Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 10"
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== Solution == | == 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 <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> | |
+ | |||
+ | \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 <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> | ||
== See also == | == See also == |
Revision as of 15:04, 26 May 2017
Problem
The scalene triangle has side lengths is perpendicular to
(a) Determine the length of
(b) Determine the area of triangle
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 and the area is given by We plug in
Since we know that Solving, we get Remembering our 8-15-17 Pythagorean triple, we see that
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 |