Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

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Kerala State Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Area of a Triangle Text Book Questions and Answers

Dot Math Textbook Page No. 68

In the picture below, the horizontal and vertical dots are one centimetre apart.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 1
What are the areas of the coloured figures below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 2
In the above picture, draw other figures by joining dots and find out their areas.
Answer:
Figure 1 is 13 cm,
Figure 2 is 6 cm
Figure 3 is 12 cm.

Explanation:
The area of figure 1 is 7+6 =13 cm,
figure 2 is 2+2+2 = 6 cm,
figure 3 is 3+3+3+3 = 12 cm.

This can be done using the Grid option in GeoGebra. Select the Polygon tool and click at various intersections of the grid lines to draw different figures.

Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 1-1

Calculate the area of these. To check the answers, select the Area tool and click within each figure.
Answer:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 1-2

Different halves
A rectangle can be halved by cutting vertically or horizontally along the middle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 9
We can also cut corner to corner to make two triangles of half the area.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 10
What happens when we draw a slanted line through the middle’?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 11

Answer:
We will get two quadrilaterals.

Explanation:
By drawing a slanted line through the middle of the rectangle we will get two quadrilaterals.

Didn’t we get two quadrilaterals of half the area?
Answer:
Such a quadrilateral with only one pair of parallel sides is called a trapezium.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Parallelogram and rectangle Textbook Page No. 70

How do we compute the area of this parallelogram?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 12
Answer:
We can compute the area of this parallelogram by multiplying the base of the perpendicular by its height.

Cut out a right angled triangle from the left as shown below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 13
And place it on the right like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 14
Now we get a rectangle of the same area.

Rectangle and triangle
In this picture below, what part of the rectangle is the red triangle?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 19

Answer:
Here, the red triangle is half part of the rectangle as the rectangle surrounds the triangle.

Rectangle and triangle Textbook Page No. 72
How about cutting the rectangle into two smaller rectangles?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 20
The area of the red triangle in each of these pieces is half the area of that piece. So the sum of their areas is equal to half the area of the original rectangle.
The original red triangle is made up of these two smaller red triangles.
Thus the area of the original red triangle is half the area of the original rectangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 21
What about the red triangle above?
Answer:
Kerala-Syllabus-7th-Standard-Maths-Solutions-Chapter-5-Area-of-a-Triangle-21
Explanation:
When the red triangle in the image above is divided in two, a square and a rectangle are formed.
The square and the rectangle are made up of these two smaller red triangles.
The area of square and rectangle will be the sum of two formed half triangles. The area of original triangle will be the half the area of square and rectangle.

Draw a rectangle in GeoGebra and mark a point on the top side. Select the Polygon tool and draw a triangle as in the problem. Colour it red. Select the Area tool and find the area of the triangle.
Drag the point on top and see what happens to the area.

Answer:
Here, we have drawn a rectangle in GeoGebra and marked a point on the top side and selected the Polygon tool and drawn a triangle, and selected the area tool.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 1-3

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 1-2

Rectangles within rectangle
Look at this rectangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 29
Do you see any relation between the areas of the red rectangles?
Think for a moment before turning the page for the answer.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Rectangles within rectangle Textbook Page No. 74
The diagonal of the large rectangle divides it into two right angled triangles of equal area. Each of these triangles is made up of the red rectangle within it and two small right angled triangles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 30
The areas of right angled triangles of the same colour are equal.
So, the area of the red rectangles are also equal.
What if we draw rectangle through some other point on the diagonal?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 31
Answer:
The areas of right angled triangles with the same colour are equal even if we draw a rectangle via another point on the diagonal.
So, the area of the red rectangles will also be equal.

In GeoGebra, draw a pair of horizontal lines, 8 units apart. On the bottom line, mark two points D and F 4 units apart. Mark a point B on the top line and draw ΔDFB using the Polygon tool. What is the area of this triangle? Check the answer using the Area tool. Now drag B and see what happens to the area.

Area of a Triangle

Square division
Draw a square and mark the mid points of its sides.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 37
Join these to the corners of the square as shown below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 38
We get a square at the centre:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 39
What part of the original square is this?

Square division Textbook Page No. 76

Cut out a figure like this in paper.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 40
Rearrange the triangular pieces like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 41
Now we get five squares of equal size.
So, the small square is \(\frac{1}{5}\) of the orignal square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Trapezium Textbook Page No. 77

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 45
In the figure ABCD is a rectangle and EFG is a right angled triangle.
What are the areas of the trapeziums AFED, and ECBF?
Answer:

Explanation:
Given that the figure ABCD is a rectangle and EFG is a right-angled triangle. So the areas of the trapeziums AFED, and ECBF is
Here, the total area will be
= AB × AD
= 20 × 10
= 200 sq cm.
The area of the rectangle is ECBG
= EC × BC
= 6 × 10
= 60 sq cm.
The area of the triangle EGF is
= \(\frac{1}{2}\) × 4 × 10
= 2 × 10
= 20 sq cm.
So the area of the trapezium ECBF is
= area of the rectangle is ECBG + area of the triangle EGF
= 60 + 20
= 80 sq cm.

Halving Textbook Page No. 68

Cut out a paper rectangle 4 centimetres wide and 3 centimetres high.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 3
Draw a line down the middle as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 4
Now we have two rectangles. What is the area of each?
To see that it’s half the big rectangle, we need only fold it across.
So the area of a small rectangle is half the area of the large rectangle. That is,
\(\frac{1}{2}\) × 12 = 6 square centimetres
Can you halve the area in any other way?
Answer:
Kerala State Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle img_1
The image above shows another method for halving the area.
The diagonal of the rectangle divides it into two triangles of equal area.
Area of triangle will be \(\frac{1}{2}\)×12 = 6 square centimetres.
Therefore, the area of original rectangle will be twice of the triangle , 2×6 = 12 square centimetres.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Another half
Cut out a paper rectangle 10 centimetres wide and 8 centimetres high.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 5

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 6
Draw a line from corner to corner as shown.
The rectangle is split into two triangles.
Are their areas equal?
Can we fold and check as before?
How about cutting them out?
Then put one on top of the other.
So, what is the area of each triangle?
The area of a triangle is half the area of the rectangle. That is,
\(\frac{1}{2}\) × 10 × 8 = 40 sq. cm.
Do you notice anything about the angles of these triangles?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 7
A triangle with a right angle at one corner is called a right-angled triangle.
What is the area of the right-angled triangle shown on the right?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 8
Answer:
The area of the right-angled triangle is 10 sq cm.

Explanation:
The area of the right-angled triangle is
= \(\frac{1}{2}\) × 4 × 5
= 2 × 5
= 10 sq cm.

Cut out two right angled triangles like this and place them together as shown below.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 15
What is the area of this rectangle?
The area of the right-angled triangle is half of this, right?
Area of the triangle = \(\frac{1}{2}\) × 4 × 5
= 10 sq. cm.
In this, 4 and 5 are the lengths of the perpendicular sides of the right-angled triangle.
Thus we have the method to compute the area of a right-angled triangle.

The area of a right-angled triangle is half the product of the perpendicular sides.

Now calculate the area of the figures shown below:

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 16

Answer:
The area of the figures is 30 cm sq, 8.75 cm sq, 13.5 cm sq.

Explanation:
The area of the first figure is
= \(\frac{1}{2}\) × 10 × 6
= 5 × 6
= 30 cm sq.
The area of the second figure is
= \(\frac{1}{2}\) × 3.5 × 5
= \(\frac{17.5}{2}\)
= 8.75 cm sq.
The area of the third figure is
= \(\frac{1}{2}\) × 4.5 × 6
= 4.5 × 3
= 13.5 cm sq.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 17

Answer:
The total area is 40 cm sq.

Explanation:
The area of the rectangle is length × breadth
= 6 × 5
= 30 cm sq.
And the area of the triangle is
= \(\frac{1}{2}\) × 2 × 5
= 5 cm sq.
The total area is the area of the rectangle + the area of the triangle
= 30 + 5 + 5
= 40 cm sq.

Answer:
The total area is 16.5 cm sq.

Explanation:
The area of the rectangle is length × breadth
= 4 × 3
= 12 cm sq.
The area of the first triangle is
= \(\frac{1}{2}\) × 1 × 3
= \(\frac{3}{2}\)
= 1.5 cm sq.
The area of the second triangle is
= \(\frac{1}{2}\) × 2 × 3
= 3 cm sq.
The total area is the area of the rectangle + the area of the triangle
= 12 + 1.5 + 3
=16.5 cm sq.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 18

Answer:
The total area of the triangle is 15 cm sq.

Explanation:
The area of the first triangle is
= \(\frac{1}{2}\) × 3 × 6
= 3 × 3
= 9 cm sq.
The area of the second triangle is
= \(\frac{1}{2}\) × 2 × 6
= 6 cm sq.
The total area of the triangle is
= 9 + 6
=15 cm sq.

  • The area of a right-angled triangle is 96 square centimeters. One of the perpendicular sides is 16 centimeters long. What is the length of the other?

Answer:
The length of the other side is 6 sq cm.

Explanation:
Given that the area of a right-angled triangle is 96 square centimeters and one side of the perpendicular sides is 16 centimeters. So the length of the other side is
96 = 16 × L
L = 96 ÷ 16
= 6 sq cm.

  • The perpendicular sides of a right-angled triangle are 12 and 15 centimeters long. Another right angled triangle of the same area has one of the perpendicular sides 18 centimeters long. What is the length of the other?

Answer:
The area of the first right-angled triangle is 90 sq cm and the length of the other side is 10 cm.

Explanation:
Given the perpendicular sides of a right-angled triangle are 12 and 15 centimeters long. So the area of the first triangle is
= \(\frac{1}{2}\) × 12 × 15
= 90 sq cm.
So the area will be
90 = \(\frac{1}{2}\) × 18 × h
90 = 9 × h
h = \(\frac{90}{9}\)
= 10 cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Other triangles Textbook Page No. 72

Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 22
No angle of it is right.
How do we calculate the area?
Can we cut it into two right-angled triangles?
Look at the earlier problems you have done.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 23
So to compute the area, what all lengths are to be measured?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 24
Area = (\(\frac{1}{2}\) × 2 × 3) + (\(\frac{1}{2}\) × 4 × 3)
= 3 + 6
= 9 sq. cm.
We can calculate the area of any triangle like this.
What is the general method to calculate the area of a triangle?
Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 25
To find the area, first, draw a perpendicular from the top comer to divide it into two right-angled triangles.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 26
Now some lengths are to be measured.
Let’s write letters for these for the time being.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 27
So, how do we write the area?
The sum of the areas of two right-angled triangles is
(\(\frac{1}{2}\) × x × z) + (\(\frac{1}{2}\) × y × z)
= \(\frac{1}{2}\)xz + \(\frac{1}{2}\)yz
= \(\frac{1}{2}\)(x + y)z
In this, x + y is the length of the bottom side.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 28

So, how do we compute the area of a triangle?
The area of a triangle is half the product of one side with the perpendicular from the opposite side.

Now compute the area of these figures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 32

Answer:
The area of the first triangle is 25 sq cm.
The area of the second triangle is 11 sq cm.

