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

Revision as of 19:50, 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

Using the Pythagorean Theorem, the height of the first triangle is $\sqrt{25^2-15^2}=20$ . Using this, its area is $A=\frac{20\cdot 30}{2}=300$ . Similarly, the height of the second triangle is $\sqrt{25^2-20^2}=15$ , and the area is $B=\frac{15\cdot 40}{2}=300$ . Therefore, $A=B$ and the answer is $\boxed{\textbf{(C)}}$ .

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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 ==Solution==
+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>.
 ==See Also==
 {{AMC8 box|year=2011|num-b=15|num-a=17}}

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

Revision as of 19:50, 25 November 2011

Solution

See Also