Area of Composite Shapes worksheet with 12 problems for calculating shaded areas.

I will solve the problem by finding the area of each shaded composite shape. I'll go through each one step-by-step, breaking down the composite shapes into simpler geometric figures (rectangles, triangles, circles, etc.) and then summing or subtracting their areas as needed.

Let me begin with problem 1.

Problem 1

This shape is a rectangle with a semicircle on the right side.
- Rectangle: 5 m × 4 m = 20 m²
- Semicircle: radius = 4 m / 2 = 2 m, so area = (π × r²) / 2 = (π × 2²) / 2 = (4π) / 2 = 2π m² ≈ 6.28 m²
- Total area = 20 + 2π ≈ 26.28 m²

Problem 2

This is an L-shaped figure that can be divided into two rectangles.
- Bottom rectangle: 10 ft × 4 ft = 40 ft²
- Top rectangle: 4 ft × 3 ft = 12 ft² (since the total height is 7 ft and bottom is 4 ft, the top part is 3 ft high)
- Total area = 40 + 12 = 52 ft²

Problem 3

This is a rectangle with a triangle on the right side.
- Rectangle: 12 cm × 8 cm = 96 cm²
- Triangle: base = 15 cm - 12 cm = 3 cm, height = 8 cm, so area = (1/2) × 3 × 8 = 12 cm²
- Total area = 96 + 12 = 108 cm²

Problem 4

This is an L-shaped figure that can be divided into two rectangles.
- Bottom rectangle: 15 mi × 2 mi = 30 mi²
- Left rectangle: 12 mi × 6 mi = 72 mi² (since total height is 8 mi and bottom is 2 mi, the left part is 6 mi high)
- Total area = 30 + 72 = 102 mi²

Problem 5

This is a rectangle with a smaller rectangle cut out from the top.
- Outer rectangle: 10 km × 5 km = 50 km²
- Inner rectangle (cutout): 8 km × 3 km = 24 km²
- Total area = 50 - 24 = 26 km²

Problem 6

This is a T-shaped figure that can be divided into two rectangles.
- Top rectangle: 7 yd × 2 yd = 14 yd²
- Bottom rectangle: 5 yd × 2 yd = 10 yd²
- Total area = 14 + 10 = 24 yd²

Problem 7

This is a rectangle with a triangle cut out from the bottom.
- Rectangle: 5 in × 10 in = 50 in²
- Triangle: base = 5 in, height = 10 in - 8 in = 2 in, so area = (1/2) × 5 × 2 = 5 in²
- Total area = 50 - 5 = 45 in²

Problem 8

This is a rectangle with a trapezoid cut out from the right side.
- Rectangle: 10 m × 8 m = 80 m²
- Trapezoid: bases are 8 m and 1 m, height is 10 m - 8 m = 2 m? Wait, let me reconsider. Actually, looking at the diagram, the cutout appears to be a parallelogram or trapezoid with parallel sides of 8 m and 1 m, but the height is not clearly specified. Let me re-examine.

Actually, looking more carefully, the cutout is a parallelogram with base 1 m and height 8 m (the vertical dimension). So area = base × height = 1 m × 8 m = 8 m².
- Total area = 80 - 8 = 72 m²

Problem 9

This is a ring (annulus) formed by two concentric circles.
- Outer circle: radius = 3 m, area = π × 3² = 9π m²
- Inner circle: radius = 2 m, area = π × 2² = 4π m²
- Total area = 9π - 4π = 5π m² ≈ 15.71 m²

Problem 10

This is a rectangle with a triangle cut out from the bottom left corner.
- Rectangle: 10 in × 10 in = 100 in²
- Triangle: base = 10 in - 9 in = 1 in, height = 10 in - 9 in = 1 in, so area = (1/2) × 1 × 1 = 0.5 in²
- Total area = 100 - 0.5 = 99.5 in²

Problem 11

This is a hexagon-like shape that can be seen as a rectangle with two triangles on the ends.
- Middle rectangle: 13 ft × 4 ft = 52 ft²
- Two triangles: each has base = 2 ft, height = 4 ft, so area of one triangle = (1/2) × 2 × 4 = 4 ft², total for two = 8 ft²
- Total area = 52 + 8 = 60 ft²

Problem 12

This is a square with a circle cut out from the center.
- Square: 4 yd × 4 yd = 16 yd²
- Circle: diameter = 4 yd, so radius = 2 yd, area = π × 2² = 4π yd² ≈ 12.57 yd²
- Total area = 16 - 4π ≈ 3.43 yd²

Now I'll summarize all the answers:

1. A = 20 + 2π ≈ 26.28 m²
2. A = 52 ft²
3. A = 108 cm²
4. A = 102 mi²
5. A = 26 km²
6. A = 24 yd²
7. A = 45 in²
8. A = 72 m²
9. A = 5π ≈ 15.71 m²
10. A = 99.5 in²
11. A = 60 ft²
12. A = 16 - 4π ≈ 3.43 yd²