Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 11 Squares and Right Triangles Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 7 Maths Chapter 11 Solutions Squares and Right Triangles
Class 7 Maths Chapter 11 Squares and Right Triangles Questions and Answers Kerala State Syllabus
Squares and Right Triangles Class 7 Questions and Answers Kerala Syllabus
Page 149
Question 1.
Now compute the lengths of the sides of squares with area given below Write the answers using square roots.
i) 49 square centimetres
ii) 169 square centimetres
iii) 400 square centimetres
iv) 225 square centimetres
v) 1.69 square centimetres
vi) 6 \(\frac{1}{4}\) square metres
Answer:
i) 49 square centimetres
Area = 49 square centimetres
49 = 7 × 7 = 7²
By reversing, \(\sqrt{49}\) = 7
So the length of the sides of square is 7 cm
ii) 169 square centimetres
Area =169 square centimetres
169 = 13 × 13 = 13²
By reversing, \(\sqrt{169}\) = 13
So the length of the sides of square is 13 cm
iii) 400 square centimetres Area = 400 square centimetres
400 = 20 × 20 = 20²
By reversing, \(\sqrt{400}\) = 20
So the length of the side of square is 20 cm
iv) 225 square centimetres
Area = 225 square centimetres
225 = 15 × 15 = 15²
By reversing, \(\sqrt{225}\) = 15
So the length of the side of square is 15 cm
v) 1.69 square centimetres Area =1.69 square centimetres
1.69 = 1.3 × 1.3 = (1.3)²
By reversing, \(\sqrt{169}\) = 1.3
So the length of the side of square is 1.3 cm
vi) 6 \(\frac{1}{4}\) square metres
By reversing, \(\sqrt{6.25}\) = 2.5
6.25 = 2.5 × 2.5 = (2.5)²
Area = 6\(\frac{1}{4}\) square metres = \(\frac{25}{4}\) = 6.25
So the length of the side of square is 2.5 m
Page 152
Question 1.
Now can’t you draw squares of areas as below:
i) 32 sq. cm
ii) 50 sq. cm
iii) 12. 5sq. cm
iv) 24\(\frac{1}{2}\) sq. cm
Answer:
i) 32 cm
Here we want to draw a square of area 32 sq. cm
So first we want to find half its area, ie \(\frac{32}{2}\) = 16 cm
Next we want to find square root of 16 ie, \(\sqrt{16}\) = 4 cm
So we have to draw a square of length 4 cm
Next we have to draw its diagonal and measure the length of its diagonal.
Then extend any side of square, and mark the length of the diagonal on the extended side of the square.
ii) 50 sq. cm
We want to draw a square of area 50 sq. cm
So first we want to find half its area, ie \(\frac{50}{2}\) = 25cm
Next we want to find square root of 25 ie, \(\sqrt{25}\) = 5 cm
So we have to draw a square of length 5 cm
Next we have to draw its diagonal and measure the length of its diagonal.
Then extend any side of square, and mark the length of the diagonal on the extended side of the square.
Thus we get a square of Area 50 sq. cm
iii) 12.5 sq. cm
First we want to find half its area , ie \(\frac{12.5}{2}\)= 6.25cm
Next we want to find square root of 6.25 ie, \(\sqrt{6.25}\) =2. 5 cm
So we have to draw a square of length 2.5 cm
Next we have to draw its diagonal and measure the length of its diagonal.
Then extend any side of square, and mark the length of the diagonal on the extended side of the square.
Thus we get a square of Area 12.5 sq. cm
iv) 24 \(\frac{1}{2}\)sq. cm
Here 24\(\frac{1}{2}\) = \(\frac{49}{2}\) = 24.5
First we want to find half its area ie \(\frac{24.5}{2}\) =12.25 cm
Next we want to find square root of 12.25 ie, \(\sqrt{12.25}\) = 3. 5 cm
So we have to draw a square of length 3.5 cm
Next we have to draw its diagonal and measure the length of its diagonal.
Then extend any side of square, and mark the length of the diagonal on the extended side of the square.
Thus, we get a square of area 24 \(\frac{1}{2}\)sq. cm
Page 157
Question 1.
Draw squares of areas given below
i) 17 square centimetres
ii) 18 square centimetres
iii) 19 square centimetres
Answer:
i) 17 square centimetres
We can split 17 as 17= 16 + 1 = 4² + 1²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 4 and 1 cm, then the area of the square on the hypotenuse is 17 sq.cm.
ii) 18 square centimetres We can split 18 as 18 = 9 + 9 = 3² + 3²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 3 and 3 cm, then the area of the square on the hypotenuse is 18 sq.cm.
iii) 19 square centimetres
We can split 19 as difference of two squares:
19 = 100 – 81 = 10² – 9²
So, if we draw a right triangle with a hypotenuse of 10 centimetres and another side of 9 centimetres then the area of the square on the third side would be 19 square centimetres.
