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Difference between revisions of "2011 AMC 8 Problems/Problem 16"

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(→‎Solution)
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'''25-25-30'''
 
'''25-25-30'''
  
−
We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the [[Pythagorean Theorem]], we have \[ \begin{eqnarray*}
+
We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the [[Pythagorean Theorem]], we have  
−
15^2 + x^2 &=& 25^2 \\
+
<math><cmath> 15^2 + x^2 =25^2 </cmath></math>
−
x^2 &=& 25^2 - 15^2 \\
+
<math><cmath> \x^2 = 25^2 - 15^2</cmath> </math>
−
&=& (25 + 15)(25-15) \\
+
<math><cmath>x^2 = (25 + 15)(25-15)</cmath> </math>
−
&=& 40\cdot 10 \\
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<math><cmath>x^2= 40\cdot 10</cmath></math>
−
&=& 400\\
+
<math><cmath>x^2= 400</cmath></math>
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x &=& \sqrt{400} \\
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<math><cmath>x = \sqrt{400}</cmath></math>
−
&=& 20
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<math><cmath>x= 20</cmath></math>
−
\end{eqnarray*} \]
+
 
  
 
Thus we have two 15-20-25 right triangles.
 
Thus we have two 15-20-25 right triangles.
Line 22: Line 22:
 
We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles.
 
We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles.
  
−
Let the area of a 15-20-25 right triangle be <math>x</math>.
+
Let the area of a 15-20-25 right triangle be <cmath>x</cmath>.
−
<cmath> a = 2x \\ b = 2x \\ \boxed{a = b} </cmath>
+
<math><cmath>a = 2x</cmath></math>
−
 
+
<math><cmath> b = 2x</cmath></math>
−
Using the [[Pythagorean Theorem]], the height of the first triangle is <math>\sqrt{25^2-15^2}=20</math>. Using this, its area is <math>A=\frac{20\cdot 30}{2}=300</math>. Similarly, the height of the second triangle is <math>\sqrt{25^2-20^2}=15</math>, and the area is <math>B=\frac{15\cdot 40}{2}=300</math>. Therefore, <math>A=B</math> and the answer is <math>\boxed{\textbf{(C)}}</math>.
+
<math><cmath>\textbf{C) }\boxed{a = b} </cmath></math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}

Revision as of 09:53, 27 November 2011

Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$?

$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \\ \\ \textbf{(E) }A = \dfrac{16}9B$

Solution

25-25-30

We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have $<cmath> 15^2 + x^2 =25^2 </cmath>$ $<cmath> \x^2 = 25^2 - 15^2</cmath>$ (Error compiling LaTeX. Unknown error_msg) $<cmath>x^2 = (25 + 15)(25-15)</cmath>$ $<cmath>x^2= 40\cdot 10</cmath>$ $<cmath>x^2= 400</cmath>$ $<cmath>x = \sqrt{400}</cmath>$ $<cmath>x= 20</cmath>$


Thus we have two 15-20-25 right triangles.

25-25-40

We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles.

Let the area of a 15-20-25 right triangle be \[x\]. $<cmath>a = 2x</cmath>$ $<cmath> b = 2x</cmath>$ $<cmath>\textbf{C) }\boxed{a = b} </cmath>$

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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