# 1950 AHSME Problems/Problem 47

## Problem

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad \textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad \textbf{(E)}\ x=\dfrac{1}{2}b$

## Solution 1:Similarity

Draw the triangle.The small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle.Hence, the proportions of the heights is equal to the proportions of the sides.

In particular, we get $$\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}$$ The answer is $\boxed{\textbf{(C)}}$.

## Solution 2: Area

The smaller triangle at the top has an area of $$\dfrac{2x(h-x)}{2}=x(h-x)$$

Because the two other pieces are complementary triangles, we can add them together to create one triangle with area $$\dfrac{x(b-2x)}{2}$$

Lastly the area of the rectangle is $2x^2$. These areas together sum to the area of the big rectangle, which is $\dfrac{bh}{2}$. Hence, $$\dfrac{bh}{2}=2x^2 + \dfrac{x(b-2x)}{2}+x(h-x)$$

Solving we get that $$x = \dfrac{bh}{2h + b}$$ The answer is $\boxed{\textbf{(C)}}$.

## Video Solution

~MathProblemSolvingSkills.com