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Difference between revisions of "1999 AMC 8 Problems/Problem 23"
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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} = \boxed{\text{(C)}\ \sqrt{13}}</math>.
 
Since <math>\triangle CBM</math> is a right triangle, <math>CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}</math>.
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=AH5_Gol2GCM&t=21s
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~CosineMethod
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=22|num-a=24}}
 
{{AMC8 box|year=1999|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:53, 23 January 2024

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, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}$.


Video Solution

https://www.youtube.com/watch?v=AH5_Gol2GCM&t=21s ~CosineMethod

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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