Question 234099
The area of the rectangle is:
{{{A[r] = (4x+3)(3x-2)}}}
{{{A[r] = 12x^2+x-6}}}
The area of the triangle is:
{{{A[t] = (1/2)(4x)(5x)}}}
{{{A[t] = 10x^2}}}
{{{A[r] = A[t]}}} so...
{{{12x^2+x-6 = 10x^2}}} Subtract{{{10x^2}}} from both sides.
{{{highlight(2x^2+x-6 = 0)}}} Solve by factoring.
{{{(2x-3)(x+2) =0}}} 
{{{2x-3 = 0}}} or {{{x+2 = 0}}} so that...
{{{x = 3/2}}} or {{{x = -2}}} Discard the negative solution as lengths are positive.
{{{highlight(x = 3/2)}}}
{{{PQ = 5x}}}
{{{PQ = 5(3/2)}}}
{{{highlight(PQ = 7.5)}}}cm.