Question

1. the area of the shape shown to right is closest to

a) 62 cm²
b) 52 cm²
c) 38 cm²
d) 90 cm²
(the shape is a combination of a right triangle and a semicircle. the right triangle has one leg 8 cm, hypotenuse 10 cm, and the diameter of the semicircle is equal to the other leg of the triangle.)

Explanation:

Step1: Find base of right triangle

Use Pythagorean theorem: $a^2 + b^2 = c^2$. Here, $a=8$, $c=10$, solve for $b$:
$$b = \sqrt{10^2 - 8^2} = \sqrt{100-64} = \sqrt{36} = 6 \text{ cm}$$
This base is the diameter of the semicircle, so radius $r = \frac{6}{2}=3$ cm.

Step2: Calculate area of triangle

Area formula: $\frac{1}{2} \times \text{base} \times \text{height}$
$$A_{\text{triangle}} = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$$

Step3: Calculate area of semicircle

Area formula: $\frac{1}{2} \pi r^2$
$$A_{\text{semicircle}} = \frac{1}{2} \times \pi \times 3^2 = \frac{9}{2}\pi \approx 14.14 \text{ cm}^2$$

Step4: Calculate total area

Add the two areas:
$$A_{\text{total}} = 24 + 14.14 = 38.14 \text{ cm}^2$$

Answer:

C) $38 \text{ cm}^2$