Question 2.
In the picture below, dots are marked 1 centimetre apart, horizontally and vertically and some of these are joined to make squares:
Calculate their areas.
Answer:
Square 1:
Let’s extend the two sides of square which creates a triangle.
Now the sides of the triangle will be 3 cm and 2cm with respect to the size of the dots given.
So, The area of the square 1 = 3² + 2² = 9 + 4 = 13 square centimetres
Square 2:
Let’s extend the two sides of square which creates a triangle.
Now the sides of the triangle will be 1cm and 1cm with respect to the size of the dots given.
So, The area of the square 2 = 1² + 1² = 1 + 1 = 2 square centimetres
Square 3:
Let’s extend the two sides of square which creates a triangle.
Now the sides of the triangle will be 1cm and 2cm with respect to the size of the dots given.
So, The area of the square 3 = 1² + 2² = 1 + 4 = 5 square centimetres
Square 4:
Let’s extend the two sides of square which creates a triangle.
Now the sides of the triangle will be 4cm and 2cm with respect to the size of the dots given.
So, The area of the square 1 = 4² + 2² = 16 + 4 = 20 square centimetres
Page 159
Question 1.
The lengths of two sides of some right triangles are given below. Calculate the length of the third side.
i) Perpendicular sides 6 centimetres, 8 centimetres
ii) Perpendicular sides 9 centimetres, 12 centimetres
iii) Perpendicular sides 7 centimetres, 24 centimetres
iv) Perpendicular sides 14 centimetres, 48 centimetres
v) Hypotenuse 17 centimetres, another side 15 centimetres
vi) Hypotenuse 34 centimetres, another side 30 centimetres
Answer:
i) Perpendicular sides 6 centimetres, 8 centimetres
Length of the third side = \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
= 10 centimetres
ii) Perpendicular sides 9 centimetres, 12 centimetres
Length of the third side = \(\sqrt{9^2+12^2}\)
= \(\sqrt{81+144}\)
= \(\sqrt{225}\)
= 15 centimetres
iii) Perpendicular sides 7 centimetres, 24 centimetres
Length of the third side = \(\sqrt{7^2+24^2}\)
= \(\sqrt{49+576}\)
= \(\sqrt{625}\)
= 25 centimetres
iv) Perpendicular sides 14 centimetres, 48 centimetres
Length of the third side = \(\sqrt{14^2+48^2}\)
= \(\sqrt{196+2304}\)
= \(\sqrt{2500}\)
= 50 centimetres
v) Hypotenuse 17 centimetres, another side 15 centimetres
Length of the third side = \(\sqrt{17^2-15^2}\)
= \(\sqrt{289-225}\)
= \(\sqrt{64}\)
= 8 centimetres
vi) Hypotenuse 34 centimetres, another side 30 centimetres
Length of the third side = \(\sqrt{34^2-30^2}\)
= \(\sqrt{1156-900}\)
= \(\sqrt{256}\)
= 16 centimetres
Question 2.
Two pillars of heights 2 metres and 5 metres stand 4 metres apart:
What is the distance between their tops?
Answer:
We are given that two pillars of heights 2 metres and 5 metres stand 4 metres apart
We want to find the distance between their tops.
When we draw a line parallel to 4 metre in the figure, we got a right triangle whose sides are 4 metre, 3 metre (ie 5 – 2 =3)
So the third side is the hypotenuse of the right triangle
We know that hypotenuse2 = Base² + Altitude²
Here Hypotenuse² = 4² + 3²
= 16 + 9
= 25
Hypotenuse = \(\sqrt{25}\)
= 5 cm
So the distance between their tops = 5 cm
Question 3.
Find the length of the bottom side of the triangle drawn below:
Answer:
In the given figure there are 2 right triangles,
Here hypotenuse and altitudes of both the triangles are given,
So, Base² = hypotenuse² – altitude²
= 13² – 12²
= 169 – 144
= 25m
Base = \(\sqrt{25}\) = 5 m
So the base of both the triangle = 5m
So Base of the large triangle = 5 + 5 = 10 cm
Question 4.
In the pictures below, O is the centre of the circle and A, B, P, Q are points on the circle: Calculate the lengths of the lines AB and PQ.
Answer:
Join OB and OQ in the figure;
So we get two right triangles in the 1st circle
OB = 5 cm(radius)
Let C be the Point where O meets the line AB
So OC = 4 m
AO = 5 m
So AC² = AO² – OC²
= 5² – 4²
= 25 – 16
= 9
So AC = √9 = 3 cm
CB = 3 cm
Thus AB = AC + CB
= 3 + 3
= 6 cm
In the second figure,
Let R be the point where O meets PQ.
