Grade 7 math worksheet on perimeter and area of a circle with real-world application problems.

Let's solve each problem step by step.

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Problem 1:

The rope connecting a goat to a pole is 8 m long. What area of grass can the goat eat?

#### Solution:
The goat can graze in a circular area with the rope as the radius. The formula for the area of a circle is:
\[
\text{Area} = \pi r^2
\]
where \( r \) is the radius.

Here, the radius \( r = 8 \) m. Substituting the value:
\[
\text{Area} = \pi (8)^2 = \pi \times 64
\]
Using \( \pi \approx 3.14 \):
\[
\text{Area} \approx 3.14 \times 64 = 200.96 \, \text{m}^2
\]

#### Final Answer:
\[
\boxed{200.96 \, \text{m}^2}
\]

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Problem 2:

A circular hoop has a radius of 40 cm. Find the length of tubing needed to make the hoop.

#### Solution:
The length of tubing needed to make the hoop is the circumference of the circle. The formula for the circumference of a circle is:
\[
\text{Circumference} = 2 \pi r
\]
where \( r \) is the radius.

Here, the radius \( r = 40 \) cm. Substituting the value:
\[
\text{Circumference} = 2 \pi (40) = 80 \pi
\]
Using \( \pi \approx 3.14 \):
\[
\text{Circumference} \approx 80 \times 3.14 = 251.2 \, \text{cm}
\]

#### Final Answer:
\[
\boxed{251.2 \, \text{cm}}
\]

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Problem 3:

A cylindrical tank has a radius of 1.5 m. What is the circumference of its base?

#### Solution:
The circumference of the base of the cylindrical tank is the same as the circumference of a circle with radius 1.5 m. The formula for the circumference of a circle is:
\[
\text{Circumference} = 2 \pi r
\]
where \( r \) is the radius.

Here, the radius \( r = 1.5 \) m. Substituting the value:
\[
\text{Circumference} = 2 \pi (1.5) = 3 \pi
\]
Using \( \pi \approx 3.14 \):
\[
\text{Circumference} \approx 3 \times 3.14 = 9.42 \, \text{m}
\]

#### Final Answer:
\[
\boxed{9.42 \, \text{m}}
\]

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Problem 4:

A circular garden plot has a circumference of 12.65 m. Find its radius in meters, correct to the nearest centimetre.

#### Solution:
The formula for the circumference of a circle is:
\[
\text{Circumference} = 2 \pi r
\]
where \( r \) is the radius. We are given the circumference as 12.65 m. Rearranging the formula to solve for \( r \):
\[
r = \frac{\text{Circumference}}{2 \pi}
\]
Substituting the given circumference:
\[
r = \frac{12.65}{2 \pi}
\]
Using \( \pi \approx 3.14 \):
\[
r = \frac{12.65}{2 \times 3.14} = \frac{12.65}{6.28} \approx 2.014 \, \text{m}
\]
To convert to centimetres and round to the nearest centimetre:
\[
r \approx 2.014 \, \text{m} = 201.4 \, \text{cm} \approx 201 \, \text{cm}
\]

#### Final Answer:
\[
\boxed{2.01 \, \text{m}}
\]

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Final Answers:

1. \(\boxed{200.96 \, \text{m}^2}\)
2. \(\boxed{251.2 \, \text{cm}}\)
3. \(\boxed{9.42 \, \text{m}}\)
4. \(\boxed{2.01 \, \text{m}}\)