Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 2"
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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> | 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 == |
Latest revision as of 13:04, 2 January 2020
Problem
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 be called , be called , and called . Since is a right triangle, we can write a system of equations.
Solving the second equation, we get or . Since has to be positive, . Plugging into the first equation, we get that or . Once again, side lengths must be positive, so . Now we have an altitude and a base, so the area of is
~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 |