Difference between revisions of "2013 AMC 10A Problems/Problem 15"
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<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math> | <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math> | ||
| − | ==Solution 1== | + | ==Solution 1 (cheap way)== |
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, | The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, | ||
| − | therefore the length of the side perpendicular to that altitude will be between 10 and 15. The only answer choice that meets this requirement is 12 ( | + | therefore the length of the side perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is 12, <math>\textbf{(D)}</math>. |
| + | |||
| + | ==Solution 2 (actually solving)== | ||
| + | |||
| + | Let the height to the side of length 15 be <math>h_{1}</math>, the height to the side of length 10 be <math>h_{2}</math>, the area be <math>A</math>, and the height to the unknown side be <math>h_{3}</math>. | ||
| + | |||
| + | Because the area of a triangle is <math>\frac{bh}{2}</math>, we get that <math>15(h_{1}) = 2A</math> and <math>10(h_{2}) = 2A</math>, so, setting them equal, <math>h_{2} = \frac{3h_{1}}{2}</math>. Now, we know that <math>2h_{3} = h_{1} + h_{2}</math>. Substituting, we get that | ||
| + | |||
| + | <math>h_{3} = 1.25h_{1}</math>. | ||
| + | |||
| + | Thus, the side length is going to be <math>\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = 12</math>, <math>textbf{(D)}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2013|ab=A|num-b=14|num-a=16}} | ||
Revision as of 21:33, 7 February 2013
Problem
Two sides of a triangle have lengths 
and 
. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
Solution 1 (cheap way)
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between and . The only answer choice that meets this requirement is 12, 
.
Solution 2 (actually solving)
Let the height to the side of length 15 be 
, the height to the side of length 10 be 
, the area be 
, and the height to the unknown side be 
.
Because the area of a triangle is 
, we get that 
and 
, so, setting them equal, 
. Now, we know that 
. Substituting, we get that
.
Thus, the side length is going to be 
, 
.
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
