Ratio of areas determined by a square inscribed in the corner of a right triangle

Question

I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)

A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.

Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?

• (A) $1:1$
• (B) $c:(a+b)$
• (C) $a b: c^2$
• (D) $( a + b )^2 : 2 c^2$
• (E) $c^2 : 2 a b$

Apparently the answer is $c^2 : 2 a b$ (choice E), but how?

Your help is greatly appreciated! Thank you in advance.

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(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)

Answer 1

Let the side of the square is $x$. Then, using similarity of triangles, $\frac{b-x}{x}=\frac{x}{a-x}$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=\frac{ab}{a+b}$. Thus, the shaded area is $\frac{(ab)^2}{(a+b)^2}$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5\left(\frac{ab}{a+b}(b-\frac{ab}{a+b})+\frac{ab}{a+b}(a-\frac{ab}{a+b}\right)=\frac{ab(b^2+a^2)}{2(a+b)^2}=\frac{abc^2}{2(a+b)^2}$$ so the final ratio is $$\frac{2ab}{c^2}$$

Answer 2

Let $x$ be the side length of the square.

By similarity,

• The hypotenuse of the lower small right triangle is $\bigl({\large{\frac{c}{b}}}\bigr)x$.$\\[4pt]$
• The hypotenuse of the upper small right triangle is $\bigl({\large{\frac{c}{a}}}\bigr)x$.

Hence we get $$\left(\frac{c}{b}\right)x+\left(\frac{c}{a}\right)x=c$$ which yields $$x=\frac{ab}{a+b}$$ If $S,U$ are the respective areas of the shaded and unshaded regions, then

• $S=x^2$$\\[4pt]$
• $U=\bigl({\large{\frac{1}{2}}}\bigr)ab-x^2$

hence, the required ratio can be expressed as \begin{align*} \frac{U}{S} &=\frac{\left(\frac{1}{2}\right)ab-x^2}{x^2}\\[4pt] &=\frac{\left(\frac{1}{2}\right)ab}{x^2}-1\\[4pt] &=\left( \left({\small{\frac{1}{2}}}\right)ab \right) \left({\small{\frac{1}{x}}}\right)^2 -1 \\[4pt] &=\left( \left({\small{\frac{1}{2}}}\right)ab \right) \left({\small{\frac{a+b}{ab}}}\right)^2 -1 \\[4pt] &=\frac{(a+b)^2}{2ab}-1\\[4pt] &=\frac{(a+b)^2-2ab}{2ab}\\[4pt] &=\frac{a^2+b^2}{2ab}\\[4pt] &=\frac{c^2}{2ab}\\[4pt] \end{align*}