Difference between revisions of "1960 AHSME Problems/Problem 37"
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<cmath>n = \frac{b(h-x)}{h}</cmath> | <cmath>n = \frac{b(h-x)}{h}</cmath> | ||
Thus, the area of the rectangle is <math>\frac{bx}{h}(h-x)</math>, which is answer choice <math>\boxed{\textbf{(A)}}</math>. | Thus, the area of the rectangle is <math>\frac{bx}{h}(h-x)</math>, which is answer choice <math>\boxed{\textbf{(A)}}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/ZdM2ou5Gsuw?t=172 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
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==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}} | ||
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+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 21:26, 28 December 2023
Contents
[hide]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
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 Thus, the area of the rectangle is , which is answer choice .
Video Solution
https://youtu.be/ZdM2ou5Gsuw?t=172
~MathProblemSolvingSkills.com
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |