Mathematics Semester i Solutions Solutions for Class 7 Math Chapter 7 Geometry are provided here with simple step-by-step explanations. These solutions for Geometry are extremely popular among class 7 students for Math Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Semester i Solutions Book of class 7 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Semester i Solutions Solutions. All Mathematics Semester i Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 91:

Question O1:

What is a circle?

Answer:

A circle is a path traced by a moving point, which is equidistant from a fixed point. This fixed point is called a centre.

Page No 91:

Question O2:

What is the diameter of a circle whose radius is 7 cm?

Answer:

14 cm

Page No 91:

Question O3:

What is the radius of a circle whose diameter is 4.8 cm?

Answer:

2.4 cm

 



Page No 92:

Question W1:

Fill in the blanks with suitable words:

1. Diameter of a circle is _______ its radius.

2. Radius of a circle is ______ its diameter.

3. The diameters of a circle intersect at the ______.

4. The chord passing through the centre of a circle is _________.

5. The biggest chord of a circle is ______.

6. The lengths of the radii of a circle are _______.

Answer:

1. Diameter of a circle is twice its radius.

2. Radius of a circle is half its diameter.

3. The diameters of a circle intersect at the centre.

4. The chord passing through the centre of a circle is diameter.

5. The biggest chord of a circle is diameter.

6. The lengths of the radii of a circle are equal.

Page No 92:

Question W2:

The diameters of circles are given below. Find their radii.

(1) 8 cm

(2) 6 cm

(3) 8.6 cm

(4) 5.4 cm

Answer:

(1) r =

=

= 4 cm

(2) r =

=

= 3 cm

(3) r =

=

= 4.3 cm

(4) r =

=

= 2.7 cm

Page No 92:

Question W3:

Find the diameters of the circles whose radii are given below.

(1) 5 cm

(2) 7 cm

(3) 2.4 cm

(4) 3.7 cm

Answer:

Diameter (d) = 2r, r being the radius

(1) d = 2r = (2 × 5) cm = 10 cm

(2) d = 2r = (2 × 7) cm = 14 cm

(3) d = 2r = (2 × 2.4) cm = 4.8 cm

(4) d = 2r = (2 × 3.7) cm = 7.4 cm

Page No 92:

Question W4:

Using ruler and compass draw circles with the radii given below.

Mark their centres and draw the radii.

(1) 4 cm

(2) 2.5 cm

Answer:

(1)

(2)

Page No 92:

Question W5:

Draw circles with the following radii. Draw and name the diameters. Also measure their lengths.

(1) 3 cm

(2) 3.8 cm

Answer:

(1)

Length of diameter AB = 6 cm

(2)

Length of diameter PQ = 7.6 cm

Page No 92:

Question W6:

Draw circles of the given radii and draw chords of the given lengths.

Sl. No.

Radius (cm)

Length of chord (cm)

1.

2.

2.5

4.0

3.0

3.5

Answer:

(1) Construction:

Taking O as centre and radius OR = 2.5 cm, draw a circle.

With the help of a ruler, draw a chord of 3cm length as shown below.

(2) Construction:

Taking O as centre and radius OR = 4.0cm, draw a circle.

With the help of a ruler, draw a chord of 3.5 cm length as shown below.

Page No 92:

Question W7:

Draw a circle with radius 4.3 cm. Draw a segment with the length of its chord as 6.5 cm and name the segment.

Answer:

Taking O as centre and radius OR = 4.3 cm, draw a circle.

With the help of a ruler, draw a chord of 6.5cm length as shown below.

And the area between the chord AB and arc AB is the required segment APB.



Page No 95:

Question O1:

What is the approximate value of π?

Answer:

The approximate value of π is .

Page No 95:

Question W1:

Find the circumference of circles whose radii are

(1) 21 cm

(2) 56 cm

(3) 4.2 cm

(4) 10.5 cm

Answer:

(1) C = 2πr

= 2 × × 21

= 2 × 22 × 3

= 132 cm

(2) C = 2πr

= 2 × × 56

= 2 × 22 × 8

= 352 cm

(3) C = 2πr

= 2 × × 4.2

= 2 × 22 × 0.6

= 26.4 cm

(4) C = 2πr

= 2 × × 10.5

= 2 × 22 × 1.5

= 66 cm

Page No 95:

Question O2:

What is the formula to find the circumference of a circle?

Answer:

The circumference of a circle is the product of π and its diameter. Thus, the formula to find the circumference of a circle is .

Page No 95:

Question W2:

Find the circumference of circles whose diameters are

(1) 7 cm

(2) 14 cm

(3) 3.5 cm

(4) 5.6 cm

Answer:

(1) C = πd

= × 7

= 22 cm

(2) C = πd

= × 14

= 44 cm

(3) C = πd

= × 3.5

= 11 cm

(4) C = πd

= × 5.8

= 17.6 cm

Page No 95:

Question W3:

Find the diameter of circles whose circumference is

(1) 22 cm

(2) 44 cm

(3) 8.8 cm

(4) 13.2 cm

Answer:

We know that, C = πd. So, d =

(1)

(2)

(3)

(4)

Page No 95:

Question O3:

If C = πd, then what is the value of d?

Answer:

If , then .



Page No 97:

Question W1:

The diameter of a wheel is 15.4 cm. Calculate the distance travelled when it completes 150 revolutions.

Answer:

C = πd = ×15.4 cm = 48.4 cm

Distance travelled when the wheel completes one revolution = 48.4 cm

Distance travelled when the wheel completes 150 revolution = 48.4 × 150 cm = 7260 cm

Page No 97:

Question W2:

The diameter of a bicycle wheel is 56 cm. Find the distance travelled when it completes 250 revolutions.

