Geometry Worksheets | Area Worksheets - Free Printable
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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 problem one by one. We’ll find the area of each compound shape by breaking it into simpler shapes (like triangles, rectangles, circles, semicircles) and then adding or subtracting areas as needed.
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Problem 1:
Shape: Triangle with a circle cut out from inside.
- Triangle base = 29 ft, height = 29 ft → Area = (1/2) × base × height = (1/2) × 29 × 29 = 420.5 ft²
- Circle radius = 9 ft → Area = π × r² = 3.1416 × 81 ≈ 254.47 ft²
- Shaded area = Triangle - Circle = 420.5 - 254.47 ≈ 166.0 ft²
✔ Check: 29×29=841, half is 420.5. π×81≈254.469. Difference ≈166.031 → rounded to 1 decimal: 166.0
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Problem 2:
Shape: Right triangle + semicircle
- Triangle: base = 5 in, height = 11 in → Area = (1/2) × 5 × 11 = 27.5 in²
- Semicircle: radius = 5 in → Area = (1/2) × π × r² = 0.5 × 3.1416 × 25 ≈ 39.27 in²
- Total area = 27.5 + 39.27 ≈ 66.8 in²
✔ Check: 5×11/2=27.5; π×25/2≈39.2699 → sum ≈66.7699 → round to 66.8
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Problem 3:
Shape: Square with circle cut out
- Square side = 16 in → Area = 16 × 16 = 256 in²
- Circle radius = 8 in → Area = π × 64 ≈ 201.06 in²
- Shaded area = 256 - 201.06 ≈ 54.9 in²
✔ Check: 16²=256; π×64≈201.0619 → difference ≈54.938 → round to 54.9
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Problem 4:
Shape: “C” shaped rectangle — think of it as big rectangle minus small rectangle.
- Big rectangle: width = 11 m, height = 14 m → Area = 11 × 14 = 154 m²
- Cut-out part: width = 7 m, height = 7 m → Area = 7 × 7 = 49 m²
- But wait — looking at diagram: The “C” has outer dimensions 14m tall, 11m wide. Inner empty space is 7m wide and 7m high? Actually, let's re-analyze:
Actually, better way: Break into 3 rectangles:
- Left vertical bar: 7m wide × 14m tall? No — look again.
Wait — standard approach for C-shape:
Total area = area of full rectangle minus the missing middle rectangle.
Full rectangle if no hole: 14m × 11m = 154 m²
Missing rectangle: 7m (width) × 7m (height) = 49 m²? But that would leave two sides.
Actually, looking at labels:
The figure shows:
- Overall height: 14 m
- Bottom width: 11 m
- Top horizontal part: 7 m long
- Vertical gap in middle: 7 m high
So actually, we can compute as:
Left rectangle: 7m wide × 14m tall? Not quite.
Better: Think of it as:
- Bottom rectangle: 11m × ?
Actually, let’s use subtraction:
Imagine full rectangle 14m × 11m = 154 m²
Now subtract the inner rectangle that’s missing: which is 7m wide and 7m high? But where?
From diagram: The “hole” is centered? Actually, the top arm is 7m, bottom is 11m, so the missing part is on the right side? Wait — perhaps it’s easier to add parts.
Alternative method:
Break into three rectangles:
1. Left vertical: 7m wide × 14m tall → 98 m²? Too big.
Wait — correct interpretation:
Looking at typical such problems:
The shape is like a rectangle 14m high and 11m wide, but with a rectangular notch taken out from the right side, starting from top.
Actually, from the diagram description:
It says:
- Height total: 14 m
- Width at bottom: 11 m
- Top horizontal segment: 7 m
- And there’s a 7 m vertical drop labeled inside.
This suggests:
The shape consists of:
- A left rectangle: 7m wide × 14m tall → 98 m²
- Plus a bottom rectangle extending to the right: (11 - 7)=4m wide × ? height.
But the vertical label “7m” is inside the notch — meaning the notch is 7m deep vertically.
Actually, standard solution for this type:
Area = area of large rectangle minus area of small rectangle cut out.
Large rectangle: 14m × 11m = 154 m²
Cut-out rectangle: 7m (height) × (11 - 7)m = 7m × 4m = 28 m²? Wait, no.
Wait — if the top part is only 7m wide, and bottom is 11m, then the missing part is on the right, from top down 7m, and width = 11 - 7 = 4m.
Yes! So:
Cut-out rectangle: width = 4m, height = 7m → area = 28 m²
Shaded area = 154 - 28 = 126 m²
✔ Confirm: Yes, that matches common textbook problems. So 126.0 m²
---
Problem 5:
Shape: Rectangle with semicircle cut out from right side.
- Rectangle: 11 cm × ? height not given? Wait — diagram shows height same as diameter of semicircle.
