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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 11"

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==Problem==
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Consider Square <math>ABCD</math>, a square with side length <math>10</math>. Let Points <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math> be the midpoints of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. Find the area of the square formed by the four line segments <math>AG</math>, <math>BH</math>, <math>CE</math>, and <math>DF</math>.
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<center><asy> draw((0,0)--(10,0)); draw((0,0)--(0, 10)); draw((10,0)--(10, 10)); draw((10,10)--(0, 10)); draw((0,10)--(5, 0)); draw((0,0)--(10, 5)); draw((10,0)--(5, 10)); draw((10,10)--(0, 5)); </asy></center>
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<math>\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50</math>
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==Solution==
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Note that the area of square ABCD is 100. We can break the large square into 20 right triangles, all of which are congruent. Thus, the area of one triangle is 5, and because the smaller square is composed of 4 of these triangles, the area of the smaller square is 20, or <math>\boxed{\textbf B}</math>
 
Note that the area of square ABCD is 100. We can break the large square into 20 right triangles, all of which are congruent. Thus, the area of one triangle is 5, and because the smaller square is composed of 4 of these triangles, the area of the smaller square is 20, or <math>\boxed{\textbf B}</math>

Latest revision as of 20:17, 8 October 2025

Problem

Consider Square $ABCD$, a square with side length $10$. Let Points $E$, $F$, $G$, $H$ be the midpoints of sides $AB$, $BC$, $CD$, and $DA$, respectively. Find the area of the square formed by the four line segments $AG$, $BH$, $CE$, and $DF$.

[asy] draw((0,0)--(10,0)); draw((0,0)--(0, 10)); draw((10,0)--(10, 10)); draw((10,10)--(0, 10)); draw((0,10)--(5, 0)); draw((0,0)--(10, 5)); draw((10,0)--(5, 10)); draw((10,10)--(0, 5)); [/asy]

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$

Solution

Note that the area of square ABCD is 100. We can break the large square into 20 right triangles, all of which are congruent. Thus, the area of one triangle is 5, and because the smaller square is composed of 4 of these triangles, the area of the smaller square is 20, or $\boxed{\textbf B}$

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