Question

what is the area of the shaded region?
21 mm²
24 mm²
42 mm²
48 mm² (diagram labels: 6 mm, 4 mm, 3 mm, 2 mm, 5 mm)

Explanation:

Answer:

To find the area of the shaded region, we can break it into two parts: a trapezoid (or a combination of a rectangle and a triangle) and a rectangle, or use the method of subtracting the unshaded triangle from the large triangle.

First, let's consider the large triangle:

• The total height of the large triangle is \( 6 + 4 + 2 = 12 \) mm.
• The base of the large triangle is \( 5 \) mm.
• Area of large triangle: \( \frac{1}{2} \times 5 \times 12 = 30 \) mm². Wait, that doesn't match. Wait, maybe another approach.

Wait, the shaded region can be seen as a rectangle at the bottom and a trapezoid above, or subtract the unshaded triangle.

Unshaded triangle:

• Height: \( 4 \) mm, base: \( 3 \) mm.
• Area: \( \frac{1}{2} \times 3 \times 4 = 6 \) mm².

Now, the large figure (before subtracting the unshaded triangle) can be considered as a trapezoid or a combination. Wait, maybe the total area of the shaded region is the area of the large triangle minus the unshaded triangle, plus the rectangle? Wait, no. Let's re-examine the diagram.

Wait, the vertical segments: 6 mm, 4 mm, 2 mm. Horizontal: 5 mm (bottom), 3 mm (middle).

Alternative approach: Split the shaded region into two parts:

1. The bottom rectangle: height \( 2 \) mm, width \( 5 \) mm. Area: \( 2 \times 5 = 10 \) mm².
2. The upper trapezoid: the two parallel sides are \( 6 \) mm (left) and \( (5 - 3) = 2 \) mm? Wait, no. Wait, the upper part is a triangle? Wait, maybe the correct way is:

Total height of the shaded region's vertical side: \( 6 + 2 = 8 \) mm? No, the 4 mm is the height of the unshaded triangle.

Wait, let's calculate the area of the shaded region by adding the area of the rectangle (2 mm height, 5 mm base) and the area of the trapezoid (with bases 6 mm and (5 - 3) mm? No, maybe the upper part is a triangle with base 5 mm and height 6 mm, plus the rectangle (2 mm height, 5 mm base), minus the unshaded triangle (base 3 mm, height 4 mm).

Wait:

• Area of triangle (top): \( \frac{1}{2} \times 5 \times 6 = 15 \) mm².
• Area of rectangle (bottom): \( 2 \times 5 = 10 \) mm².
• Total before subtracting unshaded: \( 15 + 10 = 25 \) mm². No, that's not right.

Wait, maybe the unshaded triangle has height 4 mm and base 3 mm. The large triangle (including the shaded and unshaded) has height \( 6 + 4 + 2 = 12 \) mm and base 5 mm. Area: \( \frac{1}{2} \times 5 \times 12 = 30 \) mm². Then subtract the unshaded triangle: \( 30 - 6 = 24 \) mm²? Wait, no, 30 - 6 = 24? Wait, 30 - 6 is 24, which is one of the options. Wait, maybe that's the way.

Wait, let's check again:

• Large triangle: height = 6 + 4 + 2 = 12 mm, base = 5 mm. Area = 0.5 5 12 = 30 mm².
• Unshaded triangle: height = 4 mm, base = 3 mm. Area = 0.5 3 4 = 6 mm².
• Shaded area = 30 - 6 = 24 mm².

Yes, that matches the option 24 mm².

So the area of the shaded region is \( 24 \, \text{mm}^2 \).

\boxed{24}