Semicircle radius = 4 cm → diameter = 8 cm → so rectangle height = 8 cm
Rectangle area = 11 × 8 = 88 cm²
Semicircle area = (1/2) × π × 4² = 0.5 × 3.1416 × 16 ≈ 25.13 cm²
Shaded area = 88 - 25.13 ≈ 62.9 cm²
✔ Check: 11×8=88; π×16/2≈25.1327 → difference ≈62.867 → round to 62.9
---
Problem 6:
Shape: Rectangle + triangle on right.
- Rectangle: 12m × 12m = 144 m²
- Triangle: base = 6m, height = 12m → Area = (1/2) × 6 × 12 = 36 m²
- Total = 144 + 36 = 180.0 m²
✔ Simple addition: 144+36=180 → 180.0
---
Problem 7:
Shape: Two triangles on top of a semicircle.
- Each triangle: base = ? , height = 5 yd. From diagram, they share the diameter of the semicircle.
Semicircle radius = 2 yd → diameter = 4 yd → so each triangle has base = 2 yd? Or together?
Actually, the two triangles are symmetric, sitting on the diameter. So combined, their bases make up the diameter = 4 yd.
Each triangle: base = 2 yd, height = 5 yd → area per triangle = (1/2) × 2 × 5 = 5 yd² → two triangles = 10 yd²
Semicircle: radius = 2 yd → area = (1/2) × π × 4 = 2π ≈ 6.283 yd²
Total shaded area = 10 + 6.283 ≈ 16.3 yd²
✔ Check: Triangles: 2 * (0.5*2*5) = 10; semicircle: 0.5*π*4≈6.2832 → sum≈16.2832 → round to 16.3
---
Problem 8:
Shape: Large square with small square cut out.
- Large square: 6 ft × 6 ft = 36 ft²
- Small square: 3 ft × 3 ft = 9 ft²
- Shaded area = 36 - 9 = 27.0 ft²
✔ Straightforward: 36-9=27 → 27.0
---
Problem 9:
Shape: Rectangle + triangle on top (house shape).
- Rectangle: 20 yd × 15 yd = 300 yd²
- Triangle: base = 20 yd, height = 10 yd → Area = (1/2) × 20 × 10 = 100 yd²
- Total = 300 + 100 = 400.0 yd²
✔ 300+100=400 → 400.0
---
Final Answers:
1) 166.0
2) 66.8
3) 54.9
4) 126.0
5) 62.9
6) 180.0
7) 16.3
8) 27.0
9) 400.0
──────────────────────────────────────
Final Answer:
1) 166.0
2) 66.8
3) 54.9
4) 126.0
5) 62.9
6) 180.0
7) 16.3
8) 27.0
9) 400.0
---
Problem 1:
Shape: Triangle with a circle cut out from inside.
- Triangle base = 29 ft, height = 29 ft → Area = (1/2) × base × height = (1/2) × 29 × 29 = 420.5 ft²
- Circle radius = 9 ft → Area = π × r² = 3.1416 × 81 ≈ 254.47 ft²
- Shaded area = Triangle - Circle = 420.5 - 254.47 ≈ 166.0 ft²
✔ Check: 29×29=841, half is 420.5. π×81≈254.469. Difference ≈166.031 → rounded to 1 decimal: 166.0
---
Problem 2:
Shape: Right triangle + semicircle
- Triangle: base = 5 in, height = 11 in → Area = (1/2) × 5 × 11 = 27.5 in²
- Semicircle: radius = 5 in → Area = (1/2) × π × r² = 0.5 × 3.1416 × 25 ≈ 39.27 in²
- Total area = 27.5 + 39.27 ≈ 66.8 in²
✔ Check: 5×11/2=27.5; π×25/2≈39.2699 → sum ≈66.7699 → round to 66.8
---
Problem 3:
Shape: Square with circle cut out
- Square side = 16 in → Area = 16 × 16 = 256 in²
- Circle radius = 8 in → Area = π × 64 ≈ 201.06 in²
- Shaded area = 256 - 201.06 ≈ 54.9 in²
✔ Check: 16²=256; π×64≈201.0619 → difference ≈54.938 → round to 54.9
---
Problem 4:
Shape: “C” shaped rectangle — think of it as big rectangle minus small rectangle.
- Big rectangle: width = 11 m, height = 14 m → Area = 11 × 14 = 154 m²
- Cut-out part: width = 7 m, height = 7 m → Area = 7 × 7 = 49 m²
- But wait — looking at diagram: The “C” has outer dimensions 14m tall, 11m wide. Inner empty space is 7m wide and 7m high? Actually, let's re-analyze:
Actually, better way: Break into 3 rectangles:
- Left vertical bar: 7m wide × 14m tall? No — look again.
Wait — standard approach for C-shape:
Total area = area of full rectangle minus the missing middle rectangle.
Full rectangle if no hole: 14m × 11m = 154 m²
Missing rectangle: 7m (width) × 7m (height) = 49 m²? But that would leave two sides.
