The DAV Maths Class 7 Solutions and **DAV Class 7 Maths Chapter 11 Worksheet 5** Solutions of Perimeter and Area offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 11 WS 5 Solutions

Question 1.

Find the circumference of a circle whose radius is 7 cm.

Answer:

Circumference of a circle = 2πr

= 2 × \(\frac{22}{7}\) × 7

= 44 cm

Hence the required circumference = 44 cm.

Question 2.

Find the diameter of a circle whose circumference is 66 m.

Answer:

Circumference of a circle = 2πr

66 = 2 × \(\frac{22}{7}\) × r

r = \(\frac{66 \times 7}{2 \times 22}=\frac{21}{2}\) m

∴ Diameter = 2 × radius = 2× \(\frac{21}{2}\) = 21 m

Hence the required diameter = 21 m

Question 3.

If circumference of a circle is 176 cm, find its radius.

Answer:

Circumference of the circle = 2πr

176 = 2 × \(\frac{22}{7}\) × r

∴ r = \(\frac{176 \times 7}{22 \times 2}\) = 28 cm

Hence the required radius = 28 cm.

Question 4.

The diameter of a wheel is 1.4 m, find its circumference.

Answer:

Diameter = 1.4 m

∴ its radius = \(\frac{1.4}{2}\) = 0.7 m

Circumference = 2πr = 2 × \(\frac{22}{7}\) × 0.7 = 4.4 m

Hence the required circumference = 4.4 m.

Question 5.

The radii of two circles are in the ratio 2 : 3. What is the ratio of their circumference?

Answer:

Let the radii of the two circles be 2x cm and 3x cm and C_{1} and C_{2} be their radii respectively.

∵ \(\frac{C_1}{C_2}=\frac{2 \pi r_1}{2 \pi r_2}\)

\(\frac{C_1}{C_2}=\frac{r_1}{r_2}=\frac{2 x}{3 x}=\frac{2}{3}\)

Hence the ratio of their circumference is 2 : 3.

Question 6.

Tire moon is nearly 385000 km away from the earth. It takes a round of the earth every month. How much distance does it travel in one month?

Answer:

Radius of the circular path = 385000 km.

∴ Its circumference = 2πr

= 2 × \(\frac{22}{7}\) × 385000

= 2 × 22 × 55000

= 2420000 km.

Hence the distance covered by moon in one month

= 2420000 km.

Question 7.

The diameter of the wheel of car is 35 cm. How much distance will it cover in 1000 revolutions?

Answer:

The diameter of the wheel = 35 cm

∴ Its radius = \(\frac{35}{2}\) cm

Circumference = 2πr

= 2 × \(\frac{22}{7} \times \frac{35}{2}\) = 110 cm

∴ Distance covered by the wheel in 1000 revolutions

= 1000 × 110 cm

= 110000 cm = 1.1 km.

Hence the required distance covered = 1.1 km.

Question 8.

The radius of the wheel of a bus is 0.70 m. How many revolutions will it make in covering 22 km?

Answer:

Circumference of the wheel = 2πr

= 2 × \(\frac{22}{7}\) × 0.70 = 4.4 m

Hence the required number of revolutions = 5000 m.

Question 9.

A wire is in the form of a circle with radius 42 cm. It is bent into a square. Find the sides of the square.

Answer:

Circumference of the circle = 2πr = 2 × \(\frac{22}{7}\) × 42 = 264 cm

Perimeter of the square = 4 × side

Now 4 × side = 264

∴ Side = \(\frac{264}{4}\) = 66 cm

Hence the side of the square = 66 cm.

Question 10.

The diameter of a circular park is 140 m. Around it on the outside, a path having the width of 7 m is constructed. If the path has to be fenced from inside and outside at the rate of ₹ 7 per meter, find its total cost.

Answer:

Radius of the inner park

Inner circumference = 2πr

= 2 × \(\frac{22}{7}\) × 70

= 440 m

Radius of the outer park = 70 + 7 = 77 m

Outer circumference = 2π × r

= 2 × \(\frac{22}{7}\) × 77 = 484 m

Total length of the two circumference

= 440 m + 484 m = 924 m

Cost of fencing = 7 × 924 = ₹ 6468

Hence the required cost = ₹ 6468.

Question 11.

An athlete runs around a circular park 10 times. If the diameter of park is 280 m, find the distance covered by the athlete in km.

Answer:

Diameter of the circular park = 280 m

Its radius = \(\frac{280}{2}\) = 140 m

Circumference = 2πr

= 2 × \(\frac{22}{7}\) × 140 = 880 m

Total distance travelled by the athlete in 10 rounds = 880 × 10

= 8800 m

= 8.8 km.

Hence the required distance = 8.8 km.

Question 12.

A circular piece of the wire is converted into a rhombus of side 11 cm. Find the diameter of the circular piece.

Answer:

Perimeter of the rhombus = 11 × 4 = 44 cm

∴ Circumference of the circular piece of wire = 44 cm

Circumference = 2πr

44 = 2 × \(\frac{22}{7}\) × r

∴ r = \(\frac{44 \times 7}{2 \times 22}\) = 7cm

∴ Its diameter = 2 × 7 = 14 cm

Hence the required diameter = 14 cm.

Question 13.

A race track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track.

Answer:

Let the radii of the outer and inner circles be r_{1} and r_{2} respectively.

∴ Inner circumference = 2πr_{2}

352 = 2 × \(\frac{22}{7}\) × r_{2}

r_{2} = \(\frac{352 \times 7}{2 \times 22}\) = 56 m

Outer circumference = 2πr_{1}

396 = 2 × \(\frac{22}{7}\) × r_{1}

r_{1} = \(\frac{352 \times 7}{2 \times 22}\) = 56 m

∴ Width of the track = r_{1} – r_{2} = 63 m – 56 m = 7 m

Hence the required width = 7 m.

Question 14.

Radius of a circular region is 63 cm. Find the least length of rope which is sufficient to encircle the circular region.

Answer:

Radius = 63 cm

Circumference = 2πr = 2 × \(\frac{22}{7}\) × 63 = 396 cm

Here the least length of the rope = 396 cm or 3.96 m.