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 C1 and C2 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 r1 and r2 respectively.
∴ Inner circumference = 2πr2
352 = 2 × \(\frac{22}{7}\) × r2
r2 = \(\frac{352 \times 7}{2 \times 22}\) = 56 m
Outer circumference = 2πr1
396 = 2 × \(\frac{22}{7}\) × r1
r1 = \(\frac{352 \times 7}{2 \times 22}\) = 56 m
∴ Width of the track = r1 – r2 = 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.