Explanation:
The area of the first triangle is
= \(\frac{1}{2}\) × 5 × 10
= 5 × 5
= 25 sq cm.
The area of the second triangle is
= \(\frac{1}{2}\) × 5.5 × 4
= 5.5 × 2
= 11 sq cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 33

Answer:
The area of the first figure is 9 sq cm.
The area of the second figure is 16 sq cm.

Explanation:
The area of the first figure is
= \(\frac{1}{2}\) × 3 × 6
= 3 × 3
= 9 sq cm.
The height of the second triangle is
= 2+6
= 8 cm and the base is 4 cm.
The area of the second figure is
= \(\frac{1}{2}\) × 8 × 4
= 4 × 4
= 16 sq cm.

Answer:
The area of the first figure is 9 sq cm.
The area of the second figure is 16 sq cm.
The area of the third figure is 21 sq cm.

Explanation:
The area of the first figure is
= \(\frac{1}{2}\) × 3 × 5
= \(\frac{15}{2}\)
= 7.5 sq cm.
The area of the second figure is
= \(\frac{1}{2}\) × 4 × 5
= 2 × 5
= 10 sq cm.
The area of the third figure is
= \(\frac{1}{2}\) × 6 × 7
= 3 × 7
= 21 sq cm.
So the total area of the figure is 7.5+10+21 = 38.5 sq cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 34
Answer:
The area of the first figure is 24 sq cm.
The area of the second figure is 24 sq cm.
The area of the rectangle is 120 sq cm.
The total area is 168 sq cm.

Explanation:
The area of the first figure is
= \(\frac{1}{2}\) × 4 × 12
= 2 × 12
= 24 sq cm.
The area of the second figure is
= \(\frac{1}{2}\) × 4 × 12
= 2 × 12
= 24 sq cm.
The area of the rectangle is
= l × b
= 10 × 12
= 120 sq cm.
So the total area is
= 24 + 24 + 120
= 168 sq cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Another triangle Textbook Page No. 75
Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 35
How do we compute its area?
How do we draw a perpendicular from A to BC?
How about extending BC to the right’?
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 36
Now how do we calculate the area of ΔABC?
We get ΔABC by removing ΔACD from ΔABD.

ΔABD is a right-angled triangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 42
Area of ΔABD = \(\frac{1}{2}\) × BD × AD
ΔACD also is a right-angled triangle.
Area of ΔACD = \(\frac{1}{2}\) × CD × AD
Now we can find the area of ΔABC
Area of ΔABC = Area of ΔABD – Area of ΔACD
= \(\frac{1}{2}\) × BD × AD – \(\frac{1}{2}\) × CD × AD
= \(\frac{1}{2}\) × (BD – CD) × AD
From the picture,
BD – CD = BC
Thus we have
Area of ΔABC = \(\frac{1}{2}\) × (BD – CD) × AD
= \(\frac{1}{2}\) × BC × AD
Measure BC, and AD and compute the area.
Here AD is the height measured from BC.
So for triangles of this kind also, the area is half the product of a side and the height from it.

Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 43
Compute the area of the triangle by measuring out the needed lengths.
Answer:
Let the base of triangle be 6cm wide and the height be 4cm long.
Thus, the area of triangle will be \(\frac{1}{2}\) × 6 × 4 = 12 square cm.

Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle

Let’s do it! Textbook Page No. 77

Question 1.
A rectangular plot is 30 metres long and 10 metres wide. Within it, a triangular part is marked for plant¬ing plantain.
Kerala Syllabus 7th Standard Maths Solutions Chapter 5 Area of a Triangle 44

(i). What is the area of this part?
Answer:
Area = 300 m sq.

Explanation:
Given that a rectangular plot is 30 meters long and 10 meters wide. So the area of the land will be 30×10 = 300 m sq.
The area of part 1 is
= 10 × 10
= 100 sq m.
The area of part 2 is
= \(\frac{1}{2}\) × 4 × 10
= 2 × 10
= 20 sq m.
The area of part 3 is
= \(\frac{1}{2}\) × 16 × 10
= 8 × 10
= 80 sq m.

(ii). What is the area of the triangular part to the right of the area for plantain?
Answer:
The area of triangular part to the right of the area for plantain is 20 sq m.

Explanation:
The area of triangular part to the right of the area for plantain is
= \(\frac{1}{2}\) × 4 × 10
= 2 × 10
= 20 sq m.

(iii). What is an area of the trapezium to the left of the plantain area?
Answer:
The area of the trapezium to the left of the plantain area is 120 sq m.

Explanation:
The area of the trapezium to the left of the plantain area is
= area of part 1 + area of part 2
= 100 + 20
= 120 sq m.

Question 2.
In ΔABC, the angle at B is right. Its area is 48 square centimeters and the length of BC is 8 centimeters. The side of BC is extended by 6 centimeters to D. What is the area of ΔADC?
Answer:
The length of AB is 12 cm and the area of ADC is 84 sq cm.

Explanation:
Given that the area of ΔABC is 48 sq cm and the length of BC is 8 cm, so
ΔABC = \(\frac{1}{2}\) × BC × AB
48 =  \(\frac{1}{2}\) × 8 × AB
48 = 4 × AB
AB = \(\frac{48}{4}\)
= 12 cm.
And the area of ADC is
ΔADC = \(\frac{1}{2}\) × AD × AB
= \(\frac{1}{2}\) × 14 × 12
= 7 × 12
= 84 sq cm.

Kerala Syllabus 7th Standard Basic Science Solutions Guide

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Kerala State Syllabus 7th Standard Basic Science Textbooks Solutions Part 1

  • Chapter 1 Reaping Gold from Soil
  • Chapter 2 Wonders of Visible Light
  • Chapter 3 Acids and Alkalis
  • Chapter 4 Through the Alimentary Canal
  • Chapter 5 When Current Flows

Kerala State Syllabus 7th Standard Basic Science Textbooks Solutions Part 2

  • Chapter 6 For a Pollution Free Nature
  • Chapter 7 Pressure in Liquids and Gases
  • Chapter 8 Breath and Blood of Life
  • Chapter 9 Paths of Heat Flow
  • Chapter 10 Safety in Food too…

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Kerala State Syllabus 7th Standard Social Science Textbooks Solutions Part 1

  • Chapter 1 Europe in Transition
  • Chapter 2 From Trade to Power
  • Chapter 3 Resistance and the First War of Independence
  • Chapter 4 India Towards A New Era
  • Chapter 5 Economic Sources
  • Chapter 6 Understanding of Maps
  • Chapter 7 Earth and Biosphere

Kerala State Syllabus 7th Standard Social Science Textbooks Solutions Part 2

  • Chapter 8 Towards a New Kerala Society
  • Chapter 9 Gandhiji and the Freedom Struggle
  • Chapter 10 Our Constitution
  • Chapter 11 Individual and Society
  • Chapter 12 Insolation and Atmospheric Condition
  • Chapter 13 A Glimpse of India

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इकाई 1

  • Chapter 1 गुलमोहर का जन्मदिन (कहानी)
  • Chapter 2 हम सब सुमन एक उपवन के (कविता)

इकाई 2

  • Chapter 1 अनमोल खज़ाना (कहानी)
  • Chapter 2 कोशिश करनेवालों की (कविता)

इकाई 3

  • Chapter 1 ज़रूरी खुराक़ (बाल एकांकी)
  • Chapter 2 एक दौड़ ऐसी भी (लेख)

इकाई 4

  • Chapter 1 राजा का दरवाज़ा (कहानी)
  • Chapter 2 तब याद तुम्हारी आती है (कविता)

इकाई 5

  • Chapter 1 आसमान के लिए (जीवनी)
  • Chapter 2 मेरा जीवन (कविता)

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Kerala Syllabus 7th Standard English Guide

Unit 1 Nature’s Plenty

  • Chapter 1 How Far is the River? (Ruskin Bond)
  • Chapter 2 The Echoing Green (William Blake)
  • Chapter 3 A Snake in the Grass (R.K.Narayan)

Unit 2 Tales and Tunes

  • Chapter 1 The Owl and the Pussy-Cat (Edward Lear)
  • Chapter 2 The Song of Songs (Swapna Dutta)

Unit 3 Man and Media

  • Chapter 1 The Story of Messages
  • Chapter 2 The Radio
  • Chapter 3 The Television
  • Chapter 4 The Newspaper
  • Chapter 5 The Computer
  • Chapter 6 Books (Eleanor Farjeon)
  • Chapter 7 Polya (Mikhail Zoschenko)

Unit 4 Rhythms of Life

  • Chapter 1 To My Mother (Christina Rossetti)
  • Chapter 2 Somebody’s Mother (Mary Dow Brine)
  • Chapter 3 The Wooden Cup (Domenico Vittorini)
  • Chapter 4 A Village Pooram (Anita Nair)

Unit 5 Light and Shade

  • Chapter 1 The Lonely Child and the Puppy
  • Chapter 2 The Boy and the Balloon (Albert Lamorisse)
  • Chapter 3 Daddy Fell into the Pond (Alfred Noyes)

Unit 6 Moments of Humour

  • Chapter 1 My Financial Career (Stephen Leacock)
  • Chapter 2 Mother and the Mouse (Faith Trekson)
  • Chapter 3 Master of the Game

We hope the given Kerala Syllabus 7th Standard English Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, English Textbook Activity Answers, Chapters Summary in Malayalam, English Study Material, Kerala Reader English Book Answers, English Teachers Hand Book will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 7th English Textbooks Answers Guide Pdf of Kerala Class 7 Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 7th Standard Textbooks Solutions Guide

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Kerala State Syllabus 7th Standard Textbooks Solutions

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Kerala Syllabus 7th Standard Maths Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 7th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 7th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 7th Standard Maths Textbooks Solutions Pdf of Kerala Class 7 Part 1 and Part 2.

