Question

The longer leg of a right triangle is 2cm more than twice the shorter leg. Find the measure of each leg og the triangle with the smallest area possible

Answer 1

hypotenuse =
\[h = \sqrt{(x^2 + (2x+2)^2)} = \sqrt{(x^2 + 4x^2 + 8x + 4)} = \sqrt{(5x^2 + 8x + 4)}\]

Answer 2

Area = height x base x half \[Area = \frac 12 \times base \times height\]\[Area = \frac 12 \times x \times (2x + 2) \]\[Area = \frac 12 \times 2x^2 + 2x\]\[Area = x^2 + x \]