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

(Created page with "The side lengths of square WXYZ must be 5 cm, since the area is 25 cm ^2. First, you should determine the height of triangle ABC. The distance from O to line WZ must be 2.5 cm, s...")
 
 
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The side lengths of square WXYZ must be 5 cm, since the area is 25 cm ^2. First, you should determine the height of triangle ABC. The distance from O to line WZ must be 2.5 cm, since line WX = 5 cm, and the distance from O to Z is half of that. The distance from line WZ to line BC must be 2, since the side lengths of the small squares are 1, and there are two squares from line WZ to line BC. So, the height of ABC must be 4.5, which is 2.5 + 2. The length of BC can be determined by subtracting 2 from 5, since the length of WZ is 5, and the two squares in the corners give us 2 together. This gives us the base for ABC, which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of (C) - 27/4.
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==Problem==
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In the figure, the area of square <math>WXYZ</math> is <math>25 \text{ cm}^2</math>. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In <math>\triangle ABC</math>, <math>AB = AC</math>, and when <math>\triangle ABC</math> is folded over side <math>\overline{BC}</math>, point <math>A</math> coincides with <math>O</math>, the center of square <math>WXYZ</math>. What is the area of <math>\triangle ABC</math>, in square centimeters?
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<asy>
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defaultpen(fontsize(8));
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size(225);
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pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);
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draw((-4,0)--Y--X--(-4,10)--cycle);
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draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);
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dot(O);
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label("$A$", A, NW);
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label("$O$", O, NE);
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label("$B$", B, SW);
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label("$C$", C, NW);
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label("$W$",W , NE);
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label("$X$", X, N);
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label("$Y$", Y, S);
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label("$Z$", Z, SE);
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</asy>
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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>
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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
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==Video Solution==
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https://www.youtube.com/watch?v=4RBCH1rUcSw
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 +
~David
 +
 
 +
==See Also==
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{{AMC8 box|year=2003|num-b=24|after=Last Problem}}
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{{MAA Notice}}

Latest revision as of 16:52, 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
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
All AJHSME/AMC 8 Problems and Solutions

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