Kerala State Syllabus 7th Standard Maths Textbooks Solutions

Kerala Syllabus 7th Standard Maths Guide

Kerala State Syllabus 7th Standard Maths Textbooks Solutions Part 1

Kerala State Syllabus 7th Standard Maths Textbooks Solutions Part 2

We hope the given Kerala Syllabus 7th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 7th Maths Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

You can Download Square and Square Root Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square and Square Root Text Book Questions and Answers

Triangular numbers Textbook Page No. 80

See the dots arranged in triangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 1
How many dots are there in each?
1, 3, 6
How many dots would be there in the next triangle?
Such numbers as 1, 3, 6, 10, … are called triangular numbers.
The first triangular number is 1.
The second is 1 + 2 = 3.
The third is 1 + 2 + 3 = 6.
What is the tenth triangular number?
Answer:
From the above logic we need to add 10 numbers sum to obtain the 10th triangular number.
To find the 10th triangular number T10 we need to add 1+ 2+3+4+5+6+7+8+9+10 = 55
Therefore, 10th triangular number T10 is 55.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Squares and triangles Textbook Page No. 81

Look at these pictures:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 4
Each square is divided into two triangles.
Let’s translate this into numbers:
4 = 1 + 3
9 = 3 + 6
16 = 6 + 10
Check whether the same pattern continues. What do we see?
All perfect squares after 1 are the sums of two consecutive triangular numbers.
What is the sum of the seventh and eighth triangular numbers?
Answer:
Yes adding consecutive triangular numbers gives us perfect squares. So, we get the next pattern 9+16 = 25 which in turn is a perfect square.
7th Triangular number can be obtained as T7 = 1+2+3+4+5+6+7 = 28
8th Triangular number can be obtained as T8 = 1+2+3+4+5+6+7+8 = 36
Sum of Seventh and Eighth Triangular Numbers = T7+T8
= 28+36
= 64
Thus, the sum of 7th, 8th Triangular Numbers is 64

Increase and decrease Textbook Page No. 82

Look at this number pattern:
1 = 1
4 = 1 + 2 + 1
9 = 1 + 2 + 3 + 2 + 1
16 = 1 + 2 + 3 + 4 + 3 + 2 + 1
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 5
Can you split some more perfect squares like this?
Answer:
Writing some more perfect squares we have 25 = 1+2+3+4+5+4+3+2+1
36 = 1+2+3+4+5+6+5+4+3+2+1
49 = 1+2+3+4+5+6+7+6+5+4+3+2+1
64 = 1+2+3+4+5+6+7+8+7+6+5+4+3+2+1

Square difference

We have see that
22 = 12 + (1 + 2)
32 = 22 + (2 + 3)
42 = 32 + (3 + 4) and so on.
We can write these in another manner also:
22 – 12 = 1 + 2
32 – 22 = 2 + 3
42 – 32 = 3 + 4
In general, the difference of the squares of two consecutive natural numbers is their sum. Now look at these:
32 – 12 = 9 – 1 = 8
42 – 22 = 16 – 4 = 12
52 – 32 = 25 – 9 = 16
What is the relation between the difference of the squares of alternative natural numbers and their sum?
Answer:
32 – 12 = (3+1) x 2 = 8
42 – 22 = (4+2) x 2 = 12
52 – 32 =  = (5+3) x 2 = 16

The difference between squares of two alternate natural numbers is always even i.e. twice the sum of two numbers that are squared.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Project Textbook Page No. 84

Last digit
Look at the last digit of squares of natural numbers from 1 to 10:
1, 4, 9, 6, 5, 6, 9, 4, 1, 0
Now, look at the last digits of squares of natural numbers from 11 to 20.
Do we have the same pattern?
Let’s look at another thing: Does any perfect square end in 2?
Which are the digits which do not occur at the end of perfect squares?
Is 2637 then a perfect square?
Answer:
Last digits of squares of natural numbers from 11 to 20 are 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. Yes they have the same pattern
As we observe the digits 2, 3, 7, 8 doesn’t occur at the end of perfect squares.
As we know 2637 ends with digit 7 at the end it is not a perfect square.

To decide that a number is not a perfect square, we need only look at the last digit.
Can we decide that a number is a perfect square from its last digit alone?
Answer:
Yes, we can decide if a number is perfect square or not by seeing the last digit alone as you can see above.

Rectangle and square

Look at this picture.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 6
Dots in a rectangle.
Can you rearrange the dots to make another rectangle?
Can you rearrange the dots to make a square? Start like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 7
How many more are needed to make a square?
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 8
How many dots were there in the original rectangle? How many in this square?
What do we see here?
42 = (3 × 5) + 1
Can we do this for all rectangular arrangements?
The numbers here are 3, 4, 5.
So, for this trick to work, what should be the relation between the number of dots in each row and column of the rectangle?
We can write this in numbers as
22 = (1 × 3) + 1
32 = (2 × 4) + 1
42 = (3 × 5) + 1
Try to continue this
Answer:
52 = (4 x 6) +1
62 = (5 x 7) +1
72 = (6 x 8) +1
82 = (7 x 9) +1
92 = (8 x 10)+1

Square root of a perfect square

784 is a perfect square. What is its square root? 784 is between the perfect squares 400 and 900; and we know that their square roots are 20 and 30. So \(\sqrt{784}\) is between 20 and 30. Since last digit of 784 is 4, its square root should have 2 or 8 as the last digit. So \(\sqrt{784}\) is either 22 or 28.
784 is near to 900 than 400. So \(\sqrt{784}\) must be 28. Now calculate 282 and check.
Given that 1369, 2116, 2209 are perfect squares, find their square roots like this.

Project Textbook Page No. 87

Digit sum

16 is a perfect square and the sum of its digits is 7.
The next perfect square 25 also has digit sum 7.
The digit sum of 36 is 9.
The sum of the digits of the next perfect square 49 is 13. If we add the digits again, the sum is 4. Find the sum of the digit sums (reduced to a single digit number) of perfect squares starting from 1.
Do you see any pattern?
Is 3324 is perfect square?
Answer:
Given number is 3324
Now Split the number and add each number 3 + 3 + 2 + 4  = 12
As the result is more than 1 number we should add it again 1 + 2 = 3
All possible numbers that are perfect square have roots of either 1, 4, 7, 9
As 3 is not in the list 3324 is not a Perfect Square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Rows and columns Textbook Page No. 80

Look this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 2
Dots in rows and columns make a rectangle.
How many dots in all?
Did you count the dots one by one?
Can you make other rectangles with 24 dots?
Is any one of these a square?
How many more dots do we need to make a square? Can you remove some dots and make a square? How many?
Can you remove some dots and make a square? How many?
Numbers which can be arranged in squares are called square numbers.
Do you see anything special about of the number of dots making a square?
Answer:
There are total 4×6=24 dots in all.
No, the dots were counted by multiplying the number of dots in rows and number of dots in column.
None of these is a square.
We need more 12 dots to make a square, 24+12=36 or 62.
Yes, we can remove some dots and make a square.
We need to remove 8 dots to make a square, 24-8=16 or 42.
The number of dots making a square is the sqaure of that number.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root img_1

Squares

What are the ways in which we can write 36 as the product of two numbers?
2 × 18, 3 × 12, 4 × 9
We can also write
36 = 6 × 6
And we have seen that it can also be shortened as 36 = 62.
36 is 6 multiplied by 6 itself; that is, the second power of 6.
There is another name for this:
36 is the square of 6.
Then what is the square of 5?
What is the square of \(\frac{1}{2}\)?
Answer:
Square of 5 = 52 = 25
\(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

Perfect squares

1, 4, 9, 16,… are the squares of the natural numbers. They are called perfect squares.
What is the perfect square after 16?
Why is 20 not a perfect square?
Answer:
Prime Factorization of 20 can be written as 22 x 51
As 5 is not in pair 20 is not a perfect square.
The next perfect square after 16 is 25.

Let us look at the succession of perfect squares in another way.
To reach 4 from 1, we must add 3.
To reach 9 from 4?
We can state these as
4 – 1 = 3
9 – 4 = 5
16 – 9 = 7
All these differences are odd numbers, right?
So, the difference of two consecutive perfect squares is an odd number.
Let’s write this as,
4=1 + 3
9 = 4 + 5 = 1 + 3 + 5
16 = 9 + 7 = 1 + 3 + 5 + 7
What do we see here?
When we add consecutive odd numbers starting from 1, we get the perfect squares.
This can be seen from these pictures also.
Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 3
Can you write down the squares of natural numbers upto 20, by adding odd numbers? You can proceed like this
12 = 1
22 = 1 + 3 = 4
32 = 4 + 5 = 9
42 = 9 + 7 = 16

What is the relation between the number of consecutive odd numbers from 1 and their sum?
What is the sum of 30 consecutive odd numbers starting from 1?
Answer:
Let us assume the arithmetic series 1, 3, 5, 7, 9, 11, 13…..
we know the formula for sum of arithmetic series Sn = n/2[2a+(n-1)d]
Here d = 2
first term a = 1
Substituting the inputs we have Sn= 30/2[2×1+(30-1)2]
= 30/2[2+58]
= 30/2[60]
= 900
Therefore, the sum of 30 consecutive odd numbers starting from 1 is 900.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Tricks with ten Textbook Page No. 82

The square of 10 is 100. What is the square of 100?
In the square of 1000, how many zeros are there after 1?
What about the square of 10000?
What happens to the number of zeros on squaring?
So how do we spot the perfect squares among 10, 100, 1000, 10000 and so on?
Is one lakh a perfect square?
What about ten lakhs (million)?
Now find out the squares of 20, 200 and 2000.
Is 400000000 a perfect square?
What if we put in one more zero?
Answer:
Square of a number containing x zeros will become 2 times number of zeros.
1 lakh is not a perfect square as the number of zeros is not even.
Ten Lakhs is a perfect square as it comes with 6 zeros that are even.
20 = 20 x 20 = 400
200 = 200 x 200 = 40000
2000 = 2000 x 2000 = 4000000
400000000 is a perfect square as the number of zeros 8 is even.

Now some problems. Do them all in your head.

• Find out the squares of these numbers.

• 30
Answer:
302 = 30 x 30
= 900
The square of 30 is 900.

• 400
Answer:
4002 = 400 x 400
= 160000
The square of 400 is 160000

• 7000
Answer:
70002 = 7000 x 7000
= 49000000
The Square of 7000 is 49000000

• 6 × 1025
Answer:
(6 × 1025)= (6 × 1025) x (6 × 1025)
= 36 x (1025)2
= 36 x 1050

• Find out the perfect squares among these numbers.

• 2500
Answer:
The given number 2500 can be written as 502
Hence the 2500 is a perfect square number.

• 36000
Answer:
We cannot write the given number 36000 as square of two equal numbers. Hence it is not a perfect square.

• 1500
Answer:
We cannot write the given number 1500 as square of two equal numbers. Hence it is not a perfect square.

• 9 × 107
Answer:
We cannot write the given number 9 × 107 as square of two equal numbers. Hence it is not a perfect square.

• 16 × 1024
Answer:
The given number 2500 can be written as (4x 1012)2
Hence the given number 16 × 1024 is a perfect square.

Next square

What is the square of 21?
Wait a bit before you start multiplying.
The square of 20 is 400, isn’t it? So to get the square of 21, we need only add an odd number.
Which odd number?
Let’s start from the beginning. We can write
22 = 12 + 3 = 12 + (1 + 2)
32 = 22 + 5 = 22 + (2 + 3)
42 = 32 + 7 = 32 + (3 + 4)
52 = 42 + 9 = 42 + (4 + 5)
and so on. Continuing like this, how do we write 212?
212 = 202 + (20 + 21)
That is,
212 = 400 + 41 = 441
Now we can continue as before with
222 = 441 +43 = 484
and so on.
How do we find out the square of 101?
1002 = 10000
What more should we add?
100 + 101 = 201
So, 1012 = 10000 + 201 = 10201

• Find out the squares of these numbers using the above idea.
• 51
Answer:
Using the above process we write the 512 = 502+(50+51)
= 2500+(50+51)
= 2500+101
= 2601

• 61
Answer:
It can be written as 612 = 602+(60+61)
= 3600+(60+61)
= 3600+121
= 3721

• 121
Answer:
The given number is written as 1212 = 1202+(120+121)
= 14400+(120+121)
= 14400+241
= 14641

• 1001
Answer:
It is written as 10012 = 10002+(1000+1001)
= 10002+(1000+1001)
= 1000000+2001
= 1002001

• Compute the squares of natural numbers from 90 to 100.
Answer:
902
= 90 x 90
=8100
912
= 91 x 91
= 8281
922
= 92 x 92
= 8464
932
= 93 x 93
= 8649
942
= 94 x 94
= 8836
952
= 95 x 95
= 8025
962
= 96 x96
= 8928
972
= 97 x 97
= 9409
982
= 98 x 98
= 9310
992
= 99 x 99
= 9801
1002
= 100 x 100
= 10,000

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Fraction squares Textbook Page No. 83

A fraction multiplied by itself is also a square.
What is the square of \(\frac{3}{4}\) ?
(\(\frac{3}{4}\))2 = \(\frac{3}{4}\) × \(\frac{3}{4}\) = \(\frac{3 \times 3}{4 \times 4}\) = \(\frac{9}{16}\)
That is,
(\(\frac{3}{4}\))2 = \(\frac{9}{16}\) = \(\frac{3^{2}}{4^{2}}\)
So to square a fraction, we need only square the numerator and denominator separately.

