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Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 1 Solutions Parallel Lines

Class 7 Maths Chapter 1 Parallel Lines Questions and Answers Kerala State Syllabus

Parallel Lines Class 7 Questions and Answers Kerala Syllabus

Page 17

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
Draw a horizontal line segment AB that is 5 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 60 degrees from line AB.

From point A, draw a line segment AC that is 3 cm long, making sure it forms the 60° angle with AB.

From point B, draw a line segment BD that is 3 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.
From point C, draw a line segment CD that is 5 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Now finding the remaining angles,
We know that, ∠A + ∠C = 180°
∠C = 180° – 60° = 120°
Similarly, ∠A + ∠B = 180°
∠B = 180° – 60° = 120°
And, ∠B + ∠D = 180°
∠D = 180° – 120° = 60°

Question 2.
The top and bottom blue lines in the figure are parallel. Find the angle between the green lines.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 40°
∠DBC – ∠BGF = 50°
:. ∠ABC = ∠ABD + ∠DBC
= 40° + 50° = 90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.
Draw this figure:

Answer:
Draw a line segment AB measuring 2 cm.

Draw a perpendicular line from point A.

Measure 20° on the protector, draw two lines on both sides of the perpendicular line, and mark C and D.

Using a set square, draw a line parallel to line AC starting from point B

Using a set square, draw a line parallel to line AD starting from point B and mark the intersection point as E.

Page 21

Question 1.
Draw the triangle with the given measures.

Answer:
First off all we have to find the third angle of the given triangle
⇒ Third angle = 180° – (60° + 40°) = 180° – 100° = 80°
Now we can start drawing the triangle, for that
Draw a horizontal line segment that is 5 cm long.

At left end of the line, use a protractor to measure and mark an angle of 40° from the line

At right end of the line, use a protractor to measure and mark an angle of 40°from the line

Construct line segments forming angles of 40° and 80°, extending each until the lines intersect

Question 2.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure

Since, ABCD is rectangle ∠A, ∠B, ∠C, ∠D equals to 90°
Now, it is given that one part of ∠A is 40°, so the remaining part of A is 50°
Similarly, one part of ∠D is 25°, so the remaining part of ∠D is 65°
It implies that two angles of the triangle are 50° and 65°.
Therefore, the third angle of the triangle is = 180° – (50° + 65°)
= 180° – (105°)
= 75°

So, the angles of triangles are 50°, 65°, 75o

Question 3.
The top and bottom lines in the figure are parallel. Calculate the third angle of the bottom triangle and all angles of the top triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel lines
The two given angles of the bottom triangle are 35° and 45°
Therefore, third angle of bottom triangle = 180° – (35° +45°)
= 180° – (80°)
= 100°
In bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, ∠A and ∠D are equal. ∠D of top triangle is = 35°
Similarly, in bottom triangle, ∠B and in top triangle, ∠C are Alternate angles
Therefore, B and C are equal, ∠C of top triangle is = 45°
So, the third angle of top triangle = 180° – (35° + 45o)
= 180° – (80°)
= 100°
All angles of bottom triangle = 35°, 45°,
All angles of top triangle = 35°, 45°, 100°.

Question 4.
The left and right sides of the large triangle are parallel to the left and right sides of the small triangle. Calculate the other two angles of the large triangle and all angles of the small triangle.

Answer:

Consider the above figure,
Given that AB and CD are parallel, and AC and DE are also parallel,
It’s given that the middle angle is 70°.
Considering the parallel lines AB and CD, AC is a slanting line intersecting the parallel lines.
Therefore, the middle angle and ∠A of the larger triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠A = 70°
Now, we can calculate the third angle of the larger triangle
i.e., ∠B = 180° – (70° +60°)
= 180° – (130°) = 50°

All three angles of larger triangle are 50°, 60°, 70°
Similarly, considering the parallel lines AC and DE, CD is a slanting line intersecting the parallel lines.
Considering the parallel lines AC and DE, CD is slanting line through the parallel lines.
Here, middle angle and the ∠D of smaller triangle are alternate angles,
Therefore, these two angles are equal
i.e., ∠D = 70°
Now, ZC of larger triangle, middle angle.and ∠C of the smaller triangle form a straight line.
Therefore, ∠C of smaller triangle is = 180° − (60° + 70°)
= 180° – 130°
= 50°
Now we can find the third angle of smaller triangle (∠E) = 180° – (50° + 70°)
= 180° – (120°)
= 60°
All three angle of smaller triangle are 50°, 60°, 70°

Question 5.
A triangle is drawn inside a parallelogram.
Calculate the angles of the triangle.
Answer:

Consider the above figure,
∠D and ∠B are opposite angles of parallelogram.
Therefore, D = ∠B = 110°
But, one part of ∠D is 60o, so remaining part of ∠D = 110° – 60° = 50°
Now, considering the angles A and <D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 110° = 70°
But, one part of ∠A is 30°, so remaining part of ∠A = 70° – 30° = 40°
It means, we get two angles of the triangle which are 50° and 40°
So, third angle of the triangle = 180° – (50° + 40°)
= 180° – (90°)
= 90°

Hence, all three angles of the triangles are 50°, 40°, 90°

Intext Questions and Answers

Question 1.
Find the unknown angle in the following figures?

Answer:
i. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 45°

ii. By extending the parallel lines and slanting line we get the figure below

From the figure, it is clear that two small angles are formed when the slanting line intersects the top and bottom parallel lines. Since these angles are equal in measure.
Therefore, the unknown angle is 40°.

iii. By drawing a vertical line, as shown in the figure below, we can observe that this vertical line is parallel to the other parallel lines and bisects the middle angle. Therefore, the angles on the left side of the vertical line are 30°

Similarly, since the angle on the right side is also 30°, then the unknown angle will also be 30o.

When two parallel lines are cut by a third line (called a slanting line), different pairs of angles are formed.

Question 2.
One angle of a triangle is 72°. The other two angles are of equal measure. What are their measures?
Answer:
Sum of two other angles = 180° – 72° = 108°
Since the other two angles are equal, each angle is = \(\frac{108^0}{2}\) = 54

Question 3.
When a line crosses another line, how many angles are formed between them ?

Answer:
When a line crosses another line, then 4 angles are formed between them.

When two lines cross perpendicular, then all angles are 90°.

Consider the figure below,

Here, 4 angles are 21, 22, 23, 24, where 21 and 22 are small angles, and 23 and 24 are large angles Then we can say that,
The two small angles are of the same measure.
i.e., ∠1 = ∠2

The two large angles are of the same measure.
i.e., ∠3 = ∠4

The sum of a small angle and a large angle is 180°.
i.e., ∠1 + ∠4 = 180° and ∠3 + ∠4 = 180°.

Consider the figure below, which shows two parallel lines intersected by another line, forming eight angles.

Then we can say that,
∠1 = ∠5
∠2 = ∠6
∠3 = ∠7
∠4 = ∠8
For example,
In the given figure, one of the angles is given as 50°

Then, the remaining angles are shown below

A line intersects two parallel lines at angles of the same measure

Question 4.
Can you calculate the other seven angles which the parallel lines make with the slanting line?

Answer:

Class 7 Maths Chapter 1 Kerala Syllabus Parallel Lines Questions and Answers

Question 1.
Draw the parallelogram below with the given measures.

Answer:
Draw a horizontal line segment AB that is 6 cm long. This will be one side of the parallelogram.

At point A, use a protractor to measure and mark an angle of 50° from line AB.

From point A, draw a line segment AC that is 4 cm long, making sure it forms the 50° angle with AB.

From point B, draw a line segment BD that is 4 cm long, parallel to AC. You can use a ruler and a set square to ensure parallelism.

From point C, draw a line segment CD that is 6 cm long, parallel to AB. The points C and D should connect to complete the parallelogram.

Question 2.
The top and bottom lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Draw a horizontal line parallel to the other two parallel lines

From the figure, we get,
∠EAB = ∠ABD = 30°
∠DBC = ∠BCF = 60°
∴ ∠ABC = ∠ABD + ∠DBC
= 30° + 60° = 90°

Question 3.
Two vertical lines in the figure are parallel. Find the unknown angle shown in the figure.

Answer:
Consider the figure below

Given that AB and CD are parallel line
Draw a line PQ parallel to both AB and CD
∠A and ∠AQP are alternate angles
Therefore, ∠A = ∠AQP = 30°

Similarly,
∠C and ∠CQP are alternate angles
Therefore, ∠C = ∠CQP = 40°
But, ∠Q = ∠AQP + ∠CQP
= 30° + 40° = 70°

Question 4.
The figure shows a triangle drawn in a rectangle. Calculate the angles of the triangle.

Answer:
Consider the figure below,

Since, ABCD is rectangle ∠P, ∠Q, ∠R, ∠S equals to 90°
Now, it is given that one part of ∠Q is 50°, so the remaining part of∠A is 40°
Similarly, one part of ∠R is 30°, so the remaining part of ∠R is 60°
It implies that two angles of the triangle are 40° and 60o.
Therefore, the third angle of the triangle is = 180° – (40° + 60°)
= 180° — (100°) = 80°
So, the angles of triangles are 40°, 60°, 80°

Question 5.
A triangle is drawn inside a parallelogram. Calculate the angles of the triangle.

Answer:
Consider the figure below

∠D and ∠B are opposite angles of parallelogram.
Therefore, ∠D = ∠B = 105°
But, one part of ∠D is 50°, so remaining part of ∠D = 105° – 50° = 55°
Now, considering the angles ∠A and ∠D, their sum is equals to 180°
i.e., ∠A + ∠D = 180°
∠A = 180° – ∠D
= 180° – 105° = 75o
But, one part of ∠A is 20°, so remaining part of ∠A = 75° – 20o = 55°
It means, we get two angles of the triangle which are 55° and 55°
So, third angle of the triangle = 180° – (55° + 55°)
= 180° – (110°) = 70°
Hence, all three angles of the triangles are 55°, 55°, 70°.

Practice Questions

Question 1.
Draw the parallelogram below with the given measures.

Calculate the other three angles.
Answer:
The other three angles of Parallelogram 60°, 60°, 120°

Question 2.
Given that PQ and QR are parallel lines, find the angle shown in the figure?

Answer:
90°

Question 3.
In the figure, the pair of lines slanted to the left are parallel; and also the pair of lines slanted to the right.

Draw this figure:
Answer:
third angle of the bottom triangle = 90°

Question 4.
The top and bottom lines in the figure are parallel.

Calculate the third angle of the bottom triangle and all angles of the top triangle.
Answer:
all angles of the top triangle = 90°, 40o, 50°,

Question 5.
The left and right lines in the figure are parallel.

Calculate the third angle of the larger triangle and all angles of the smaller triangle.
Answer:
Third angle of the larger triangle = 80°

Question 6.
A triangle is drawn inside a parallelogram.

Calculate the angles of the triangle.
Answer:
All angles of smaller triangle = 80°, 55°, 45°

Question 7.
The figure shows a triangle drawn in a rectangle.

Calculate the angles of the triangle.
Answer:
All angles of triangle = 70°, 70°, 40°

Question 8.
Draw the triangle with the given measures.

Answer:
90°, 55°, 35° 70°.

Class 7 Maths Chapter 1 Notes Kerala Syllabus Parallel Lines

Understanding geometry begins with recognising the relationships between lines and angles. In this chapter, we delve into the fascinating world of parallel lines and the angles they create when intersected by other lines. This exploration will lay the foundation for more advanced geometric concepts and develop your ability to visualise and solve geometric problems.

Lines and Angles
We start by examining lines and the angles formed when they intersect. A special focus is given to pairs of angles that are formed on the same side of the slanting line, both above and below the parallel lines. These pairs of angles are equal in measure, helping us understand the special properties of angles created when parallel lines are intersected by a slanting line.

Matching Angles
Next, we explore the idea of matching angles. This section introduces alternate angles and co-interior angles more explicitly. Alternate angles are pairs of angles that lie on opposite sides of the transversal but inside the two lines. These angles are crucial in identifying and proving the parallel nature of lines. Co-interior angles are also revisited here, reinforcing their properties and significance.

Triangle Sum
Finally, we turn our attention to triangles, a fundamental shape in geometry. One of the most important properties of triangles is the sum of their interior angles. You will learn and prove that the total sum of all three angles in any triangle is always 180°. This section will include various exercises to solidify your understanding and application of this essential geometric principle.

By the end of this chapter, you will have a solid understanding of how parallel lines interact with transversals, the relationships between the resulting angles, and the fundamental properties of triangles. These concepts are not only crucial for your current studies but also form the bedrock of many advanced topics in geometry.

Matching Angles
Of the angles made when two parallel lines are cut by a slanting line,

• • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°
• If the intersecting line is perpendicular to one of the parallel lines, it would be perpendicular to the other line too, and all angles would be right angles.

Interior Angles:
These are the angles on the inside of the parallel lines.
When you look at the pairs of interior angles on the same side of the slanting line, they are called co-interior angles.
The sum of each pair of co-interior angles is always 180°.

Exterior Angles:
These are the angles on the outside of the parallel lines.
Similarly, when you look at the pairs of exterior angles on the same side of the slanting line, they are called co-exterior angles.
The sum of each pair of co-exterior angles is also 180°.

Alternate Interior Angles:
Alternate interior angles are pairs of angles that lie between the two parallel lines and on opposite sides of the slanting line.
These angles are same measure

Alternate Exterior Angles:
Alternate exterior angles are pairs of angles that lie outside the two parallel lines and on opposite sides of the slanting line.
Like alternate interior angles, alternate exterior angles are also equal.

Corresponding Angles:
These angles are in the same relative position at each intersection where a slanting line crosses the parallel lines.
Each pair of corresponding angles has the same measure.

Find out the other angles in the given parallelogram?

Answer:
First, consider the given angle (55° angle). To determine other angles, observe the intersections made by the left side with the top and bottom parallel lines.
The 55° angle and the angle above it form a pair of angles that add up to 180°. Therefore, the top angle is:
180° – 55° = 125°
Next, look at the angle to the right of the marked 55° angle.
To find this angle, examine the angles formed by the left and right parallel sides with the bottom line. The 55° angle and the angle to its right also form a pair of angles that add up to 180°.
Thus, this angle is also 125°, as calculated earlier.

Triangle Sum
The sum of all angles of a triangle is 180°

• A line intersects two parallel lines at angles of the same measure
• When two parallel lines are cut by a slanting line
  • the small angles are of the same measure.
  • the large angles are of the same measure.
  • a small angle and a large angle add up to 180°.
• If the intersecting line is perpendicular to one of the parallel lines, would be perpendicular to the other line too, and all angles would be right angles.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 4 Reciprocals Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 4 Solutions Reciprocals

Class 7 Maths Chapter 4 Reciprocals Questions and Answers Kerala State Syllabus

Reciprocals Class 7 Questions and Answers Kerala Syllabus

Page 64

Question 1.
Suma has 16 rupees with her. Safeer has 4 rupees.
(i) What part of Suma’s money does Safeer have?
(ii) How many times Safeer’s money does Suma have?
Answer:
Suma’s amount = 16
Safeer’s amount = 4
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\)
So, Safeer has of Suma’s money.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{16}{4}\)
= 4
So, Suma has 4 times that of Safeer’s money.

Question 2.
A large bag contains 9 kilograms of sugar. A small bag contains 6 kilograms.
(i) The weight of sugar in the heavier bag is how much times that in the lighter bag?
(ii) The weight of sugar in the lighter bag is what part of that in the heavier bag?
Answer:
(i) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{9}{6}\)
= \(\frac{3}{2}\)
So, the weight of sugar in the heavier bag is \(\frac{3}{2}\) times that in the lighter bag.

(ii) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{9}\)
= \(\frac{2}{3}\)
So, the weight of sugar in the lighter bag is \(\frac{2}{3}\) part of that in the heavier bag.

Question 3.
The weight of an iron block is 6 kilograms. The weight of another block is 26 kilograms.
(i) The weight of the lighter block is what fraction of that of the heavier block?
(ii) The weight of the heavier block is how much times that of the lighter block?
Answer:
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{26}\)
= \(\frac{3}{13}\)
So, the weight of the lighter block is \(\frac{3}{13}\) fraction of that of the heavier block.

(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{26}{6}\)
= \(\frac{13}{3}\)
So, the weight of the heavier block is \(\frac{13}{3}\) times that of the lighter block.

Question 4.
The length of a ribbon is 2 times the length of a smaller ribbon. What part of the length of the large ribbon is the length of the small ribbon?
Answer:
Times = \(\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
2\(\frac{2}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
\(\frac{8}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)

8 × length of the smaller ribbon = 3 × length of the large ribbon
\(\frac{\text { length of the smaller ribbon }}{\text { length of the large ribbon }}=\frac{3}{8}\)
Part = \(\frac{3}{8}\)
So, length of the smaller ribbon is \(\frac{3}{8}\) part of the length of the large ribbon.

Page 67

Question 1.
27 students of a class got A plus in Maths. They form of the entire class. How many students are there in the class?
Answer:
\(\frac{3}{4}\) × entire class = 27
entire class = 27 ÷ \(\frac{3}{4}\)
= 27 × \(\frac{4}{3}\)
= 9 × 4
= 36
Thus, there are 36 students in the class.

Question 2.
\(\frac{2}{3}\) of a bottle was filled with \(\frac{1}{2}\) litre of water. How many litres of water will the bottle hold?
Answer:
\(\frac{2}{3}\) x entire bottle = \(\frac{1}{2}\) litres
entire bottle = \(\frac{1}{2} \div \frac{2}{3}\)
= \(\frac{1}{2} \times \frac{3}{2}\)
= \(\frac{3}{4}\)
Thus, the bottle will hold – litres of water.

Question 3.
\(\frac{3}{4}\) of a vessel holds 1\(\frac{1}{2}\) litres of water. What is the capacity of the vessel in litres if it is completely filled with water?
Answer:
\(\frac{3}{4}\) × entire vessel = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) litres
entire vessel = \(\frac{3}{2} \div \frac{3}{4}\)
= \(\frac{3}{2} \times \frac{4}{3}\)
= \(\frac{4}{2}\)
= 2
Thus, the capacity of the vessel if it is completely filled with water is 2 litres.

