1993 UNCO Math Contest II Problems/Problem 10

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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 $a = 52, b = 53, c = 51.$

s=51+52+532=3522=326=78[ABC]=78(7851)(7852)(7853)=78(27)(26)(25)=5(326)(33)(213)=5332133213=152232132=15(2313)2=1170

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 $BD = \boxed{24}.$ $\square$

See also

1993 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Last Question
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All UNCO Math Contest Problems and Solutions

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