Now do these problems without pen and paper.
• Find out the squares of these numbers.
• \(\frac{2}{3}\)
Answer:
\(\frac{2}{3}\)2  = \(\frac{2}{3}\)*\(\frac{2}{3}\)
= \(\frac{2×2}{3×3}\)
= \(\frac{4}{9}\)

• \(\frac{1}{5}\) 
Answer:
\(\frac{1}{5}\)2  = \(\frac{1}{5}\)*\(\frac{1}{5}\)
= \(\frac{1×1}{5×5}\)
= \(\frac{1}{25}\)

• \(\frac{7}{3}\)
Answer:
\(\frac{7}{3}\)2  = \(\frac{7}{3}\)*\(\frac{7}{3}\)
= \(\frac{7×7}{3×3}\)
= \(\frac{49}{9}\)

• 1\(\frac{1}{2}\)
Answer:
1\(\frac{1}{2}\)2  = \(\frac{1}{2}\)*\(\frac{1}{2}\)
=1 \(\frac{1×1}{2×2}\)
=1 \(\frac{1}{4}\)
= \(\frac{5}{4}\)

• Which of the fractions below are squares?

• \(\frac{4}{15}\)
Answer:
Numerator is 4 and Denominator is 15 .
Both are not perfect Square numbers so it is not possible to write the given fraction as squares.

• \(\frac{8}{9}\)
Answer:
Numerator is 8 and Denominator is 9.
Here 8 is not a perfect Square number and Denominator is a perfect square number. so it is not possible to write the given fraction as squares.

• \(\frac{16}{25}\)
Answer:
In a given fraction, numerator is 16 and Denominator is 25 which were square numbers of 4 and 5.Hence we can Write as \(\frac{4}{5}\)2

• 2\(\frac{1}{4}\)
Answer:
When the given 2\(\frac{1}{4}\) is converted to improper fraction, we get \(\frac{9}{4}\).
Here the Numerator is 9 and Denominator is 4
We can write the numerator as perfect square i.e. (3)2
Denominator 4 can be written as (2)2
Therefore, 2\(\frac{1}{4}\) can be written as \(\frac{3}{2}\)2

• 4\(\frac{1}{9}\)
Answer:
When the given 4\(\frac{1}{9}\) is converted to improper fraction, we get \(\frac{1}{9}\).
Here the Numerator is 1 and but the Denominator is 9 as a square of number (3)2
So, we cannot write the given fraction as Squares.

• \(\frac{8}{18}\)
Answer:
Numerator is 8 and Denominator is 18.
Here numerator and Denominator are not perfect square numbers so there is no chance to write the given fraction as square of numbers.

Decimal squares

What is the square of 0.5?
We know that 52 = 25. How many decimal places would be there in the product 0.5 × 0.5?
Why?
0.5 = \(\frac{5}{10}\), right?
Can you find out the square of 0.05?
Answer:
0.052 =0.05×0.05
= 0.0025 Hence four decimal places are there.
0.0025 = \(\frac{25}{1000}\)

You have computed the squares of many natural numbers. Using that table, can you find out the square of 0.15?
Do these problems also in your head.

• Find out the squares of these numbers.
Answer:
0.152 = 0.15 x 0.15
= 0.0225

• 1.2
Answer:
1.22 = 1.2 x 1.2
= 1.44

• 0.12
Answer:
0.122 = 0.12 x 0.12
= 0.0144

• 0.013
Answer:
0.0132 = 0.013 x 0.013
= 0.000169

• Which of the following numbers are squares?

• 2.5
Answer:
We cannot write the number 2.5 as a square.

• 0.25
Answer:
0.25 = 0.5 x 0.5.
Hence the number 0.25 is written as 0.52
• 0.0016
Answer:
0.0016 = 0.04 x 0.04.
Hence the number 0.0016 is written as 0.042

• 14.4
Answer:
We cannot write the number 14.4 as square.

• 1.44
Answer:
1.44= 1.2×1.2
Hence the number 1.44 is written as 1.22

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Square product Textbook Page No. 84

What is 52 × 42?
52 × 42 = 25 × 16 = ……..
There is an easier way:
52 × 42 = 5 × 5 × 4 × 4
= (5 × 4) × (5 × 4)
= 20 × 20
= 400
= (5 x 4) x (5 x 4)
= 20 x 20
= 400

Can you find out the products below like this, without pen and paper?

• 52 × 82
Answer:
52 × 82 = 5 x 5 x 8 x 8
= (5 x 8) x (5 x 8)
= 40 x 40
= 1600

• 2.52 × 42
Answer:
2.52 × 42 = 2.5 x 2.5 x 4 x 4
= (2.5 x 4) x (2.5 x 4)
= 10 x 10
= 100

• (1.5)2 × (0.2)2
Answer:
(1.5)2 × (0.2)2 = 1.5 x 1.5 x 0.2 x 0.2
= (1.5 x 0.2) x (1.5 x 0.2)
= 0.3 x 0.3
= 0.009

What general rule did we use in all these?
The product of the squares of two numbers is equal to the square of their product.
How do we say this in algebra?
x2y2 = (xy)2, for any numbers x, y
What about for three numbers?
Answer:
let the three numbers be x, y, and z
x2y2z2 = (xyz)2, for any numbers x, y, and z.

Square factors

How do we write 30 as a product of prime numbers?
30 = 2 × 3 × 5
So how do we factorize 900?
900 = 302 = (2 × 3 × 5)2 = 22 × 32 × 52
Similarly, using the facts that 24 = 23 × 3 and 242 = 576, we get
576 = 242 = (23 × 3)2 = (23)2 × 32 = 26 × 32

Can you write each number below and its square as a product of prime powers?
• 35
Answer:
35 = 5 x 7
Thus 35 can be written as product of prime powers 51 x 71

• 45
Answer:
45 = 5 x 9
= 5 x 3 x 3
= 51 x 32
Thus, 45 can be written as 51 x 32

• 72
Answer:
72 = 24 x 3
= 8 x 3 x 3
= 2 x 2 x 2 x 3 x 3
= 23  x 32
Thus, 72 can be written as 23  x 32

• 36
Answer:
36 = 9 x 4
= 9 x 2 x 2
= 3 x 3 x 2 x 2
= 32 x 22
Thus, 36 can be written as 32 x 22

• 49
Answer:
49 = 7 x 7
= 72
Thus, 49 can be written as 72

Did you note any peculiarity of the exponents of the factors of the squares?
Answer:

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Reverse computation

We have to draw a square; and its area must be 9 square centimetres.
How do we do it?
The area of a square is the square of the side.

So if the area is to be 9 square centimetres, what should be the side?
To draw a square of area 169 square centimetres, what should be the length of a side?
For that, we must find out which number squared gives 169. Looking up our table of squares, we find 132 = 169. So we must draw a square of side 13 centimetres.

Here, given a number we found out which number it is the square of. This operation is called extracting the square root.
That is, instead of saying the square of 13 is 169, we can say in reverse that the square root of 169 is 13.
Just as we write
132 = 169
as shorthand for statement “the square of 13 is 169”, we write the statement “the square root of 169 is 13” in shorthand form as
\(\sqrt{169}\) = 13
(the extraction of square root is indicated by the symbol Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root 9)
Similarly, the fact that the square of 5 is 25 can also be stated, the square root of 25 is 5. In short hand form,
52 = 25
\(\sqrt{25}\) = 5
In general
For numbers x and y, if x2 = y, then \(\sqrt{y}\) = x

Now find out the square root of these numbers:

• 100
Answer:
10 = 100
\(\sqrt{100}\) = 10
Therefore, square root of 100 is 10

• 256
Answer:
162 = 256
\(\sqrt{256}\) = 16
Thus, the square root of 256 is 16.

• \(\frac{1}{4}\)
Answer:
\(\frac{1}{4}\) = \(\frac{1}{2}\)2
Thus, the Square root of \(\frac{1}{4}\) = \(\frac{1}{2}\)

• \(\frac{16}{25}\)
Answer:
\(\frac{16}{25}\) = \(\frac{4}{5}\)2
Thus, the square root of \(\frac{16}{25}\) is \(\frac{4}{5}\).

• 1.44
Answer:
1.44 = (1.2)2
Thus, the square root of 1.44 is 1.2

• 0.01
Answer:
0.01 = (0.1)2
Thus, the square root of 0.01 is 0.1

Square root factors Textbook Page No. 87

How do we find the square root of 1225?
Since a product of squares is the square of the product, we need only write 1225 as a product of squares.
First factorize 1225 into primes:
1225 = 52 × 72
And we can write
52 × 72 = (5 × 7)2 = 352
So, 1225 = 352
From this, we get \(\sqrt{1225}\) = 35
Let’s take another example. What is the \(\sqrt{3969}\) ?
As before, we first factorize 3969 into primes.
3969 = 32 × 32 × 72
= (3 × 3 × 7)2
From this, we get \(\sqrt{3969}\) = 3 × 3 × 7 = 63

Now compute the square roots of these.
• 256
Answer:
Given number is 256
Firstly factorizing it into primes we have 256 = 2 x 128
= 2 x 2 x 64
= 2 x 2 x 2 x 32
= 2 x 2 x 2 x 2 x 16
= 2 x 2 x 2 x 2 x 2 x 8
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= (2 x 2) x (2 x 2)  x (2 x 2) x (2 x 2)
= 4 x 4 x 4 x 4
= 16 x 16
= (16)2
From this, we get \(\sqrt{256}\) = 16

• 2025
Answer:
Given number is 2025
Writing the factorization of 2025 we have
= 3 x 3 x 3 x 3 x 5 x 5
= 9 x 9 x 5 x 5
= (9 x 5)2
= (45)2
Therefore, the square root of \(\sqrt{2025}\) is 45

• 441
Answer:
Given number is 441
Writing the factorization of 441 we have
= 3 x 3 x 7 x 7
= 32 x 72
= (3 x 7)2
= (21)2

Therefore, the square root of 441 is 21.

• 921
Answer:
Writing the factorization of 921 we have 3 x 307
Thus, 921 can’t be written as perfect square and the square root of 921 isn’t a natural number.

• 1089
Answer:
Given number is 1089
Writing the factorization of it we have 1089 = 3 x 3 x 11 x 11
= 32 x 112
= (3 x 11)2
= (33)2
Therefore, square root of 1089 is 33.

