Difference between revisions of "1987 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
− | + | We are given that the area of the inscribed square is <math>441</math>, so the side length of that square is <math>21</math>. Since the square divides the <math>45-45-90</math> larger triangle into 2 smaller congruent <math>45-45-90</math>, then the legs of the larger isosceles right triangle (<math>BC</math> and <math>AB</math>) are equal to <math>42</math>. | |
− | <math>\boxed{\mathrm{(B)}\ 392}</math> | + | <asy> |
+ | draw((0,0)--(10,0)--(0,10)--cycle); | ||
+ | draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); | ||
+ | label("A", (0,10), W); | ||
+ | label("B", (0,0), W); | ||
+ | label("C", (10,0), E); | ||
+ | label("S", (25/3,11/6), E); | ||
+ | label("S", (11/6,25/3), E); | ||
+ | label("S", (5,5), NE); | ||
+ | </asy> | ||
+ | |||
+ | We now have that <math>3S=42\sqrt{2}</math>, so <math>S=14\sqrt{2}</math>. But we want the area of the square which is <math>S^2=(14\sqrt{2})^2= \boxed{\mathrm{(B)}\ 392}</math> | ||
== See also == | == See also == |
Latest revision as of 15:54, 4 January 2016
Problem
There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure 1 below, then one finds that the area of the square is . What is the area (in ) of the square inscribed in the same as shown in Figure 2 below?
Solution
We are given that the area of the inscribed square is , so the side length of that square is . Since the square divides the larger triangle into 2 smaller congruent , then the legs of the larger isosceles right triangle ( and ) are equal to .
We now have that , so . But we want the area of the square which is
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 | ||
All AHSME Problems and Solutions |
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