Answer:

C = πd = ×56 cm = 176 cm

Distance travelled when the wheel completes one revolution = 176 cm

Distance travelled when the wheel completes 250 revolution = 176 × 250 cm = 44000 cm



Page No 98:

Question W3:

The circumference of a bicycle wheel is 132 cm. How many revolutions does it complete when it move a distance of 264 m?

Answer:

C = 132 cm

Distance Travelled = 264 m = 26400 cm

Number of revolutions = = = 200

Page No 98:

Question W4:

The radius of the wheel of a car is 98 cm. Find the number of revolutions it completes, when it travels a distance of 3.850 m.

Answer:

C = 2πr

= 2 × × 98

= 2 × 22 × 14

= 616 cm

Distance Travelled = 3850 m = 385000 cm

Number of revolutions = = = 625

Page No 98:

Question W5:

The distance covered by a wheel when it completes 20 revolutions is 110 m. Find the diameter of the wheel.

Answer:

Number of revolutions =

20 =

C =

C = 5.5 m = 550 cm

C = πd

550 = × d

d =

d = 175 cm = 1.75 m

Page No 98:

Question W6:

A wheel covers a distance of 82.5 m is 15 revolutions. Find its diameter.

Answer:

Number of revolutions =

15 =

C =

C = 5.5 m = 550 cm

C = πd

550 = × d

d =

d = 175 cm = 1.75 m



Page No 100:

Question O1:

What is the formula to find the area of a circle?

Answer:

The formula to find the area of a circle is .

Page No 100:

Question W1:

Find the areas of circles whose radii are

(1) 14 cm

(2) 21 cm

(3) 10.5 cm

(4) 3.5 cm

Answer:

(1) Area, A = πr2 = π × r × r = × 14 × 14 = 616 cm2

(2) Area, A = πr2 = π × r × r =× 21 × 21 = 1386 cm2

(3) Area, A = πr2 = π × r × r = × 10.5 × 10.5 = 346.5 cm2

(4) Area, A = πr2 = π × r × r =× 3.5 × 3.5 = 38.5 cm2

Page No 100:

Question O2:

Which of the following is area of a circle?

(a) 154 m

(b) 154 sq. m

(c) 154 cm

Answer:

(a) The unit of the quantity 154 m is m (meter). Thus, it is not the area of a circle.

(b) The unit of the quantity 154 sq. m is sq. m (square meter). Thus, it is the area of a circle.

(c) The unit of the quantity 154 cm is cm (centimeter). Thus, it is not the area of a circle.



Page No 101:

Question W2:

Find the areas of circles whose diameters are

(1) 28 cm

(2) 42 cm

(3) 7 cm

(4) 70 cm

Answer:

(1) r =

=

= 14 cm

A = πr2

A = π × r × r

= × 14 × 14

= 616 cm2

(2) r =

=

= 21 cm

A = πr2

A = π × r × r

= × 21 × 21

= 1386 cm2

(3) r =

=

= 3.5 cm

A = πr2

A = π × r × r

= × 3.5 × 3.5

= 38.5 cm2

(4) r =

=

= 35 cm

A = πr2

A = π × r × r

= × 35 × 35

= 3850 cm2

Page No 101:

Question W3:

Find the radii of circles whose areas are

(1) 1.386 cm2

(2) 15,400 sq. cm

Answer:

(1) A = πr2

i.e. 1386 = × r2

i.e. × r2 = 1386

r2 = 1386 ×

r2 = 441 = 212

r = 21 cm

(2) A = πr2

i.e. 15400 = × r2

i.e. × r2 = 15400

r2 = 15400 ×

r2 = 4900 = 702

r = 70 cm

Page No 101:

Question W4:

Find the area of circles whose circumference are

(1) 88 cm

(2) 7π cm

Answer:

(1) C = 2πr

88 = 2 × × r

r =

r = 14 cm

A = πr2

A = π × r × r

= × 14 × 14

= 616 cm2

(2) C = 2πr

7π = 2 × π× r

Cancelling π on both the sides of the equation:

r =

r = 3.5 cm

A = πr2

A = π × r × r

= × 3.5 × 3.5

= 38.5 cm2



Page No 102:

Question W1:

Find the area of the shaded potion

Answer:

Diameter of the circle, d = 14 cm

Radius of the circle, r =

=

= 7 cm

Area of the circular region, A = πr2

= π × r × r

= × 7 × 7

= 154 cm2

Area of the rectangular region = l × b

= 14 × 14

= 196 cm2

Area of the shaded portion = Area of the rectangular region − Area of the circular region

= 196 − 154

= 42 cm2

Page No 102:

Question W2:

Find the areas of sectors A, B and C

Answer:

We know:

Area of a circle =

= r r

= 14 14

= 44 14

= 616

Area of sector A = Area of the circle

= 616

= 154

Area of sector B = Area of the circle

= 616

= 308

Area of sector C = Area of the circle

= 616

= 154

Page No 102:

Question W3:

The inner and outer radii of a circular ring are 14 cm and 21 cm respectively. Find the area of the shaded portion.

Answer:

We know:

Area of the circle =

Area of the inner circle =

= r r

= 14 14

= 44 14

= 616

Area of the outer circle =

= r r

= 21 21

= 66 21

= 1386

Area of the shaded portion = Area of the outer circle − Area of the inner circle

= 1386 − 616

= 770



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