Difference between revisions of "1950 AHSME Problems/Problem 47"
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==Solution== | ==Solution== | ||
| − | {{ | + | Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get <math>\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}</math>. The answer is <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 22:54, 24 April 2013
Problem
A rectangle inscribed in a triangle has its base coinciding with the base 
of the triangle. If the altitude of the triangle is 
, and the altitude 
of the rectangle is half the base of the rectangle, then:
Solution
Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get 
. The answer is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
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| All AHSME Problems and Solutions | ||
