# Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 2"

## Problem

$[asy] pair D=(0,0),B=(0,12),C=(16,0),A=(-9,0); draw(A--B--C--A,dot); draw(D--B,dot); MP("D",D,S);MP("A",A,S);MP("C",C,S);MP("B",B,N); MP("16",(8,0),S);MP("15",(A+B)/2,NW); [/asy]$

Segment AB is perpendicular to segment BC and segment AC is perpendicular to segment BD. If segment AB has length 15 and segment DC has length 16, then what is the area of triangle ABC?

## Solution

Let the altitude $BD$ be called $h$, $AD$ be called $x$, and $BC$ called $y$. Since $ABC$ is a right triangle, we can write a system of equations. $$h^2=16x$$ $$15^2=x(x+16)$$ $$y^2=16(16+x)$$

Solving the second equation, we get $x^2+16x-225=0 \rightarrow (x-9)(x+25)=0 \rightarrow x=9$ or $-25$. Since $x$ has to be positive, $x=9$. Plugging $9$ into the first equation, we get that $h^2=4^2\cdot3^2 \rightarrow h=12$ or $-12$. Once again, side lengths must be positive, so $h=12$. Now we have an altitude and a base, so the area of $ABC$ is $\frac{(9+16)\cdot12}{2} = \boxed{150}$