Question

The altitude of a triangle is 2 m longer than its base. What are the dimensions of the altitude and the base if thearea of the triangle is 40 m^2?

Answer 1

Step-by-step explanation:

Let base of triangle be x m.

Therefore, altitude =( x + 2)m

$area \: of \: \triangle = \frac{1}{2} \times base \times altitude \\ \\ 40 = \frac{1}{2} \times x \times (x + 2) \\ \\ 80 = {x }^{2} +2x \\ \\ {x}^{2} + 2x - 80 = 0 \\ {x}^{2} + 10x - 8x - 80 \\ x(x + 10) - 8(x + 10) = 0\\ (x + 10)(x - 8) = 0 \\ x + 10 = 0 \: \: or \: \: x - 8 = 0 \\ x = - 10 \: \: or \: \: x = 8 \\ base \: can \: not \: be \: - ve \\ x \neq - 10 \\ x = 8 \\ x + 2 = 8 + 2 = 10$

Therefore

Altitude = 10 m

Base = 8 m