• 15625
Answer:
Given number is 15625
Writing the factorization of it we have 15625 = 5 x 5 x 5 x 5 x 5 x 5
= 52 x 52 x 52
= (5 x 5 x 5)2
= (125)2
Therefore, the square root of 15625 is 125

• 1936
Answer:
Given number is 1936
Writing the factorization of 1936 we have 2 x 2 x 2 x 2 x 11 x 11
= (22 x 22 x 112)
= (2 x 2 x 11)2
= 442
Therefore, the square root of 1936 is 44.

• 3025
Answer:
Given number is 3025
Writing the factorization of 3025 we have 5 x 5 x 11 x 11
= 52 x 112
= (5 x 11)2
= 552
Therefore, the square root of 3025 is 55.

• 12544
Answer:
Given number is 12544
Writing the factorization of 12544 we have 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7
= 28 x 72
= (112)2
Therefore, Square Root of 12544 is 112.

Kerala Syllabus 7th Standard Maths Solutions Chapter 6 Square and Square Root

Let’s do it!

Question 1.
The area of a square plot is 1024 square meters. What is the length of its sides?
Answer:
Given area of Square plot = 1024 Square Meters
Imagine ‘S’ is side of a Square. As we know Area of Square = Side x Side
1024 = Side x Side
1024 = Side2
Side = √1024
Side = 2 5
Hence Length of Square Plot = 32 meters

Question 2.
In a hall, 625 chairs are arranged in rows and columns, with the number of rows equal to the number of columns. The chairs in one row and one column are removed. How many chairs remain?
Answer:
Total number of chairs = 625
Number of Rows and Columns present are Equal i.e. Rows =  Columns , Let it be y.
Rows .Columns=625
y. y =625
y 2= 625
y = √625
y = √(25)2
y = 25
so Number of Rows and Number of Columns =25
When 1 Row and 1 Column is removed as below
Number of Rows =25-1=24
Number of Rows =25-1=24
When the 24 Rows and 24 Columns are arranged = 24 x 24
= 576

Question 3.
The sum of a certain number of consecutive odd numbers, starting with 1, is 5184. How many odd numbers are added?
Answer:
We know the formula of sum of arithmetic series Sn = n/2[2a+(n-1)d]
The arithmetic series 1, 3, 5, 7, 9, 11, 13 ……
Here the first term a = 1
Number of odd numbers to be added = n
Common difference d = 2
Substituting in the formula above we have
5184 = n/2[2×1+(n-1)2]
5184 = n/2[2+2n-2]
5184 = n/2[2n]
5184 = n2
n = 72
Thus, the number of odd numbers to be added is 72.

Question 4.
The sum of two consecutive natural numbers and the square of the first is 5329. What are the numbers?
Answer:
Suppose x and x+1 are two Consecutive natural numbers
As per the Given data, Square of First number x2  = 5329
When they are added it gives as  x+(x+1) +x2 = 5329
x + x+1+x2 = 5329
x + x+1+x = 5329
2x +1+x = 5329
(x+1)= 5329
(x+1) = √5329
x+1 = √(73)2
x+1 = 73
x = 73-1
x= 72
Therefore, the numbers are 72 and 73.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

You can Download Ratio Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Ratio Text Book Questions and Answers

Same shape Textbook Page No. 116

In both the rectangles below, the length is 1 centimetre more than the breadth.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 1
But not only the length of these are different, they look different also. In the larger rectangle, the sides look almost the same. Draw a rectangle of length 50 centimetres and breadth 49 centimetres in a larger sheet of paper. It looks almost a square, doesn’t it?
In the first rectangle above, length is double the breadth. Now look at this rectangle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 2
Again the length is double the breadth. Even though it is larger than the first, they have the same shape, right?
Answer:
As the length is double the breadth in the second rectangle. The first and the second rectangle will have the same shape.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Changing scale

Look at this photo:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 7
The shorter side is 2 centimetres and the longer side, 3 centimetres; that is, the longer side is 1\(\frac{1}{2}\) times shorter side.
Suppose we make the shorter side 3 centimetres and longer side 4.5 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 8
Still, the longer side is 1\(\frac{1}{2}\) times the shorter side.
Now suppose we change the shorter side to 3 centimetres and increase the longer side also by 1 centimetre, making it 4 centimeters.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 9
Does the picture look right?
Answer:
Yes, the picture looks right.
Suppose if we change the shorter side to 3 centimetres and the longer side to 4 centimetres. The longer side is 1\(\frac{1}{3}\) times the shorter side.

TV Math

The sizes of TV sets are usually given as 14 inch, 17 inch, 20 inch and so on. What does it mean?
The TV screen is a rectangle: and these are lengths of the diagonals of the screen.
Does it determine the size of the screen? Rectangles of different width and lengths can have the same diagonal:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 10
In the modern TV sets, the ratio of length to width is 16 : 9. In the earlier days, it was 4 : 3.
See their difference in two TV screens of the same diagonal length.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 11

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Flags Textbook Page No. 119

When we draw our National Flag, not only should the colours be right, the ratio of width to length should also be correct. This ratio is 2 : 3.
That is, in drawing our flag, if the length is taken as 3 centimetres, the width should be 2 centimetres.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 14
This ratio is different for flags of other countries. For example, in the flag of Australia, this ratio is 1 : 2.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 15
And in the flag of Germany, it is 3 : 5.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 16

Without fractions Textbook Page No. 120

When quantities like length are measured using a definite unit, we may not always get counting numbers; and it is this fact which led to the idea of fractions.

In comparing the sizes of two quantities, one question is whether both can be given as counting numbers, using a suitably small unit of measurement. It is this question that leads to the idea of ratio.

For example, suppose the length of two objects are found as \(\frac{2}{5}\) and \(\frac{3}{5}\), when measured using a string. If \(\frac{1}{5}\) of the string is taken as the unit, we can say the length of the first is 2 and that of the second is 3. This is the meaning of
saying the ratio of the lengths is 2 : 3.
Suppose the length of two objects are \(\frac{1}{3}\) and \(\frac{1}{5}\) of the string.
To get both lengths as counting numbers, what fraction of the string can be taken as a unit of measurement?

Circle relations

Look at the pieces of circles below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 19

The smaller piece is \(\frac{1}{4}\) of the circle and the larger piece is \(\frac{1}{2}\) of the circle.
So, the larger piece is twice the size of the smaller. That is, the ratio of the sizes of small and large pieces is 1 : 2.
Now look at these pieces:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 20
What is the ratio of their sizes?
Let’s measure each using \(\frac{1}{4}\) of circles.
The smaller figure has two such pieces. What about the larger?
Answer:
Let’s measure each using \(\frac{1}{4}\) of circles.
The larger figure has three such pieces.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 21
So what is the ratio of the sizes of these two?
Answer:
Let’s measure each using \(\frac{1}{4}\) of circles.
The smaller piece has two such pieces and the larger piece has three such pieces.
2×\(\frac{1}{4}\) : 3×\(\frac{1}{4}\)
\(\frac{2}{4}\) : \(\frac{3}{4}\)
Therefore, the ratio will be 2 : 3

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Moving Ratio Textbook Page No. 122
Have you taken apart toy cars or old clocks? There are many toothed wheels in such mechanisms. See this picture:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 22
It is a part of a machine. In the two whole wheels we see, the smaller has 13 teeth and the larger has 21. So during the time that the smaller circle makes 21 revolutions, the larger one would have made only 13 revolutions.
The speed of rotation of machines is controlled by arranging such wheels with different number of teeth.

Sand and Cement

In construction of a building, sand and cement are used in definite ratios, but the ratios are different depending on the purpose. When one bowl of cement and five bowls of sand are mixed, the ratio of cement to sand is 1 : 5

When one sack of cement is mixed with five sacks of sand, the ratio is the same. But to set a brick wall, this much cement may not be needed. In this case, the ratio may be 1 : 10 or 1 : 12.

Ratio of Parts Textbook Page No. 124

We can use ratio to compare parts of a whole also. For example, in the picture below, the lighter part is \(\frac{3}{8}\) of the circle, and the darker part is \(\frac{5}{8}\) of the circle.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 23
These two parts together make up the whole circle. The ratio of the sizes of these parts is 3 : 5.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 24
So here, the ratio 3 : 5 indicate the two fractions \(\frac{3}{8}\) and \(\frac{5}{8}\).

Generally in such instances, a ratio of the numbers indicates fractions of equal denominators adding up to 1.

Meaning of ratio Textbook Page No. 125

If we know only the ratio of two quantities, we can’t say exactly how much they are; but they can be compared in several ways.
For examples, suppose the volumes of two pots are said to be in the ratio 2 : 3. We can interpret this in the following ways:

  • To fill the smaller pot, we need \(\frac{2}{3}\) of the larger pot.
  • To fill the bigger pot, we need \(\frac{3}{2}\) = 1\(\frac{1}{2}\) times the smaller pot.
  • Whether we take \(\frac{1}{2}\) of the water in the smaller pot, or \(\frac{1}{3}\) of the water in the larger pot, we get the same amount of water.
  • If we fill both pots and pour them into a large pot, \(\frac{2}{5}\) of the total amount of water
    is from the smaller pot and \(\frac{3}{5}\) from the larger.

If the length of two pieces of rope are said to be in the ratio 3 : 5, what all interpretations like this can you make?
Answer:
If the length of two pieces of rope is 3 : 5.Then the following interpretations can be made.
The smaller rope is \(\frac{3}{5}\) of the length of the larger rope.
The smaller rope will be \(\frac{3}{8}\) of the total rope length.

Three measures

Look at this triangle:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 25
In it, the longest side is double the shortest. The side of medium length is one and a half times the shortest.
Saying this with ratios, the lengths of the shortest and the longest sides are in the ratio 1 : 2 and those of the shortest and the medium are in the ratio 2 : 3.
What is the ratio of the lengths of the medium side and the longest side?
We can say this in another way: if we use a string of length 1.5 centimetres, the shortest side is 2, the medium side is 3 and the longest side is 4.
The can be shortened by saying the sides are in the ratio 2 : 3 : 4.

Triangle Math Textbook Page No. 127

How many triangles are there with ratio of sides 2 : 3 : 4?
The lengths of sides can be 2 cm, 3 cm, 4cm.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 26
Or they can be 1 cm, 1.5 cm, 2 cm
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 27
We can take metres instead of centimetres.
Thus we have several such triangles.
In all these, what fraction of the perimeter is the shortest side?
And the side of medium length?
The longest side?
The perimeter of a triangle is 80 centimetres and its sides are in the ratio 5 : 7 : 8. Can you compute the actual lengths of sides?
Answer:
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 80 cm.
5x + 7x + 8x = 80
20x = 80
x = 4
Since the value of x is 4, the lengths will be 5×4=20cm, 7×4=28cm and 8×4=32cm.

What if the perimeter is 1 metre?
Answer:
1 metre equals 100 centimetres.
Let the ratio be x.Then the sides will be 5x, 7x and 8x.
Given that perimeter is 100 cm.
5x + 7x + 8x = 100
20x = 100
x = 5
Since the value of x is 5, the lengths will be 5×5=25cm, 7×5=35cm and 8×5=40cm.

