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

(Problem 16)
 
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Let <math>A</math> be the area of the triangle with sides of length <math>25, 25</math>, and <math>30</math>. Let <math>B</math> be the area of the triangle with sides of length <math>25, 25,</math> and <math>40</math>. What is the relationship between <math>A</math> and <math>B</math>?
 
Let <math>A</math> be the area of the triangle with sides of length <math>25, 25</math>, and <math>30</math>. Let <math>B</math> be the area of the triangle with sides of length <math>25, 25,</math> and <math>40</math>. What is the relationship between <math>A</math> and <math>B</math>?
  
<math> \text{(A) } A = \dfrac9{16}B \qquad\text{(B) } A = \dfrac34B \qquad\text{(C) } A=B \qquad\text{(D) } A = \dfrac43B \\ \\ \text{(E) }A = \dfrac{16}9B </math>
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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>
  
 
==Solution==
 
==Solution==

Revision as of 19:25, 25 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

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions