Question

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:

$x=\dfrac12h$

$x=\dfrac{bh}{b+h}$

$x=\dfrac{bh}{2h+b}$

$x=\sqrt{\dfrac{hb}{2}}$

$x=\dfrac12b$

Answer: C

Concepts:similarity rectangle altitude

Difficulty rating: 1800

Small Hint:

The segment across the triangle at height $x$ has length $b(1-\frac{x}{h})$

Big Hint:

The rectangle’s base is $2x$

Solution:

By similarity, the width of the triangle at height $x$ above its base is $b(1-\tfrac{x}{h}).$ This is the base of the inscribed rectangle. Since the rectangle’s altitude is half its base, its base is $2x.$ Hence $2x=b\left(1-\frac{x}{h}\right).$ Multiplying by $h$ and solving gives $x(2h+b)=bh,$ so $x=\dfrac{bh}{2h+b}.$

Thus, the correct answer is C.

1950 AMC 12 Problem 47

Answer: C

Difficulty rating: 1800

Small Hint:

Big Hint:

Solution:

Problem 47 in Other Years