2023 · Q73
An isosceles triangle has its base length 2a and its height is h. On each side of the triangle a square is drawn external to the triangle. What is the area of the figure thus formed?
Elementary MathematicsMensuration
- A6a² + 2h² + 2ah
- B6a² + 2h² + ah
- C4a² + 2h² + ah
- D6a² + h² + ah
Explanation
The figure consists of the isosceles triangle plus the three outward squares on its sides. The triangle's area is (1/2) × base × height = (1/2)(2a)(h) = ah. The square on the base has side 2a, area 4a². Each equal side has length √(a² + h²) (half the base and the height form a right triangle), so each of the two squares on the equal sides has area a² + h², totalling 2a² + 2h². The combined area is ah + 4a² + 2a² + 2h² = 6a² + 2h² + ah. Hence option (b). Distractor (a) doubles the triangle area, and (c) counts only one of the two leg-squares.