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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 15"

(Created page with "The area of the larger square is 144, so each side length is 12. m + n = 12. Also, Each of the smaller 4 triangles are right triangles, so m²+n²= 108. n=12-m m²+(12-m)²...")
 
 
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The area of the larger square is 144, so each side length is 12. m + n =  12. Also, Each of the smaller 4 triangles are right triangles, so m²+n²= 108.  
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The area of the larger square is 144, so each side length is 12. <math>m + n =  12</math>. Also, each of the smaller 4 triangles are right triangles, so <math>m^2+n^2= 108</math>.  
  
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n=12-m
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<math>n=12-m</math>
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m²+(12-m)²=108
 
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m²+144-24m+m²=108.
 
  
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Solving for m gives m= 12+sqrt(72)/2, m = 6+3sqrt2, a=6, b=3, c=2. a+b+c=11. (A)
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<math>m^2+(12-m)^2=108</math>
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<math>m^2+144-24m+m^2=108</math>.
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Solving for <math>m</math> gives <math>m= 12+\frac{\sqrt72}{2}</math>
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<math>m = 6+3\sqrt{2}</math>
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<math>a=6</math>, <math>b=3</math>, <math>c=2</math>. <math>a+b+c=11</math>. (A)

Latest revision as of 14:01, 11 August 2024

The area of the larger square is 144, so each side length is 12. $m + n = 12$. Also, each of the smaller 4 triangles are right triangles, so $m^2+n^2= 108$.

$n=12-m$

$m^2+(12-m)^2=108$

$m^2+144-24m+m^2=108$.

Solving for $m$ gives $m= 12+\frac{\sqrt72}{2}$

$m = 6+3\sqrt{2}$

$a=6$, $b=3$, $c=2$. $a+b+c=11$. (A)

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