Question

what is the area of the shaded region?
21 mm²
14 mm²
42 mm²
48 mm²

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, find the height of the large triangle: \( 6 + 4 + 2 = 12 \) mm, base = 5 mm. Area of large triangle: \( \frac{1}{2} \times 5 \times 12 = 30 \) mm²? Wait, no, maybe better to split the shaded region.

Wait, the shaded region can be seen as a rectangle at the bottom (2 mm height, 5 mm base: area \( 2 \times 5 = 10 \) mm²) and a trapezoid above it. The trapezoid has bases 5 mm and (5 - 3) = 2 mm? Wait, no, the vertical sides: the left side is 6 mm, and the unshaded triangle has height 4 mm and base 3 mm. Wait, maybe another approach:

Total height of the figure: 6 + 4 + 2 = 12 mm. The large triangle (if we consider the outer triangle) would have base 5 mm and height 12 mm, area \( \frac{1}{2} \times 5 \times 12 = 30 \)? No, that doesn't match. Wait, maybe the shaded region is composed of a rectangle (2 mm × 5 mm = 10) and a triangle above it. The triangle above: height 6 mm, base (5 - 3) = 2? No, wait the horizontal length: the bottom is 5 mm, the indent is 3 mm, so the top horizontal length is 5 - 3 = 2 mm? Wait, no, the left side: 6 mm (vertical), then a horizontal indent of 3 mm, then 4 mm vertical, then 2 mm vertical. Wait, maybe the shaded area is the area of the large triangle minus the unshaded triangle.

Large triangle: base 5 mm, height (6 + 4 + 2) = 12 mm? No, that can't be. Wait, the vertical segments: 6 mm, 4 mm, 2 mm. So total vertical length is 6 + 4 + 2 = 12 mm. The base is 5 mm. So area of large triangle: \( \frac{1}{2} \times 5 \times 12 = 30 \) mm². Unshaded triangle: height 4 mm, base 3 mm. Area: \( \frac{1}{2} \times 3 \times 4 = 6 \) mm². Then shaded area: 30 - 6 = 24? No, the options are 21, 14, 42, 48. Wait, maybe I misread the diagram.

Wait, maybe the figure is a large triangle with height (6 + 2) = 8 mm? No, the vertical segments are 6 mm (top), 4 mm (middle), 2 mm (bottom). Wait, the bottom rectangle is 2 mm × 5 mm = 10. The middle part: a trapezoid with height 4 mm, bases 5 mm and (5 - 3) = 2 mm? Area of trapezoid: \( \frac{1}{2} \times (5 + 2) \times 4 = 14 \). Then the top triangle: height 6 mm, base (5 - 3) = 2 mm? Area: \( \frac{1}{2} \times 2 \times 6 = 6 \). Total: 10 + 14 + 6 = 30. No. Wait, maybe the correct approach is:

The shaded region can be divided into a rectangle (2 mm × 5 mm = 10) and a triangle above it. The triangle has height (6 + 4) = 10 mm? No. Wait, the options include 21. Let's recalculate.

Wait, maybe the large triangle has base 5 mm and height (6 + 2) = 8 mm? No. Wait, the vertical length from the bottom to the top of the shaded part: 2 + 4 + 6 = 12, but maybe the base is 7? No, the base is 5. Wait, maybe the unshaded triangle has height 4 mm and base 3 mm, and the large triangle has height (6 + 4 + 2) = 12 and base 5, but that gives 30 - 6 = 24, not an option. Wait, maybe the diagram is a right triangle with height (6 + 2) = 8 mm and base 5 mm? No. Wait, the options are 21, 14, 42, 48. Let's check 21: maybe the area is calculated[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]