Here PO = 5 cm, OR = 3cm ,OQ = 5 cm
So PR² = PO² – OR²
= 5² – 3²
= 25 – 9
= 16
PR = \(\sqrt{16}\) = 4 cm
∴ RQ = 4 cm
So, PQ = 4 + 4 = 8 cm
Therefore AB = 6 cm and PQ = 8 cm
Class 7 Maths Chapter 11 Kerala Syllabus Squares and Right Triangles Questions and Answers
Question 1.
The area of a square is 196 sq.cm .Then find the length of the sides?
Answer:
It is given that area of the square = 196 Sq.cm
We know that area of the square = side × side
= 14 × 14
= 196 Sq.cm
length of the sides = \(\sqrt{196}\) = 14 cm
Next we want to find square root of 49 ie, \(\sqrt{49}\) = 7 cm
So we have to draw a square of length 7 cm
Question 2.
Draw a square of area 99 sq.cm.
Answer:
Here we want to draw a square of area 98 sq. cm
So first we want to find half its area, ie \(\frac{98}{2}\) = 49 cm
Next we want to find square root of 49 ¡e, \(\sqrt{49}\) = 7 cm
So we have to draw a square of length 7 cm
Next we have to draw its diagonal and measure the length of its diagonal.
Then extend any side of square, and mark the length of the diagonal on the extended side of the square.
Thus we get a square of Area 98sq. cm
Question 3.
Draw squares of areas 52 square centimetres and 21 square centimetres.
Answer:
For the area of square 52 square centimetres
We can split 52 as
52 = 36 + 16 = 6² + 4²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 6 and 4 cm, then the area of the square on the hypotenuse is 52 sq.cm.
For the area of square 21 square centimetres
We can split 21 as difference of two squares:
21 = 25 – 4 = 5² – 2²
So, if we draw a right triangle with a hypotenuse of 5 centimetres and another side of 2 centimetres then the area of the square on the third side would be 21 square centimetres.
Question 4.
Find out the area of the square on the diagonal of given rectangle.
Answer:
First draw the diagonal.
Now the sides of the triangle will be 8 cm and 3 cm.
So, the area of the square on the diagonal = 8² + 3² = 64 + 9 = 73 square centimetres
Question 5.
The lengths of two sides of a right triangles are given below. Calculate the length of the third side. Hypotenuse 10 centimetres, another side 8 centimetres
Answer:
The length of the third side = \(\sqrt{10^2-8^2}\)
= \(\sqrt{100-64}\)
= \(\sqrt{36}\)
= 6 cm
Question 6.
Find the sides of the triangle in the figure.
Answer:
In the given figure the one perpendicular side is 4em,
This side is parallel to the side of the rectangle and hence it is 4 cm
And the other side is 8 – 5 = 3 cm
Therefore third side of the triangle = \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5 cm
Class 7 Maths Chapter 11 Notes Kerala Syllabus Squares and Right Triangles
In this chapter we will explore two important shapes, Squares and Right Triangles. We’ll begin by exploring squares, We will learn how to calculate the area of a square and how to draw a square with double the area of another. Next, we’ll move on to right triangles. We will explore the Pythagoras theorem, which helps us find the lengths of the sides of right triangles. This is a very useful tool in both, math and real-life situations.
- In the first topic, Areas of Squares, we will explore the concept of squaring a number and the square of a number. Then, We will also look into the square root of a number.
- In the second topic, Doubling the Area, we will learn how to draw a square that has double the area of a given square.
- In the third topic, Rectangles and Squares, we will discuss the hypotenuse, perpendicular sides, and Pythagoras theorem.
- In the fourth topic, Line Math, we will examine how the Pythagorean theorem relates the sides of a right triangle and how to solve related problems.
Through this chapter, we will build a foundational understanding of squares and right triangles.
Areas Of Squares
The second power of a number is called the square of that number. And the operation of finding the square of a number is called squaring.
For example, (1.3)², (1\(\frac{1}{2}\))²
(1.3)² = (1.3) × (1.3)
= \(\frac{13}{10} \times \frac{13}{10}\)
= \(\frac{169}{100}\)
= 24
The area of a (geometrical) square is the (arithmetical) square of its sides.
Here we can also say that, to draw a square of area 25 square centimeters, what should be the length of the sides?
So to find the answer, we want to know, the square of which number is 25,
ie, which number multiplied by itself gives 25
We know that 5 × 5 = 25, so the length of the sides should be 5 cm.
In another way we can say it as :
5 is the square root of 25.