Length and width Textbook Page No. 116

Look at these rectangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 3
Is there any common relation between the lengths and widths of all these?
In all these, the length is twice the width, right? (We can also say width is half the length)
In the language of mathematics, we state this fact like this:
In all these rectangles, the width and length are in the ratio one to two.
In writing, we shorten the phrase “one to two” as 1 : 2; that is
In all these rectangles, the width and length are in the ratio 1 : 2.

In the rectangle of width 1 centimetre and length 2 centimetres, the length is twice the width. In a rectangle of width 1 metre and length 2 metres also, we have the same relation. So, in both these rectangles, width and length are in the ratio one to two (1 : 2).
We can state this in reverse: in all these rectangles, length and width are in the ratio two to one (2 : 1).
What is the width to length ratio of the rectangle below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 4
Answer:
The length of the given rectangle is 6cm and width is 2cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.

And what about this rectangle?
Answer:
The length of the given rectangle is 4.5cm and width is 1.5cm.
Therefore, the width to length ratio for the above rectangle will be 1 : 3.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 5
In both, the length is 3 times the width. So what is the ratio of the width to the length?
Answer:
In the above two rectangles, the length is thrice the width.
The ratio of the width to the length is 1 : 3.

How do we say this as a ratio?
Answer:
The width and length are in the ratio one to three.

1 metre means loo centimetres. So in such a rectangle, the ratio of the width to the length is 1: 50.
Now look at these rectangles:
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 6

In both, the length is one and a half times the width.
How do we say this as a ratio?
We can say one to one and a half; but usually we avoid fractions when we state ratios.
Suppose we take the width as 2 centimetres.
What is 1\(\frac{1}{2}\) times 2?
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 12
So, we say that in rectangles of this type, the ratio of width to length is two to three and write 2 : 3.
Can’t we say here that the ratio is 4 : 6 ?
Nothing wrong in it; but usually ratios are stated using the least possible counting numbers.
So how do we state using ratios, the fact that the length of a rectangle is two and a half times the width?

If the width is 1 centimetre, then the length is 2\(\frac{1}{2}\) centimetres.
What if the width is 2 centimetres?
Length would be 5 centimetres.
So, we say that width and length are in the ratio 2 to 5.

What if the length is one and a quarter times the width?
If the width is 1 centimetre, the length is 1\(\frac{1}{4}\) centimetre.
If the width is 2 centimetres then the length is 2\(\frac{1}{2}\) centimetres.
Still we haven’t got rid of fractions.
Now if width is 4 centimetres, what would be the length?

So, in all such rectangles, the width and length are in the ratio 4 : 5.
Do you notice another thing in all these?
If we stretch or shrink both width and length by the same factor, the ratio is not changed. For example, look at this table.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 13
In all these, the length is 3 times the width; or the reverse, the length is \(\frac{1}{3}\) of the width.
In terms of ratio, we say width and length are in the ratio 1 : 3 or length and width are in the ratio 3 : 1.
In terms of ratio, we say width and length are in the ratio 1: 3 or length and width are in the ratio 3: 1.

• For all rectangles of dimensions given below, state the ratio of width to length, using the least possible counting numbers:

• width 8 centimetres, length 10 centimetres
Answer:
The simplest form for the given numbers 8:10 is 4 : 5
Therefore, the ratio of width to length will be 4 : 5

• width 8 metres length 12 metres
Answer:
The simplest form for the given numbers 8:12 is 4 : 3
Therefore, the ratio of width to length will be 2 : 3

• width 20 centimetres length 1 metre
Answer:
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 20 : 1

• width 40 centimetres length 1 metre
Answer:
The given numbers cannot be written in simplest form.
Therefore, the ratio of width to length will be 40 : 1

• width 1.5 centimetres length 2 centimetres
Answer:
The simplest form for the given numbers 1.5:2 is 3 : 4
Therefore, the ratio of width to length will be 3 : 4

• In the table below two of the width, length and their ratio of some rectangles are given. Calculate the third and fill up the table.
Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 17
Answer:
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_1

• What does it mean to say that the width to length ratio of a rectangle is 1: 1 ? What sort of a rectangle is it?
Answer:
The width to length ratio of a rectangle is 1 : 1.
Here the width of the rectangle and length of the rectangle is same.
If the width and the length of rectangle is same, then it is a square.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Other quantities Textbook Page No. 120

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio 18

There are two ropes, the shorter \(\frac{1}{3}\) metre long and the longer, \(\frac{1}{2}\) metre long. What is the ratio of their lengths?

We can do this in different ways.
We can check how much times \(\frac{1}{3}\) is \(\frac{1}{2}\).
\(\frac{1}{2}\) ÷ \(\frac{1}{2}\) = \(\frac{3}{2}\)
So, the length of the longer rope is \(\frac{3}{2}\) times that of the shorter.
That is, 1\(\frac{1}{2}\) times.
If the length of the shorter rope is taken as 1, the length of the longer is 1\(\frac{1}{2}\); if 2, then 3.
So the length of the shorter and longer rope are in the ratio 2 : 3.

We can also think in another manner. As in the case of the width and length of rectangles, we can imagine the length to be stretched by the same factor; and the ratio won’t change.
Suppose we double the length of each piece of rope.
Then the length of the shorter is \(\frac{2}{3}\) metres and that of the longer 1 metre. This doesn’t remove fractions.

By what factor should we stretch to get rid of fractions? How about 6 ?
6 times \(\frac{1}{3}\) is 2.
6 times \(\frac{1}{2}\) is 3.
The shorter is now 2 metres and the longer is 3 metres. So the ratio is 2 : 3.
There is yet another way; we can write
\(\frac{1}{3}\) = \(\frac{2}{6}\) \(\frac{1}{2}\) = \(\frac{3}{6}\)
That is, we can think of the shorter rope as made up of 2 pieces of \(\frac{1}{6}\) metres each and the longer one as 3 of the same \(\frac{1}{6}\) metres. In this way also, we can calculate the ratio as 2 to 3.
Now look at this problem : to fill a can we need only half the water in a bottle.

To fill a larger can, we need three quarters of the bottle. What is the ratio of the volume of the smaller can to the larger can ?
Here, we can write
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So, if we pour \(\frac{1}{4}\) of the bottle 2 times, we can fill the smaller can; to fill the larger can, \(\frac{1}{4}\) of the bottle should be poured 3 times. So the volumes of the smaller and larger cans are in the ratio 2 : 3.

Another problem: Raju has 200 rupees with him and Rahim has 300. What is the ratio of the money Raju and Rahim have?
If we imagine both to have the amounts in hundred rupee notes, then Raju has 2 notes and Rahim, 3 notes. So the ratio is 2 : 3.

Let’s change the problem slightly and suppose Raju has 250 rupees and Rahim, 350 rupees.
If we think of the amounts in terms of 50 rupee notes, then Raju has 5 and Rahim has 7.
The ratio is 5 : 7.
What if the amounts are 225 and 325 rupees?

Imagine each amount as packets of 25 rupees. Raju has 225 ÷ 25 = 9 packets and Rahim has 325 ÷ 25 = 13 packets. The ratio is 9 : 13.
Let’s look at one more problem: in a class, there are 25 girls and 20 boys. What is the ratio of the number of girls to the number of boys?
If we split the girls and boys separately into groups of 5, there will be 5 groups of girls and 4 groups of boys. So the ratio is 5 to 4.

In this way, calculate the required ratios and write them using the least possible counting numbers, in these problems:

• Of two pencils, the shorter is of length 6 centimetres and the longer, 9 centimetres. What is the ratio of the lengths of the shorter and the longer pencils?
Answer:
Given that the length of shorter pencil is 6 centimetres and the longer is 9 centimetres.
The simplest form of 6 and 9 is 2 and 3.
In terms of ratio, the ratio of the lengths of the shorter and the longer pencils 2 : 3.

• In a school, there are 120 boys and 140 girls. What is the ratio of the number of boys to the number of girls?
Answer:
Given that there are 120 boys and 140 girls in a school.
The simplest form of 120 and 140 is 6 and 7.
In terms of ratio, it will be 6 : 7

• 96 women and 144 men attended a meeting. Calculate the ratio of the number of women to the number of men.
Answer:
Given that 96 women and 144 men attended a meeting.
The simplest form of 96 and 144 is  and 22 and 3.
In terms of ratio, it will be 2 : 3

• When the sides of a rectangle were measured using a string, the width was \(\frac{1}{4}\) of the string and the length \(\frac{1}{3}\) of the siring. What is the ratio of the width to the length’?
Answer:
Given that the width of rectangle is \(\frac{1}{4}\) of the string and the length is \(\frac{1}{3}\) of the siring.
The ratio of the width to the length will be 3 : 4

• To fill a larger bottle, 3\(\frac{1}{2}\) glasses of water are needed and to fill a smaller bottle, 2\(\frac{1}{4}\) of glasses are needed. What is the ratio of the volumes of the large and small bottles?
Answer:
Given that to fill a larger bottle 3\(\frac{1}{2}\) glasses of water are needed and to fill a smaller bottle, 2\(\frac{1}{4}\) of glasses are needed.
3\(\frac{1}{2}\) equals \(\frac{7}{2}\) and 2\(\frac{1}{4}\) equals \(\frac{9}{4}\).
The ratio of the volumes of the large and small bottles will be 14 : 9

Ratio of mixtures Textbook Page No. 123

To make idlis , Ammu’s mother grinds two cups of rice and one cup of urad dal. When she expected some guests the next day, she took four cups of rice. How many cups of urad should she take?
To have the same consistency and taste, the amount of urad must be halfthe amount of rice.
So for four cups of rice, there should be two cups of urad.
We can say that the quantities, rice and urad must be in the ratio 2 :1
Now another problem on mixing: to paint the walls of Abu’s home, first 25 litres of green and 20 litres of white were mixed.

When this was not enough, 15 litres of green was taken. How much white should be added to this?
To get the same final colour, the ratio of green and white should not change.
In what ratio was green and white mixed first?
That is, for 5 litres of green, we should take 4 litres of white.
To maintain the same ratio, for 15 litres of green, how many litres of white should we take?
How many times 5 is 15?
So 3 times 4 litres of white should be mixed.
That is 12 litres.
To get the same shade of green, how many litres of green should be mixed with 16 litres of white?

Now try these problems:

• For 6 cups of rice, 2 cups of urad should be taken to make dosas. How many cups of urad should taken with 9 cups of rice?
Answer:
Given that 6 cups of rice, 2 cups of urad should be taken to make dosas.
In terms of ratio, the ratio of the rice to urad will be 3 : 1.
Here, the number of rice cups is thrice the number of urad cups.
Therefore, to maintain the same ratio, for 9 cups of rice, 3 cups of urad should be taken.

• To set the walls of Nizar’s house, cement and sand were used in a ratio 1:5. He bought 45 sacks of cement. How many sacks of sand should he buy?
Answer:
Given that to set the walls of Nizar’s house, cement and sand were used in a ratio 1:5.
Here, 1 resembles the number of units for cement sacks and 5 resembles the number of unit for sand sacks.
The number of sand sacks required is 5 times the number of cement sacks.
If there are 45 sacks of cement, then the number of sand sacks required will be 45×5=225.