Question 4.
Two of the three ribbons of the same length and half the third ribbon were placed end to end. It came to 1 metre. What is the length of a ribbon in centimetres?
Answer:
1 metre = 100 cm
2.5 ribbons = 100 cm
\(\frac{5}{2}\) × length of a ribbon = 100 cm
length of a ribbon = 100 ÷ \(\frac{5}{2}\)
= 100 × \(\frac{2}{5}\)
= 20 × 2
= 40 cm

Page 68

Question 1.
A 16 metres long rod is cut into pieces of length \(\frac{2}{3}\) metre. How many such pieces will be there?
Answer:
Total length of the rod = 16 m
Length of a piece = \(\frac{2}{3}\) m
Number of pieces = Total length of the rod ÷ Length of a piece
= 16 ÷ \(\frac{2}{3}\)
= 16 × \(\frac{3}{2}\)
= 8 × 3
= 24.

Question 2.
How many \(\frac{3}{4}\) litre bottles are needed to fill 5\(\frac{1}{4}\) litres of water?
Answer:
Total water = 5\(\frac{1}{4}=\frac{21}{4}\) litres
Amount of water in a bottle = \(\) litres
Number of bottles = Total water ÷ Amount of water in a bottle
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{3}\)
= 7

Question 3.
13\(\frac{1}{2}\) kilograms of sugar is to be packed into bags with 2\(\frac{1}{4}\) kilograms sugar each. How many bags are needed?
Answer:
Total sugar = 13\(\frac{1}{2}\) = \(\frac{27}{2}\) kg
Amount of sugar in one bag = 2\(\frac{1}{4}=\frac{9}{4}\)kg
Number of bags = Total sugar Amount of sugar in one bag
= \(\frac{27}{2} \div \frac{9}{4}\)
= \(\frac{27}{2} \times \frac{4}{9}\)
= 3 × 2
= 6.

Question 4.
The area of a rectangle is 22\(\frac{1}{2}\) square centimetres, and one side is 3\(\frac{3}{4}\) centimetres long. What is the length of the other side?
Answer:
Area of a rectangle = 22\(\frac{1}{2}\) = \(\frac{45}{2}\)sq.cm
length of one side = 3\(\frac{3}{4}=\frac{15}{4}\) cm
length of the other side = area of a rectangle ÷ length of one side
= \(\frac{45}{2} \div \frac{15}{4}\)
= \(\frac{45}{2} \times \frac{4}{15}\)
= 3 × 2
= 6 cm

Question 5.
How many pieces, each of length 2\(\frac{1}{2}\) metres, can be cut off from a rope of length 11\(\frac{1}{2}\) metres? How many metres of rope will be left?
Answer:
Total length of the rope = 11\(\frac{1}{2}\) = \(\frac{23}{2}\) m
Length of a piece = 2\(\frac{1}{2}=\frac{5}{2}\)m
Number of pieces = Total length of the rope ÷ Length of a piece
= \(\frac{23}{2} \div \frac{5}{2}\)
= \(\frac{23}{2} \times \frac{2}{5}\)
= \(\frac{23}{5}\)
= 4\(\frac{3}{5}\)

So, we can cut 4 pieces of length 2\(\frac{1}{2}\) m.
4 × 2\(\frac{1}{2}\)
= 4 × \(\frac{5}{2}\)
= 2 × 5
= 10 m
Length of the remaining rope =11\(\frac{1}{2}\) – 10 = \(\frac{23}{2}\) – 10 = \(\frac{3}{2}\)m

Page 68

Question 1.
A rod is 36 metres long. How many pieces each of length 2\(\frac{1}{2}\) metres can be cut off from it? What is the length of the rod left?
Appu did the problem this way.
36 ÷ 2\(\frac{1}{2}\) = 36 ÷ \(\frac{5}{2}\)
= 36 ÷ \(\frac{2}{5}\)
= \(\frac{72}{5}\)
When we divide 72 by 5, the quotient is 14 and remainder is 2. So we get 14 pieces. The remaining rod is of length 2 metres.
Ammu used another idea.
2 pieces, each of length 2\(\frac{1}{2}\) metres makes 5 metres.
7 × 5 = 35
So 7 × 2 = 14 pieces can be cut off. The remainder is 36 – 35 = 1 metre. Whose answer is right?
Solution:
In Appu’s case, he find that 14 pieces of length 2\(\frac{1}{2}\) metres can be cut off from it.
This 14 pieces together forms;
14 × 2\(\frac{1}{2}\) = 14 × \(\frac{5}{2}\)
= 7 × 5
= 35 m
So, the remaining length = 36 – 35 = 1 m.
Thus, Ammu’s answer is the right one.

Class 7 Maths Chapter 4 Kerala Syllabus Reciprocals Questions and Answers

Question 1.
Find the reciprocal of \(\frac{3}{7}\)
Answer:
reciprocal of \(\frac{3}{7}=\frac{7}{3}\)

Question 2.
Simplify \(\frac{3}{4} \div \frac{5}{6}\)
Answer:
\(\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{18}{20}=\frac{9}{10}\)

Question 3.
A recipe requires cup of sugar to make 12 cookies. How much sugar is needed to make 36 cookies?
12 കുക്കീസ് ഉണ്ടാക്കുവാനായി \(\frac{3}{4}\) കപ്പ് പഞ്ചസാര വേണം. അങ്ങനെയങ്കിൽ 36 കുക്കീസ് ഉണ്ടാക്കുവാൻ എത്ര കപ്പ് പഞ്ചസാര വേണം?
Answer:
\(\frac{3}{4}\) cup of sugar gives 12 cookies.
12 × 3 gives 36.
So, for 36 cookies we need \(\frac{3}{4}\) × 3 = \(\frac{9}{4}\) cups of sugar.

Question 4.
If of the cake is eaten, what fraction of the whole cake remains?
ഒരു കേക്കിന്റെ \(\frac{2}{5}\) ഭാഗം കഴിച്ചെങ്കിൽ ബാക്കി എത്ര ഭാഗം ഉണ്ട്?
Answer:
The whole cake is \(\frac{5}{5}\) = 1
Remaining cake = \(\frac{5}{5}-\frac{2}{5}=\frac{3}{5}\)
Thus, \(\frac{3}{5}\) part of the whole cake remains.

Question 5.
A ribbon is cut into 12 equal parts. What fraction of the ribbon is 3 parts?
ഒരു റിബൺ 12 തുല്യ ഭാഗങ്ങളായി മുറിച്ചു. അതിലെ 3 ഭാഗം ആകെയുള്ള റിബണിന്റെ എത്ര ഭാഗമാണ്?
Answer:
Part (or fraction) = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)

Practice Questions

Question 1.
Solve \(\frac{4}{5} \div \frac{5}{4}\)
Answer:
\(\frac{16}{25}\)

Question 2.
Calculate \(\frac{3}{7} \div \frac{7}{3}\)
Answer:
1

Question 3.
A company produces \(\frac{2}{5}\) of its daily output in the morning and \(\frac{3}{8}\) in the afternoon. What fraction of the total daily output is produced in the afternoon?
Answer:
\(\frac{15}{31}\)

Question 4.
Calculate the reciprocal of \(\frac{2}{5}\). Which is larger?
Answer:
\(\frac{5}{2}\), reciprocal is the largest

Question 5.
When a tank is \(\frac{1}{4}\) full, it contains 80 litres of water. What will it contain when it is \(\frac{3}{8}\) full?
Answer:
120 litres

Class 7 Maths Chapter 4 Notes Kerala Syllabus Reciprocals

We are already familiar with a number of mathematical operations, such as addition, subtraction, multiplication and division. This chapter introduces a new type of mathematical operation named “reciprocal”. Following are the topics discussed in this chapter.

Times and parts
If we are given a large number and a small number, we can say the large number is how many times the small number. Always,
Times = \(\frac{\text { large number }}{\text { small number }}\)

If we are given a large number and a small number, we can say the small number is what part of the large number. Always,
Part = \(\frac{\text { small number }}{\text { large number }}\)

Topsy – Turvy
A fraction obtained by interchanging the numerator and denominator of the given fraction is called the reciprocal of the given fraction.
If we multiply a natural number by its reciprocal, we will get one as the result.
Eg:
The reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\)

If we multiply a fraction by its reciprocal, we will get one as the result.
Eg:
Consider the fraction \(\frac{3}{4}\). Its reciprocal is \(\frac{4}{3}\). Now, \(\frac{3}{4} \times \frac{4}{3}\) = 1

Zero does not have a reciprocal because any number multiplied by zero is zero, not 1.

Fraction division
Division of natural numbers is the same as multiplication by the reciprocal.
Dividing a fraction (or a number) by another fraction is exactly same as multiplying the first fraction (or number) by the reciprocal of the second fraction.
Eg:
\(\frac{1}{2} \div \frac{3}{4}=\frac{1}{2} \times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}\)
8 ÷ \(\frac{2}{5}\) = 8 × \(\frac{5}{2}\) = 4 × 5 = 20

If we are given a large number and a small number, we can say;
the large number is how many times the small number
the small number is what part of the large number large number
Times = \(\frac{\text { large number }}{\text { small number }}\)
Part = \(\frac{\text { small number }}{\text { large number }}\)

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 2 Fractions Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 2 Solutions Fractions

Class 7 Maths Chapter 2 Fractions Questions and Answers Kerala State Syllabus

Fractions Class 7 Questions and Answers Kerala Syllabus

Page 24

Do the following problems mentally. Write each as how many times and also as a product.

Question 1.
Each piece of a pumpkin weighs a quarter kilogram. What is the weight of two pieces together? What is the weight of four such pieces? Six pieces?
Answer:
Two times quarter kilogram is half kilogram.
Four times quarter kilogram is one kilogram.
Six times quarter kilogram is one and a half kilogram.
Mathematically,
Each piece of a pumpkin weighs \(\frac{1}{4}\) kilogram.
The weight of two pieces is two times \(\frac{1}{4}\), which is \(\frac{1}{2}\) kilogram.
The weight of four pieces is 4 times \(\frac{1}{4}\) which is 1 kilogram.
The weight of six pieces is 6 times \(\frac{1}{4}\) which is 1\(\frac{1}{2}\) kilogram.

Using products,
The weight of two pieces = 2 × \(\frac{1}{4}=\frac{2}{4}\) = \(\frac{1}{2}\) kilogram
The weight of four pieces= 4 × \(\frac{1}{4}=\frac{4}{4}\) = 1 kilogram
The weight of six pieces = 6 × \(\frac{1}{4}\) = 1\(\frac{1}{2}\) kilogram.

Question 2.
We can fill a cup with one-third of a litre of milk. How much milk is needed to fill two cups? Four cups?
Answer:
Two times one-third of a litre is two-third of a litre.
Four times one-third of a litre is four-third of a litre. Mathematically,
Each cup contains one-third of a litre of milk.
So, milk needed to fill two cups is 2 times \(\frac{1}{3}\) which is \(\frac{2}{3}\) litre.
Milk needed to fill four cups is 4 times \(\frac{1}{3}\), which is 1\(\frac{1}{3}\) litre.
Using products,
Milk needed to fill two cups = 2 × \(\frac{1}{3}=\frac{2}{3}\) litre
Milk needed to fill four cups = 4 × \(\frac{1}{3}\) = 1\(\frac{1}{3}\) litre.

Question 3.
What is the total length of four pieces of ribbons, each of length three fourths of a metre? What about five pieces?
Answer:
Four times three fourth of a metre is three metre.
Five times three fourth of a metre is three and three-fourth of a metre.
Mathematically,
Length of each piece is three-fourth of a metre.
So, total length of four pieces of ribbon is 4 times \(\frac{3}{4}\), which is 3 metre.
Total length of five pieces of ribbon is 5 times \(\frac{3}{4}\), which is 3\(\frac{3}{4}\) metre.
Using products,
Total length of four pieces of ribbon = 4 × \(\frac{3}{4}\) = 3 metre
Total length of five pieces of ribbon = 5 × \(\frac{1}{2}\) = 3\(\frac{3}{4}\) metre

Question 4.
It takes hour to walk around a play ground once.
(i) How much time does it take to walk 4 times around at this speed?
(ii) What about 7 times?
Answer:
(i) 4 times \(\frac{1}{4}\) is 1 hour.
Using products,
Time taken to walk 4 time around = 4 × \(\frac{1}{4}\) = 1 hour

(ii) 7 times \(\frac{1}{4}\) is 1\(\frac{3}{4}\) hour.
Using products,
Time taken to walk 7 times around = 7 × \(\frac{1}{4}\) = 1\(\frac{3}{4}\)hour.

Page – 25,26

Question 1.
The weight of an iron block is kilogram.
(i) What is the total weight of such 15 blocks?
(ii) 16 blocks?
Answer:
(i) Weight of one block = \(\frac{1}{4}\) kilogram.
Weight of 15 blocks = 15 × \(\frac{1}{4}=\frac{15 \times 1}{4}\)
= \(\frac{15}{4}=\frac{12+3}{4}\)
= \(\frac{12}{4}+\frac{3}{4}\)
= 3\(\frac{3}{4}\)

(ii) Weight of 16 blocks = 16 × \(\frac{1}{4}=\frac{16}{4}\) = 4 kilogram.

Question 2.
Some 2 metre long rods are cut into 5 pieces of equal length.
(i) What is the length of each piece?
(ii) What is the total length of 4 pieces?
(iii) of 10 pieces?
Answer:
(i) Total length of the rod = 2 metre.
Length of each piece = \(\frac{2}{5}\) metre

(ii) Total length of 4 pieces = 4 × \(\frac{2}{5}=\frac{4 \times 2}{5}=\frac{8}{5}\)
= \(\frac{5+3}{5}=\frac{5}{5}+\frac{3}{5}\)
= 1\(\frac{1}{5}\) metre

Question 3.
5 litres of milk is filled in 6 bottles of the same size.
(i) How many litres of milk does each bottle hold?
(ii) How many litres in 3 bottles together?
(iii) In 4 bottles?
Answer:
Total milk = 5 litre
Number of bottles = 6
(i) Milk in each bottle = \(\frac{5}{6}\) litre

(ii) Milk in 3 bottles together = 3 × \(\frac{5}{6}=\frac{3 \times 5}{6}=\frac{15}{6}\)
= \(\frac{12+3}{6}=\frac{12}{6}+\frac{3}{6}\)
= 2\(\frac{1}{2}\) litre.

(iii) Milk in 4 bottles = 4 × \(\frac{5}{6}=\frac{4 \times 5}{6}=\frac{20}{6}\)
= \(\frac{18+2}{6}=\frac{18}{6}+\frac{2}{6}\)
= 3\(\frac{1}{3}\) litre

Page – 28

Do these problems in head. Then write each as a part and also as a product of numbers.

Question 1.
Nine litres of milk is divided equally among three children. How many litres will each get? What if there are four children?
Answer:
Each will get a third of nine litres, that is three litres.
As a part,
Each will get \(\frac{1}{3}\) of 9, which is 3 litres.

As a product,
Each will get \(\frac{1}{3}\) × 9 = \(\frac{9}{3}\) = 3 litres.
If there are four children,
Each will get a fourth of nine litres. A fourth of 8 litre is 2 litre and then a fourth of the remaining one litre. So, two and one-fourth of a litre.

As a part,
Each will get \(\frac{1}{4}\) of 9, which is 2\(\frac{1}{4}\) litres.

As a product,
Each will get \(\frac{1}{4}\) × 9 = \(\frac{9}{4}\) = 2\(\frac{1}{4}\) litres.

Question 2.
Six kilograms of rice was packed in five bags of the same size. How many kilograms of rice in each bag? What if it is packed in four bags?
Answer:
Each bag has one fifth of six kilograms. One fifth of 5 kilogram is 1 kilogram and then one fifth of remaining one kilogram. So, one and one-fifth of a kilogram.

As a part,
Each bag has \(\frac{1}{5}\) of 6, which is 1\(\frac{1}{5}\) kilograms.

As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{5}\) = 1\(\frac{1}{5}\) kilograms.
If it is packed in four bags,
Each bag has one-fourth of six kilograms. One-fourth of four kilograms is 1 kilogram and then one- fourth of remaining two kilograms. So, one and a half kilograms.

As a part,
Each bag has \(\frac{1}{4}\) of 6, which is 1\(\frac{1}{2}\) kilograms.
As a product,
Each bag has \(\frac{1}{4}\) × 6 = \(\frac{6}{4}\) = 1\(\frac{1}{2}\) kilograms.

Question 3.
A seven metre long string is divided into six equal pieces. What is the length of each piece? What if it is divided into three equal pieces?
Answer:
Each piece is one-sixth of seven metres. One-sixth of six metres is 1 metre and then one-sixth of the remaining one metre. So, one and one-sixth of a metre.

As a part,
Length of each piece is \(\frac{1}{6}\) of 7, which is 1\(\frac{1}{6}\) metres.

As a product,
Length of each piece = \(\frac{1}{6}\) × 7 = \(\frac{7}{6}\) = 1 \(\frac{1}{6}\) metres.
If it is divided into three equal pieces,
Each piece is one-third of seven metres. One-third of six metres is 2 metres and one-third of remaining one metre. So, two and one-third of a metre.

As a part,
Length of each piece is \(\frac{1}{3}\) of 7, which is 2\(\frac{1}{3}\) metres.
As a product,
Length of each piece \(\frac{1}{3}\) × 7 = \(\frac{7}{3}\) = 2\(\frac{1}{3}\) metres.
The calculations of the types of problems above can be done as follows:

Question 4.
We have to cut off \(\frac{3}{5}\) of a 7 metre long string. How long is this piece?
Answer:
Here we have to calculate \(\frac{3}{5}\) of 7.
\(\frac{3}{5}\) x 7 = \(\frac{3 \times 7}{5}=\frac{21}{5}=\frac{20+1}{5}\)
\(\frac{20}{5}+\frac{1}{5}\) = 4 + \(\frac{1}{5}\)
= 4\(\frac{1}{5}\)
So, we need to cut off 4\(\frac{1}{5}\) metres.

Page – 29

Question 1.
There are 35 children in a class. of them are girls. How many girls are there in the class?
Answer:
Number of girls = \(\frac{3}{5}\) × 35
= 3 × \(\frac{35}{5}\)
= 3 × 7
= 21
Or,
Number of girls = \(\frac{3}{5}\) × 35
= \(\frac{3 \times 35}{5}\)
= \(\frac{105}{5}\)
= 21

Question 2.
10 kilograms of rice is filled equally in 8 bags. If the rice in 3 such bags are taken together, how many kilograms would that be?
Answer:
Rice in 3 bags are taken together.
That is, we have to find- of 10 kilograms.
So, amount of rice = \(\frac{3}{8}\) × 10
= \(\frac{30}{8}=\frac{24+6}{8}\)
= \(\frac{24}{8}+\frac{6}{8}\)
= 3\(\frac{3}{4}\) kilograms.

Question 3.
The area of the rectangle in the figure is 27 square centimetres. It is divided into 9 equal parts.

