Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1987 AHSME Problems/Problem 21"
(→‎Solution)
 
Line 28: Line 28:
 
    
 
    
 
== 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 14:54, 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 $441 \text{cm}^2$. What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?

[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]

$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$

Solution

We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$. [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 $3S=42\sqrt{2}$, so $S=14\sqrt{2}$. But we want the area of the square which is $S^2=(14\sqrt{2})^2= \boxed{\mathrm{(B)}\ 392}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_AHSME_Problems/Problem_21&oldid=74314"