• To paint a house, 24 litres of paint was mixed with 3 litres of turpentine. How many litres of turpentine should be mixed with 32 litres of paint?
Answer:
Given that to paint a house, 24 litres of paint was mixed with 3 litres of turpentine.
In terms of ratio, the ratio of the paint to turpentine will be 8 : 1.
Let x be the number of litres of turpentine that should be mixed with 32 litres of paint.
\(\frac{8}{1}\) = \(\frac{32}{x}\)
8x = 32
x = \(\frac{32}{8}\)
x = 4
Therefore, 4 litres of turpentine should be mixed with 32 litres of paint.

• In the first ward of a panchayat, the male to female ratio is 10 : 11. There are 3311 women. How many men are there? What is the total population in that panchayat?
Answer:
Given that In the first ward of a panchayat, the male to female ratio is 10 : 11.
Let the number of men be m and there are 3311 women.
\(\frac{10}{11}\) = \(\frac{m}{3311}\)
11 × m  = 10 × 3311
m = \(\frac{10 × 3311}{11}\)
m = 3010
Therefore, the number of mens will be 3010.
The total population in that panchayat will be the sum of number of mens and womens, which is 3010+3311=6321

• In a school, the number of female and male teachers are in the ratio 5 : 1. There are 6 male teachers. How many female teachers are there?
Answer:
Given that, In a school, the number of female and male teachers are in the ratio 5 : 1.
Here, the number of female teachers is five times the number of male teachers.
If there are 6 male teachers.Then the female teachers will be 5×6=30.
Thus, there are 30 female teachers.

• Ali and Ajayan set up a shop together. Ali invested 5000 rupees and Ajayan, 3000 rupees. They divided the profit for a month in the ratio of their investments and Ali got 2000 rupees. How much did Ajayan get? What is the total profit?
Answer:
Given that, Ali invested 5000 rupees and Ajayan, 3000 rupees.
In terms of ratio, the ratio of Ali to Ajayan will be 5 : 3
Ali got 2000 rupees.Let m be the amount Ajayan should get.
\(\frac{5}{3}\) = \(\frac{2000}{m}\)
5×m = 2000×3
m = \(\frac{2000×3}{5}\)
m = 1200
Ajayan will get 1200 rupees.
The total profit will be the sum of maount got by Ali and Ajayan, which is 2000+1200=3200 rupees.

Kerala Syllabus 7th Standard Maths Solutions Chapter 9 Ratio

Division Problem Textbook Page No. 125

We have seen that to make idlis, rice and urad are to be taken in the ratio 2:1. In 9 cups of such a mixture of rice and urad, how many cups of rice are taken?
2 cups of rice and 1 cup of urad together make 3 cups of mixture.
Here we have 9 cups of mixture.
How many times 3 is 9 ?
To maintain the same ratio, both rice and urad must be taken 3 fold.
So 6 cups of rice and 3 cups of urad.

Another problem: In a co-operative society there are 600 members are male and 400 are female. An executive committee of 30 members is to be formed with the same male to female ratio as in the society. How many male and female members are to be there in the committee?
In the society, the male to female ratio is 3 : 2.
3 men and 2 women make 5 in all.
Here we need a total of 30.
How many times 5 to 30?
So there should be 3 × 6 = 18 men and 2 × 6 = 12 women in the committee.

One more problem: a rectangular piece of land is to be marked on the school ground for a vegetable garden. Hari and Mary started making a rectangle with a 24 metre long rope. Vimala Teacher said it would be nice, if the sides are in the ratio 3 : 5. What should be the length and width of the rectangle?
The length of the rope is 24 metres and this is the perimeter of the rectangle.
Answer:
Given that, Hari and Mary started making a rectangle with a 24 metre long rope and their teacher suggested to make it in 3 : 5 ratio.
Let the ratio be x. Then the sides will be 3x and 5x.
Perimeter will be the total length of the rope which is 24 metre long.
3x + 5x = 24
8x = 24
x = \(\frac{24}{8}\)
x = 3
Therefore, the ength and width of the rectangle will be 9 metre and 15 metre.

If we take the length and width as 3 meters and 5 metres, what would be the perimeter?
Answer:
If the length and width is 3 meters and 5 metres respectively.
Then the perimeter will be 3+5=8 metres.

How much of 16 is 24?
\(\frac{24}{16}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
So width should be 1\(\frac{1}{2}\) times 3 metres; that is
3 × 1\(\frac{1}{2}\) = 4\(\frac{1}{2}\) metres.
And length should be 1\(\frac{1}{2}\) times 5 metres, that is,
5 × 1\(\frac{1}{2}\) = 7\(\frac{1}{2}\) metres

Now try these problems:

• Suhra and Sita started a business together. Suhra invested 40000 rupees and Sita, 30000 rupees. They made a profit of 7000 rupees which they divided in the ratio of their investments. How much did each get?
Answer:
Suhra invested 40000 rupees and Sita, 30000 rupees in a business together.
In terms of ratio, the ratio of Suhra to Sita will be 4 : 3.
They made a profit of 7000 rupees.
Let the ratio be x, then the investment will be 4x and 3x.
4x + 3x = 7000
7x = 7000
x = 1000
Therefore, Suhra’s profit will be 4×1000=4000 rupees
Sita’s profit will be 3×1000=3000 rupees.

• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of their investments. How much did each get?
Answer:
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = \(\frac{6500}{13}\)
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees

• John and Ramesh took up a job on contract. John worked 7 days and Ramesh, 6 days. They got 6500 rupees as wages which they divided in the ratio of the numbers of days each worked. How much did each get?
Answer:
Given that John worked 7 days and Ramesh, 6 days.
In terms of ratio, the ratio for the number of days worked by John to Ramesh will be 7 : 6
Let the ratio be x, then the investment will be 7x and 6x.
7x + 6x = 6500
13x = 6500
x = \(\frac{6500}{13}\)
x = 500
Therefore, John will get 7×500 = 3500 rupees
Ramesh will get 6×500 = 3000 rupees

• Angles of a linear pair are in the ratio 4 : 5. What is the measure of each angle?
Answer:
Angles of a linear pair are in the ratio 4 : 5.
Let the ratio be x,then it will be 4x and 5x.
4x + 5x = 90
9x = 90
x =10
4x = 4 × 10 = 40
5x = 5 × 10 = 50
Thus, the angles measurement will be 40 degrees and 50 degrees.

• Draw a line AB of length 9 centimetres. A point P is to be marked on it, such that the lengths of AP and PB are in the ratio 1 : 2. How far from A should P be marked? Compute this and mark the point.
Answer:
Length of AB is 9 centimetres.
Point P is marked in 1 : 2 ratio.
Let the ratio be x, then it will be 1x and 2x, the total is 3x.
\(\frac{1x}{3x}\) × 9 = 3 centimetres
\(\frac{2x}{3x}\) × 9 = 6 centimetres
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_2
P should be marked 3 cm away from A.

• Draw a line 15 centimetres long. A point is to be marked on it, dividing the length in the ratio 2 : 3. Compute the distances and mark the point.
Answer:
Length of AB is 15 centimetres.
Point P is marked in 2 : 3 ratio.
Let the ratio be x, then it will be 2x and 3x, the total is 5x.
\(\frac{2x}{5x}\) × 15 = 6 centimetres
\(\frac{3x}{5x}\) × 15 = 9 centimetres
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_3
P should be marked 6 cm away from A.

• Sita and Soby divided some money m the ratio 3 : 2 and Sita got 480 rupees. What is the total amount they divided?
Answer:
Sita and Soby divided some money ‘m’ in the ratio 3 : 2 and Sita got 480 rupees.
Let the ratio be m, then the ratio of money Sita got will be 3m.
3m = 480
m = \(\frac{480}{3}\)
m = 160
Money Soby will get 2×160 = 320 rupees.

• In a right triangle, the two smaller angles are in the ratio 1 : 4. Compute these angles.
Answer:
In a right triangle, the two smaller angles are in the ratio 1 : 4.
Right triangle means 90 degrees.
Let the ratio be x, total will be 1x + 4x = 5x.
\(\frac{1x}{5x}\)×90 = 18 degrees
\(\frac{4x}{5x}\)×90 = 72 degrees
Then the angles will be 18 degrees and 72 degrees.

• Draw a rectangle of perimeter 30 centimetres and lengths of sides in the ratio 1 : 2. Draw two more rectangles of the same perimeter, with lengths of sides in the ratio 2 : 3 and 3 : 7. Compute the areas of all three rectangles.
Answer:
A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 1 : 2
Let the ratio be x, the total will be 1x + 2x = 3x.
One of the side will be \(\frac{1x}{3x}\)×30 = 10 centimetres
The other side will be \(\frac{2x}{3x}\)×30 = 20 centimetres
Therefore, the rectangle with 10 centimetres and 20 centimetres is drawn as shown below.
Area of this rectangle will be 10×5=50 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_7

A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 2 : 3
Let the ratio be x, the total will be 2x + 3x = 5x.
One of the side will be \(\frac{2x}{5x}\)×30 = 12 centimetres
The other side will be \(\frac{3x}{5x}\)×30 = 18 centimetres
Therefore, the rectangle with 12 centimetres and 18 centimetres is drawn as shown below.
Area of this rectangle will be 9×6=54 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_8

A rectangle of perimeter 30 centimetres is drawn with lengths of sides in the ratio 3 : 7
Let the ratio be x, the total will be 3x + 7x = 10x.
One of the side will be \(\frac{3x}{10x}\)×30 = 9 centimetres
The other side will be \(\frac{7x}{10x}\)×30 = 21 centimetres
Therefore, the rectangle with 9 centimetres and 21 centimetres is drawn as shown below.
Area of this rectangle will be 10.5×4.5=47.25 square centimetres.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 9 Ratio img_9

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles

You can Download Adding Angles Questions and Answers, Activity, Notes, Kerala Syllabus 7th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles

Adding Angles Text Book Questions and Answers

Joining angles Textbook Page No. 8

Can you draw this angle?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 1

Draw another angle on top like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 2

How many angles at A now?
∠CAB = _________
∠DAC = _________
Answer: ∠CAB = 30°
∠DAC = 45°

Explanation: Given the two figures, one figure is 30° Angles.
The other figure is 45° and 30°.
Now, we will find the angles of ∠CAB and ∠DAC.
Observe the figure properly, then we find it easily.
The angle between C, A, and B is 30°.
Next, the angle between D, A, and C is 45°.
So, the ∠CAB = 30°, and ∠DAC = 45°

Do you see one more large angle?
How much is it?
∠DAB = _______
How did you compute it?
∠DAB = 45° + 30° = 75°
In the figures below, the measures of two angles are shown. Write the third angle as a sum or difference and compute its measure:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 3
∠DAB = ____ + _____ = ______
Answer:∠DAB = 100°.

Explanation: Given the figure with two angles.
In between A, B, C, and D, we have two angles.
One angle is 60° and the other angle is 40°.
In a given question in between the angles, we have an addition symbol.
So, we find the third angle. We can add two angles we get the third angle.
so, the angles is 60° + 40° , ∠DAB = 100°.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 4
∠DAB = ____ + _____ = ______
Answer: ∠DAB = 90°.

