# 1993 UNCO Math Contest II Problems/Problem 10

## Problem

The scalene triangle $ABC$ has side lengths $51, 52, 53.$ $AD$ is perpendicular to $BC.$ $[asy] draw((0,0)--(52,0)--(24,sqrt(3)*26)--cycle); draw((24,0)--(24,sqrt(3)*26)); draw((0,-8)--(52,-8),arrow=Arrow()); draw((52,-8)--(0,-8),arrow=Arrow()); draw((24,3)--(21,3)--(21,0),black); MP("B",(0,0),SW);MP("A",(24,sqrt(3)*26),N);MP("C",(52,0),SE);MP("D",(24,0),S); MP("52",(26,-8),S);MP("53",(38,sqrt(3)*13),NE);MP("51",(12,sqrt(3)*13),NW); [/asy]$

(a) Determine the length of $BD.$

(b) Determine the area of triangle $ABC.$

## Solution

We will solve both parts at once since it is easy to get the two answers from this all-inclusive solution.

Heron's Formula states that in a triangle with sides $a, b, c$ and $s = \frac{a + b + c}{2},$ the area is given by $$\sqrt{s(s - a)(s - b)(s - c)}.$$ We plug in $a = 52, b = 53, c = 51.$ \begin{align*} s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\ [ABC] &= \sqrt{78(78 - 51)(78 - 52)(78 - 53)} = \sqrt{78(27)(26)(25)} = 5 \sqrt{(3 \cdot 26)\left(3^3\right)(2 \cdot 13)} \\ &= 5 \cdot 3 \sqrt{3 \cdot 2 \cdot 13 \cdot 3 \cdot 2 \cdot 13} = 15 \sqrt{2^2 \cdot 3^2 \cdot 13^2} = 15 \sqrt{(2 \cdot 3 \cdot 13)}^2 \\ &= \boxed{1170} \end{align*}

Since $[ABC] = \frac{bh}{2},$ we know that $1170 = \frac{AD \cdot 52}{2} = 26 \cdot AD.$ Solving, we get $AD = 45.$ Remembering our 8-15-17 Pythagorean triple, we see that $BD = \boxed{24}.$ $\square$