Actually, looking at labels:
The figure shows:
- Overall height: 14 m
- Bottom width: 11 m
- Top horizontal part: 7 m long
- Vertical gap in middle: 7 m high
So actually, we can compute as:
Left rectangle: 7m wide × 14m tall? Not quite.
Better: Think of it as:
- Bottom rectangle: 11m × ?
Actually, let’s use subtraction:
Imagine full rectangle 14m × 11m = 154 m²
Now subtract the inner rectangle that’s missing: which is 7m wide and 7m high? But where?
From diagram: The “hole” is centered? Actually, the top arm is 7m, bottom is 11m, so the missing part is on the right side? Wait — perhaps it’s easier to add parts.
Alternative method:
Break into three rectangles:
1. Left vertical: 7m wide × 14m tall → 98 m²? Too big.
Wait — correct interpretation:
Looking at typical such problems:
The shape is like a rectangle 14m high and 11m wide, but with a rectangular notch taken out from the right side, starting from top.
Actually, from the diagram description:
It says:
- Height total: 14 m
- Width at bottom: 11 m
- Top horizontal segment: 7 m
- And there’s a 7 m vertical drop labeled inside.
This suggests:
The shape consists of:
- A left rectangle: 7m wide × 14m tall → 98 m²
- Plus a bottom rectangle extending to the right: (11 - 7)=4m wide × ? height.
But the vertical label “7m” is inside the notch — meaning the notch is 7m deep vertically.
Actually, standard solution for this type:
Area = area of large rectangle minus area of small rectangle cut out.
Large rectangle: 14m × 11m = 154 m²
Cut-out rectangle: 7m (height) × (11 - 7)m = 7m × 4m = 28 m²? Wait, no.
Wait — if the top part is only 7m wide, and bottom is 11m, then the missing part is on the right, from top down 7m, and width = 11 - 7 = 4m.
Yes! So:
Cut-out rectangle: width = 4m, height = 7m → area = 28 m²
Shaded area = 154 - 28 = 126 m²
✔ Confirm: Yes, that matches common textbook problems. So 126.0 m²
---
Problem 5:
Shape: Rectangle with semicircle cut out from right side.
- Rectangle: 11 cm × ? height not given? Wait — diagram shows height same as diameter of semicircle.
Semicircle radius = 4 cm → diameter = 8 cm → so rectangle height = 8 cm
Rectangle area = 11 × 8 = 88 cm²
Semicircle area = (1/2) × π × 4² = 0.5 × 3.1416 × 16 ≈ 25.13 cm²
Shaded area = 88 - 25.13 ≈ 62.9 cm²
✔ Check: 11×8=88; π×16/2≈25.1327 → difference ≈62.867 → round to 62.9
---
Problem 6:
Shape: Rectangle + triangle on right.
- Rectangle: 12m × 12m = 144 m²
- Triangle: base = 6m, height = 12m → Area = (1/2) × 6 × 12 = 36 m²
- Total = 144 + 36 = 180.0 m²
✔ Simple addition: 144+36=180 → 180.0
---
Problem 7:
Shape: Two triangles on top of a semicircle.
- Each triangle: base = ? , height = 5 yd. From diagram, they share the diameter of the semicircle.
Semicircle radius = 2 yd → diameter = 4 yd → so each triangle has base = 2 yd? Or together?
Actually, the two triangles are symmetric, sitting on the diameter. So combined, their bases make up the diameter = 4 yd.
Each triangle: base = 2 yd, height = 5 yd → area per triangle = (1/2) × 2 × 5 = 5 yd² → two triangles = 10 yd²
Semicircle: radius = 2 yd → area = (1/2) × π × 4 = 2π ≈ 6.283 yd²
Total shaded area = 10 + 6.283 ≈ 16.3 yd²
✔ Check: Triangles: 2 * (0.5*2*5) = 10; semicircle: 0.5*π*4≈6.2832 → sum≈16.2832 → round to 16.3
---
Problem 8:
Shape: Large square with small square cut out.
- Large square: 6 ft × 6 ft = 36 ft²
- Small square: 3 ft × 3 ft = 9 ft²
- Shaded area = 36 - 9 = 27.0 ft²
✔ Straightforward: 36-9=27 → 27.0
---
Problem 9:
Shape: Rectangle + triangle on top (house shape).
- Rectangle: 20 yd × 15 yd = 300 yd²
- Triangle: base = 20 yd, height = 10 yd → Area = (1/2) × 20 × 10 = 100 yd²
- Total = 300 + 100 = 400.0 yd²
✔ 300+100=400 → 400.0
---
Final Answers:
1) 166.0
2) 66.8
3) 54.9
4) 126.0
5) 62.9
6) 180.0
7) 16.3
8) 27.0
9) 400.0
──────────────────────────────────────
Final Answer:
1) 166.0
2) 66.8
3) 54.9
4) 126.0
5) 62.9
6) 180.0
7) 16.3
8) 27.0
9) 400.0
Parent Tip: Review the logic above to help your child master the concept of worksheet for area.
