Perimeter and Area of Semicircle and Quadrants worksheet with problems and space for answers.
Worksheet with eight diagrams of semicircles and quarter circles, each labeled with dimensions, asking to calculate perimeter and area.
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Step-by-step solution for: Perimeter & Area of Semi and Quadrant Circle worksheet | Live ...
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Show Answer Key & Explanations
Step-by-step solution for: Perimeter & Area of Semi and Quadrant Circle worksheet | Live ...
To solve the problem, we need to calculate the perimeter and area of the given semicircles and quarter circles. Let's break it down step by step for each part.
1. Circumference of a full circle: \( C = 2\pi r \)
2. Area of a full circle: \( A = \pi r^2 \)
#### For Semicircles:
- Perimeter: The perimeter of a semicircle is half the circumference of the full circle plus the diameter.
\[
P_{\text{semicircle}} = \pi r + 2r
\]
- Area: The area of a semicircle is half the area of the full circle.
\[
A_{\text{semicircle}} = \frac{1}{2} \pi r^2
\]
#### For Quarter Circles:
- Perimeter: The perimeter of a quarter circle is one-fourth the circumference of the full circle plus twice the radius.
\[
P_{\text{quarter circle}} = \frac{\pi r}{2} + 2r
\]
- Area: The area of a quarter circle is one-fourth the area of the full circle.
\[
A_{\text{quarter circle}} = \frac{1}{4} \pi r^2
\]
- Use \( \pi = 3.142 \)
---
#### Part (a): Semicircle with radius \( r = 3 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3 + 2 \times 3 = 9.426 + 6 = 15.426 \approx 15.4 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3^2 = \frac{1}{2} \times 3.142 \times 9 = 14.139 \approx 14.1 \, \text{cm}^2
\]
#### Part (b): Quarter Circle with radius \( r = 4 \) cm
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 4}{2} + 2 \times 4 = 6.284 + 8 = 14.284 \approx 14.3 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 4^2 = \frac{1}{4} \times 3.142 \times 16 = 12.568 \approx 12.6 \, \text{cm}^2
\]
#### Part (c): Semicircle with diameter \( d = 6 \) cm
- Radius \( r = \frac{d}{2} = \frac{6}{2} = 3 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3 + 2 \times 3 = 9.426 + 6 = 15.426 \approx 15.4 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3^2 = \frac{1}{2} \times 3.142 \times 9 = 14.139 \approx 14.1 \, \text{cm}^2
\]
#### Part (d): Semicircle with radius \( r = 3.5 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3.5 + 2 \times 3.5 = 10.997 + 7 = 17.997 \approx 18.0 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3.5^2 = \frac{1}{2} \times 3.142 \times 12.25 = 19.23875 \approx 19.2 \, \text{cm}^2
\]
#### Part (e): Quarter Circle with radius \( r = 2 \) mm
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 2}{2} + 2 \times 2 = 3.142 + 4 = 7.142 \approx 7.1 \, \text{mm}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 2^2 = \frac{1}{4} \times 3.142 \times 4 = 3.142 \approx 3.1 \, \text{mm}^2
\]
#### Part (f): Quarter Circle with radius \( r = 1.5 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 1.5}{2} + 2 \times 1.5 = 2.3565 + 3 = 5.3565 \approx 5.4 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 1.5^2 = \frac{1}{4} \times 3.142 \times 2.25 = 1.769625 \approx 1.8 \, \text{m}^2
\]
#### Part (g): Quarter Circle with radius \( r = 7 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 7}{2} + 2 \times 7 = 10.997 + 14 = 24.997 \approx 25.0 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 7^2 = \frac{1}{4} \times 3.142 \times 49 = 38.4845 \approx 38.5 \, \text{m}^2
\]
#### Part (h): Quarter Circle with radius \( r = 13 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 13}{2} + 2 \times 13 = 20.423 + 26 = 46.423 \approx 46.4 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 13^2 = \frac{1}{4} \times 3.142 \times 169 = 132.69875 \approx 132.7 \, \text{m}^2
\]
---
\[
\boxed{
\begin{array}{ll}
\text{(a)} & \text{Perimeter} = 15.4 \, \text{cm}, \, \text{Area} = 14.1 \, \text{cm}^2 \\
\text{(b)} & \text{Perimeter} = 14.3 \, \text{cm}, \, \text{Area} = 12.6 \, \text{cm}^2 \\
\text{(c)} & \text{Perimeter} = 15.4 \, \text{cm}, \, \text{Area} = 14.1 \, \text{cm}^2 \\
\text{(d)} & \text{Perimeter} = 18.0 \, \text{cm}, \, \text{Area} = 19.2 \, \text{cm}^2 \\
\text{(e)} & \text{Perimeter} = 7.1 \, \text{mm}, \, \text{Area} = 3.1 \, \text{mm}^2 \\
\text{(f)} & \text{Perimeter} = 5.4 \, \text{m}, \, \text{Area} = 1.8 \, \text{m}^2 \\
\text{(g)} & \text{Perimeter} = 25.0 \, \text{m}, \, \text{Area} = 38.5 \, \text{m}^2 \\
\text{(h)} & \text{Perimeter} = 46.4 \, \text{m}, \, \text{Area} = 132.7 \, \text{m}^2 \\
\end{array}
}
\]
Formulas Needed:
1. Circumference of a full circle: \( C = 2\pi r \)
2. Area of a full circle: \( A = \pi r^2 \)
#### For Semicircles:
- Perimeter: The perimeter of a semicircle is half the circumference of the full circle plus the diameter.