What is the area of the darker part in square centimetres?
Answer:
5 of the 9 equal parts are darker.
So, we have to find \(\frac{5}{9}\) of 27 square centimetres.

Area of darker part = \(\frac{5}{9}\) × 27 = 5 × \(\frac{27}{9}\)
= 5 × 3 = 15 square centimeres.

Area of darker part = \(\frac{5}{9}\) × 27 = \(\frac{5 \times 27}{9}=\frac{135}{9}\)
= 15 square centimetres.

Page – 32

Question 1.
Draw rectangles and find these products.
(i) \(\frac{1}{2} \times \frac{1}{4}\)
(ii) \(\frac{1}{3} \times \frac{1}{6}\)
(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Answer:
(i) \(\frac{1}{2} \times \frac{1}{4}\)
Draw a rectangle a divide it into 4 equal parts. Then divide one part into half.

Now extend the horizontal line.

So, \(\frac{1}{2} \times \frac{1}{4}\) = \(\frac{1}{8}\)

(ii) \(\frac{1}{3} \times \frac{1}{6}\)
Draw a rectangle and divide it into 6 equal parts. Then divide each part into 3 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{3} \times \frac{1}{6}=\frac{1}{18}\)

(iii) \(\frac{1}{5} \times \frac{1}{8}\)
Draw a rectangle a divide it into 8 equal parts. Then divide one part into 5 equal parts.

Now extend the horizontal line.

So, \(\frac{1}{5} \times \frac{1}{8}=\frac{1}{40}\)

Question 2.
A one metre long string is divided into five equal parts. How long is half of each part in metres? In centimetres?
Answer:
When one metre long string divided into five equal parts each part is one- fifth of one metre.
Now half of each part is \(\frac{1}{2}\) of \(\frac{1}{5}\)of one metre.
∴ Length of each part = \(\frac{1}{2} \times \frac{1}{5}\) = \(\frac{1}{10}\) metre
= \(\frac{1}{10}\) × 100
= 10 centimetre.

Question 3.
One litre of milk is filled in two bottles of equal size. A quarter of the milk in one bottle was used to make tea. How many litres of milk were used for tea? In millilitres?
Answer:
When one litre of milk is filled in two bottles of equal size, each bottle contain half of a litre.
A quarter of the milk in one bottle was used to make tea, which is of of one litre.
Milk used for tea \(\frac{1}{4}\) of \(\frac{1}{2}\) one litre
= \(\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}\) litre
= \(\frac{1}{8}\) × 1000 = 125 millilitre.

Consider another type of problem:
Find of \(\frac{4}{5}\) of \(\frac{2}{3}\).
Answer:

Page – 34

Question 1.
A rope 2 metres long is cut into 5 equal pieces. What is the length of three quarters of one of the pieces in metres? In centimetres?
Answer:
When 2 metre long rope cut into 5 equal pieces, length of each piece is \(\frac{1}{5}\) of 2 metre, which is \(\frac{2}{5}\) metre.
Length of three quarters of one of the pieces is \(\frac{3}{4}\) of \(\frac{2}{5}\)
Required length of the piece = \(\frac{3}{4} \times \frac{2}{5}\)
= 3 × \(\frac{1}{4} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{20}\) × 2
= \(\frac{6}{20}=\frac{3}{10}\)

Question 2.
4 bottles of the same size were filled with 3 litres of water. One of these was used to fill 5 cups of the same size. How much water is there in one such cup, in litres? And in millilitres?
Answer:
When 4 bottles of the same size were filled with 3 litres of water, each bottle has of 3 litre, which is \(\frac{3}{4}\) litre.
One of these was used to fill 5 cups of the same size. Then, amount of water in one cup is \(\frac{1}{5}\) of \(\frac{3}{4}\).
Amount of water in one cup = \(\frac{1}{5} \times \frac{3}{4}\)
= \(\frac{1}{5} \times \frac{1}{4}\) × 3
= \(\frac{1}{20}\) × 3
= \(\frac{3}{20}\)

Question 3.
A watermelon weighing four kilograms was cut into five equal pieces. One piece was again halved. What is the weight of each of these two pieces in kilograms? And in grams?
Answer:
When watermelon weighing 4 kg cut into five equal pieces, weight of each piece is \(\frac{1}{5}\) of 4 kg, which is \(\frac{4}{5}\)kg.
Each piece is again halved.
Then weight of each of these two pieces is \(\frac{1}{2}\) of \(\frac{4}{5}\)
Required weight = \(=\frac{1}{2} \times \frac{4}{5}\)
= \(=\frac{1}{2} \times \frac{1}{5}\) × 4
= \(\frac{1}{10}\) × 4
= \(\frac{4}{10}=\frac{2}{5}\) kilograms

Question 4.
A vessel full of milk is used to fill three bottles of the same size. Then the milk in each bottle was used to fill four cups of the same size. What fraction of the milk in the first vessel does each cup contain?
Answer:
When a vessel full of milk used to fill three bottles of same size, each bottle has one-third of milk.
When milk in each bottle was used to fill four cups of same size, each cup has \(\frac{1}{4}\) of \(\frac{1}{3}\) of the milk.
∴ fraction of milk in each cup = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)

Question 5.
Draw a line AB of length 12 centimetres. Mark AC as \(\frac{2}{3}\) of AB. Mark AD as \(\frac{1}{4}\) of AC. What part of AB is AD?
Answer:

AD = \(\frac{1}{4}\) of AC
= \(\frac{1}{4}\) of \(\frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{2}{3}\) of AB
= \(\frac{1}{4} \times \frac{1}{3}\) × 2 of AB
= \(\frac{1}{12}\) × 2 of AB
= \(\frac{2}{12}=\frac{1}{6}\) of AB

Question 6.
Calculate the following using multiplication:
(i) \(\frac{3}{7}\) of \(\frac{2}{5}\)
Answer:
\(\frac{3}{7}\) of \(\frac{2}{5}\) = \(\frac{3}{7} \times \frac{2}{5}\)
= 3 × \(\frac{1}{7} \times \frac{1}{5}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(ii) \(\frac{2}{3}\) of \(\frac{3}{4}\)
Answer:
\(\frac{2}{3}\) of \(\frac{3}{4}\) = \(\frac{2}{3} \times \frac{3}{4}\)
= 2 × \(\frac{1}{3} \times \frac{1}{4}\) × 3
= 2 × \(\frac{1}{12}\) × 3
= 6 × \(\frac{1}{12}=\frac{6}{12}=\frac{1}{2}\)

(iii) \(\frac{3}{5}\) of \(\frac{2}{7}\)
Answer:
\(\frac{3}{5}\) of \(\frac{2}{7}\) = \(\frac{3}{5} \times \frac{2}{7}\)
= 3 × \(\frac{1}{5} \times \frac{1}{7}\) × 2
= 3 × \(\frac{1}{35}\) × 2
= \(\frac{6}{35}\)

(iv) \(\frac{5}{6}\) of \(\frac{3}{10}\)
Answer:
\(\frac{5}{6}\) of \(\frac{3}{10}\) = \(\frac{5}{6} \times \frac{3}{10}\)
= 5 × \(\frac{1}{6} \times \frac{1}{10}\) × 3
= 5 × \(\frac{1}{60}\) × 3
= 15 × \(\frac{1}{60}=\frac{15}{60}=\frac{1}{4}\)

Page – 37

Question 1.
One and a half metres of cloth is needed for a shirt. How much cloth is required for five such shirts?
Answer:
Cloth needed for a shirt = 1 metres
Cloth needed for 5 shirts = 5 × 1\(\frac{1}{2}\)
= 5 × (1 + \(\frac{1}{2}\))
= (5 × 1) + (5 × \(\frac{1}{2}\))
= 5 + 2\(\frac{1}{2}\)
= 7\(\frac{1}{2}\) metres.
Or,
5 × 1\(\frac{1}{2}\) = 5 × \(\frac{3}{2}\)
= \(\frac{15}{2}\) = 7\(\frac{1}{2}\)

Question 2.
The price of one kilogram of okra is thirty rupees. What is the price of two and a half kilograms?
Answer:
Price of one kilogram of okra = 30 rupees
Price of two and a half kilograms of okra = 30 × 2\(\frac{1}{2}\)
= 30 × (2 + \(\frac{1}{2}\))
= (30 × 2) + (30 × \(\frac{1}{2}\))
= 60 + 15
= 75 rupees
Or,
30 × 2\(\frac{1}{2}\) = 30 × \(\frac{5}{2}\)
= \(\frac{30}{2}\) × 5
= 15 × 575 rupees

Question 3.
A person walks two and a half kilometres in an hour. At the same speed, how far will he walk in one and a half hours?
Answer:
Distance he walks in one hour = 2 \(\frac{1}{2}\) km
Distance he walks in one and a half hours = 1 \(\frac{1}{2}\) x 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{5}{2}\)
= \(\frac{3 \times 5}{2 \times 2}=\frac{15}{4}\)
= 3 \(\frac{3}{4}\) km

Question 4.
Roni has 36 stamps with her. Sahira says she has 2 times this. How many stamps does Sahira have?
Answer:
Number of stamps with Roni = 36
Number of stamps with Sahira = 2\(\frac{1}{2}\) × 36
= (2 + \(\frac{1}{2}\)) × 36
= (2 × 36) +(\(\frac{1}{2}\) × 36)
= 72 + 18
= 90
Or
2\(\frac{1}{2}\) × 36 = \(\frac{5}{2}\) × 36
= 5 × \(\frac{36}{2}\)
= 5 × 18
= 90

Question 5.
Joji works 4\(\frac{1}{2}\) hours each day. How many hours does he work in 6 days?
Answer:
Number of hours Joji works each day = 4 hours
Number of hours Joji work in 6 days = 6 × 4\(\frac{1}{2}\)
= 6 × (4 + \(\frac{1}{2}\))
= (6 × 4) + (6 × \(\frac{1}{2}\))
= 24 + 3
= 27 hours

Question 6.
Calculate the following:
(i) 4 times 5\(\frac{1}{3}\)
Answer:
4 times 5\(\frac{1}{3}\) = 4 × 5\(\frac{1}{3}\)
= 4 × (5 + \(\frac{1}{3}\))
= (4 × 5) + (4 × \(\frac{1}{3}\))
= 20 + 1\(\frac{1}{3}\)
= 21 \(\frac{1}{3}\)

(ii) 4\(\frac{1}{3}\) times 5
Answer:
4\(\frac{1}{3}\) times 5 = 4\(\frac{1}{3}\) × 5
= (4 + \(\frac{1}{3}\)) × 5
= (4 × 5) + (\(\frac{1}{3}\) × 5)
= 20 + 1\(\frac{2}{3}\)
= 21\(\frac{2}{3}\)

(iii) 1\(\frac{1}{2}\) times \(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) times \(\frac{2}{3}\) = 1\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(iv) \(\frac{2}{5}\) times 2\(\frac{1}{2}\)
Answer:
\(\frac{2}{5}\) times 2\(\frac{1}{2}\) = \(\frac{2}{5}\) × 2\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{2}{3}\)
= \(\frac{3 \times 2}{2 \times 3}\)
= 1

(v) 2\(\frac{1}{2}\) times 5\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) times 5\(\frac{1}{2}\) = 2\(\frac{1}{2}\) × 5\(\frac{1}{2}\)
= \(\frac{5}{2} \times \frac{11}{2}\)
= \(\frac{5 \times 11}{2 \times 2}\)
= \(\frac{55}{4}\)
= 13 \(\frac{3}{4}\)

Page – 42

Question 1.
The length and breadth of some rectangles are given below. Find the area of each:
(i) 3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Answer:
3\(\frac{1}{4}\) centimetres, 4\(\frac{1}{2}\) centimetres
Area of the rectangle = 3\(\frac{1}{4}\) × 4\(\frac{1}{2}\)
= \(\frac{13}{4} \times \frac{9}{2}\)
= \(\frac{13 \times 9}{4 \times 2}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) square centimetres

(ii) 5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Answer:
5\(\frac{1}{3}\) metres, 6\(\frac{3}{4}\) metres
Area of the rectangle = 5\(\frac{1}{3}\) × 6\(\frac{3}{4}\)
= \(\frac{16}{3} \times \frac{27}{4}\)
= \(\frac{16}{4} \times \frac{27}{3}\)
= 4 × 3 = 12

Question 2.
What is the area of a square of side 1 metres?
Answer:
Area of the square = 1\(\frac{1}{2}\) × 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{3 \times 3}{2 \times 2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) square metre

Question 3.
The perimeter of a square is 14 metres. What is its area?
Answer:
Perimeter of the square = 14 metres
4 × side = 14
∴ Side = \(\frac{14}{4}\)= 3\(\frac{1}{2}\) metre
Area = 3\(\frac{1}{2}\) × 3\(\frac{1}{2}\)
= \(\frac{7}{2} \times \frac{7}{2}\)
= \(\frac{7 \times 7}{2 \times 2}\)
= \(\frac{49}{4}\)
= 12\(\frac{1}{4}\) square metres

Intext Questions and Answers

Question 1.
A bottle holds a quarter litre of water. How much water is needed to fill three such bottles?
Answer:
Each bottle holds a quarter litre of water. So, water needed to fill three such bottles is three times a quarter litre, which is three-quarter litres.
This can be calculated as,
3 times \(\frac{1}{4}\) is \(\frac{3}{4}\) is \(\frac{3}{4}\)
As a product,
3 × \(\frac{1}{4}=\frac{3}{4}\)

Question 2.
The calculations in the type of problems above can be done easily as follows:
\(\frac{3}{4}\) litres of milk in a bottle; how many litres in 7 such bottles?
Answer:
Amount of milk in one bottle = \(\frac{3}{4}\) litres
Amount of milk in 7 such bottles = 7 times \(\frac{3}{4}\)
7 × \(\frac{3}{4}=\frac{7 \times 3}{4}=\frac{21}{4}\)

Split 21 as a multiple of 4.
\(\frac{21}{4}=\frac{20+1}{4}=\frac{20}{4}+\frac{1}{4}\)
= 5 + \(\frac{1}{4}\)
= 5\(\frac{1}{4}\)

Question 3.
A five metre long string is cut into three equal pieces. What is the length of each piece?
Answer:
3 metres cut into three equal pieces, each piece is 1 metre. The remaining 2 metre cut into three equal pieces, each piece must be two-third of a metre. So, the length of each piece is 1 metres.
In terms of number alone,
\(\frac{1}{3}\) of 5 is 1\(\frac{2}{3}\)
Writing it as a product,
\(\frac{1}{3}\) × 5 = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Question 4.
Draw rectangle and find \(\frac{1}{3} \times \frac{1}{2}\)
Answer:
First, draw a rectangle and divide it into two equal part. Then each part is \(\frac{1}{2}\)

Then divide one part into three equal parts. Here, each part is \(\frac{1}{3} \times \frac{1}{2}\)

Now extend the horizontal line.

The rectangle is divided into 6 equal parts and so each part is
Hence, \(\frac{1}{3} \times \frac{1}{2}\) = \(\frac{1}{6}\)

Question 5.
Find 3 × 2\(\frac{1}{4}\)
Answer:
3 × 2\(\frac{1}{4}\) = 3 × (2 + \(\frac{1}{4}\))
= (3 × 2) + (3 × \(\frac{1}{4}\))
= 6 + \(\frac{3}{4}\)
= 6\(\frac{3}{4}\)

Another way,
3 × 2\(\frac{1}{4}\) = 3 × \(\frac{9}{4}\)
= \(\frac{27}{4}\)
= 6\(\frac{3}{4}\)
Consider another problem.

Question 6.
Find 3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
Answer:
3\(\frac{1}{2}\) × 2\(\frac{1}{4}\)
= \(\frac{7}{2} \times \frac{9}{4}\)
= \(\frac{7 \times 9}{2 \times 4}\)
= \(\frac{63}{8}=\frac{56+7}{8}=\frac{56}{8}+\frac{7}{8}\)
= 7\(\frac{7}{8}\)

Question 7.
Find the area of the rectangle having length 5\(\frac{1}{2}\) cm and breadth 3\(\frac{1}{3}\) cm.
Answer:

Now, divide the length 5\(\frac{1}{2}\) cm into 11 equal parts of length \(\frac{1}{2}\) cm.
Divide the breadth 3\(\frac{1}{3}\) cm into 10 equal parts of length \(\frac{1}{3}\) cm.