In shorthand notation, “5 squared us 25”
ie, 5² = 25
The reverse statement is 5 is the square root of 25. We √ use to write this in shorthand that is,
√25 = 5.
It is not easy to compute the square roots of numbers. For small natural numbers, we can check the squares one by one and find the square root .For slightly larger natural numbers, we can try to factorize them and compute the square root.
For example consider the number 196,where we can factorize 196 and writeas:
196 = 2 × 2 × 7 × 7 = 2² × 7²
But we know that product of powers is the power of the product ie,
xnyn =(xy)n
So, 196 = 2² × 7² = (2 × 7)² = 14²
Writing this in reverse, we get \(\sqrt{196}\) = 14
Doubling The Area
How do you make a square of double the area of the given figure?
For this we want to measure the length of its sides. So that we can calculate its area as the square of this number. If we double this number and calculate its square root, we get the side of the square of double the area. Without any measurement or computation we can find the same in another way,
For that,
First cut out another square of the same size and cut both sides along their diagonals
Now arrange the four triangles as below:
So that we get this square,
We used all the pieces of two small squares to make this large square. So its area is double that of a small square
If each of the small squares is made up of two right triangles of the same size; the large square is made up of four such right triangles.
Thus we can conclude that, if we want to just draw a square with double the area of another, we need just draw the square with the diagonal as side:
Here the side of the large square is equal to the length to the diagonal of the smaller square.
If we want to see the squares separately, we just need to one side of the original square and mark the length of the diagonal on it.
Rectangles And Squares
Now let’s discuss how to draw the square on the diagonal of a rectangle with unequal sides?
Here, the area of this square also depends on the squares on the sides.
To understand this, draw a rectangle, and squares on its sides and a diagonal as in the above picture, on a thick sheet of paper. Then,
Draw the diagonals of the bottom square and mark the point where they intersect.
Then erase the diagonals and draw lines through the point marked, lines parallel to the sides of the largest square:
Now cut out the squares; also cut the grey square along the lines drawn inside it:
Arrange the pieces of the grey square and the whole white square within the black square as shown below:
Here the square on the diagonal to the rectangle can be exactly filled with squares on the sides of the rectangle. So, we can say that,
The area of the square on the diagonal of a rectangle is equal to the sum of the areas of the squares on the adjacent sides.
The longest side of a right triangle is called its hypotenuse.
So the above result can also be stated as:
The area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of . the squares on the perpendicular sides.
This results is known as the Pythagoras’ Theorem.
For example:
If the sides of the rectangle are 7 cm and 4cm.
If the areas of the squares on sides of the rectangle are :
7² = 49 square centimetres
4² = 16 square centimetres
So, the area of the square on the diagonal = 49 + 16 = 65 sq.cm
Let see how to draw squares of specified areas using Phythagor’s theorem,
For example,
Draw a square of area 25 square centimetres.
For that first split 25 as
25 = 16 + 9 = 4² + 3²
So, by Pythagoras’ Theorem, if we draw a right triangle with perpendicular sides 4 and 3 cm, then the area of the square on the hypotenuse is 25 sq.cm.
What about a square of area 16 square centimetres?
Here 16 cannot be written as sum of squares of two natural numbers. So we cannot draw the square using this method.
So let’s consider another method,
According to Pythagoras’ Theorem, if the area of the square on the hypotenuse of a right triangle is subtracted from the square on another side, the area on the third side is obtained.
So, for this we can split 16 as difference of two squares:
16= 25 – 9 = 5² – 3²
So, if we draw a right triangle with a hypotenuse of 5 centimetres and another side of 3 centimetres then the area of the square on the third side would be 16 square centimetres
To draw such a triangle follow the steps given below:
- Draw a line 3 centimetres long and the perpendicular at one end.
- Draw a piece of the circle centered at the other end of the line, with radius 5 centimetres.
The point where this circle meets the perpendicular is the third vertex of the triangle.
Line Math
The area of a square is the square of the length of its side; so, Pythagoras’ Theorem can also be stated as a relation between the sides of a right triangle:
The square of the hypotenuse of a right triangle is the sum of the squares of its perpendicular sides.
For example,
If the perpendicular sides of a right triangle are 3 centimetres and 4 centimetres, then the square of its hypotenuse is
3² + 4² = 25
So, the hypotenuse of this triangle is 5 centimetres
- The area of a (geometrical) square is the (arithmetical) square of its sides.
- If we want to just draw a square with double the area of another, we need just draw the square with the diagonal as side.
- The longest side of a right triangle is called its hypotenuse.
- The area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of the squares on the perpendicular sides. This results is known as the Pythagoras’ Theorem.
- The square of the hypotenuse of a right triangle is the sum of the squares of its perpendicular sides.