We are given a composite shape made of a right-angled triangle $ABC$ and two half-circles. The sides $AC = 10 \, \text{cm}$ and $CB = 10 \, \text{cm}$ are the diameters of the half-circles. We are tasked with finding the area of the composite shape.

Step 1: Area of the Right-Angled Triangle

The base and height of the triangle are $AC = 10 \, \text{cm}$ and $CB = 10 \, \text{cm}$. The formula for the area of a right-angled triangle is: $A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height}$ $A_{\text{triangle}} = \frac{1}{2} \times 10 \times 10 = 50 \, \text{cm}^2$

Step 2: Area of the Two Half-Circles

The diameters of both half-circles are 10 cm, so their radii are: $r = \frac{10}{2} = 5 \, \text{cm}$ The formula for the area of a circle is $A_{\text{circle}} = \pi r^2$, and since we have two half-circles, their combined area will be the same as the area of a full circle: $A_{\text{two half-circles}} = \pi r^2 = \pi \times 5^2 = 25\pi \, \text{cm}^2$ Using $\pi \approx 3.1416$: $A_{\text{two half-circles}} \approx 25 \times 3.1416 = 78.54 \, \text{cm}^2$

Step 3: Total Area of the Composite Shape

Now, we add the area of the triangle and the area of the two half-circles: $A_{\text{total}} = A_{\text{triangle}} + A_{\text{two half-circles}}$ $A_{\text{total}} = 50 + 78.54 = 128.54 \, \text{cm}^2$

The closest answer is: $\boxed{C. \, 128.54 \, \text{cm}^2}$

Would you like a detailed explanation or have any additional questions?

Here are 5 related questions for further exploration:

1. How would the area change if the diameters of the circles were doubled?
2. What would the perimeter of the composite shape be?
3. How can you derive the area of a sector of a circle if the angle is not 180°?
4. How does changing one of the triangle's legs affect the total area?
5. Can you calculate the centroid of the composite shape?

Tip: Always break down composite shapes into simpler parts like triangles, circles, or rectangles to simplify area calculations.