So, 11 × 10 = 110 rectangles in all, each of area \(\frac{1}{6}\) square centimetres.
∴ Area of rectangle = 110 × \(\frac{1}{6}\) = 18\(\frac{1}{3}\)
Now, 5\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
= \(\frac{11}{2} \times \frac{10}{3}\)
= 11 × \(\frac{1}{2}\) × 10 × \(\frac{1}{3}\)
= 110 × \(\frac{1}{6}\)
= 18\(\frac{1}{3}\)
So, even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Class 7 Maths Chapter 2 Kerala Syllabus Fractions Questions and Answers

Question 1.
Calculate the following:
(i) \(\frac{2}{3}\) of 16
Answer:
\(\frac{2}{3}\) of 16
= \(\frac{2}{3}\) × 16
= \(\frac{2 \times 16}{3}=\frac{32}{3}\)
= 10\(\frac{2}{3}\)

(ii) \(\frac{4}{7}\) of 25
Answer:
\(\frac{4}{7}\) of 25
= \(\frac{4}{7}\) × 25
= \(\frac{4 \times 25}{7}=\frac{100}{7}\)
= 14\(\frac{2}{3}\)

(iii) \(\frac{2}{7}\) of \(\frac{1}{4}\)
Answer:
\(\frac{2}{7}\) of \(\frac{1}{4}\)
= \(\frac{2}{7}\) × \(\frac{1}{4}\)
= \(\frac{2}{7} \times \frac{1}{4}\)
= \(\frac{2 \times 1}{7 \times 4}=\frac{2}{28}\)
= \(\frac{1}{14}\)

(iv) 1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) × 6\(\frac{2}{3}\)
= \(\frac{3}{2} \times \frac{20}{3}=\frac{3 \times 20}{2 \times 3}\)
= \(\frac{60}{6}\)
= 10

(v) 2\(\frac{3}{4}\) × \(\frac{5}{8}\)
Answer:
2\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{11}{4} \times \frac{5}{8}\)
= \(\frac{11 \times 5}{4 \times 8}=\frac{55}{32}\)
= 1\(\frac{23}{32}\)

(vi) \(\frac{4}{7}\) of \(\frac{3}{5}\)
Answer:
\(\frac{4}{7}\) of \(\frac{3}{5}\)
= \(\frac{4}{7}\) × \(\frac{3}{5}\)
= \(\frac{4 \times 3}{7 \times 5}\)
= \(\frac{12}{35}\)

Question 2.
12 metre long rope cut into 4 equal pieces.
(i) What is the length of each piece?
(ii) What if it is cut into 5 equal pieces?
Answer:
(i) Length of each piece = \(\frac{12}{4}\) = 3 metres

(ii) If it is cut into 5 equal pieces,
Length of each piece = \(\frac{12}{5}\) = 2\(\frac{2}{5}\) metres

Question 3.
Each piece of rope is \(\frac{7}{4}\) metres long. What is the total length of 8 such pieces?
Answer:
Lenth of each piece = \(\frac{7}{4}\) metres
Length of 8 pieces = \(\frac{7}{4}\) × 8 = 7 × \(\frac{8}{4}\)
= 7 × 2
= 14 metres

Question 4.
If 4 strings of length \(\frac{1}{3}\) metre were laid end to end, what would be the total length?
Answer:
Total length = \(\frac{1}{3}\) × 4 = \(\frac{4}{3}\)
= 1\(\frac{1}{3}\) metres

Question 5.
Suhara has 1 metre long silk ribbon. She gave half of it to Soumya. She in turn gave half of this to Reena. What is the length of the piece Reena got?
Answer:
Length of the piece Reena got = \(\frac{1}{2}\) of \(\frac{1}{2}\) of 1 metre
= \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) metre

Question 6.
Half the children in a class are girls. A third of them are in the Math Club. What fraction of the total children are they?
Answer:
Number of girls in Math club = \(\frac{1}{3}\) of \(\frac{1}{2}\) of total children
= \(\frac{1}{3} \times \frac{1}{2}\) of total children
= \(\frac{1}{6}\) of total children

Question 7.
The length and breadth of some rectangles are given below. Calculate their areas.
(i) 4\(\frac{1}{2}\) cm, 3\(\frac{1}{4}\) cm
Answer:
Area = 4\(\frac{1}{2}\) × 3\(\frac{1}{4}\)
= \(\frac{9}{2} \times \frac{13}{4}=\frac{9 \times 13}{2 \times 4}\)
= \(\frac{117}{8}\)
= 14\(\frac{5}{8}\) cm²

(ii) 6\(\frac{3}{4}\) cm, 5\(\frac{1}{3}\) cm
Answer:
Area = 6\(\frac{3}{4}\) × 5\(\frac{1}{3}\)
= \(\frac{27}{4} \times \frac{16}{3}\)
= \(\frac{27}{3} \times \frac{16}{4}\)
= 9 × 4
= 36 cm²

Question 8.
What is the area of a square of side 1 metre?
Answer:
Area 1\(\frac{1}{2}\) x 1\(\frac{1}{2}\)
= \(\frac{3}{2} \times \frac{3}{2}\)
= \(\frac{9}{4}\)
= 2\(\frac{1}{4}\) m²

Practice Questions

Question 1.
Each bottle has \(\frac{3}{4}\) litres of water. What is the quantity of water in 14 such bottles?
Answer:
\(\frac{21}{2}\) = 10\(\frac{1}{2}\) litres

Question 2.
A farmer plants the sapling of a plant at a uniform distance of cm. If he plants 27 such saplings in a row, find the total distance between the first and the last sapling.
Answer:
45 cm

Question 3.
In a class of 60 students, two-thirds are boys. How many girls are there in the class?
Answer:
20

Question 4.
The weight of an iron block is \(\frac{3}{4}\) kilogram. What is the total weight of such 17 blocks?
Answer:
\(\frac{51}{4}\) = 12\(\frac{3}{4}\) kilogram

Question 5.
There are some cans, each containing 3 litres of milk. The milk in each vessel is used to fill 5 identical bottles.
(i) How much milk is there in each bottle?
Answer:
\(\frac{3}{5}\) litres

(ii) How much milk in 3 such bottles?
Answer:
\(\frac{9}{5}\) = 1 \(\frac{4}{5}\) litres

(iii) In 10 bottles?
Answer:
6 litres

Question 6.
Calculate the following:
(i) \(\frac{1}{8} \times \frac{1}{5}\)
Answer:
\(\frac{1}{40}\)

(ii) \(\frac{1}{6} \times \frac{1}{7}\)
Answer:
\(\frac{1}{42}\)

(iii) \(\frac{2}{5} \times \frac{7}{9}\)
Answer:
\(\frac{14}{45}\)

(iv) \(\frac{4}{5}\) of \(\frac{2}{3}\)
Answer:
\(\frac{8}{15}\)

(v) \(\frac{2}{9}\) of \(\frac{3}{2}\)
Answer:
\(\frac{1}{3}\)

(vi) 1\(\frac{3}{4}\) of 4
Answer:
7

(vii) \(\frac{3}{8}\) of 2\(\frac{1}{2}\)
Answer:
\(\frac{15}{16}\)

(viii) 3\(\frac{1}{4}\) × 5\(\frac{2}{9}\)
Answer:
\(\frac{611}{36}\)

(ix) 4\(\frac{1}{7}\) × 3\(\frac{1}{8}\)
Answer:
\(\frac{725}{56}\)

Question 7.
If three litres of milk is equally divided among four persons, how much would each get?
Answer:
\(\frac{3}{4}\) litres

Question 8.
Six kilogram of rice is packed into four identical bags.
(i) How much rice is in each bag?
Answer:
\(\frac{3}{2}\) Kilograms

(ii) What if it is packed into two bags?
Answer:
3 Kilograms

Class 7 Maths Chapter 2 Notes Kerala Syllabus Fractions

Fractions are a way to represent parts of a whole. A fraction consists of a numerator and a denominator. The numerator is the number written above the fraction line, which tells how many equal parts you consider, and the denominator is the number written below the fraction line, which tells the total number of equal parts the whole thing has been cut into.

Multifold multiplication
If we have to find 4 times three-quarter, we calculate it in following way:
\(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\) = 3
This can be easily calculated as:
4 × \(\frac{3}{4}=\frac{12}{4}\) = 3

Part and multiplication
If three litres of milk divided among four people, each will get \(\frac{1}{4}\) of 3 litres.
i.e, \(\frac{1}{4}\) of 3 = \(\frac{1}{4}\) × 3 = \(\frac{3}{4}\) litres.

Part of part
If we have to calculate \(\frac{1}{5}\) of \(\frac{1}{4}\), it can be done as as follows:
\(\frac{1}{5} \times \frac{1}{4}=\frac{1}{5 \times 4}=\frac{1}{20}\)
Now, if have to calculate \(\frac{3}{5}\) of \(\frac{2}{7}\),
\(\frac{3}{5} \times \frac{2}{7}=3 \times \frac{1}{5} \times \frac{1}{7} \times 2=\frac{3 \times 2}{5 \times 7}=\frac{6}{35}\)

To calculate 2\(\frac{1}{4}\) × 5,
2\(\frac{1}{4}\) × 5 = (2 + \(\frac{1}{4}\)) × 5
= (2 × 5) + (\(\frac{1}{4}\) × 5)
= 10 + 1\(\frac{1}{4}\)
= 11\(\frac{1}{4}\)

To calculate 1\(\frac{1}{2}\) × 3\(\frac{3}{4}\)
1\(\frac{1}{2}\) × 3\(\frac{3}{4}\) = \(\frac{3}{2} \times \frac{15}{4}\)
= \(\frac{3 \times 15}{2 \times 4}\)
= \(\frac{45}{8}=\frac{40+5}{8}=\frac{40}{8}+\frac{5}{8}\)
= 5 \(\frac{5}{8}\)
Even if the lengths are in fractions, the area of a rectangle is still the product of the lengths of sides.

Share And Fraction
If 4 litres of milk is divided equally among 3 persons, how much milk will each one get? First give 1 litre to each one. If the remaining 1 litre is divided among 3 persons, each will get \(\frac{1}{3}\) litre. So, in total each get 1\(\frac{1}{3}\) litres.
Numerically,
4 ÷ 3 = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 3 Triangles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 3 Solutions Triangles

Class 7 Maths Chapter 3 Triangles Questions and Answers Kerala State Syllabus

Triangles Class 7 Questions and Answers Kerala Syllabus

Page 52

Question 1.
The sides of a triangle are natural numbers. If the lengths of two sides are 5 centimetres and 8 centimetres, what are the possible numbers which can be the length of the third side?
Answer:
We know that the sum of the lengths of any two sides is greater than the length of the third side.
So, 5 + 8 should be greater than the third side.
That is, 13 is greater than the third side.

So, the length of the third side should be less than 13.
Also, we know that the difference of the lengths of any two sides is smaller than the length of the third side.

So, 8 – 5 should be less than the third side.
3 should be lesser than the third side.

So, the length of the third side should be something greater than 3.
The natural numbers greater than 3 and lesser than 13 are; 4, 5, 6, 7, 8, 9, 10, 11 and 12.
These are the possible numbers that can be the length of the third side.

Question 2.
The lengths of the sides of a triangle are all natural numbers and two of the sides are 1 centimetre and 99 centimetres. What is the length of the third side?
Answer:
We know that the sum of the lengths of any two sides is greater than the length of the third side.
So, 1 + 99 should be greater than the third side.
That is, 100 is greater than the third side.
So, the length of the third side should be less than 100.
Also, we know that the difference of the lengths of any two sides is smaller than the length of the third side.
So, 99 – 1 should be less than the third side.
98 should be less than the third side.
So, the length of the third side should be something greater than 98.
The natural number greater than 98 and lesser than 100 is 99.
Thus the length of the third side is 99 cm.

Question 3.
Which of the following sets of three lengths can be used to draw a triangle?
(i) 4 centimetres, 6 centimetres, 10 centimetres
(ii) 3 centimetres, 4 centimetres, 5 centimetres
(iii) 10 centimetres, 5 centimetres, 4 centimetres
Answer:
(i) 4 + 6 = 10, which is not greater than the third side. So, it is not possible to draw a triangle with these lengths.
(ii) Here, the sum of any two sides is greater than the third side. So, we can draw a triangle with this set of lengths.
(iii) 5 + 4 = 9, which is not greater than the third side. So, it is not possible to draw a triangle with these lengths.

Question 4.
Draw these pictures:
(i)

Answer:
Steps to draw:

• Draw a square of side length 4 cm. (You can use the compass and the protractor)
• Draw the triangles of length 6 cm, 6 cm and 4 cm on all four sides of the square. (Measure 6 cm on your compass and draw an arc from each end of the sides of the square. join the meeting point of the arc’ to both ends of the sides)
• Join the third vertices of all the triangles.

(ii)

Answer:
Steps to draw:

• Draw a triangle of sides 12 cm, 10 cm, and 8 cm.
• Draw a triangle of sides 6 cm, 5 cm, and 4 cm. (Make sure that the midpoint of the side of length 6 cm is the upper vertex of the first triangle)
• Draw a triangle of sides 3 cm, 2.5 cm, and 2 cm. (Make sure that the midpoint of the side of length 3 cm is the upper vertex of the second triangle)
• Draw a line passing through the left vertices of all the triangles.
• Draw a line passing through the right vertices of all the triangles.

Intext Questions and Answers

Question 1.
Try to draw the pictures below:
(i)

Answer:
Step 1: Draw a triangle with equal side lengths as discussed above. (You are free to choose the side length)

Step 2: Extend the bottom side of the triangle to the right.

Step 3: Draw an equilateral triangle with this new line as the base.

Step 4: Draw the third triangle as shown in the figure.

Step 5: Delete the circles.

(ii)

Step 1: Draw a triangle with equal side lengths as discussed. (You are free to choose the side length)

Step 2: Draw a line of length the same as the side length of the triangle as shown in the figure.

Step 3: Draw an equilateral triangle with this new line as the base.

Step 4: Draw a line of length the same as the side length of the triangle as shown in the figure.

Step 5: Draw an equilateral triangle with this new line as the base.

Step 6: Delete the circles.

Question 2.
How do we draw a triangle with all its sides are equal?
Answer:
Let’s do it with the help of an example. Suppose we want to draw a triangle with all its sides equal to 3 cm. Following are the steps to draw it.
Step 1: Draw a line of length 3 cm

Step 2: Draw a circle with a radius of 3 cm (length of the given line) from the right end of the line.

Step 3: Draw a circle with a radius of 3 cm (length of the given line) from the left end of the. line.

Step 4: Both the circles intersect at two points. Choose any of them as the third vertex and complete the triangle.

Step 5: Delete the circles. Thus, we get the required triangle.

Question 3.
Draw the following triangle using circles.

Answer:
Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Question 4.
Can you draw these pictures by joining more and more small triangles?

Answer:
Draw two equilateral triangles of the same size as shown below.

Join all the outer vertices.

Draw the lines as shown.

Erase the unnecessary lines.

Consider the marked points.

Join the opposite points.

Shade it as shown in the text.

Question 5.
Draw the following figures using the methods we have used in this lesson.

Answer:
(i) Step 1:

Step 2:

Step 3:

Step 4:

(ii) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

(iii) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

Step 7:

(iv) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

Class 7 Maths Chapter 3 Kerala Syllabus Triangles Questions and Answers

Question 1.
Which one of the following measures can be the sides of a triangle
(a) 3 cm, 4 cm, 7 cm
(b) 6 cm, 7 cm, 14 cm
(c) 4 cm, 5 cm, 10 cm
(d) 4 cm, 5 cm, 8 cm
Answer:
a) 3+4=7. It is not greater than the third side. So, it is not possible to draw such a triangle.
b) 6+7=13. It is not greater than the third side. So, it is not possible to draw such a triangle.
c) 5+49. It is not greater than the third side. So, it is not possible to draw such a triangle.
d) The sum of the lengths of any two sides is greater than the third side. So, it is possible to draw such a triangle.

Question 2.
Zeenath doing a project to find out the measures which can be used to draw a triangle. Which among the following measures can be used to draw a triangle?
(a) 8 cm 4 cm 3 cm
(b) 9 cm 5 cm 12 cm
(c) 5 cm 5 cm 10 cm
(d) 6 cm 7 cm 15 cm
Answer:
a) 4+37. It is not greater than the third side. So, it is not possible to draw such a triangle.
b) The sum of the lengths of any two sides is greater than the third side. So, it is possible to draw such a triangle.
c) 5+5 10. It is not greater than the third side. So, it is not possible to draw such a triangle.
d) 6+7=13. It is not greater than the third side. So, it is not possible to draw such a triangle.

Question 3.
Can a triangle have sides with lengths 6 cm, 5 cm and 9 cm?
Answer:
6 + 5 = 11
6 + 9 = 15
5 + 9 = 14
Here, sum of any two sides is greater than the third side.
Thus, a triangle can have sides with lengths 6 cm, 5 cm and 9 cm.

Question 4.
Is it possible to have a triangle with the following sides?
(i) 3 cm, 4 cm and 7 cm.
(ii) 7 cm, 7 cm and 7 cm.
(iii) 2 cm, 4 cm and 2 cm.
(iv) 3 cm, 5 cm and 7 cm.
Answer:
(i) 3 + 4 = 7, which is not greater than the third side. So, triangle is not possible.
(ii) All sides equal means equilateral triangle. An equilateral triangle can be of any length. So, triangle is possible.
(iii) 2 + 2 = 4, which is not greater than the third side. So, triangle is not possible.
(iv) 3 + 5 = 8
3 + 7 = 10
5 + 7 = 12
Here, sum of any two sides is greater than the third side. So, triangle is possible.

Question 4.
Draw the right triangle whose sides are 4 cm, 12 cm and 10 cm.
Answer:

Class 7 Maths Chapter 3 Notes Kerala Syllabus Triangles

We are already familiar with the concept of triangles. This chapter introduces a new idea called an equilateral triangle. We will also discuss how to draw different triangles using circles.
Following are the main topics discussed in this chapter.

Steps to draw an equilateral triangle using circles

• Draw a line of the given length
• Draw a circle with the same radius from the right end of the line.
• Draw a circle with the same radius from the left end of the line.
• Choose the intersecting point as the third vertex of the triangle

Steps to draw a non-equilateral triangle using circles

• Draw a line of length same as the first side of the triangle.
• Draw a semicircle with the length of the second side as the radius from one end of the line.
• Draw a semicircle with the length of the third side as the radius radius from other end of the line.
• Choose the intersecting point of these two circles as the third vertex of the triangle.

Relation between sides of the triangles
In any triangle;
the sum of the lengths of any two sides greater than the length of the third side.
the difference of the lengths of any two sides is smaller than the length of the third side.

Other ideas
A triangle having all three sides of the same length is called an equilateral triangle.
Sum of all the angles in a triangle is 180°
If only two angles are given we can draw more than one triangle with these.
If the length of two sides and the angle between them is specified, then there is only one triangle with this measure.

Lines And Math
Triangles with all sides equal are called equilateral triangles.
Eg:

In an equilateral triangles, all angles are equal to 60°

How can we draw triangles with unequal sides using circles?
Step 1: Draw a line of length same as the first side of the triangle.
Step 2: Draw a semicircle with the length of the second side as the radius from one end of the first line.
Step 3: Draw a semicircle with the length of the third side as the radius from the other end of the line.
Step 4: Choose the intersecting point of these two semi circles as the third vertex draw a triangle.

Relation between the length of sides of a triangle.
In any triangle;
the sum of the lengths of any two sides is greater than the length of the third side.
the difference of the lengths of any two sides is smaller than the length of the third side.

Angle Math
Sum of all the angles in a triangle is 180°
The total measure of the three angles of a triangle is 180°. This is called the angle sum property of a traingles

We can draw several triangles with two specified angles.
Eg:

To draw a specific triangle, we have to specify not only two angles, but the length of the side on which they stand also.
To draw more than one triangle with the same angles it is enough to draw lines parallel to the sides of the first triangle.
Eg:

Sides And Angles
If we specify the lengths of two sides of a triangle and the angle between them, we have the triangle. In a triangle with angles 30°, 60°, and 90°, the larger side is two times the smaller sides.
Eg:

How do we draw a triangle with all its sides are equal?

• Triangles with all sides equal are called equilateral triangles.
• How do we draw triangles with unequal sides using circles?
• In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Relation between the length of sides of a triangle.

In any triangle;

• the sum of the lengths of any two sides is greater than the length of the third side.
• the difference of the lengths of any two sides is smaller than the length of the third side.
• Sum of all the angles in a triangle is 180°
• We can draw several triangles with two specified angles.
• To draw a specific triangle, we have to specify not only two angles, but the length of the side on which they stand also.
• If we specify the lengths of two sides of a triangle and the angle between them, we have the triangle.
  In a triangle with angles 30°, 60°, and 90o, the larger side is two times the smaller sides.

Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 5 Decimal Methods Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 7 Maths Chapter 5 Solutions Decimal Methods

Class 7 Maths Chapter 5 Decimal Methods Questions and Answers Kerala State Syllabus

Decimal Methods Class 7 Questions and Answers Kerala Syllabus

Page 71

Question 1.
Write the following numbers as fractions.
(i) 45.6
(ii) 45.06
(iii) 45.67
(iv) 4.506
(v) 456.07
Answer:
(i) Here, the numerator is 456.
There is only one digit after the decimal point. So, the denominator is 10. Thus, the fractional form is \(\frac{456}{10}\)

(ii) Here, the numerator is 4506.
There are two digits after the decimal point. So, the denominator is 100. Thus, the fractional form is \(\frac{4506}{100}\)

(iii) 45.67 = \(\frac{4567}{100}\)
(iv) 4.506 = \(\frac{4506}{1000}\)
(v) 456.07 = \(\frac{45607}{100}\)

Page 73

Question 1.
The figure below shows a regular pentagon.

Find the perimeter of the pentagon.
Answer:
Length of a side = 2.35 m
Number of sides = 5
Perimeter = Number of sides × Length of a side
= 5 × 2.35
= 5 × \(\frac{235}{100}\)
= \(\frac{1175}{100}\)
= 11.75 metres

Question 2.
A kid needs 1.45 metres of cloth for a shirt. How many metres of cloth is needed for 4 shirts?
Answer:
Amount of cloth needed for one shirt = 1.45 m
Number of shirts = 4
The amount of cloth needed for 4 shirts.
= 4 × Amount of cloth for one shirt
= 4 × 1.45
= 4 × \(\frac{145}{100}\)
= \(\frac{580}{100}\)
= 5.80 m

Question 3.
A bag holds 4.75 kilograms of rice. How much rice can 8 such bags hold?
Answer:
Amount of rice in one bag = 4.75 kg
Amount of rice in 8 bags = 8 × amount of rice in one bag
= 8 × 4.75
= 8 × \(\frac{475}{100}\)
= \(\frac{8 \times 475}{100}\)
= \(\frac{3800}{100}\)
= 38 kg

Question 4.
A vessel full of oil was used to fill in 6 bottles. Each bottle holds 0.75 litres. How much oil was there in the vessel?
Answer:
Amount of oil in one bottle = 0.75 litres
Amount of oil in the vessel = Amount of oil in six bottles
= = 6 × amount of oil in one bottle
= 6 × 0.75
= 6 × \(\frac{75}{100}\)
= \(\frac{6 \times 75}{100}\)
= \(\frac{450}{100}\)
= 4.50 litres

Page 74

Question 1.
Find the area in square metres of a rectangle of length 6.25 metres and width 4.2 metres.
Answer:
length = 6.25 m
width = 4.2 m
Area = length × width
= 6.25 × 4.2
= \(\frac{625}{100} \times \frac{42}{10}\)
= \(\frac{26250}{1000}\)
= 26.250 square metres

Question 2.
The weight of 1 millilitre of coconut oil is 0.91 grams. What is the weight of 10.5 millilitres of coconut oil?
Answer:
Weight of 1 millilitre of coconut oil = 0.91 g
Weight of 10.5 millilitres of coconut oil
= 10.5 × Weight of 1 millilitre of coconut oil
= 10.5 × 0.91
= \(\frac{9555}{1000}\)
= 9.555 g

Question 3.
The price of 1 litre of petrol is 110.12 rupees. What is the price of 2.5 litres of petrol?
Answer:
Price of 1 litre of petrol = 110.12 rupees
Price of 2.5 litres of petrol = 2.5 × Price of 1 litre of petrol
= 2.5 × 110.12
= \(\frac{25}{10} \times \frac{11012}{100}\)
= \(\frac{275300}{1000}\)
= 275.300 rupees

Page 76

Question 1.
Given that 1234 × 56 = 69104
Answer:
(i) Find the answers to the following problems without actual multiplication.
(i) 1.234 × 56
(ii) 12.34 × 5.6
(iii)123.4 × 0.56
(iv) 1234 × 0.056
Answer:
(i) 1.234 × 56
= 69.104

(ii) 12.34 × 5.6
= 69.104

(iii) 123.4 × 0.56
= 69.104

(iv) 1234 × 0.056
= 69.104

(ii) Like this, how many products can you find which gives 6.9104?
Answer:
0.1234 × 56 = 6.9104
1.234 × 5.6 = 6.9104
12.34 × 0.56 = 6.9104
123.4 × 0.056 = 6.9104
1234 × 0.0056 = 6.9104
So, 5 products gives 6.9104

Question 2.
In the following products, how many of them give the same product as 1.234 × 5.67?
(i) 12.34 × 0.567
(ii) 1.234 × 567
(iii) 0.1234 × 5.67
(iv) 1.234 × 56.7
(v) 123.4 × 0.0567
Answer:
1.234 × 5.67 has 5 digits after the decimal point.
(i) number of digits after the decimal point in 12.34 is 2. number of digits after the decimal point in 0.567 is 3.
Thus, the number of digits after the decimal point in the product = 2 + 3
= 5
∴same.

(ii) number of digits after the decimal point in the product = 3 + 0
= 3
∴ different.

(iii)number of digits after the decimal point in the product = 4 + 2
= 6
∴ different.

(iv)number of digits after the decimal point in the product = 3 + 1
= 4
∴ different.

(v) number of digits after the decimal point in the product = 1 + 4
= 5
∴ same.

Question 3.
Find the greatest and least products from the following:
(i) 0.11 × 0.11
(ii) 1.1 × 1.1
(iii) 1.01 × 1.01
(iv) 0.101 × 1.1
(v) 10.1 × 0.101
Answer:
(i) 0.11 × 0.11 = \(\frac{11}{100} \times \frac{11}{100}\)
= \(\frac{121}{10000}\)
= 0.0121

(ii) 1.1 × 1.1 = \(\frac{11}{10} \times \frac{11}{10}\)
= \(\frac{121}{100}\)
= 1.21

(iii) 1.01 × 1.01 = \(\frac{101}{100} \times \frac{101}{100}\)
= \(\frac{10201}{10000}\)
= 0.1111

(iv) 0.101 × 1.1 = \(\frac{101}{1000} \times \frac{11}{10}\)
= \(\frac{1111}{10000}\)
= 1.0201

(v) 10.1 × 0.101 = \(\frac{101}{10} \times \frac{101}{1000}\)
= \(\frac{10201}{10000}\)
= 1.0201
Greatest value = 1.21 ⇒ greatest product = 1.1 × 1.1
Least value = 0.0121 ⇒ least product 0.11 × 0.11

Page 78

Question 1.
The perimeter of an equilateral triangle is 12.9 centimetres. What is the length of each side?
Answer:
Perimerter = 12.9 cm
Length of one side =Perimeter ÷ number of sides
= 12.9 ÷ 3
= \(\frac{129}{10} \times \frac{1}{3}\)
= \(\frac{129}{3} \times \frac{1}{10}\)
= 43 × \(\frac{1}{10}\)
= 4.3 cm

Question 2.
16.5 kilograms of rice was divided equally among 5 people. How many kilograms did each get?
Answer:
Total amount of rice = 16.5 kg
Amount of rice each gets = total rice ÷ number of people
= 16.5 ÷ 5
= \(\frac{165}{10} \times \frac{1}{5}\)
= \(\frac{165}{5} \times \frac{1}{10}\)
= 33 × \(\frac{1}{10}\)
= 3.3 kg

Question 3.
A large vessel contains 25.2 litres of coconut oil. It was used to fill 6 small vessels of the same size. How much does each small vessel contain?
Answer:
Amount of coconut oil in the large vessel = 25.2 litres
Amount of coconut oil in each of the small vessel
= oil in large vessel ÷ number of small vessels
= 25.2 ÷ 6
= \(\frac{252}{10} \times \frac{1}{6}\)
= \(\frac{252}{6} \times \frac{1}{10}\)
= 42 × \(\frac{1}{10}\)
= 4.2 litres

Question 4.
33.6 kilograms of rice was divided equally among 8 people. Sujatha divided what she got into three equal parts and gave one part to Razia. How much did Razia get?
Answer:
Total rice = 33.6 kg
Amount of rice Sujatha gets = total rice ÷ 8
= 33.6 ÷ 8
= \(\frac{336}{10} \times \frac{1}{8}\)
= \(\frac{336}{8} \times \frac{1}{10}\)
= 42 × \(\frac{1}{10}\)
= 4.2 kg

Amount of rice Razia gets = Amount of rice Sujatha gets ÷ 3
= 4.2 ÷ 3
= \(\frac{42}{10} \times \frac{1}{3}\)
= \(\frac{42}{3} \times \frac{1}{10}\)
= 14 × \(\frac{1}{10}\)
= 1.4 kg

Question 5.
We have 7407 ÷ 6 = 1234.5
Use this result to find answers to the following questions without actual division:
(i) 740.7 ÷ 6
(ii) 74.07 ÷ 6
(iii) 7.407 ÷ 6
Answer:
(i) 740.7 ÷ 6 = \(\frac{7407}{10} \times \frac{1}{6}\)
= \(\frac{7407}{6} \times \frac{1}{10}\)
= 1234.5 × \(\frac{1}{10}\)
= 123.45

OR

Here, the decimal point in the numerator is shifted one place to the left. So, the decimal point in the answer moves one place to the left.
∴ 740.7 ÷ 6 = 123.45

(ii) Here, the decimal point in the numerator is shifted two places to the left. So, the decimal point in the answer moves two places to the left.
∴ 74.07 ÷ 6 = 12.345

(iii)Here, the decimal point in the numerator is shifted two places to the left. So, the decimal point in the answer moves three places to the left.
∴ 7.407 ÷ 6 = 1.2345

Page 80

Question 1.
Find the decimal forms of the following fractions.
(i) \(\frac{3}{5}\)
Answer:
\(\frac{3}{5}=\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)
= 0.6

(ii) \(\frac{4}{5}\)
Answer:
\(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\)
= 0.8

(iii) \(\frac{1}{20}\)
Answer:
\(\frac{1}{20}=\frac{1 \times 5}{20 \times 5}=\frac{5}{100}\)

(iv) \(\frac{7}{8}\)
Answer:
\(\frac{7}{8}=\frac{7 \times 125}{8 \times 125}=\frac{875}{1000}\)
=0.875

Question 2.
3 litres of milk is used to fill in 8 bottles of the same size. How many litres does each bottle hold?
Answer:
Total milk = 3 litres
Number of bottles = 8.
Amount of milk in one bottle = \(\frac{\text { Total milk }}{\text { number of bottles }}\)
= \(\frac{3}{8}\)
= \(\frac{3 \times 125}{8 \times 125}\)
= \(\frac{375}{1000}\)
= 0.375 litres

Question 3.
A rope 17 metres long is cut into 25 equal pieces. What is the length of each piece in metres?
Answer:
Total length = 17 m
Number of pieces = 25
Length of each piece = \(\frac{\text { Total length }}{\text { number of pieces }}\)
= \(\frac{17}{25}\)
= \(\frac{17 \times 4}{25 \times 4}\)
= \(\frac{68}{100}\)
= 0.68 m

Question 4.
19 kilograms of rice was equally divided among 20 people. How much kilograms did each get?
Answer:
Total rice = 19 kg
Number of peoples = 20
Amount of rice each get = \(\frac{\text { Total rice }}{\text { number of people }}\)
= \(\frac{19}{20}\)
= \(\frac{19 \times 5}{20 \times 5}\)
= \(\frac{95}{100}\)
= 0.95 kg

Page – 81

Question 1.
A ribbon 14.5 centimetres long is cut into two equal pieces. What is the length of each piece in centimetres?
Answer:
Length of the ribbon = 14.5 cm
Number of pieces = 2
Length of each piece = \(\frac{1}{\text { Number of pieces }}\) × Length of the ribbon

Question 2.
What is the length of a side of a square of perimeter 20.5 metres?
Answer:
Perimeter of the square = 20.5 m
Length of a side = \(\frac{1}{4}\) × 20.5

Question 3.
The price of 6 pens is 40.50 rupees. What is the price of one pen?
Answer:
Price of 6 pens = 40.50
Price of one pen = \(\frac{1}{6}\) × 40.50

Page 82

Question 1.
A vessel contains 4.05 litres of coconut oil. It is to be used to fill 0.45 litre bottles. How many bottles are needed?
Answer:
Total oil = 4.05 = \(\frac{405}{100}\) litres
Capacity of one bottle = 0.45 = \(\frac{45}{100}\) litres
Number of bottles =
Here, \(\frac{45}{100}\) is the dividing fraction. Its reciprocal is \(\frac{100}{45}\)
Number of bottles = \(\frac{405}{100} \times \frac{100}{45}\)
= \(\frac{405}{45}\)
= \(\frac{81 \times 5}{9 \times 5}\)
= \(\frac{81}{9}\)
= 9

Question 2.
An iron rod 17.5 metres long is cut into pieces of length 2.5 metres each. How many pieces are there?
Answer:
Length of the iron rod = 17.5 = \(\frac{175}{10}\)m
Length of one piece = 2.5 = \(\frac{25}{10}\) m
Number of pieces = \(\frac{\text { Length of the iron rod }}{\text { Length of one piece }}\)
Here, \(\frac{25}{10}\) is the dividing fraction. Its reciprocal is \(\frac{10}{25}\)
Therefore, number of pieces = \(\frac{175}{10} \times \frac{10}{25}\)
= \(\frac{175}{25}\)
= 7

Question 3.
6.5 kilograms of chilli powder was packed in 0.25 kilogram packets. How many packets are there?
Answer:
Total chilli powder = 6.5 = \(\frac{65}{10}\)kg
Amount of chilli powder in a small packet = 0.25 = \(\frac{25}{100}\) kg
Number of small packets = \(\frac{\text { Total chilli powder }}{\text { Amount of chilli powder in a small packet }}\)
Here, \(\frac{25}{100}\) is the dividing fraction. Its reciprocal is \(\frac{100}{25}\)
Therefore, the number of packets = \(\frac{65}{10} \times \frac{100}{25}\)
= \(\frac{650}{25}\)
= 26

Intext Questions And Answers

Question 1.
Write the decimal forms of the following fractions.
(i) \(\frac{325}{10}\)
Answer:
\(\frac{325}{10}\)
= 32.5

(ii) \(\frac{325}{100}\)
Answer:
\(\frac{325}{100}\)
= 3.25

(iii) \(\frac{325}{1000}\)
Answer:
\(\frac{325}{1000}\)
= 0.325

Question 2.
Complete the following table.

Express these in decimal forms also.
\(\frac{476}{10}\) =
\(\frac{476}{100}\) =
\(\frac{476}{1000}\) =
\(\frac{476}{10000}\)
Answer:

In decimal form;
\(\frac{476}{10}\)
= 47.6

\(\frac{476}{100}\)
= 4.76

\(\frac{476}{1000}\)
= 0.476

\(\frac{476}{10000}\)
= 0.0476

Question 3.
What is the fractional form of 327.045?
Answer:
Here, the numerator is 327045.
In the given decimal number,we have three digits after the decimal point. So, the denominator is 1000.
Thus, the fractional form is \(\frac{327045}{1000}\)

Class 7 Maths Chapter 5 Kerala Syllabus Decimal Methods Questions and Answers

Question 1.
What is the fraction form of 3.05?
Answer:
3.05 = \(\frac{305}{100}\)

Question 2.
\(\frac{1.234}{0.01234}\) = ?
Answer:

Question 3.
\(\frac{2.3 \times 3.2}{0.4}\) = ?
Answer:

Question 4.
\(\frac{0.013 \times 0.013}{0.0169}\) = ?
Answer:

Question 5.
If \(\frac{37}{1.5} \times \frac{1.2}{4}\) = 7.4, What is \(\frac{3.7}{0.15} \times \frac{12}{0.4}\)
Answer:
In both the cases the total number of decimal places in the numerator is same.
In the first case the total number of decimal places in the denominator is one.
In the second case the total number of decimal places in the denominator is three.
The increase in the decimal places in the denominator is two.
So, the new answer is; old answer × 100. That is,
7.4 × 100 = 740

Practice Questions

Question 1.
If the breadth of a rectangle is 2.5 metres and the area is 5.6 square metres, what is its length?
Answer:
2.24 m

Question 2.
What is the decimal form of \(\frac{3.2}{16}\)
Answer:
0.2

Question 3.
If the length of a box is 1.2 metres, breadth is 0.8 m and heigtht is 0.2 m, what is the volume of that box?
Answer:
0.192 cubic metres

Question 4.
The total weight of a box containing 5 detergent packets of the same size is the weight of each packet?
Answer:
1.68 kg

Question 5.
How many packets of \(\frac{1}{12}\) kg salt can be made from \(\frac{15}{2}\) kg of salt?
Answer:
90

Class 7 Maths Chapter 5 Notes Kerala Syllabus Decimal Methods

Decimal numbers are one of the main and commonly used concepts in mathematics. This chapter deals with some of the main ideas of decimal numbers. Following are the concepts discussed in this chapter.

Converting a fraction to a decimal

• To convert a fraction with denominator 10 and numerator a natural number to a decimal, put the decimal point just before the last digit of the numerator.
• To convert a fraction with denominator 100 and numerator a natural number to a decimal, put the decimal point just before the second last digit of the numerator.
• To convert a fraction with denominator 1000 and numerator a natural number to a decimal, put the decimal point just before the third last digit of the numerator.

In other cases, to convert a fraction with numerator a natural number to a decimal, first convert the denominator of the fraction to any of the forms 10,100, 1000, … by multiplying both the numerator and the denominator by a suitable number. Then, write the decimal form of that fraction.

Converting a decimal to a fraction
In this case, the numerator will be the given number itself (ignore the decimal point).
To write the denominator, count the number of digits after the decimal point.
If there is only one digit after the decimal point, then write 10 in the denominator.
If there are two digits after the decimal point, then write 100 in the denominator.
If there are three digits after the decimal point, then write 1000 in the denominator. And so on.

Multiplication of a decimal with a natural number
To multiply a decimal with a natural number, first convert the decimal to its fractional form and then multiply with the natural number.

Multiplication of two decimals
To multiply two decimal numbers, first of all, convert the two decimals to fractions then take their products.
The number of digits after the decimal point in the product is the sum of the number of digits after the decimal point in the numbers being multiplied.

