Area of Compound Shapes | Revision Worksheets - Level 1 | Shapes ... - Free Printable

Let's solve each of the compound shape problems step by step, calculating the area of the shaded region in each figure. We'll use:

- $ \pi = 3.14 $
- Round answers to 2 decimal places if needed.

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Problem 1)

Shape: A rectangle with a semicircle on top.

- Rectangle: 13 in (width) × 12 in (height)
- Semicircle: diameter = 13 in → radius = 6.5 in
- The semicircle is attached to the top of the rectangle.

Area of rectangle:
$$
A_{\text{rect}} = 13 \times 12 = 156 \text{ in}^2
$$

Area of semicircle:
$$
A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times (6.5)^2 = \frac{1}{2} \times 3.14 \times 42.25 = 66.3325 \text{ in}^2
$$

Total shaded area:
$$
A = 156 + 66.3325 = 222.3325 \approx \boxed{222.33} \text{ in}^2
$$

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Problem 2)

Shape: A large right triangle with a smaller triangle cut out from it.

We are given:
- Large triangle: base = 15 ft, height = 10 ft
- Cut-out triangle: base = 7 ft, height = 7 ft (from the diagram)

Area of large triangle:
$$
A_{\text{large}} = \frac{1}{2} \times 15 \times 10 = 75 \text{ ft}^2
$$

Area of small triangle (cut out):
$$
A_{\text{small}} = \frac{1}{2} \times 7 \times 7 = 24.5 \text{ ft}^2
$$

Shaded area:
$$
A = 75 - 24.5 = \boxed{50.50} \text{ ft}^2
$$

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Problem 3)

Shape: A circle with a triangle inscribed inside it (triangle connects two ends of a diameter and the top of the circle).

Given:
- Circle radius = 12 yd → diameter = 24 yd
- Triangle has base = 24 yd (diameter), height = 12 yd (radius, since it goes from center to top)

But note: The shaded region is the circle minus the triangle.

Area of circle:
$$
A_{\text{circle}} = \pi r^2 = 3.14 \times (12)^2 = 3.14 \times 144 = 452.16 \text{ yd}^2
$$

Area of triangle:
$$
A_{\text{tri}} = \frac{1}{2} \times 24 \times 12 = 144 \text{ yd}^2
$$

Shaded area:
$$
A = 452.16 - 144 = \boxed{308.16} \text{ yd}^2
$$

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Problem 4)

Shape: A rectangle with a right triangle removed from the bottom-left corner.

Given:
- Rectangle: 11 m (width) × 13 m (height)
- Cut-out triangle: base = 11 m, height = 11 m? Wait — let’s check.

Wait: The figure shows a trapezoid-like shape with a "bite" taken out. But actually, it looks like a rectangle with a right triangle cut from the bottom-left.

From the diagram:
- The full height is 13 m.
- The cut-out triangle has base = 11 m (same as width), and height = 13 - 11 = 2 m?

Wait — no. Actually, the vertical side on the left is 11 m, but the total height is 13 m. So the bottom part is missing a triangle.

So:
- The shaded region is a rectangle (11 × 13) minus a right triangle at the bottom-left.

Triangle:
- Base = 11 m
- Height = 13 - 11 = 2 m

Rectangle area:
$$
A_{\text{rect}} = 11 \times 13 = 143 \text{ m}^2
$$

Triangle area:
$$
A_{\text{tri}} = \frac{1}{2} \times 11 \times 2 = 11 \text{ m}^2
$$

Shaded area:
$$
A = 143 - 11 = \boxed{132.00} \text{ m}^2
$$

---

Problem 5)

Shape: A rectangle with a semicircle on top.

Given:
- Rectangle: width = 22 ft, height = 11 ft
- Semicircle: diameter = 22 ft → radius = 11 ft

Area of rectangle:
$$
A_{\text{rect}} = 22 \times 11 = 242 \text{ ft}^2
$$

Area of semicircle:
$$
A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times (11)^2 = \frac{1}{2} \times 3.14 \times 121 = 190.07 \text{ ft}^2
$$

Total shaded area:
$$
A = 242 + 190.07 = \boxed{432.07} \text{ ft}^2
$$

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Problem 6)

Shape: A composite shape made of a rectangle and a trapezoid or triangle?

Looking at the figure:
- Left side: a rectangle 5 yd wide × 12 yd tall
- Right side: a triangle with base = 26 - 5 = 21 yd, height = 7 yd? Wait.

Wait — the total base is 26 yd. The left rectangle is 5 yd wide. The right part is a trapezoid?

Actually, from the diagram:
- The left vertical segment is 12 yd high.
- The right side rises from 0 to 7 yd over a horizontal distance of 26 - 5 = 21 yd.

Wait — the shaded region is composed of:
1. A rectangle: 5 yd × 12 yd
2. A trapezoid: bases = 12 yd and 7 yd, height = 21 yd

Yes, that makes sense.

So:
- Rectangle: $ 5 \times 12 = 60 $ yd²
- Trapezoid: $ A = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (12 + 7) \times 21 = \frac{1}{2} \times 19 \times 21 = 199.5 $ yd²

Total shaded area:
$$
A = 60 + 199.5 = \boxed{259.50} \text{ yd}^2
$$

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✔ Final Answers:

| Problem | Area (rounded to 2 decimals) |
|--------|-------------------------------|
| 1) | 222.33 in² |
| 2) | 50.50 ft² |
| 3) | 308.16 yd² |
| 4) | 132.00 m² |
| 5) | 432.07 ft² |
| 6) | 259.50 yd² |

Let me know if you'd like a visual explanation or need help with any specific one!