\[
P_{\text{semicircle}} = \pi r + 2r
\]
- Area: The area of a semicircle is half the area of the full circle.
\[
A_{\text{semicircle}} = \frac{1}{2} \pi r^2
\]
#### For Quarter Circles:
- Perimeter: The perimeter of a quarter circle is one-fourth the circumference of the full circle plus twice the radius.
\[
P_{\text{quarter circle}} = \frac{\pi r}{2} + 2r
\]
- Area: The area of a quarter circle is one-fourth the area of the full circle.
\[
A_{\text{quarter circle}} = \frac{1}{4} \pi r^2
\]
Given Data:
- Use \( \pi = 3.142 \)
---
Solutions:
#### Part (a): Semicircle with radius \( r = 3 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3 + 2 \times 3 = 9.426 + 6 = 15.426 \approx 15.4 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3^2 = \frac{1}{2} \times 3.142 \times 9 = 14.139 \approx 14.1 \, \text{cm}^2
\]
#### Part (b): Quarter Circle with radius \( r = 4 \) cm
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 4}{2} + 2 \times 4 = 6.284 + 8 = 14.284 \approx 14.3 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 4^2 = \frac{1}{4} \times 3.142 \times 16 = 12.568 \approx 12.6 \, \text{cm}^2
\]
#### Part (c): Semicircle with diameter \( d = 6 \) cm
- Radius \( r = \frac{d}{2} = \frac{6}{2} = 3 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3 + 2 \times 3 = 9.426 + 6 = 15.426 \approx 15.4 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3^2 = \frac{1}{2} \times 3.142 \times 9 = 14.139 \approx 14.1 \, \text{cm}^2
\]
#### Part (d): Semicircle with radius \( r = 3.5 \) cm
- Perimeter:
\[
P = \pi r + 2r = 3.142 \times 3.5 + 2 \times 3.5 = 10.997 + 7 = 17.997 \approx 18.0 \, \text{cm}
\]
- Area:
\[
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 3.5^2 = \frac{1}{2} \times 3.142 \times 12.25 = 19.23875 \approx 19.2 \, \text{cm}^2
\]
#### Part (e): Quarter Circle with radius \( r = 2 \) mm
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 2}{2} + 2 \times 2 = 3.142 + 4 = 7.142 \approx 7.1 \, \text{mm}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 2^2 = \frac{1}{4} \times 3.142 \times 4 = 3.142 \approx 3.1 \, \text{mm}^2
\]
#### Part (f): Quarter Circle with radius \( r = 1.5 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 1.5}{2} + 2 \times 1.5 = 2.3565 + 3 = 5.3565 \approx 5.4 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 1.5^2 = \frac{1}{4} \times 3.142 \times 2.25 = 1.769625 \approx 1.8 \, \text{m}^2
\]
#### Part (g): Quarter Circle with radius \( r = 7 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 7}{2} + 2 \times 7 = 10.997 + 14 = 24.997 \approx 25.0 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 7^2 = \frac{1}{4} \times 3.142 \times 49 = 38.4845 \approx 38.5 \, \text{m}^2
\]
#### Part (h): Quarter Circle with radius \( r = 13 \) m
- Perimeter:
\[
P = \frac{\pi r}{2} + 2r = \frac{3.142 \times 13}{2} + 2 \times 13 = 20.423 + 26 = 46.423 \approx 46.4 \, \text{m}
\]
- Area:
\[
A = \frac{1}{4} \pi r^2 = \frac{1}{4} \times 3.142 \times 13^2 = \frac{1}{4} \times 3.142 \times 169 = 132.69875 \approx 132.7 \, \text{m}^2
\]
---
Final Answers:
\[
\boxed{
\begin{array}{ll}
\text{(a)} & \text{Perimeter} = 15.4 \, \text{cm}, \, \text{Area} = 14.1 \, \text{cm}^2 \\
\text{(b)} & \text{Perimeter} = 14.3 \, \text{cm}, \, \text{Area} = 12.6 \, \text{cm}^2 \\
\text{(c)} & \text{Perimeter} = 15.4 \, \text{cm}, \, \text{Area} = 14.1 \, \text{cm}^2 \\
\text{(d)} & \text{Perimeter} = 18.0 \, \text{cm}, \, \text{Area} = 19.2 \, \text{cm}^2 \\
\text{(e)} & \text{Perimeter} = 7.1 \, \text{mm}, \, \text{Area} = 3.1 \, \text{mm}^2 \\
\text{(f)} & \text{Perimeter} = 5.4 \, \text{m}, \, \text{Area} = 1.8 \, \text{m}^2 \\
\text{(g)} & \text{Perimeter} = 25.0 \, \text{m}, \, \text{Area} = 38.5 \, \text{m}^2 \\
\text{(h)} & \text{Perimeter} = 46.4 \, \text{m}, \, \text{Area} = 132.7 \, \text{m}^2 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of semicircle worksheet.