Dividing a decimal by a natural number
To divide a decimal by a natural number, first write the decimal as a fraction and then divide it by the natural number.

Dividing two decimals
To divide two decimals, first convert both the decimals to fractions. Then, take the reciprocal of the dividing fraction. Now multiply the fractions and then write the decimal form of the resulting fraction.

(i) How to convert a fraction with denominator 10 and numerator a natural number to a decimal?
To convert a fraction with denominator 10 to a decimal, put the decimal point just before the last digit of the numerator.
Eg:
\(\frac{157}{10}\) = 15.7
\(\frac{15}{10}\) = 1.5
\(\frac{5}{10}\) = 0.5

(ii) How to convert a fraction with denominator 100 and numerator a natural number to a decimal?
To convert a fraction with denominator 100 to a decimal, put the decimal point just before the second last digit of the numerator.
Eg:
\(\frac{482}{100}\) = 4.82, \(\frac{48}{100}\) = 0.48, \(\frac{4}{100}\) = 0.04

(iii) How to convert a fraction with denominator 1000 and numerator a natural number to a decimal?
To convert a fraction with denominator 1000 to a decimal, put the decimal point just before the third last digit of the numerator.
Eg:
\(\frac{690}{1000}\) = 0.690, \(\frac{69}{1000}\) = 0.069, \(\frac{6}{1000}\) = 0.006

What is the reason for the change in the position of the decimal point when we divide a number each time by 10?
It is because each time we divide a number by 10 (or multiply by the places shift one place to the right. As a result, the decimal point moves one point to the left when we divide a number each time by 10.
Eg:
Consider the number 498. According to the place value, this number can be written as follows:

How to convert a decimal to a fraction?
The numerator will be the given number without the decimal point.
To write the denominator, count the number of digits after the decimal point.
If there is only one digit after the decimal point, then write 10 in the denominator.
If there are two digits after the decimal point, then write 100 in the denominator.
If there are three digits after the decimal point, then write 1000 in the denominator. And so on.

Multiples
How to multiply a decimal with a natural number?
To multiply a decimal with a natural number, first convert the decimal to its fractional form and then multiply with the natural number.
Eg:
5 × 1.35 = 5 × \(\frac{135}{100}\)
= \(\frac{675}{100}\)
= 6.75

Decimal Multiplication
How to multiply two decimal numbers?
First of all, convert the two decimals to fractions then take their products. Then write the decimal form of this product.
Eg:
0.15 × 3.2 = \(\frac{15}{100} \times \frac{32}{10}\)
= \(=\frac{15 \times 32}{100 \times 10}\)
= \(\frac{480}{1000}\)
= 0.480

Multiplication Operations
The number of digits after the decimal point in the product is the sum of the number of digits after the decimal point in the numbers being multiplied.
OR
The number of decimal places in the product is equal to the sum of the decimal places in the numbers being multiplied.
Eg:
We can find that 3.14 × 1.2 = 3.768
Here,
number of digits after the decimal point in 3.14 is 2.
number of digits after the decimal point in 1.2 is 1.
number of digits after the decimal point in the product 3.768
= 3
= 2 + 1
= number of digits after the decimal point in 3.14
+ number of digits after the decimal point in 1.2

The number of digits after the decimal point is exactly same as the number of zeroes in the denominator of the fractional form. (Only if the denominator is 10, 100, 1000, …)
Eg:

Parts
How to divide a decimal by a natural number?
To divide a decimal by a natural number, first write the decimal as a fraction and then divide it by the natural number. Then write the decimal form of the result.
Example 1:
10.2 ÷ 2 = \(\frac{102}{10}\) ÷ 2
= \(\frac{102}{10} \times \frac{1}{2}\)
= \(\frac{102}{2} \times \frac{1}{10}\)
= 51 × \(\frac{1}{10}\)
= 5.1

Example 2:

23.2 ÷ 4 = \(\frac{1}{4}\) × 23.2
= \(\frac{1}{4} \times \frac{232}{10}\)
= \(\frac{232}{40}\)
= \(\frac{58 \times 4}{10 \times 4}\)
= \(\frac{58}{10}\)
= 5.8

Fraction And Decimal
We get various forms of a fraction by multiplying the numerator and the denominator by the same number.
Eg:
Consider the fraction \(\frac{1}{2}\)
When we multiply both the numerator and the denominator by 5 it becomes;
\(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\)
When we multiply both the numerator and the denominator by 8 it becomes;
\(\frac{1}{2}=\frac{1 \times 8}{2 \times 8}=\frac{8}{16}\)
Thus, \(\frac{5}{10}\) and \(\frac{8}{16}\) are different forms of the fraction \(\frac{1}{2}\)

How to convert a fraction to a decimal?
To convert a fraction to a decimal, first convert the denominator of the fraction to any of the forms 10, 100, 1000, by multiplying both the numerator and the denominator by a suitable number. Then, write the decimal form of that fraction.
Eg:
Decimal form of \(\frac{1}{4}=\frac{1 \times 25}{4 \times 25}=\frac{25}{100}\) = 0.25
Decimal form of \(\frac{1}{8}=\frac{1 \times 125}{8 \times 125}=\frac{125}{1000}\) = 0.125

Decimal Division
How to divide two decimals?
To divide two decimals first convert both the decimals to fractions. Then take the reciprocal of the dividing fraction. Now multiply the fractions and then write the decimal form of the resulting fraction.
Eg:
3.25 ÷ 2.5 ?
3.25 =
2.5 =
Dividing fraction is \(\frac{25}{10}\).Its reciprocal is \(\frac{10}{25}\)
3.25 ÷ 2.5 = \(\frac{325}{100} \times \frac{10}{25}\)
= \(\frac{325}{10} \times \frac{1}{25}\)
= \(\frac{325}{10} \times \frac{4}{100}\)
= \(\frac{1300}{1000}\)
= 1.300

Expert Teachers at HSSLive.Guru have created Kerala Syllabus 5th Standard Textbooks Solutions Notes Pdf Free Download all Subjects in English Medium and Malayalam Medium of Chapter wise Questions and Answers, SCERT Textbooks for Class 5 Solutions. Here we have given SCERT Solutions for Class 5 Part 1 and Part 2.

SCERT Textbooks for Class 5 Solutions

SCERT Textbooks Solutions for Class 5

SCERT Solutions for Class 5

• SCERT Class 5 Maths Solutions
• Kerala Syllabus 5th Standard Basic Science Notes Pdf
• Kerala Syllabus 5th Standard Social Science Notes Pdf
• Kerala Syllabus 5th Standard English Textbook Solutions
• Kerala Syllabus 5th Standard Hindi Textbook Solutions
• Class 5 Malayalam Kerala Padavali Notes Pdf
• Class 5 Malayalam Adisthana Padavali Notes Pdf

We hope the given SCERT Textbooks Solutions for Class 5 Pdf Free Download all Subjects in English Medium and Malayalam Medium of Kerala Syllabus 5th Standard Notes will help you. If you have any queries regarding SCERT Textbooks for Class 5 Solutions of Kerala Syllabus 5th Standard Textbook Solutions Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 7th Standard Social Science Notes Pdf Download English Medium and Malayalam Medium of SCERT Class 7 Social Science Textbook Solutions Notes Pdf, Class 7 Social Science Notes Kerala Syllabus, are part of Kerala Syllabus 7th Standard Textbooks Solutions. Here we have given 7th Standard Social Science Notes Question Answer Pdf of Class 7 Social Science Notes Pdf Kerala State Syllabus Part 1 and Part 2.

Class 7 Social Science Notes Kerala Syllabus

SCERT Class 7 Social Science Textbook Solutions

Class 7 Social Science Notes Kerala Syllabus English Medium Part 1

• Chapter 1 Medieval India Class 7 Notes
• Chapter 2 Medieval India: Cultural Movements Class 7 Notes
• Chapter 3 Constitution: Path and Guiding Light Class 7 Notes
• Chapter 4 From Injustice to Justice Class 7 Notes
• Chapter 5 Our Earth Class 7 Notes
• Chapter 6 Indian Subcontinent Class 7 Notes
• Chapter 7 From Food Production to Food Security Class 7 Notes

SCERT Class 7 Social Science Solutions Notes Pdf Part 2

• Chapter 8 Power to the People Class 7 Notes
• Chapter 9 Maps and Technology to Know the Earth Class 7 Notes
• Chapter 10 Budget: The True Record of Development Class 7 Notes
• Chapter 11 Against Discrimination Class 7 Notes
• Chapter 12 The Foundation Stones of History Class 7 Notes

7th Standard Social Science Notes Pdf Malayalam Medium

Std 7 Social Science Malayalam Medium Notes Pdf Part 1

• Chapter 1 മധ്യകാല ഇന്ത്യ Notes
• Chapter 2 മധ്യകാല ഇന്ത്യ: സാംസ്കാരിക പ്രസ്ഥാനങ്ങൾ Notes
• Chapter 3 ഭരണഘടന: വഴിയും വഴികാട്ടിയും Notes
• Chapter 4 അനീതിയിൽ നിന്ന് നീതിയിലേക്ക് Notes
• Chapter 5 നമ്മുടെ ഭൂമി Notes
• Chapter 6 ഇന്ത്യൻ ഉപഭൂഖണ്ഡം Notes
• Chapter 7 ഭക്ഷ്യോൽപാദനം മുതൽ ഭക്ഷ്യസുരക്ഷ വരെ Notes

Class 7 Social Science Notes Kerala Syllabus Malayalam Medium Part 2

• Chapter 8 അധികാരം ജനങ്ങൾക്ക് Notes
• Chapter 9 ഭൂമിയെ അറിയാൻ ഭൂപടങ്ങളും ആധുനികസങ്കേതങ്ങളും Notes
• Chapter 10 ബജറ്റ്: വികസനത്തിന്റെ നേർരേഖ Notes
• Chapter 11 വിവേചനത്തിനെതിരെ Notes
• Chapter 12 ചരിത്രത്തിന്റെ ആധാര ശിലകൾ Notes

We hope the given Kerala SCERT Class 7 Social Science Question Answer, Kerala Syllabus 7th Standard Social Science Notes Pdf Malayalam Medium English Medium of Class 7 Social Science Notes State Syllabus will help you. If you have any queries regarding Class 7 Social Science Notes Pdf Kerala Syllabus, SCERT Class 7 Social Science Solutions Part 1 & 2, drop a comment below and we will get back to you at the earliest.

By reviewing Std 6 Social Science Notes Pdf Malayalam Medium and മധ്യകാല ഇന്ത്യ അധികാരകേന്ദ്രങ്ങൾ Class 6 Social Science Chapter 1 Question Answer Notes Malayalam Medium, students can improve their conceptual understanding.

Class 6 Social Science Chapter 1 Notes Malayalam Medium മധ്യകാല ഇന്ത്യ അധികാരകേന്ദ്രങ്ങൾ

Medieval India: The Centres of Power Class 6 Notes Malayalam Medium

Question 1.
അധികാര കേന്ദ്രമായി മാറാൻ ഡൽഹിയെ സഹായിച്ച് ഭൂമിശാസ്ത്ര ഘടകങ്ങൾ എന്തെല്ലാം ?
Answer:
സിന്ധു – ഗംഗാ സമതലത്തിലാണ് ഡൽഹി സ്ഥിതി ചെയ്യുന്നത്. ഈ പ്രദേശത്തെ ഫല ഭൂയിഷ്ഠിത കാർഷിക പുരോഗതിക്ക് സഹായ കമായി. ആരവല്ലി പർവ്വത നിരകൾ ശത്രുക്ക ളുടെ ആക്രമണങ്ങളെ പ്രതിരോധിക്കാൻ ഡൽഹിയെ സഹായിച്ചു.

ആരവല്ലി പർവ്വത നിരകളിലെ പാറക്കൂട്ടങ്ങളിൽ നിന്ന് കോട്ടകളു ടെയും കെട്ടിടങ്ങളുടെയും നിർമ്മാണത്തിനാ വശ്യമായ കല്ലുകളെ ശേഖരിക്കുവാൻ കഴിഞ്ഞു. യമുനാനദി ജലഗതാഗതത്തെ സഹാ യിച്ചതിനൊപ്പം ഡൽഹിക്ക് ആവശ്യമായ ജലം ഉറപ്പു വരുത്തി. ഈ ഭൂമിശാസ്ത്ര പ്രത്യേകത കളാണ് ഭരണാധികാരികളെ ഡൽഹിയിലേക്ക് ആകർഷിച്ചത്.

Question 2.
ഇന്ത്യ ചരിത്രത്തിൽ നിർണ്ണായകമായ ഒന്നായി രുന്ന പാനിപ്പത്ത് യുദ്ധം സാധൂകരിക്കുക.
Answer:
ഡൽഹിക്കടുത്തുളള പാനിപ്പത്തിൽ കാബൂ ളിലെ’ (അഫ്ഗാനിസ്ഥാൻ) ഭരണാധികാരി ബാബറുടെയും സൽത്തനത്ത് ഭരണാധികാരി ഇബ്രാഹിം ലോധിയുടെയും സൈന്യങ്ങൾ ഏറ്റുമുട്ടി. ഇബ്രാഹിം ലോധിയെ പരാജയ പ്പെടുത്തി ബാബർ ഡൽഹി പിടിച്ചെടുത്തു.

പീരങ്കിയും വെടിമരുന്നും ഉപയോഗിച്ചതിനാൽ ബാബറുടെ സൈന്യത്തിനെ ഇബ്രാഹിം ലോധിയുടെ സൈന്യത്തെ എളുപ്പത്തിൽ പരാ ജയപ്പെടുത്താനായി. സൽത്തനത്ത് ഭരണം അവസാനിപ്പിച്ച് ഡൽഹി കേന്ദ്രമായി ബാബർ തുടക്കം കുറിച്ച് ഭരണം മുഗൾ ഭരണം എന്നറിയപ്പെടുന്നു.

Question 3.
മുഗൾ സാമ്രാജ്യത്തിന്റെ വികസനത്തിൽ അക്ബർ സ്വീകരിച്ച നയം വിശകലനം ചെയ്യുക.
Answer:
മുഗൾ സാമ്രാജ്യം വിസ്തൃതമാക്കുന്നതിൽ അക്ബർ പ്രധാന പങ്ക് വഹിച്ചു. അദ്ദേഹം വലിയൊരു സൈന്യത്തെ രൂപകരിച്ചു. ഇതിനു വേണ്ടി വ്യത്യസ്തമായ ഒരു രീതി സ്വീകരിച്ചു. മൻസബ്ദാരി സമ്പ്രദായം എന്നാണ് ഇതറി യപ്പെടുന്നത്. ഈ സമ്പ്രദായത്തിൽ ഓരോ ഉദ്യോഗസ്ഥനും തങ്ങളുടെ കീഴിൽ ഒരു നിശ്ചിത എണ്ണം സൈനികരെ നിലനിർത്താൻ ബാധ്യസ്ഥനായിരുന്നു.

Question 4.
“റർ പ്രദേശത്തിന് വേണ്ടി വിജയനഗര ത്തിലെയും ബാഹ്മിനിയിലെയും ഭരണാധികാ രികൾ നിരന്തരം യുദ്ധം ചെയ്തു. ഈ പ്രസ്താവനയുടെ പശ്ചാത്തലത്തിൽ റെർ പ്രദേശത്തിന്റെ പ്രത്യേകതകൾ പരിശോധിക്കുക.
Answer:
കൃഷ്ണ തുംഗഭദ്ര എന്നീ നദികൾക്കിടയിൽ കിടക്കുന്ന റെർ പ്രദേശം ഫലഭൂയിഷ്ടമാ യിരുന്നു. ദക്ഷിണേന്ത്യയിലെ നെല്ലറ എന്നാണ് ഇതറിയപ്പെടുന്നത്. ഈ പ്രദേശത്ത് ആധി പത്യം നോടാൻ വിജയനഗര ഭരണാധികാരി കളും ബാഹ്മിനി രാജാക്കന്മാരും നിരന്തരം യുദ്ധങ്ങൾ നടത്തിയിരുന്നു.

Question 5.
മറാത്ത രാജ്യത്തിന്റെ വളർച്ചയെ സഹായിച്ച ഭൂമിശാസ്ത്ര ഘടകങ്ങൾ ഏതൊക്കെയാ യിരുന്നു.
Answer:
സി. ഇ പതിനോഴാം നൂറ്റാണ്ടിലാണ് മറാത്ത കൾ പ്രബല ശക്തിയായി മാറിയത്. ഭൂമി ശാസ്ത്ര സവിശേഷതകൾ മറാത്തകളുടെ വളർച്ചയെ ഏറെ സ്വാധീനിച്ചു. വിന്ധ്യ സത്പുര പർവ്വതങ്ങളും നർമ്മദ താപ്തി നദികളും മറാത്ത പ്രദേശത്തെ സമീപ പ്രദേശങ്ങളിൽ നിന്നും വേർതിരിച്ചു. മറാത്ത പ്രദേശത്തിന് പ്രകൃതിദത്തമായ സംരക്ഷണം നൽകി.

Question 6.
(എ) വിഭാഗത്തിന് യോജിക്കുന്നവ (ബി) വിഭാ ഗത്തിൽ നിന്നു കണ്ടെത്തി എഴുതുക

Answer:

Question 7.
താഴെ തന്നിരിക്കുന്ന ഭൂപടം നിരീക്ഷിക്കൂ. എന്തെല്ലാം വിവരങ്ങൾ കണ്ടെത്താം ?

Answer:

• സിന്ധു ഗംഗാ സമതലം
• ആരവല്ലി പർവ്വതം
• യമുനാ നദി
• ഡൽഹി

Question 8.
ഡൽഹിയെ അധികാര കേന്ദ്രമായി തിരഞ്ഞെടു ക്കുവാൻ ഭരണാധികാരികളെ പ്രേരിപ്പിച്ച ഭൂമി ശാസ്ത്ര ഘടകങ്ങൾ എന്തെല്ലാമായിരുന്നു ?
Answer:
സിന്ധു ഗംഗാ സമതലത്തിലാണ് ഡൽഹി സ്ഥിതി ചെയ്യുന്നത്. ഈ പ്രദേശത്തെ ഫല ഭൂയിഷ്ഠത കാർഷിക പുരോഗതിക്ക് സഹാ യകരമായി. ആരവല്ലി പർവ്വത നിരകൾ ശത്രു ക്കളുടെ ആക്രമണങ്ങളെ പ്രതിരോധിക്കാൻ ഡൽഹിയെ സഹായിച്ചു. ആരവല്ലി പർവ്വത നിരകളിലെ പാറക്കൂട്ടങ്ങളിൽ നിന്ന് കോട്ട കളുടെയും കെട്ടിടങ്ങളുടെയും നിർമ്മാണ ത്തിനാവശ്യമായ കല്ലുകൾ ശേഖരിക്കുവാൻ കഴിഞ്ഞു. യമുനാ നദി ജലഗതാഗതത്തെ സഹായിച്ചതിനൊപ്പം ഡൽഹിക്ക് ആവശ്യമായ ജലം ഉറപ്പു വരുത്തി ഈ ഭൂമിശാസ്ത്ര പ്രത്യേ കതകളാണ് ഭരണാധികാരികളെ ഡൽഹിയി ലേക്ക് ആകർഷിച്ചത്.

