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?

  1. A. 6a^{2}+2h^{2}+2ah
  2. B. 6a^{2}+2h^{2}+ah ✓
  3. C. 4a^{2}+2h^{2}+ah
  4. D. 6a^{2}+h^{2}+ah

Correct Answer: B. 6a^{2}+2h^{2}+ah

Explanation

Area of the triangle is \frac{1}{2} \times 2a \times h = ah. The length of the equal sides is \sqrt{a^2+h^2}. The area of the square on the base is (2a)^2 = 4a^2. The areas of the two squares on the equal sides are 2 \times (\sqrt{a^2+h^2})^2 = 2a^2+2h^2. The total area of the figure formed by the triangle and the three squares is ah + 4a^2 + 2a^2 + 2h^2 = 6a^2+2h^2+ah.

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