Difference between revisions of "1960 AHSME Problems/Problem 37"
Rockmanex3 (talk | contribs) (WIP Solution) |
Rockmanex3 (talk | contribs) (→Solution (WIP)) |
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Since <math>CD</math> is perpendicular to <math>AB</math>, <math>ND = WX</math>. That means <math>CN = h-x</math>. | Since <math>CD</math> is perpendicular to <math>AB</math>, <math>ND = WX</math>. That means <math>CN = h-x</math>. | ||
The sides of the rectangle are parallel, so <math>XY \parallel WZ</math>. That means by AA Similarity, <math>\triangle CXY \sim \triangle CAB</math>. Letting <math>n</math> be the length of the base of the rectangle, that means | The sides of the rectangle are parallel, so <math>XY \parallel WZ</math>. That means by AA Similarity, <math>\triangle CXY \sim \triangle CAB</math>. Letting <math>n</math> be the length of the base of the rectangle, that means | ||
− | <cmath>\frac{h-x}{n} = \frac{}{}</cmath> | + | <cmath>\frac{h-x}{n} = \frac{h}{b}</cmath> |
+ | <cmath>n = \frac{b(h-x)}{h}</cmath> | ||
+ | That means the area of the rectangle is | ||
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=36|num-a=38}} | {{AHSME 40p box|year=1960|num-b=36|num-a=38}} |
Revision as of 19:16, 15 May 2018
Problem
The base of a triangle is of length , and the altitude is of length . A rectangle of height is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
Solution (WIP)
Let , , and . Since is perpendicular to , . That means . The sides of the rectangle are parallel, so . That means by AA Similarity, . Letting be the length of the base of the rectangle, that means That means the area of the rectangle is
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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All AHSME Problems and Solutions |