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

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== Solution ==
 
== Solution ==
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Let the altitude <math>BD</math> be called <math>h</math>, <math>AD</math> be called <math>x</math>, and <math>BC</math> called <math>y</math>. Since <math>ABC</math> is a right triangle, we can write a system of equations.
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<cmath>h^2=16x</cmath>
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<cmath>15^2=x(x+16)</cmath>
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<cmath>y^2=16(16+x)</cmath>
  
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Solving the second equation, we get <math>x^2+16x-225=0 \rightarrow (x-9)(x+25)=0 \rightarrow x=9</math> or <math>-25</math>. Since <math>x</math> has to be positive, <math>x=9</math>. Plugging <math>9</math> into the first equation, we get that <math>h^2=4^2\cdot3^2 \rightarrow h=12</math> or <math>-12</math>. Once again, side lengths must be positive, so <math>h=12</math>. Now we have an altitude and a base, so the area of <math>ABC</math> is <math>\frac{(9+16)\cdot12}{2} = \boxed{150}</math>
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~Ultraman
  
 
== See also ==
 
== See also ==
 
{{UNCO Math Contest box|year=2018|n=II|num-b=1|num-a=3}}
 
{{UNCO Math Contest box|year=2018|n=II|num-b=1|num-a=3}}
  
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[[Category:]]
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[[Category:Introductory Geometry Problems]]

Latest revision as of 14:04, 2 January 2020

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 $\frac{(9+16)\cdot12}{2} = \boxed{150}$

~Ultraman

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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