Question 9.
ചുവടെ കൊടുത്തിരിക്കുന്ന ഫ്ളോചാർട്ടിൽ സൽത്തനത്ത് കാലത്ത് ഡൽഹി ഭരിച്ച രാജ വംശങ്ങളെയും പ്രധാന ഭരണാധികാരി കളെയും ക്രമമായി നൽകിയിരിക്കുന്നു.

Question 10.
ഫ്ളോ ചാർട്ടിലെ വിവരങ്ങളുടെ അടിസ്ഥാന ത്തിൽ താഴെ പറയുന്ന പട്ടിക പൂർത്തിയാ ക്കുക.
Answer:

Question 11.
ഗുജറാത്തിന്റെ ഭൂമിശാസ്ത്രപരമായ പ്രത്യേക തകൾ എന്തെല്ലാം ?
Answer:
നീണ്ട സമുദ്രം ഗുജറാത്തിന്റെ പ്രധാന സവി ശേഷതയാണ്. അതു കൊണ്ടുതന്നെ നിരവധി തുറമുഖങ്ങൾ ഗുജറാത്തിലുണ്ട്.

Question 12.
ഗുജറാത്തിന്മേലുളള ആധിപത്യം അലാ വുദ്ദീൻ ഖൽജിയുടെ സൈന്യത്തെ എങ്ങനെ യാണ് ശക്തിപ്പെടുത്തിയത് ?
Answer:
ഗുജറാത്ത് കീഴടക്കിയതോടെ തുറമുഖങ്ങൾ അലാവുദ്ദീൻ ഖൽജിയുടെ നിയന്ത്രണത്തി ലായി. ഇറാഖിൽ നിന്ന് ഈ തുറമുഖങ്ങൾ വഴി മികച്ച ഇനം കുതിരകളെ ഇറക്കുമതി ചെയ്യാൻ കഴിഞ്ഞു. ഇത് അദ്ദേഹത്തിന്റെ സൈന്യത്തെ ശക്തിപ്പെടുത്തി. ഈ സൈന്യ ത്തിന്റെ സഹായത്തോടെ തെക്കെ ഇന്ത്യയിലും പടിഞ്ഞാറെ ഇന്ത്യയിലുമുളള പ്രദേശങ്ങൾ അലാവുദ്ദീൻ ഖൽജി തന്റെ നിയന്ത്രണ ത്തിലാക്കി.

Question 13.
ഭൂപടം നിരീക്ഷിച്ച് അക്ബറിന്റെ അധീനത യിൽ ഉൾപ്പെട്ട പ്രദേശങ്ങൾ കണ്ടെത്തി എഴുതുക

Answer:
അക്ബറുടെ അധീനതയിൽ ഉൾപ്പെട്ട പ്രദേശ

• ങ്ങൾ
• ഖാൻഡേഷ്
• ഗുജറാത്ത്
• ബംഗാൾ
• ബീഹാർ
• ഡൽഹി
• മുൾത്താൻ
• കാബൂൾ
• കാശ്മീർ

Question 14.
ഭൂപടം നിരീക്ഷിച്ച് ഇന്ത്യക്ക് പുറത്ത് ചോളന്മാ രുടെ സ്വാധീനം വ്യാപിച്ച രാജ്യങ്ങൾ കണ്ട ത്തുക.

Answer:

• ശ്രീലങ്ക
• മലേഷ്യ
• ഇന്തൊനേഷ്യൻ ദ്വീപുകൾ

Question 15.

വിജയനഗര രാജ്യത്തെക്കുറിച്ച് പേർഷ്യൻ സഞ്ചാരിയായിരുന്ന അബ്ദുൾ റസാഖിന്റെ വിവരണത്തിലെ ഒരു ഭാഗമാണ് മുകളിൽ കൊടുത്തിരിക്കുന്നത്. ഇതിൽ നിന്നും എന്തെല്ലാം കാര്യങ്ങൾ കണ്ടെത്താം
Answer:

• വിജയനഗര രാജ്യത്തിൽ ധാരാളം തുറമുഖ ങ്ങൾ ഉണ്ടായിരുന്നു.
• രാജ്യത്തിന്റെ ഭൂരിഭാഗം പ്രദേശങ്ങളും ഫല ഭൂയിഷ്ഠവും പതിവായി കൃഷിയിറക്കുന്നതു മാണ്.
• രാജാവിന്റെ ഭരണത്തിന് കീഴിലുള്ള പ്രദേശ ങ്ങളിൽ യാത്ര ചെയ്യണമെങ്കിൽ കുറഞ്ഞത് മൂന്നു മാസത്തെ
• സമയമെങ്കിലും വേണം. സൈന്യത്തിന്റെ അംഗബലം 11 ലക്ഷം വരും.

Question 16.
റെയ്ച്ചൂർ പ്രദേശങ്ങൾക്ക് വേണ്ടി യുദ്ധം നടത്താൻ വിജയനഗര – ബാഹ്മിനി രാജാ ക്കന്മാരെ പ്രേരിപ്പിച്ച ഘടകങ്ങൾ എന്തൊക്കെ യായിരുന്നു ?
Answer:
കൃഷ്ണ, തുംഗഭദ്ര എന്നീ നദികൾക്കിടയിൽ കിടക്കുന്ന റെയ്ച്ചൂർ പ്രദേശം ഫലഭൂയിഷ്ഠ മായിരുന്നു. ദക്ഷിണേന്ത്യയിലെ നെല്ലറ എന്നാണ് ഇതറിയപ്പെടുന്നത്. ഈ പ്രദേശത്ത് ആധിപത്യം നേടാൻ വിജയനഗര ഭരണാധികാ രികളും ബാഹ്മിനി രാജാക്കന്മാരും നിരന്തരം യുദ്ധങ്ങൾ നടത്തിയിരുന്നു.

Question 17.
മറാത്തകളുടെ വളർച്ചയെ സഹായിച്ച ഘടകങ്ങൾ എന്തൊക്കെയായിരുന്നു ?
Answer:
വിന്ധ്യ- സത്പുര പർവ്വതങ്ങളും നർമ്മദ താപ്തി നദികളും മറാത്ത പ്രദേശത്തെ സമീപ പ്രദേശങ്ങളിൽ നിന്നും വേർതിരിച്ചു. ഇത് മറാത്ത പ്രദേശത്തിന് പ്രകൃതിദത്തമായ സംരക്ഷണം നൽകി. മറാത്ത ഭാഷയും സാഹിത്യവും ജനങ്ങളിൽ ഐക്യബോധം വളർത്തി.

മധ്യകാല ഇന്ത്യ അധികാരകേന്ദ്രങ്ങൾ Class 6 Notes Questions and Answers

Question 1.
ഏത് രാജാക്കാൻമാരുടെ കാലത്താണ് ഡൽഹി ആദ്യമായി അധികാര കേന്ദ്രമാകുന്നത് ?
Answer:
താമര രാജാക്കന്മാരുടെ

Question 2.
തൊമര രാജാക്കൻമാരുടെ കാലത്ത് ഡൽഹി അറിയപ്പെട്ടിരുന്നത് ?
Answer:
ദില്ലിക

Question 3.
തൊമര രാജാക്കൻമാരെ തുടർന്ന് ഡൽഹിയിൽ അധികാരത്തിലെത്തിയ രാജവംശം ?
Answer:
ചൗഹാൻ രാജവംശം

Question 4.
ചൗഹാൻ വംശത്തിലെ അവസാന ഭരണാധി കാരി ?
Answer:
പൃഥ്വിരാജ് ചൗഹാൻ

Question 5.
പൃഥ്വിരാജ് ചൗഹാനെ കീഴടക്കി ഡൽഹിയിൽ ആധിപത്യം സ്ഥാപിച്ച ഘോറിലെ (ഇന്നത്തെ അഫ്ഗാനിസ്ഥാൻ) ഭരണാധികാരി?
Answer:
മുഹമ്മദ്

Question 6.
മുഹമ്മദിന്റെ സേനനായകൻ ?
Answer:
കുത്ബുദ്ദീൻ ഐബക്

Question 7.
കുത്ബുദ്ദീൻ ഐബകിന്റെ രാജവംശം അറിയ പ്പെടുന്നത് ?
Answer:
മംലൂക്ക് വംശം (അടിമവംശം)

Question 8.
കുത്ബുദ്ദീൻ ഐബകിന് ശേഷം അധികാരത്തി ലെത്തിയത് ?
Answer:
ഇൽതുത്മിഷ്

Question 9.
ഇൽത്തുമിഷ് കീഴടക്കിയ പ്രദേശങ്ങൾ ?
Answer:
മുൾട്ടാൻ, ലാഹോർ, ബംഗാൾ

Question 10.
ഇൽത്തുമിഷ് പ്രചാരത്തിലാക്കിയ നാണയ ങ്ങൾ ?
Answer:
തങ്ക, ജിതൽ

Question 11.
ഡൽഹി സൽത്തനത്തിലെ ഏക വനിതാ ഭരണാധികാരി ?
Answer:
സുൽത്താന റസിയ

Question 12.
ഇൽത്തുമിഷിന് ശേഷം അധികാരത്തിലെ ത്തിയ പ്രധാന ഭരണാധികാരി ?
Answer:
ബാൽബൻ

Question 13.
മംലൂക്ക് വംശത്തെ തുടർന്ന് അധികാരത്തിലെ ത്തിയ രാജവംശം ?
Answer:
ഖൽജി രാജവംശം (ഖിൽജി രാജവംശം)

Question 14.
അലാവുദ്ദീൻ ഖൽജിയുടെ അധികാരത്തിൻ കീഴിലായ ആദ്യ പ്രദേശം ?
Answer:
ഗുജറാത്ത്

Question 15.
ഖൽജി വംശത്തെ തുടർന്ന് അധികാരത്തിൽ എത്തിയത് ?
Answer:
തുഗ്ലക്ക് വംശം

Question 16.
ഭരണം കാര്യക്ഷമമാക്കുവാൻ തലസ്ഥാനം ഡൽഹിയിൽ നിന്ന് ദൗലത്താബാദിലേക്ക്മാറ്റിയ ഭരണാധികാരി ?
Answer:
മുഹമ്മദ് ബിൻ തുഗ്ലക്

Question 17.
ദൗലത്താബാദിന്റെ മുൻകാല പേര് ?
Answer:
ദേവഗിരി

Question 18.
പ്രധാന ചോള ഭരണാധികാരികൾ ആരെല്ലാം ?
Answer:
രാജ രാജ ചോളൻ, രാജേന്ദ്ര ചോളൻ

Question 19.
വിജയനഗര രാജ്യത്തിന്റെ സ്ഥാപകർ ?
Answer:
ഹരിഹരൻ, ബുക്കൻ

Question 20.
വിജയനഗരത്തിലെ പ്രധാന ഭരണാധികാരി ?
Answer:
കൃഷ്ണദേവരായർ

Question 21.
ബാഹ്മിനി രാജ്യത്തിന്റെ സ്ഥാപകൻ ?
Answer:
അലാവുദ്ദീൻ ഹസ്സൻ ബാൻഷാ

Question 22.
മറാത്ത രാജ്യത്തിലെ പ്രധാന ഭരണാധികാരി ?
Answer:
ശിവജി

Question 23.
മറാത്ത രാജ്യത്തിന്റെ ആസ്ഥാനം ?
Answer:
പൂനെ

Question 24.

മുകളിൽ നൽകിയിരിക്കുന്ന വിവരണത്തിൽ നിന്നും യമുനാനിദയെക്കുറിച്ചുള്ള എന്തെല്ലാം വിവരങ്ങളാണ് ലഭ്യമാവുന്നത് ?
Answer:

• ഇന്ത്യയിലെ ഏറ്റവും വലിയ നദിയായ ഗംഗയുടെ പോഷകനദികളിൽ ഒന്നാണ് യമുനാനദി.
• യമുനാനദി ഒഴുകുന്ന വഴിയിലെ പ്രധാന നഗരങ്ങളിൽ ഒന്നാണ് ഡൽഹി.
• ഉത്തരാഖണ്ഡിലെ യമുനോത്രിയിൽ നിന്നാണ് യമുനാനദിയുടെ ഉത്ഭവം.

Question 25.
ഡൽഹിയുടെ ഭൂമിശാസ്ത്രപരമായ പ്രത്യേക തകൾ എന്തെല്ലാം ?
Answer:

• സിന്ധു – ഗംഗാ സമതലത്തിലാണ് ഡൽഹി സ്ഥിതി ചെയ്യുന്നത്.
• ആരവല്ലി പർവ്വത നിരകൾ ശത്രുക്കളുടെ ആക്രമണങ്ങളെ പ്രതിരോധിക്കാൻ ഡൽ ഹിയെ സഹായിച്ചു.
• യമുനാനദി ഒഴുകുന്ന വഴിയിലെ പ്രധാന നഗരങ്ങളിൽ ഒന്നാണ് ഡൽഹി.

Question 26.
സൽത്തനത്ത് ഭരണം അവസാനിപ്പിച്ച് ഡൽഹി കേന്ദ്രമായി ബാബർ തുടക്കം കുറിച്ച മുഗൾഭരണം വരുവാനിടയായ സാഹചര്യങ്ങൾ എന്തെല്ലാം ?
Answer:

• 1526- ൽ, ഡൽഹിക്കടുത്തുള്ള പാനിപ്പ ത്തിൽ കാബൂളിലെ (അഫ്ഗാനിസ്ഥാൻ) ഭരണാധികാരി ബാബറുടെയും സൽത്ത നത്ത് ഭരണാധികാരി ഇബ്രാഹിം ലോദിയു ടെയും സൈന്യങ്ങൾ ഏറ്റുമുട്ടി.
• പീരങ്കിയും വെടിമരുന്നും ഉപയോഗിച്ചതി നാൽ ബാബറുടെ സൈന്യത്തിന് ഇബ്രാഹിം ലോദിയുടെ സൈന്യത്തെ എളുപ്പത്തിൽ പരാജയപ്പെടുത്താനായി.
• അങ്ങനെ സൽത്തനത്ത് ഭരണം അവ സാനിക്കുകയും ഡൽഹി കേന്ദ്രമായി ബാബർ തുടക്കം കുറിച്ച് മുഗൾ ഭരണം നിലവിൽ വരികയും ചെയ്തു.

Question 27.
അക്ബർ നാമയെക്കുറിച്ച് ചെറുവിവരണം തയ്യാറാക്കുക ?
Answer:
അക്ബറുടെ കൊട്ടാരത്തിൽ ജീവിച്ചിരുന്ന അബുൾ ഫസൽ എഴുതിയ ചരിത്ര കൃതി യാണ് “അക്ബർ നാമ”. ഈ പുസ്തകത്തിന് മൂന്ന് ഭാഗങ്ങളുണ്ട്. ആദ്യ ഭാഗം അക്ബറുടെ മുൻഗാമികളെയും രണ്ടാം ഭാഗം അക്ബറുടെ ഭരണകാലത്തെയും മൂന്നാം ഭാഗം അക്ബറുടെ ഭരണ സംവിധാനത്തെയും പ്രതിപാദിക്കുന്നു.

Question 28.
മൻസബ്ദാരി സമ്പ്രദായം എന്നാൽ എന്ത് ?
Answer:
മുഗൾ ഭരണാധികാരി അക്ബർ തുടക്കം കുറിച്ച് സമ്പ്രദായമാണ് മൻസബ്ദാരി സമ്പ ദായം. ഈ സമ്പ്രദായത്തിൽ ഓരോ ഉദ്യോഗ സ്ഥനും തങ്ങളുടെ കീഴിൽ ഒരു നിശ്ചിത എണ്ണം സൈനികരെ നിലനിർത്തുവാൻ ബാധ്യസ്ഥരായിരുന്നു.

Question 29.
മധ്യകാല ഇന്ത്യയുമായി ബന്ധപ്പെട്ട ഫ്ളോ ചാർട്ട് പൂർത്തീകരിക്കുക.

