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

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<math> \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 </math>
 
<math> \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 </math>
  
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==Solution==
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==Solution 1==
 
'''25-25-30'''
 
'''25-25-30'''
  
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<cmath> b = 2x</cmath>
 
<cmath> b = 2x</cmath>
 
<cmath>\boxed{\textbf{(C) } A = B} </cmath>
 
<cmath>\boxed{\textbf{(C) } A = B} </cmath>
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 +
==Solution 2==
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 +
Using Heron's formula, we can calculate the area of the two triangles. The formula states that
 +
<cmath>\begin{center} A = \sqrt{s(s - a)(s - b)(s - c)} \end{center}</cmath>
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where <math>s</math> is the semiperimeter of a triangle with side lengths <math>a</math>, <math>b</math>, and <math>c</math>.
 +
 +
For the <math>25-25-30</math> triangle, <cmath>s = \frac{25 + 25 + 30}{2} = 40</cmath>. Therefore,
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<cmath>\begin{center} A = \sqrt{40 \cdot 15 \cdot 15 \cdot 10} = 300 \end{center}</cmath>
 +
 +
For the <cmath>25-25-40</cmath> triangle, <cmath>s = \frac{25 + 25 + 40}{2} = 45</cmath>. Therefore,
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<cmath>\begin{center} B = \sqrt{45 \cdot 20 \cdot 20 \cdot 5} = 300 \end{center}</cmath>
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 +
Hence, <cmath>A = B \longrightarrow \boxed{textbf{(C) }</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}
 
{{AMC8 box|year=2011|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:40, 1 December 2017

Problem

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 1

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 \[15^2 + x^2 =25^2\] \[x^2 = 25^2 - 15^2\] \[x^2 = (25 + 15)(25-15)\] \[x^2= 40\cdot 10\] \[x^2= 400\] \[x = \sqrt{400}\] \[x= 20\]


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 $x$. \[a = 2x\] \[b = 2x\] \[\boxed{\textbf{(C) } A = B}\]

Solution 2

Using Heron's formula, we can calculate the area of the two triangles. The formula states that 

\[\begin{center} A = \sqrt{s(s - a)(s - b)(s - c)} \end{center}\] (Error compiling LaTeX. Unknown error_msg)

where $s$ is the semiperimeter of a triangle with side lengths $a$, $b$, and $c$.

For the $25-25-30$ triangle, \[s = \frac{25 + 25 + 30}{2} = 40\]. Therefore, 

\[\begin{center} A = \sqrt{40 \cdot 15 \cdot 15 \cdot 10} = 300 \end{center}\] (Error compiling LaTeX. Unknown error_msg)

For the \[25-25-40\] triangle, \[s = \frac{25 + 25 + 40}{2} = 45\]. Therefore, 

\[\begin{center} B = \sqrt{45 \cdot 20 \cdot 20 \cdot 5} = 300 \end{center}\] (Error compiling LaTeX. Unknown error_msg)

Hence, 

\[A = B \longrightarrow \boxed{textbf{(C) }\] (Error compiling LaTeX. Unknown error_msg)

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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