Math worksheet for calculating the area of compound shapes, featuring nine figures combining basic geometric forms with labeled dimensions.
A worksheet titled "Compound Shapes" with nine problems, each showing a geometric figure composed of basic shapes like triangles, rectangles, circles, and semicircles, with dimensions labeled. Students are instructed to find the area of each figure and round to one decimal place.
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Show Answer Key & Explanations
Step-by-step solution for: Geometry Worksheets | Area Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Geometry Worksheets | Area Worksheets
Let's solve each of the compound shapes step by step and find the area of each figure. We'll break down each shape into simpler geometric parts (like rectangles, triangles, circles, etc.), calculate their areas, and then add or subtract as needed.
---
- The large shape is a triangle with base = 29 ft and height = 29 ft.
- A circle of radius 9 ft is cut out from the center.
Area of triangle:
$$
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 29 \times 29 = \frac{841}{2} = 420.5 \text{ ft}^2
$$
Area of circle:
$$
A = \pi r^2 = \pi \times 9^2 = 81\pi \approx 254.5 \text{ ft}^2
$$
Total area (triangle minus circle):
$$
420.5 - 254.5 = 166.0 \text{ ft}^2
$$
✔ Answer: 166.0 ft²
---
- A right triangle with legs 11 in and 5 in.
- A semicircle with diameter = 5 in (since it's attached to the side of length 5 in).
Area of triangle:
$$
A = \frac{1}{2} \times 11 \times 5 = 27.5 \text{ in}^2
$$
Area of semicircle:
- Radius $ r = \frac{5}{2} = 2.5 $ in
$$
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2.5)^2 = \frac{1}{2} \pi (6.25) \approx 9.8 \text{ in}^2
$$
Total area:
$$
27.5 + 9.8 = 37.3 \text{ in}^2
$$
✔ Answer: 37.3 in²
---
- Square: side = 16 in
- Circle: diameter = 16 in → radius = 8 in
Area of square:
$$
A = 16 \times 16 = 256 \text{ in}^2
$$
Area of circle:
$$
A = \pi r^2 = \pi \times 8^2 = 64\pi \approx 201.1 \text{ in}^2
$$
Shaded area (square minus circle):
$$
256 - 201.1 = 54.9 \text{ in}^2
$$
✔ Answer: 54.9 in²
---
- Outer rectangle: 14 m × 11 m
- Inner missing rectangle: 7 m × 7 m (the cut-out)
Outer area:
$$
14 \times 11 = 154 \text{ m}^2
$$
Inner cut-out area:
$$
7 \times 7 = 49 \text{ m}^2
$$
Shaded area:
$$
154 - 49 = 105 \text{ m}^2
$$
✔ Answer: 105 m²
---
- Rectangle: 11 cm × 4 cm? Wait — actually, the height is 4 cm, but the width is 11 cm. But the semicircle has radius 4 cm, so the diameter is 8 cm.
Wait: The figure shows a rectangle with a semicircular hole on the right side, radius = 4 cm → diameter = 8 cm.
So:
- Rectangle: 11 cm (width) × 4 cm (height) → but wait — is the rectangle height 4 cm?
Actually, looking at the diagram:
- The rectangle is 11 cm wide and height not given, but the radius is 4 cm, and the semicircle is cut out from the right side, so the rectangle must be 4 cm tall.
But that doesn't make sense — because the curved edge is on the right, so the rectangle height should match the semicircle’s diameter → 8 cm.
Wait — the radius is 4 cm, so diameter = 8 cm.
But the rectangle is 11 cm wide, and the height appears to be 8 cm? Let's re-express:
From the image description:
- The rectangle is 11 cm wide and height = 8 cm (since semicircle has radius 4 cm → diameter 8 cm).
- Semicircle of radius 4 cm is removed from the right side.