Answer:
(B) ചൗഹാൻ രാജവംശം
(C) സൽത്തനത്ത് ഭരണം
(i) ചോളന്മാർ
(iii) ബാഹ്മിനി
(iv) മറാഠകൾ

Question 30.
സി. ഇ. 1205 ൽ കുത്ബുദ്ദീൻ ഐബക് ഡൽഹി കേന്ദ്രമാക്കി ഭരണം ആരംഭിച്ചു
(a) മുകളിൽ നൽകിയിരിക്കുന്ന പ്രസ്ഥാവന യുമായി ബന്ധപ്പെട്ട രാജവംശം ഏതാണ് ?
(b) സി.ഇ 1206 മുതൽ സി.ഇ 1526 വരെ ഡൽഹി ആസ്ഥാനമാക്കി ഭരണം നടത്തി യവർ ……… എന്ന് അറിയപ്പെടുന്നു.
Answer:
(a) മാലൂക്ക് വംശം (അടിമ വംശം)
(c) സുൽത്താന്മാർ

Question 31.
(a) മൻസബ്ദാരി സമ്പ്രദായം നടപ്പിലാക്കിയ ഭരണാധികാരി ആരാണ് ?
(b) മൻസബ്ദാരി സമ്പ്രദായം എന്നാൽ എന്ത് ?
(c) മുഗൾ ഭരണത്തിൽ പ്രധാന പദവികൾ വഹിച്ച ഏതെങ്കിലും രണ്ട് രജപുത്രരുടെ പേര് എഴുതുക ?
Answer:
(a) അക് ബർ
(b) മൻസബ്ദാരി സമ്പ്രദായത്തിൽ ഓരോ ഉദ്യോഗസ്ഥനും തങ്ങളുടെ കീഴിൽ ഒരു നിശ്ചിത എണ്ണം സൈനികരെ നിലനിർ ത്താൻ ബാധ്യസ്ഥരാണ്. മുഗൾ സാമ്രാജ്യം വിസ്തൃതമാക്കുന്നതിന് അക്ബർ ചക വർത്തി നടപ്പാക്കിയ വ്യത്യസ്തമായ ഒരു രീതിയാണ് മൻസബ്ദാരി സമ്പ്രദായം.
(c) രാജാമാൻസിംഗ്, ബീർബൽ

Question 32.
(a) മധ്യകാല ഇന്ത്യയിലെ ഒരു അധികാര കേന്ദ്രമാണ് ഡൽഹി. ഈ കാലയളവിൽ ഡൽഹിയെ കൂടാതെ ഇന്ത്യയിൽ നില നിന്നിരുന്ന മറ്റ് രാജ്യങ്ങളുടെ പേരെഴു തുക
(b) പശ്ചിമേന്ത്യയിൽ നിലനിന്നിരുന്ന രാജ്യത്തെക്കുറിച്ച് വിശദീകരിക്കുക ?
Answer:
(a) ചോള രാജ്യം, വിജയനഗരം, ബാഹ്മിനി, മറാത്ത
(b) മറാത്ത രാജ്യം
സി.ഇ പതിനേഴാം നൂറ്റാണ്ടിലാണ് മറാത്ത കൾ പ്രബലശക്തിയായി മാറിയത്. ഭൂമി ശാസ്ത്ര സവിശേഷതകൾ മറാത്തകളുടെ വളർച്ചയെ ഏറെ സ്വാധീനിച്ചു. വിന്ധ്യ സത്പുര പർവ്വതങ്ങളും നർമ്മദ താപ്തി നദികളും മറാത്ത പ്രദേശത്തെ സമീപ പ്രദേശങ്ങളിൽ നിന്നും വേർതിരിച്ചു. ഇത് മറാത്ത പ്രദേശത്തിന് പ്രകൃതിദത്തമായ സംരക്ഷണം നൽകി. മറാത്ത ഭാഷയും സാഹിത്യവും ജനങ്ങളിൽ ഐക്യബോധം വളർത്തി.

Question 33.
സിന്ധു ഗംഗാ സമതലത്തിലാണ് ഈ നഗരം സ്ഥിതി ചെയ്യുന്നത്. ഈ പ്രദേശത്തെ ഫല കാർഷിക പുരോഗതിക്ക്
സഹായകമായി.
(a) മുകളിൽ പ്രതിപാദിച്ചിരിക്കുന്ന സ്ഥലമേത് ?
(b) ഈ പ്രദേശത്തിന്റെ ഭൂമിശാസ്ത്രപരമായ സവിശേഷതകളിൽ ഏതെങ്കിലും രണ്ടെണ്ണം എഴുതുക
Answer:
(a) ഡൽഹി
(b) ആരവല്ലി പർവ്വത നിരകൾ ശത്രുക്കളുടെ ആക്രമണങ്ങളെ പ്രതിരോധിക്കാൻ ഡൽഹിയെ സഹായിച്ചു.
യമുന നദി ജലഗതാഗതത്തെ സഹായിച്ച തിനൊപ്പം ഡൽഹിക്ക് ആവശ്യമായ ജലം ഉറപ്പു വരുത്തി.
“മധ്യകാല ഇന്ത്യയിലെ സൽത്തനത്ത് – മുഗൾ ഭരണത്തിന്റെ വ്യാപനം” എന്ന വിഷയത്തെ ആസ്പദമാക്കി ഒരു സെമിനാർ പേപ്പർ തയ്യാറാക്കുക
Answer:
(1206 മുതൽ 1526 വരെയുളള കാലഘട്ടത്തിൽ സൽത്തനത്ത് ഭരണം വിക ഡൽഹിയിൽ സിച്ചു. ആ കാലയളവിൽ അഞ്ചിലധികം രാജ വംശങ്ങൾ ഡൽഹി ഭരിച്ചു. മാലൂക്ക് വംശം (അടിമവംശം), ഖൽജി വംശം, തുഗ്ലക്ക് വംശം, സയ്യിദ് വംശം, ലോദി വംശം എന്നീ രാജ വംശങ്ങളാണ് . സൽത്തനത്ത് കാലത്ത് ഡൽഹിയിൽ ഭരണം നടത്തിയിരുന്നത്. കുത്ബുദ്ദീൻ ഐബകിനാൽ സ്ഥാപിതമായ മംലൂക്ക് രാജവംശം ബാൽബൻ ഭരണത്തോടെ അവസാനിച്ചു. മാലൂക്ക് വംശത്തെ തുടർന്ന് അലാവുദ്ദീൻ ഖൽജി ഭരണാധികാരിയായ ഖൽജി വംശം നിലവിൽ വന്നു. ജലാലുദ്ദീൻ ഖൽജിയാണ് ഖൽജി വംശത്തിലെ ആദ്യ

ഭരണാധികാരി അദ്ദേഹം ജല വ്യാപാര സമ്പ്രദായം പരിഷ്കരിക്കുകയും ഇറാഖിൽ നിന്നും മികച്ച കുതിരകളെ ഇറക്കുമതി ചെയ്യാൻ ഗുജറാത്തിലെ തുറമുഖങ്ങൾ ഉപയോഗിക്കുകയും ചെയ്തു. ഇതിൽ അദ്ദേ ഹത്തിന്റെ സൈനിക ശക്തിയെ ശക്തിപ്പെ ടുത്തി.

ഖൽജി വംശത്തിന് ശേഷം തുഗ്ലക്ക് വംശമാണ് നിലവിൽ വന്നത്. ഗിയാബുദ്ദീൻ ബാൽ ബണാൽ 1320 -ൽ സ്ഥാപിക്കപ്പെട്ട തുഗ്ലക്ക് വംശം 1412 – ൽ അവസാനിച്ചു. മുഹമ്മദ് ബിൻ തുഗ്ലക്ക്, ഫിറോസ് ഷാ തുഗ്ലക്ക് എന്നിവരും തുഗ്ലക്ക് വംശത്തിലെ ഭരണാധികാരികൾ ആയിരുന്നു. തലസ്ഥാന മാറ്റം, ടോക്കൺ കറൻസി, കാർഷിക പരിഷ്ക്കാരങ്ങൾ മുതലാ യവ മുഹമ്മദ് ബിൻ തുഗ്ലക്കിന്റെ സംഭാവനകൾ ആയിരുന്നു.

തുഗ്ലക്ക് വംശത്തിന് ശേഷം ഡൽഹി ഭരിച്ചത് സയ്യിദ് വംശമാണ്. ഖിസിർ ഖാൻ തുടക്കം കുറിച്ച് സയ്യിദ് വംശം 1451- ൽ ലോദി വംശ ത്തിന്റെ ഭരണ കാലഘട്ടത്തോടെ അവ സാനിച്ചു. തുഗ്ലക്ക് വംശത്തിന് ശേഷം അധി കാരത്തിലെത്തിയ സയ്യിദ് വംശത്തിലെയും ലോദി വംശത്തിലെയും ഭരണാധികാരികൾ ദുർബലരായിരുന്നു. തുഗ്ലക്ക് വംശത്തിന് ശേഷം സൽത്തനത്ത് ഭരണം ക്ഷയിക്കാൻ തുടങ്ങി.

1526 ഓടെ സൽത്തനത്ത് ഭരണം അവസാനി പ്പിച്ച് ഡൽഹി കേന്ദ്രമായി ബാബർ തുടക്കം കുറിച്ചുളള മുഗൾ ഭരണം നിലവിലെത്തി. ബാബറിന്റെ മരണത്തിനു ശേഷം ഹുമയൂൺ അധികാരത്തിൽ കയറി. ബാബറിന്റെ മുത്ത മകനായിരുന്നു ഹുമയൂൺ.

1539 ൽ ഷേർസാസൂരി ഹുമയൂണിനെ ചൗസാ യുദ്ധത്തിൽ പരാജയപ്പെടുത്തുകയും ഇതിന്റെ ഫലമായി 15 വർഷത്തേളം ഹുമയൂണിന് അലഞ്ഞു തിരിഞ്ഞ് നടക്കേണ്ടി വരികയും ചെയ്തു. എന്നാൽ പേർഷ്യയിലെ സാവിദ്

രാജവംശത്തിന്റെ സഹായത്തോടെ 1555 ൽ തന്റെ സാമ്രാജ്യം അധിക പ്രദേശങ്ങളോടെ അദ്ദേഹം തിരിച്ചു പിടിച്ചു. ഹുമയൂണിന് ശേഷം അദ്ദേഹത്തിന്റെ മകൻ അക്ബർ തന്റെ പതിമൂന്നാം വയസ്സിൽ പഞ്ചാബിന്റെ ഗവർണ റായി ചുമതലയേറ്റു. ഇന്ത്യയുടെ ചരിത്രത്തിലെ എക്കാലത്തെയും മികച്ച ഒരു മുഗൾ ഭരണാധി കാരിയായിരുന്നു അക്ബർ.

അക്ബറിന്റെ മരണത്തിനു ശേഷം 1605-ൽ ജഹാംഗീർ സിംഹാസനസ്ഥനായി. വിദ്യാഭ്യാസവും ബുദ്ധിയുമുള്ള മികച്ച ഭരണാധികാരി യായിരുന്നു ജഹാംഗീർ, ജഹാംഗീറിനു ശേഷം ഷാജഹാൻ ചക്രവർത്തി സ്ഥാനമേറ്റു. ഷാജ ഹാൻ ചക്രവർത്തിയുടെ ഭരണകാലം “സുവർണ്ണ കാലഘട്ടം” എന്നാണ് അറിയ പ്പെടുന്നത്. ഷാജഹാന്റെ പുത്രൻ ഔറംഗ സേബ് 1658 -ൽ മുഗൾ ഭരണാധികാരിയായി സ്ഥാനമേറ്റു. ഔറംഗസേബിന്റെ ഭരണകാല ത്താണ് മുഗൾ സാമ്രാജ്യം ഏറ്റവും കൂടുതൽ വികസിച്ചത്. 1857 വരെ ഇത് നീണ്ടു നിന്നു.

Question 34.
ഡൽഹിക്കു പുറമെ മധ്യകാല ഇന്ത്യയുടെ വിവിധ ഭാഗങ്ങളിൽ നിലനിന്ന അധികാര കേന്ദ്രങ്ങളെക്കുറിച്ച് കുറിപ്പ് തയ്യാറാക്കുക
Answer:
(എ) ചോളരാജ്യം
സി.ഇ ഒമ്പതാം നൂറ്റാണ്ടോടെയാണ് ചോള രാജവശം പ്രബലമായത്. രാജ രാജ ചോളനും രാജേന്ദ്ര ചോളനും ആയിരുന്നു പ്രധാന ചോള ഭരണാധികാരികൾ. ശകതമായ നാവിക സേന ചോളൻമാർക്കുണ്ടായിരുന്നു. ഇവരുടെ സ്വാധീനം മലേഷ്യ, ഇന്തൊനേഷ്യൻ ദ്വീപു കൾ എന്നിവിടങ്ങളിലേക്ക് വ്യാപിച്ചിരുന്നു.

(ബി) വിജയനഗരം
സി.ഇ പതിനാലം നൂറ്റാണ്ടിൽ നിലവിൽ വന്ന വിജയനഗര രാജ്യത്തിന്റെ സ്ഥാപകർ ഹരിഹരൻ, ബുക്കൻ എന്നിവരായിരുന്നു. കൃഷ്ണ ദേവരായർ ആയിരുന്നു വിജയ നഗര രാജ്യത്തിന്റെ പ്രധാന ഭരണാധികാരി.

(സി) ബാഹ്മിനി രാജ്യം
അലാവുദ്ദീൻ ഹസ്സൻ ബാഹ്മൻഷാ ആയി രുന്നു ബാഹ്മിനി രാജ്യത്തിന്റെ സ്ഥാപകൻ. കൃഷ്ണാ, തുംഗഭദ്ര എന്നീ നദികൾക്കിടയിൽ കിടക്കുന്ന റെയ്ച്ചൂർ പ്രദേശം ഫലഭൂയി ഷ്ഠമായിരുന്നു. ദക്ഷിണേന്ത്യയിലെ നെല്ലറ എന്നാണ് ഇതറിയപ്പെടുന്നത്. ഈ പ്രദേശത്ത് ആധിപത്യം നേടാൻ വിജയനഗര ഭരണാധി കാരികളും ബാഹ്മിനി രാജാക്കൻമാരും നിരന്തരം യുദ്ധങ്ങൾ നടത്തിയിരുന്നു.

(ഡി) മറാത്ത രാജ്യം
മറാത്താ രാജ്യത്തിലെ പ്രധാന ഭരണാധികാരി ശിവജി ആയിരുന്നു. “ഛത്രപതി” എന്ന സ്ഥാന പേർ അദ്ദേഹം സ്വീകരിച്ചു. ശക്തമായ നാവിക സേനയും കരസേനയും മറാത്ത രാജ്യത്തിന്റെ പ്രത്യേകത ആയിരുന്നു. പ്രബലമായ രാഷ്ട്രീയ ശക്തിയായി മാറാൻ ഇത് അവരെ സഹായിച്ചു. മറാത്ത രാജ്യത്തിന്റെ ആസ്ഥാനം പൂനൈ ആയിരുന്നു.

Medieval India: The Centres of Power Class 6 Notes Pdf Malayalam Medium

സി.ഇ എട്ടാം നൂറ്റാണ്ടു മുതൽ പതിനെട്ടാം നൂറ്റാണ്ടു വരെയുളള കാലം ഇന്ത്യാചരിത്രത്തിൽ പൊതുവെ മധ്യകാലഘട്ടം എന്നറിയപ്പെടുന്നു. മുഗൾ രാജവംശവും സൽത്തനത്ത് ഭരണവുമായിരുന്നു മധ്യകാലഘട്ടത്തിൽ ഡൽഹിയിൽ നിലനിന്നിരുന്നത്. ഈ കാലയളവിൽ ദക്ഷിണേന്ത്യയിലും പശ്ചിമേന്ത്യയിലും നിലനിന്നിരുന്ന പ്രധാന രാജ്യങ്ങളാണ് ചോള രാജ്യം, വിജയനഗരം, ബാഹ്മിനി,
മറാത്ത

ഈ പാഠഭാഗത്തിൽ പ്രധാനമായും രണ്ട് ആശയങ്ങളെക്കുറിച്ചാണ് നാം പഠിക്കുന്നത്,
(a) ഡൽഹിയുടെ മധ്യകാലചരിത്രം
(b) ദക്ഷിണേന്ത്യയിലും പശ്ചിമേന്ത്യയിലും നിലനിന്നിരുന്ന വിവിധ അധികാര കേന്ദ്രങ്ങൾ പ്രധാന

• സി. ഇ എട്ടാം നൂറ്റാണ്ടു മുതൽ പതിനെട്ടാം നൂറ്റാണ്ടു വരെയുള്ള കാലം ഇന്ത്യാ ചരിത്രത്തിൽ പൊതുവെ മധ്യകാലഘട്ടം എന്നറിയപ്പെടുന്നു.
• സിന്ധു – ഗംഗ സമതലത്തിലാണ് ഡൽഹി സ്ഥിതി ചെയ്യുന്നത്.
• ആരവല്ലി പർവത നിരകൾ ശത്രുക്കളുടെ ആക്രമണങ്ങളെ പ്രതിരോധിക്കുവാൻ ഡൽഹിയെ സഹായിച്ചു.
• യമുനാ നദി ജലഗതാഗതത്തെ സഹായിച്ചതിനൊപ്പം ആവശ്യമായ ജലം ഉറപ്പു വരുത്തി.
• താമര രാജാക്കാന്മാരെ തുടർന്ന് ചൗഹാൻ രാജവംശത്തിൽപ്പെട്ടവർ ഡൽഹിയിൽ ഭരണാധി കാരികളായി.
• ഘോറിലെ മുഹമ്മദിന്റെ മരണശേഷം സി.ഇ 1206 -ൽ കുത്ബുദ്ദീൻ ഐബക് ഡൽഹി കേന്ദ്രമാക്കി ഭരണം ആരംഭിച്ചു.
• സി. ഇ 1206 മുതൽ സി.ഇ 1526 വരെ ഡൽഹി ആസ്ഥാനമാക്കി ഭരണം നടത്തിയവർ സുൽത്താൻമാർ എന്നറിയപ്പെടുന്നു.

• കുത്ബുദ്ദീൻ ഐബകിനു ശേഷം അധികാരത്തിലെത്തിയത് ഇൽതുത്മിഷ് ആയിരുന്നു.
• ഇൽതുത്മിഷിനുശേഷം അധികാരത്തിലെത്തിയ പ്രധാന ഭരണാധികാരി ബാൽബൻ ആയിരുന്നു.
• തുഗ്ലക്ക് വംശത്തിന് ശേഷം സൽത്തനത്ത് ഭരണം ക്ഷയിക്കാൻ തുടങ്ങി.
• മുഗൾ സാമ്രാജ്യം വിസ്തൃതമാക്കുന്നതിൽ അക്ബർ പ്രധാന പങ്ക് വഹിച്ചു.
• മൻസബ്ദാരി സമ്പ്രദായം എന്നാൽ, ഓരോ ഉദ്യോഗസ്ഥനും തങ്ങളുടെ കീഴിൽ ഒരു നിശ്ചിത എണ്ണം സൈനികരെ നിലനിർത്തുവാൻ ബാധ്യസ്ഥതയുളളവരാണ്.
• ചോള, വിജയനഗരം, ബാഹ്മിനി, മറാത്ത എന്നിവയാണ് മധ്യകാലഘട്ടത്തിൽ ദക്ഷിണേന്ത്യയിലും പശ്ചിമേന്ത്യയിലും നിലനിന്നിരുന്ന പ്രധാന രാജ്യങ്ങൾ.
• സി.ഇ ഒമ്പതാം നൂറ്റാണ്ടോടെയാണ് ചോള വംശം പ്രബലമായത്.
• സി.ഇ പതിനാലാം നൂറ്റാണ്ടിൽ നിലവിൽ വന്ന വിജയനഗര രാജ്യത്തിന്റെ സ്ഥാപകർ ഹരിഹരൻ, ബുക്കൻ എന്നിവരായിരുന്നു
• അലാവുദ്ദീൻ ഹസ്സൻ ബാൻഷാ ആയിരുന്നു ബാഹ്മിനി രാജ്യത്തിന്റെ സ്ഥാപകൻ.
• സി.ഇ പതിനേഴാം നൂറ്റാണ്ടിലാണ് മറാത്തകൾ പ്രബല ശക്തിയായി മാറിയത്.