Question

the height of a triangle is 9mm less than the square of its base length. The height of a similar triangle is 6mm more than twice the length of the first triangle. Write an expression for the area of the second triange in simplifies form

Answer 1

This is what we are given:

The two triangles are similar.

Answer 2

Area of left triangle:

\(A_L = \dfrac{bh}{2} \)

\(A_L = \dfrac{x(x^2 - 9)}{2} \)

The triangles are similar. The scale factor is:

\(S = \dfrac{2x + 6}{x^2 - 9} = \dfrac{2(x + 3)}{(x + 3)(x - 3)} =\dfrac{2}{x - 3}\)

The scale factor of the areas is:

\(S^2 = \left(\dfrac{2}{x - 3}\right)^2 = \dfrac{4}{(x - 3)^2}\)

The area of the right triangle is:

\(A_R = A_LS^2 = \dfrac{x(x^2 - 9)}{2}\dfrac{4}{(x - 3)^2} \)

\(A_R = \dfrac{x(x+ 3)(x - 3)} {2} \dfrac{4}{(x - 3)(x - 3)} \)

\(A_R = \dfrac{x(x+ 3)\cancel{(x - 3)}} {\cancel{2}~~1} \dfrac{\cancel{4}~~2}{\cancel{(x - 3)}(x - 3)} \)

\(A_R = \dfrac{2x(x + 3)}{(x - 3)} \)