The question on the image concerns finding the perimeter and area of an irregular figure, which includes a semicircle and a right triangle.

Problem Breakdown:

1. Semicircle: The diameter is given as 8 cm, making the radius $r = \frac{8}{2} = 4$ cm.
2. Right Triangle: The base of the triangle is 4 cm, and its height (same as the diameter of the semicircle) is 8 cm.

Solutions:

Part (a)

Determine the measures of the two legs of the right triangle:

• One leg: 4 cm (base of the triangle).
• Other leg: 8 cm (height of the triangle, same as the diameter of the semicircle).

Part (b)

Explain how to use these measures to find the hypotenuse:

• Use the Pythagorean theorem, $c = \sqrt{a^2 + b^2}$, where $a$ and $b$ are the legs of the triangle, and $c$ is the hypotenuse.

Part (c)

Calculate the hypotenuse: $c = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \approx 8.94 \text{ cm}$

Would you like further details on the area or perimeter of the combined figure?

Follow-Up Questions:

1. How would you calculate the area of the semicircle?
2. What is the perimeter of the irregular figure?
3. How is the hypotenuse used in perimeter calculations for composite shapes?
4. What is the total area of the figure, combining the semicircle and the triangle?
5. How would the calculations change if the triangle's base were altered?

Tip: Remember to use consistent units throughout calculations for accuracy.