Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 10"
(→‎Solution)
Line 23: Line 23:
 
Heron's Formula states that in a triangle with sides <math>a, b, c</math> and <math>s = \frac{a + b + c}{2},</math> the area is given by <cmath>\sqrt{s(s - a)(s - b)(s - c)}.</cmath> We plug in <math>a = 52, b = 53, c = 51.</math>
 
Heron's Formula states that in a triangle with sides <math>a, b, c</math> and <math>s = \frac{a + b + c}{2},</math> the area is given by <cmath>\sqrt{s(s - a)(s - b)(s - c)}.</cmath> We plug in <math>a = 52, b = 53, c = 51.</math>
  
 +
<cmath>
 
\begin{align*}
 
\begin{align*}
 
s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\
 
s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\
Line 29: Line 30:
 
&= \boxed{1170}
 
&= \boxed{1170}
 
\end{align*}
 
\end{align*}
 +
</cmath>
  
 
Since <math>[ABC] = \frac{bh}{2},</math> we know that <math>1170 = \frac{AD \cdot 52}{2} = 26 \cdot AD.</math> Solving, we get <math>AD = 45.</math> Remembering our 8-15-17 Pythagorean triple, we see that <math>BD = \boxed{24}.</math> <math>\square</math>
 
Since <math>[ABC] = \frac{bh}{2},</math> we know that <math>1170 = \frac{AD \cdot 52}{2} = 26 \cdot AD.</math> Solving, we get <math>AD = 45.</math> Remembering our 8-15-17 Pythagorean triple, we see that <math>BD = \boxed{24}.</math> <math>\square</math>

Revision as of 15:06, 26 May 2017

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

See also

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions

Template:Heron's Formula

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_UNCO_Math_Contest_II_Problems/Problem_10&oldid=85856"