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Difference between revisions of "1999 AMC 8 Problems/Problem 23"
(Added problem, see also box. Also cleaned up formatting of solution.)
Line 1: Line 1:
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Since the squares have side length <math>3</math>, the area of the entire square is <math>9</math>.
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==Problem==
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Square <math>ABCD</math> has sides of length 3. Segments <math>CM</math> and <math>CN</math> divide the square's area into three equal parts. How long is segment <math>CM</math>?
  
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The segments divide the square into 3 equal parts, so the area of each part is <math>3</math>.
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<asy>
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pair A,B,C,D,M,N;
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A = (0,0);
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B = (0,3);
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C = (3,3);
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D = (3,0);
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M = (0,1);
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N = (1,0);
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draw(A--B--C--D--cycle);
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draw(M--C--N);
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label("$A$",A,SW);
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label("$M$",M,W);
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label("$B$",B,NW);
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label("$C$",C,NE);
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label("$D$",D,SE);
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label("$N$",N,S);
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</asy>
  
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The base of a triangle is <math>3</math>, so the height must be <math>2</math>.
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<math>\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}</math>
  
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<math>3-2=1</math>.
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==Solution==
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Since the square has side length <math>3</math>, the area of the entire square is <math>9</math>.
  
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<math>CM=\sqrt {3^2+2^2</math>} = <math>\sqrt 13</math>
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The segments divide the square into 3 equal parts, so the area of each part is <math>9 \div 3 = 3</math>.
  
 +
Since <math>\triangle CBM</math> has area <math>3</math> and base <math>CB = 3</math>, using the area formula for a triangle:
  
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-Mrdavid445
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<math>A_{tri} = \frac{1}{2}bh</math>
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<math>3 = \frac{1}{2}3h</math>
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<math>h = 2</math>
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Thus, height <math>BM = 2</math>.
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Since <math>\triangle CBM</math> is a right triangle, <math>CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \rightarrow \boxed{C}</math>
 +
 
 +
==See also==
 +
{{AMC8 box|year=1999|num-b=22|num-a=24}}

Revision as of 16:32, 30 July 2011

Problem

Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$?

[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$N$",N,S); [/asy]

$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$

Solution

Since the square has side length $3$, the area of the entire square is $9$.

The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$.

Since $\triangle CBM$ has area $3$ and base $CB = 3$, using the area formula for a triangle:

$A_{tri} = \frac{1}{2}bh$

$3 = \frac{1}{2}3h$

$h = 2$

Thus, height $BM = 2$.

Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \rightarrow \boxed{C}$

See also

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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