Let's solve each of these composite figures step by step. We'll break down each shape into simpler parts, calculate their areas, and then add them together.
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Problem 1:
Figure: A parallelogram on top of a trapezoid.
-
Top part (Parallelogram):
- Base = 15 cm
- Height = 14 cm
- Area of parallelogram = base × height
$$
A = 15 \times 14 = 210 \text{ cm}^2
$$
-
Bottom part (Trapezoid):
- Bases: 15 cm (top) and 34 cm (bottom)
- Height = 19 cm
- Area of trapezoid = $ \frac{1}{2} \times (b_1 + b_2) \times h $
$$
A = \frac{1}{2} \times (15 + 34) \times 19 = \frac{1}{2} \times 49 \times 19 = 465.5 \text{ cm}^2
$$
-
Total area:
$$
210 + 465.5 = \boxed{675.50} \text{ cm}^2
$$
---
Problem 2:
Figure: Rectangle with two semicircles on top (forms a "heart" shape).
- The top has
two semicircles, which together make
one full circle.
- Diameter of each semicircle = 28 m → Radius = $ \frac{28}{2} = 14 $ m
- Area of full circle = $ \pi r^2 = \pi \times 14^2 = 196\pi \approx 615.75 \text{ m}^2 $
- Rectangle:
- Width = 28 m
- Height = 24 m
- Area = $ 28 \times 24 = 672 \text{ m}^2 $
-
Total area:
$$
615.75 + 672 = \boxed{1287.75} \text{ m}^2
$$
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Problem 3:
Figure: Two semicircles joined together — one large and one small.
- First semicircle (left):
- Diameter = 22 ft → Radius = 11 ft
- Area = $ \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (11)^2 = \frac{1}{2} \pi (121) = 60.5\pi $
- Second semicircle (right):
- Total length from left to right is 30 ft, and the first semicircle takes up 22 ft, so the second semicircle's diameter = $ 30 - 22 = 8 $ ft
- Radius = $ 4 $ ft
- Area = $ \frac{1}{2} \pi (4)^2 = \frac{1}{2} \pi (16) = 8\pi $
-
Total area:
$$
60.5\pi + 8\pi = 68.5\pi \approx 215.14 \text{ ft}^2
$$
So, $ \boxed{215.14} \text{ ft}^2 $
---
Problem 4:
Figure: A semicircle attached to a right triangle.
- Semicircle:
- Radius = 12.5 mm
- Area = $ \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (12.5)^2 = \frac{1}{2} \pi (156.25) = 78.125\pi \approx 245.44 \text{ mm}^2 $
- Triangle:
- Base = 32.5 mm
- Height = 12.5 mm
- Area = $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 32.5 \times 12.5 = 203.125 \text{ mm}^2 $
-
Total area:
$$
245.44 + 203.125 = \boxed{448.57} \text{ mm}^2
$$
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Problem 5:
Figure: A triangle on top of a square.
- The triangle has:
- Base = ? We need to find it.
- Height = 35 in
- Side = 37 in (hypotenuse)
But wait — the figure shows a triangle with a height of 35 in, and slanted sides of 37 in. It appears that the triangle is
isosceles, and the height splits the base into two equal parts.
We can use the Pythagorean theorem to find half the base:
$$
\text{Half base}^2 + 35^2 = 37^2 \\
x^2 + 1225 = 1369 \\
x^2 = 144 \Rightarrow x = 12
$$
So full base = $ 2 \times 12 = 24 $ in
- Area of triangle = $ \frac{1}{2} \times 24 \times 35 = 420 \text{ in}^2 $
- Square below:
- Side = 24 in (same as triangle base)
- Area = $ 24 \times 24 = 576 \text{ in}^2 $
-
Total area:
$$
420 + 576 = \boxed{996} \text{ in}^2
$$
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✔ Final Answers:
1. $ \boxed{675.50} \text{ cm}^2 $
2. $ \boxed{1287.75} \text{ m}^2 $
3. $ \boxed{215.14} \text{ ft}^2 $
4. $ \boxed{448.57} \text{ mm}^2 $
5. $ \boxed{996} \text{ in}^2 $
Let me know if you'd like the work shown in a different format!
Parent Tip: Review the logic above to help your child master the concept of surface area of composite figures worksheet.