So:
Rectangle area:
$$
11 \times 8 = 88 \text{ cm}^2
$$
Semicircle area:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (16) = 8\pi \approx 25.1 \text{ cm}^2
$$
Shaded area:
$$
88 - 25.1 = 62.9 \text{ cm}^2
$$
✔ Answer: 62.9 cm²
---
- Left part: rectangle 12 m × 12 m
- Right part: triangle with base 6 m and height 12 m (same as rectangle)
Rectangle area:
$$
12 \times 12 = 144 \text{ m}^2
$$
Triangle area:
$$
\frac{1}{2} \times 6 \times 12 = 36 \text{ m}^2
$$
Total area:
$$
144 + 36 = 180 \text{ m}^2
$$
✔ Answer: 180 m²
---
- Bottom: semicircle with radius 2 yd
- Two identical right triangles on top, each with base 3 yd and height 5 yd
Area of semicircle:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4) = 2\pi \approx 6.3 \text{ yd}^2
$$
Area of one triangle:
$$
\frac{1}{2} \times 3 \times 5 = 7.5 \text{ yd}^2
$$
Two triangles:
$$
2 \times 7.5 = 15 \text{ yd}^2
$$
Total area:
$$
6.3 + 15 = 21.3 \text{ yd}^2
$$
✔ Answer: 21.3 yd²
---
- Outer rectangle: 6 ft × 6 ft → area = 36 ft²
- Inner rectangle: 3 ft × 3 ft → area = 9 ft²
Shaded area:
$$
36 - 9 = 27 \text{ ft}^2
$$
✔ Answer: 27 ft²
---
The figure consists of:
- A trapezoid at the bottom: height = 15 yd, bases = 20 yd (bottom), and 10 yd (top)?
Wait — no. Actually, it's a house-shaped figure: a rectangle and a triangle on top.
Looking closely:
- Bottom: rectangle 20 yd × 15 yd
- Top: triangle with base 20 yd and height 10 yd
Rectangle area:
$$
20 \times 15 = 300 \text{ yd}^2
$$
Triangle area:
$$
\frac{1}{2} \times 20 \times 10 = 100 \text{ yd}^2
$$
Total area:
$$
300 + 100 = 400 \text{ yd}^2
$$
✔ Answer: 400 yd²
---
| Problem | Area |
|--------|------|
| 1) | 166.0 ft² |
| 2) | 37.3 in² |
| 3) | 54.9 in² |
| 4) | 105 m² |
| 5) | 62.9 cm² |
| 6) | 180 m² |
| 7) | 21.3 yd² |
| 8) | 27 ft² |
| 9) | 400 yd² |
Let me know if you'd like a printable version or explanations for any specific steps!
---
1) Triangle with a circle inside
- The large shape is a triangle with base = 29 ft and height = 29 ft.
- A circle of radius 9 ft is cut out from the center.
Area of triangle:
$$
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 29 \times 29 = \frac{841}{2} = 420.5 \text{ ft}^2
$$
Area of circle:
$$
A = \pi r^2 = \pi \times 9^2 = 81\pi \approx 254.5 \text{ ft}^2
$$
Total area (triangle minus circle):
$$
420.5 - 254.5 = 166.0 \text{ ft}^2
$$
✔ Answer: 166.0 ft²
---
2) Right triangle + semicircle
- A right triangle with legs 11 in and 5 in.
- A semicircle with diameter = 5 in (since it's attached to the side of length 5 in).
Area of triangle:
$$
A = \frac{1}{2} \times 11 \times 5 = 27.5 \text{ in}^2
$$
Area of semicircle:
- Radius $ r = \frac{5}{2} = 2.5 $ in
$$
A = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2.5)^2 = \frac{1}{2} \pi (6.25) \approx 9.8 \text{ in}^2
$$
Total area:
$$
27.5 + 9.8 = 37.3 \text{ in}^2
$$
✔ Answer: 37.3 in²
---
3) Square with a circle inscribed
- Square: side = 16 in
- Circle: diameter = 16 in → radius = 8 in
Area of square:
$$
A = 16 \times 16 = 256 \text{ in}^2
$$
Area of circle:
$$
A = \pi r^2 = \pi \times 8^2 = 64\pi \approx 201.1 \text{ in}^2
$$
Shaded area (square minus circle):
$$
256 - 201.1 = 54.9 \text{ in}^2
$$
✔ Answer: 54.9 in²
---
4) "C"-shaped figure
- Outer rectangle: 14 m × 11 m
- Inner missing rectangle: 7 m × 7 m (the cut-out)
Outer area:
$$
14 \times 11 = 154 \text{ m}^2
$$
Inner cut-out area:
$$
7 \times 7 = 49 \text{ m}^2
$$
Shaded area:
$$
154 - 49 = 105 \text{ m}^2
$$
✔ Answer: 105 m²
---
5) Rectangle with a semicircle removed
- Rectangle: 11 cm × 4 cm? Wait — actually, the height is 4 cm, but the width is 11 cm. But the semicircle has radius 4 cm, so the diameter is 8 cm.
