Math worksheet for calculating the area of compound shapes with various geometric figures and dimensions.

Let’s solve each compound shape one by one. We’ll break them into simpler shapes (like rectangles, triangles, semicircles), find their areas, and add or subtract as needed. Remember:
- Area of rectangle = length × width
- Area of triangle = ½ × base × height
- Area of circle = πr² → so semicircle = ½πr²
Use π ≈ 3.14 if needed.

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Problem 1: Trapezoid-like shape (actually a right trapezoid)
It has two parallel sides: 14 in and 7 in, with height 14 in (the bottom side).
Area of trapezoid = ½ × (sum of parallel sides) × height
= ½ × (14 + 7) × 14
= ½ × 21 × 14
= 10.5 × 14 = 147.0 in²

✔ Check: Yes, that’s correct.

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Problem 2: Rectangle + Semicircle on top
Rectangle: 18 cm × 18 cm → area = 324 cm²
Semicircle: diameter = 9 cm → radius = 4.5 cm
Area of semicircle = ½ × π × r² = ½ × 3.14 × (4.5)²
= ½ × 3.14 × 20.25 = 1.57 × 20.25 ≈ 31.7925
Total area = 324 + 31.7925 ≈ 355.8 cm²

✔ Rounded to one decimal: 355.8

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Problem 3: Rectangle + Semicircle on top
Rectangle: 20 yd × 15 yd = 300 yd²
Semicircle: diameter = 6 yd → radius = 3 yd
Area = ½ × π × 3² = ½ × 3.14 × 9 = 1.57 × 9 = 14.13
Total = 300 + 14.13 = 314.1 yd²

✔ Rounded: 314.1

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Problem 4: Triangle + Semicircle below
Triangle: base = ? Wait — the diagram shows a triangle with height 12 m, and the base is the same as the diameter of the semicircle? Actually, looking at the figure: it's a triangle sitting on top of a semicircle. The dashed line is 8 m — that’s the radius of the semicircle? No — wait, the label says “8 m” with an arrow pointing to the radius of the semicircle. So radius = 8 m → diameter = 16 m → that’s the base of the triangle too.

So:
- Triangle: base = 16 m, height = 12 m → area = ½ × 16 × 12 = 96 m²
- Semicircle: radius = 8 m → area = ½ × π × 64 = 32π ≈ 32 × 3.14 = 100.48 m²
Total = 96 + 100.48 = 196.5 m²

✔ Rounded: 196.5

Wait — let me double-check the diagram description. It says “12 m” for the triangle height, and “8 m” with an arrow to the radius of the semicircle. Yes, so base of triangle = diameter = 16 m. Correct.

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Problem 5: T-shaped figure — can be split into two rectangles.
Top rectangle: 12 in wide × 12 in tall → area = 144 in²
Bottom rectangle: 20 in wide × 12 in tall → area = 240 in²
But wait — they overlap? No, actually, the top rectangle sits centered on the bottom one. But since we’re just adding areas, and there’s no overlap in the shaded region, total area = 144 + 240 = 384 in²

✔ Simple addition — yes.

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Problem 6: Rectangle + Triangle on the right
Left part: rectangle 20 cm × 20 cm = 400 cm²
Right part: triangle with base 10 cm and height 20 cm → area = ½ × 10 × 20 = 100 cm²
Total = 400 + 100 = 500.0 cm²

✔ Easy.

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Problem 7: Triangle + Semicircle
Triangle: right triangle with legs 13 ft and 6 ft? Wait — the diagram shows a vertical side 13 ft, horizontal side 6 ft, and a right angle between them? Actually, looking: it’s a right triangle with base 6 ft and height 13 ft? But then attached to the base is a semicircle with radius 6 ft? That doesn’t make sense — because the semicircle is drawn under the 6 ft side.

Actually, re-examining: the triangle has height 13 ft, and the base is 6 ft (same as the diameter of the semicircle? Or radius?) — the label says “6 ft” with an arrow to the radius of the semicircle. So radius = 6 ft → diameter = 12 ft. But the triangle’s base is only 6 ft? That doesn’t match.

Wait — perhaps the triangle is attached to the flat side of the semicircle. If the semicircle has radius 6 ft, its diameter is 12 ft. But the triangle’s base is labeled 6 ft? That seems inconsistent.

Looking again: In problem 7, the figure is a right triangle with vertical leg 13 ft, horizontal leg 6 ft, and then a semicircle attached to the horizontal leg — but the semicircle’s radius is labeled 6 ft? That would mean the diameter is 12 ft, but the leg is only 6 ft. That can’t be.

Perhaps the “6 ft” is the radius, and the triangle’s base is also 6 ft? Then the semicircle is drawn with radius 6 ft, meaning it extends beyond the triangle? That might be possible.

Assume:
- Triangle: base = 6 ft, height = 13 ft → area = ½ × 6 × 13 = 39 ft²
- Semicircle: radius = 6 ft → area = ½ × π × 36 = 18π ≈ 18 × 3.14 = 56.52 ft²
Total = 39 + 56.52 = 95.5 ft²

✔ Rounded: 95.5

Even though the geometry looks odd, based on labels, this is what we have.

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Problem 8: House-shaped figure — rectangle + triangle on top
Rectangle: 6 yd wide × 8 yd tall → area = 48 yd²
Triangle: base = 6 yd, height = 9 yd → area = ½ × 6 × 9 = 27 yd²
Total = 48 + 27 = 75.0 yd²

✔ Straightforward.

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Problem 9: L-shaped figure — can be split into two rectangles.
Option 1: Left rectangle: 11 ft high × 10 ft wide = 110 ft²
Right rectangle: 8 ft high × 8 ft wide = 64 ft²
But wait — do they overlap? The total width is 10 + 8 = 18 ft, and heights are different. Actually, the left part is 11 ft tall, right part is 8 ft tall, and they share a common side? Better to think:

The whole shape can be seen as a big rectangle minus a missing corner, but easier to split vertically:

- Left rectangle: width 10 ft, height 11 ft → 110 ft²
- Right rectangle: width 8 ft, height 8 ft → 64 ft²
But the right rectangle starts at the bottom, so no overlap. Total = 110 + 64 = 174.0 ft²

Alternatively, split horizontally:
- Bottom rectangle: full width 18 ft × 8 ft = 144 ft²
- Top rectangle: only left part, 10 ft wide × (11-8)=3 ft tall = 30 ft²
Total = 144 + 30 = 174 ft² — same answer.

✔ Confirmed.

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Now, compiling all answers:

1) 147.0
2) 355.8
3) 314.1
4) 196.5
5) 384.0
6) 500.0
7) 95.5
8) 75.0
9) 174.0

All rounded to one decimal place as instructed.

Final Answer:
1) 147.0 in²
2) 355.8 cm²
3) 314.1 yd²
4) 196.5 m²
5) 384.0 in²
6) 500.0 cm²
7) 95.5 ft²
8) 75.0 yd²
9) 174.0 ft²