Andhra Pradesh AP Board 4th Class Maths Solutions 7th Lesson Geometry Textbook Exercise Questions and Answers.

## AP State Syllabus 4th Class Maths Solutions Chapter 7 Geometry

Textbook Page No. 86

Think and Discuss

Question 1.

Does a cone or cylinder have any edges and corners ?

Solution:

Yes, cone has 1 edge and 1 comer.

Yes, cylinder has 2 edges and no comers.

Do this

a) Draw a line to divide the adjacent, rectangles into two equal parts.

Solution:

b) Name some objects which are in rectangle shape.

Solution:

Book, Brick, Cell phone, Exam pad.

Textbook Page No. 87

Do this

Draw a line to divide the following squares into two equal parts.

Solution:

Textbook Page No. 89

Do this

a) How many triangles are formed when a square or rectangle is cut diagonally ?

Solution:

When a square or rectangle is cut diagonally two triangles are formed.

b) In a figure, the four sides are 20cm, 16 cm, 20 cm, 16 cm then what is the shape of the object ?

Solution:

Given measurements 20 cm, 16 cm, 20 cm, and 16 cm. Here two sides are equal in measurements. So it is a rectangle.

c) In a figure, measurements of four sides are 15 cm each and the adjacent sides are vertical to each other. What is the shape of the object ?

Solution:

Given measurement is 15 cm. All sides are equal and vertical to each other are equal.

∴ It is a square.

Textbook Page No. 90

Do this

Question 1.

Observe the following:

Question 2.

Observe the net shapes of the boxes and match them with their 3 – D shapes.

Solution:

1 – b

2 – d

3 – a

4 – c

Textbook Page No. 92

Try this

Make some squares, rectangles and triangles with some cool drink straws. Find the perimeter of the shapes.

Solution:

P = 6m

P = 4m

P = 3m

Exercise – 7.1

Question 1.

The length and breadth of a rectangular field are 60m and 40m respectively. If Somaiah walked around the field, then find the distance covered by him.

Solution:

Length of a rectangular field = 60m

Breadth of a rectangular field = 40 m

Somaiah walked around the field

Distance covered by him = 2(1 + b)

= 2(60 + 40)

= 2 × 100

= 200 m

Question 2.

Somulu’s site is in square shape. He wanted to construct the compound wall around it. Find the length of the compound wall.

Solution:

Length at a square shaped side = 14 cm

He wanted to construct the compound wall around it.

Total length at the compound wall

= 4 side = 4 × 14 = 56 cm

Question 3.

A park is in a triangular shape as shown below. What is the perimeter of the park ?

Solution:

Park is in a triangular shape.

Length of park sides are 30m, 40m and 50m.

Perimeter of the park = 30 + 40 + 50

= 120 m

4. Find the perimeter of the following figures.

a)

Perimeter = ………………. cm

Solution:

From figure, perimeter of figure a

= 6cm + 6cm + 6cm + 6 cm

= 24 cm

b)

Perimeter = ………………. cm

Solution:

From figure, perimeter of figure a

= 4cm + 7cm + 4cm + 7 cm

= 22 cm

c)

Perimeter = ……………… cm

Solution:

From figure, perimeter of figure

= 5cm + 5 cm + 5 cm

= 15 cm

d)

Perimeter = ……………… cm

Solution:

Perimeter of the shape a

= 3 cm + 1 cm + 3 cm + 1 cm

= 8 cm

Textbook Page No. 95

Do this

Find the perimeter of the given shapes

Solution:

∴ Perimeter of the shape

= 3 cm + 1 cm + 3 cm + 1 cm

= 8 cm

Solution:

∴ Perimeter of the shape

= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) cm

= 8 cm

Solution:

∴ Perimeter of the shape

= 2 + 2 + 1 + 1 + 1

= 7 cm

Solution:

∴ Perimeter of the shape

= (1 + 2 + 1 + 2) cm

= 6 cm

Solution:

∴ Perimeter of the shape

= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) cm

= 11 cm

Textbook Page No. 97

Do this

Calculate the area of the shapes formed by colouring three, four and five grids.

Solution:

Area of the shapes formed by colouring

Three grids =3 square units.

Four grids =4 square units.

Five grids = 5 square units.