Wait: The figure shows a rectangle with a semicircular hole on the right side, radius = 4 cm → diameter = 8 cm.
So:
- Rectangle: 11 cm (width) × 4 cm (height) → but wait — is the rectangle height 4 cm?
Actually, looking at the diagram:
- The rectangle is 11 cm wide and height not given, but the radius is 4 cm, and the semicircle is cut out from the right side, so the rectangle must be 4 cm tall.
But that doesn't make sense — because the curved edge is on the right, so the rectangle height should match the semicircle’s diameter → 8 cm.
Wait — the radius is 4 cm, so diameter = 8 cm.
But the rectangle is 11 cm wide, and the height appears to be 8 cm? Let's re-express:
From the image description:
- The rectangle is 11 cm wide and height = 8 cm (since semicircle has radius 4 cm → diameter 8 cm).
- Semicircle of radius 4 cm is removed from the right side.
So:
Rectangle area:
$$
11 \times 8 = 88 \text{ cm}^2
$$
Semicircle area:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (16) = 8\pi \approx 25.1 \text{ cm}^2
$$
Shaded area:
$$
88 - 25.1 = 62.9 \text{ cm}^2
$$
✔ Answer: 62.9 cm²
---
6) Rectangle + triangle
- Left part: rectangle 12 m × 12 m
- Right part: triangle with base 6 m and height 12 m (same as rectangle)
Rectangle area:
$$
12 \times 12 = 144 \text{ m}^2
$$
Triangle area:
$$
\frac{1}{2} \times 6 \times 12 = 36 \text{ m}^2
$$
Total area:
$$
144 + 36 = 180 \text{ m}^2
$$
✔ Answer: 180 m²
---
7) Two triangles + semicircle
- Bottom: semicircle with radius 2 yd
- Two identical right triangles on top, each with base 3 yd and height 5 yd
Area of semicircle:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4) = 2\pi \approx 6.3 \text{ yd}^2
$$
Area of one triangle:
$$
\frac{1}{2} \times 3 \times 5 = 7.5 \text{ yd}^2
$$
Two triangles:
$$
2 \times 7.5 = 15 \text{ yd}^2
$$
Total area:
$$
6.3 + 15 = 21.3 \text{ yd}^2
$$
✔ Answer: 21.3 yd²
---
8) Large rectangle with small rectangle cut out
- Outer rectangle: 6 ft × 6 ft → area = 36 ft²
- Inner rectangle: 3 ft × 3 ft → area = 9 ft²
Shaded area:
$$
36 - 9 = 27 \text{ ft}^2
$$
✔ Answer: 27 ft²
---
9) Trapezoid + triangle
The figure consists of:
- A trapezoid at the bottom: height = 15 yd, bases = 20 yd (bottom), and 10 yd (top)?
Wait — no. Actually, it's a house-shaped figure: a rectangle and a triangle on top.
Looking closely:
- Bottom: rectangle 20 yd × 15 yd
- Top: triangle with base 20 yd and height 10 yd
Rectangle area:
$$
20 \times 15 = 300 \text{ yd}^2
$$
Triangle area:
$$
\frac{1}{2} \times 20 \times 10 = 100 \text{ yd}^2
$$
Total area:
$$
300 + 100 = 400 \text{ yd}^2
$$
✔ Answer: 400 yd²
---
✔ Final Answers:
| Problem | Area |
|--------|------|
| 1) | 166.0 ft² |
| 2) | 37.3 in² |
| 3) | 54.9 in² |
| 4) | 105 m² |
| 5) | 62.9 cm² |
| 6) | 180 m² |
| 7) | 21.3 yd² |
| 8) | 27 ft² |
| 9) | 400 yd² |
Let me know if you'd like a printable version or explanations for any specific steps!
Parent Tip: Review the logic above to help your child master the concept of area of figures worksheet.
