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Difference between revisions of "2003 AMC 8 Problems/Problem 25"

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<math> \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2</math>
 
<math> \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2</math>
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==Solution==
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We see that <math>XY = 5</math>, the vertical distance between <math>B</math> and <math>X</math> is <math>1</math>, and the vertical distance between <math>C</math> and <math>Y</math> is <math>1</math>. Therefore, <math>BC = 5 - 1 - 1 = 3</math>. We are given that the length of the altitude of <math>\triangle ABC</math> is equal to the distance between <math>\overline{BC}</math> and <math>O</math>, which is <math>1 + 1 + \frac{5}{2} = \frac{9}{2}</math>. So the area of <math>\triangle ABC</math> is <math>\frac{1}{2}\left(3\cdot \frac{9}{2}\right)</math>, which is <math>\boxed{\textbf{(C)} \ \frac{27}{4}}</math>.
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~sidkris
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 17:52, 20 June 2024

Problem

In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC$, $AB = AC$, and when $\triangle ABC$ is folded over side $\overline{BC}$, point $A$ coincides with $O$, the center of square $WXYZ$. What is the area of $\triangle ABC$, in square centimeters?

[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label("$A$", A, NW); label("$O$", O, NE); label("$B$", B, SW); label("$C$", C, NW); label("$W$",W , NE); label("$X$", X, N); label("$Y$", Y, S); label("$Z$", Z, SE); [/asy]

$\textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2$

Solution

We see that $XY = 5$, the vertical distance between $B$ and $X$ is $1$, and the vertical distance between $C$ and $Y$ is $1$. Therefore, $BC = 5 - 1 - 1 = 3$. We are given that the length of the altitude of $\triangle ABC$ is equal to the distance between $\overline{BC}$ and $O$, which is $1 + 1 + \frac{5}{2} = \frac{9}{2}$. So the area of $\triangle ABC$ is $\frac{1}{2}\left(3\cdot \frac{9}{2}\right)$, which is $\boxed{\textbf{(C)} \ \frac{27}{4}}$.

~sidkris

Video Solution

https://www.youtube.com/watch?v=4RBCH1rUcSw

~David

See Also

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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