Try this

Find the area and perimeter of each shaded shape in the pictures. What do you notice?

i)

This figure occupies 4 grids. Hence its area = 4 square units.

Solution:

This figure occupied 4 grids.

Hence its area = 4 square units.

Perimeter of this figure

= (2 + 2+ 2+ 2) cm

= 8cm

ii)

This figure occupies 9 grids.

Hence shaded area …………. square units.

Solution:

This figure occupies 9 grids.

Hence shaded area =9 square units.

Perimeter of this figure

= each grid perimeter × 9

= 4cm × 9

= 36 cm

iii)

This figure occupies ………….. grids.

Hence its area = …………. square units.

Solution:

This figure occupied 8 grids.

Hence shaded area 8 square units.

Perimeter of this figure

= each grid perimeter × 8

= 4 cm × 8

= 32 cm

Exercise – 7.2

Question 1.

On a grid paper, draw different shapes totalling to an area of 8 square units.

Solution:

Question 2.

Draw a rectangle of 4 units length and 3 units breadth on a grid paper. Also, calculate its area.

Solution:

Question 3.

Draw a square of side 5 units on a grid paper and calculate them.

Solution:

Textbook Page No. 100

Do this

a) Draw a few circles using a bangle, plate, bottle cap etc.

Solution:

b) Draw a circles using one-rupee, two- rupee-five-rupee and ten-rupees coins.

Solution:

Exercise – 7.3

Question 1.

What is the perimeter of the following figures?

Solution:

Perimeter of the given figure

= 3 cm + 3 cm + 3 cm

= 12 cm

Perimeter of the given figure

= 3 cm + 6 cm + 4 cm + 3 cm

= 16 cm

Question 2.

A piece of land in Simhachalam is in the following shape. Find the length of the fencing wire required to provide fencing around the land.

Solution:

Length of the fencing wire required to provide fencing around the land

= Perimeter of figure

= 18 m + 17 m + 16 m + 14 m

= 65 m

Question 3.

The perimeter of the following shape is ……………. metres.

Solution:

Perimeter of the given shape

= 3m + 3m + 2m + 5m + 6m

= 19 m

Question 4.

What is the perimeter of the figure?

Solution:

Perimeter of the figure

= 2cm + cm + 3 cm + 2 cm + 2 cm + 5 cm + 7 cm + 5 cm

= 30 cm

Question 5.

What is the perimeter and area of figures given below.

Solution:

Perimeter of given figure – 1:

= 2 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 2 cm + 1 cm + 1 cm+ 1 cm + 1 cm

= 14 cm

Perimeter of given figure – 2:

= 1 cm + 1 cm + 1 cm + 1 cm + 1 cm+ 1 cm + 1 cm + 1 cm + 1 cm+ 1 cm + 1 cm + 1 cm+ 1 cm+ 1 cm + 1 cm + 1 cm + 1 cm + 1 cm+ 1 cm + 1 cm

= 20 cm

Multiple Choice Questions

Question 1.

Carrot has………….

A) Joker hat

B) Ball

C) Brick

D) Drum

Answer:

A) Joker hat

Question 2.

Samosa is in …………. shape.

A) Rectangle

B) Square

C) Circle

D) Triangle

Answer:

D) Triangle

Question 3.

Shape of Joker cap ……………..

A) Cuboid

B) Cube

C) Square

D) Cone

Answer:

D) Cone

Question 4.

Carrams board is in ……………….. shape.

A) Rectangle

B) Square

C) Triangle

D) Circle

Answer:

B) Square

Question 5.

A Rubic cube shape is ……………..

A) Cube

B) Cuboid

C) Cone

D) Square

Answer:

A) Cube

Question 6.

Tooth paste box is in ………….. shape.

A) Square

B) Circle

C) Rectangle

D) Triangle

Answer:

C) Rectangle

Question 7.

Area of shaded region is ……………. sq. units.

A) 9

B) 18

C) 4

D) 10

Answer:

C) 4

Question 8.

Perimeter

of is ………………….. m

A) 20 m

B) 12 m

C) 15 m

D)18 m

Answer:

A) 20 m

Question 9.

Name of shape of the object

A) Square

B) Triangle

C) Circle

D) Rectangle

Answer:

C) Circle

Question 10.

Tangram has …………… shapes.

A) 7

B) 5

C) 6

D) 4

Answer:

A) 7