Explanation: Given the figure with two angles.
In between A, B, C, and D, we have two angles.
One angle is 30° and the other angle is 60°.
In a given question in between the given angles, we have an addition symbol.
So, we find the third angle. We can add two angles we get the third angle.
so, the angles is 30° + 60° , ∠DAB = 90°.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 5
∠CAB = ____ – _____ = ______
Answer: ∠CAB = 30°.

Explanation: Given the figure with two angles.
In between A, B, C, and D, we have two angles.
The angle between D, A, and B is 80° and the angle between D, A, and C is 5`0°.
In the given question in between the values, we have a subtraction symbol.
So, we find the third angle. We can subtract two angles we get the third angle.
Thus the angle is 80° –  50° = 30°, ∠CAB = 30°.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 6
∠DAC = ____ – _____ = ______
Answer: ∠DAC = 75°.

Explanation: Given the figure with two angles.
In between A, B, C, and D, we have two angles.
The angle between D, A, and B is 120° and the angle between C, A, and B is 45°.
In the given question, we have a subtraction symbol.
So, we find the third angle. We can subtract two angles we get the third angle.
Thus, the angle is 120° –  45° = 75°, ∠DAC = 75°.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles

On both sides Textbook Page No. 9

Draw a line and a perpendicular to it as below:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 7

Now draw an angle within it like this:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 8

What is the measure of ∠DAC?
∠DAC = ____ – _____ = _____
Answer: ∠DAC = 40°.

Explanation: Given the figure,
In between D, A, and B, we have a 50°angle.
Now, we find the angle of DAC.
The given figure is the right angle which has an angle of 90°, so the angle CAB = 90°
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_image
In a given question, we have a subtraction symbol.
So, we find the third angle. We can subtract two angles we get the third angle.
Thus, the angle is 90° –  50° = 40°, ∠DAC = 40°.

Let’s stretch AB a bit to the left:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 9

How much is ∠DAE?
∠DAE = ____ + _____ = _____
Answer:∠DAE = 130°.

Explanation: Given the figure, we have two angles.
In between D, A, and B, we have a 50°angle.
In between C, A, and D we have a 40°angle.
Now, we find the angle of DAE.
The given figure is a right angle which has an angle of 90°. So, the angle CAE = 90°
In a given question, we have an addition symbol.
So, we find the third angle. We can add two angles 90° and 40°. We get the third angle.
Thus, the angle is 90°+ 40° = 130°, ∠DAE = 130°.

Do you see how ∠DAE is related to ∠DAB?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 10

Now look at this figure:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 11
Can you compute ∠DCA?
Answer: ∠DCA = 120°.

Explanation: Given the two figures, with an angle.
In between D, A, and B, we have a 50°angle.
In between D, C, and B, we have a 60°angle.
In a given question, it mentioned how the angle of DAE is related to DCA.
So, in the above question, the DAE value is 130°.
DAE + DAB = 130° + 50°= 180°
Now, we will find the value of DCA.
So, subtract the DCB value from 180°.
i.e., 180° – 60° = 120°.
The angle value of DCA = 120°

How about drawing a perpendicular at C and splitting this angle?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 12
How much is ∠DCE?
So, how much is ∠DCA?
∠DCE = 90° – 60° = 30°
∠DCA = 90° + 30° = 120°
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 13
Like this, can you compute the angle on the right in the figure below?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 14
Answer: Given the figure, with an angle.
In between D, C, and A, we have a 130°angle.
Linear pairs are angles formed in a line, their sum is equal to 180°.
Now, we will find the value of DCB.
So, subtract the DCA value from 180°.
i.e., 180° – 130° = 50°.
The angle value of DCB = 50°

In the figures below, the angles made on either side by joining two lines are shown; the measure of one angle is given. Compute the measure of the other:

(i)

Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Image 1

What do we see in all these?

Answer: Given the figure, with an angle.
The given angle is 100°
Linear pairs are angle formed in a line, and their sum is equal to 180°.
Now, we will find the value of the angle.
So, subtract 100°value from 180°.
i.e., 180° – 100° = 80°.
Therefore, the angle value is 80°.

(ii)

Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Image 2

What do we see in all these?

Answer: Given the figure, with an angle.
The given angle is 40°
Linear pairs angle formed in a line, their sum is equal to 180°.
Now, we will find the value of the angle.
So, subtract 40°value from 180°.
i.e., 180° – 40° = 140°.
Therefore, the angle value is 140°.

(iii)

Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Image 3

What do we see in all these?

Answer: Given the figure, with an angle.
The given angle is 50°
Linear pairs angle formed in a line, their sum is equal to 180°.
Now, we will find the value of the angle.
So, subtract 50° value from 180°.
i.e., 180° – 50° = 130°.
Therefore, the angle value is 130°.

(iv)
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 16
What do we see in all these?

Answer: Given the figure, with an angle.
The given angle is 80°
Linear pairs angle formed in a line, their sum is equal to 180°.
Now, we will find the value of the angle.
So, subtract the 80° value from 180°.
i.e., 180° – 80° = 100°.
Then, the angle value is 100°.

If a line is drawn from another line, then the sum of the angles on either side is 180°.
A pair of angles that got like this is called a linear pair.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles

Angle calculation Textbook Page No. 10

• How much is ∠ACE in this figure?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 17
Answer: The angle value of ∠ACE is 105°.

Explanation: Given that,
The value of ∠BCD = 25°.
The value of ∠DCE = 50°.
Now we will find the value of ∠ACE.
All these angles are in a line. So their sum is 180°.
So, ∠BCD + ∠DCE + ∠ACE = 180°.
25°+ 50° + ∠ACE = 180°.
75° + ∠ACE = 180°.
∠ACE = 180°-75°.
∠ACE = 105°.
So, the angle value of ∠ACE is 105°.

• How much is the angle between the two slanted lines in this figure?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 18
Answer: The angle between two slanted lines DBE is 75°.

Explanation: Given the figure. The below is the given with angles.
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_image(i)
Now, we will find the angle value of ∠DBE.
So, the angle value of ∠ABD = 45°
The angle value of ∠CBE = 60°.
We know that these angles are in a line. So the sum is 180°.
So, the value of ∠DBE is,
∠ABD + ∠CBE + ∠DBE = 180°.
45° + 60° + ∠DBE = 180°.
∠DBE = 180°- 45°- 60°
∠DBE = 75° .
So, the angle between two slanted lines DBE is 75°.

• In the figure below, ∠ACD = ∠BCE. How much is each?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 19
Answer: The angle of each is 105°.

Explanation: Given the figure,
The value of ∠ECD= 30°.
The value of ∠ACD and the value of ∠BCE is equal. Because both have the same angle distance.
So, ∠ACE = ∠BCD.
Here are the angles ∠ECD, ∠ACE, ∠and BCD in a line. So their sum is 180°.
∠ACE + ∠ECD + ∠BCD = 180°.
∠ACE = ∠BCD.
So, 2 ∠ACE + 30°   = 180°.
2 ∠ACE = 180°- 30°
2 ∠ACE = 150°.
∠ACE = 150°/2 = 75°
Now, the value of ∠ACE = 75°+ 30° = 105°.
So, ∠ACE = ∠BCD (∠BCD = 75°+ 30° = 105°).
Hence, the angle of each is 105°.

Cutting across Textbook Page No. 11

In the figure below, how much is the angle on the left?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 20

Suppose we extend the upper line downwards:
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 21
Now there are two more angles underneath.
How much is each?
The angles at the top and bottom, on the left of the slanted line, form a linear pair.
There is such a linear pair on the right also.
Now, can’t we compute these angles?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 22

In the figure below also, two lines cut across each other. Can you compute the other three angles marked?
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 23
What do we see from all these?

Among the four angles made by two lines cutting across each other, the sum of each pair of nearby angles is 180°. Each pair of opposite angles are equal.

Now, can’t you calculate the angles marked in each of the figures below? Write them in the figure.
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 24

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles

Let’s do it! Textbook Page No. 12

In each figure, some angles are given. Find all other angles.

Question 1.

i.
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 25
Answer: The other angles are 150°, 80°, and 30°.

Explanation: Given the figure,
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Fig(i)
The value of ∠EOD= 30°.
The opposite angles are equal. So, ∠EOD = ∠AOB = 30°.
Here are the angles ∠BOC, ∠COD, ∠and DOE in a line. So their sum is 180°.
∠BOC + ∠COD + ∠DOE  = 180°.
∠BOC + 70°+ 30°  = 180°.
∠BOC = 180°-70°- 30°
∠BOC = 80°.
Next, the angles ∠AOE = ∠BOD. Because both are opposite angles.
So,  ∠BOD = ∠BOC+∠COD
∠BOD = 80°+70° = 150°.
The other angles are 150°, 80°, and 30°.

ii.
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 27
Answer: The other angles are 50°, 130°, and 60°.

Explanation: Given the figure,

The value of ∠COD= 50°.
The opposite angles are equal. So, ∠COD = ∠AOB = 50°.
Here are the angles ∠of BOC, ∠and COD in a line. So their sum is 180°.
∠BOC + ∠COD = 180°.
∠BOC + 50°  = 180°.
∠BOC = 180°-50°
∠BOC = 130°.
Next, the angles ∠AOE, ∠EOD, ∠and DOC are in a line. So their sum is 180°.
∠AOE + ∠EOD + ∠DOC  = 180°.
∠EOD + 70°+ 50°  = 180°.
∠EOD = 180°-70°-50°
∠EOD = 60°.
The other angles are 50°, 130°, and 60°.

iii.

Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 28
Answer: The other angles are 60°, 80°, and 80°.

Explanation: Given the figure,
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Fig(iii)
The value of ∠COB= 60°.
The opposite angles are equal. So, ∠COB = ∠EOF = 60°.
The value of ∠DOE= 40°.
Here are the angles ∠BOC, ∠COD, ∠and DOE in a line. So their sum is 180°.
∠BOC + ∠COD + ∠DOE= 180°.
∠COD + 60° + 40° = 180°.
∠COD = 180°-60°- 40°
∠COD = 80°.
The angles ∠COD = ∠AOF = 80°. Because both are opposite angles.
The other angles are 60°, 80°, and 80°.

iv.
Kerala Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles 29
Answer: The other angles are 90°, 50°, 40°, and 40°.

Explanation: Given the figure,
Kerala State Syllabus 7th Standard Maths Solutions Chapter 1 Adding Angles_Fig(iv)
The value of ∠COB= 90°.
The opposite angles are equal. So, ∠COB = ∠EOF = 90°.
The value of ∠DOE= 50°.
∠DOE = ∠BOA= 50°. Because these are opposite angles.
Here are the angles ∠BOC, ∠COD, ∠and DOE in a line. So their sum is 180°.
∠BOC + ∠COD + ∠DOE= 180°.
∠COD + 90° + 50° = 180°.
∠COD = 180°-90°- 50°
∠COD = 40°.
Next, the angles ∠COD = ∠AOF  = 40°. Because both are opposite angles.
Hence, the other angles are 90°, 50°, 40°, and 40°.

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