Difference between revisions of "2010 AMC 10B Problems/Problem 7"

(Created page with 'The triangle is isosceles. The height of the triangle is therefore given by <math>h = \sqrt{10^2 - ( \dfrac{12}{2})^2} = \sqrt{64} = 8</math> Now, the area of the triangle is <…')
 
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==Problem==
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A triangle has side lengths <math>10</math>, <math>10</math>, and <math>12</math>. A rectangle has width <math>4</math> and area equal to the
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area of the triangle. What is the perimeter of this rectangle?
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<math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36</math>
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==Solution==
 
The triangle is isosceles. The height of the triangle is therefore given by <math>h = \sqrt{10^2 - ( \dfrac{12}{2})^2} =  \sqrt{64} = 8</math>
 
The triangle is isosceles. The height of the triangle is therefore given by <math>h = \sqrt{10^2 - ( \dfrac{12}{2})^2} =  \sqrt{64} = 8</math>
  
 
Now, the area of the triangle is <math>\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48 </math>
 
Now, the area of the triangle is <math>\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48 </math>
  
We have that the area of the rectangle is the same as the area of the triangle, namely 48. We also have the width of the rectangle: 4.
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We have that the area of the rectangle is the same as the area of the triangle, namely <math>48</math>. We also have the width of the rectangle: <math>4</math>.
  
The length of the rectangle therefor is:
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The length of the rectangle therefore is:
 
<math>l = \dfrac{48}{4} = 12</math>
 
<math>l = \dfrac{48}{4} = 12</math>
  
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The answer is:
 
The answer is:
  
<math> \boxed{\mathrm{(D)} = 32} </math>
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<math> \boxed{\textbf{(D)}\ 32} </math>
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==See Also==
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{{AMC10 box|year=2010|ab=B|num-b=6|num-a=8}}

Revision as of 01:40, 26 November 2011

Problem

A triangle has side lengths $10$, $10$, and $12$. A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$

Solution

The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - (	\dfrac{12}{2})^2} =  \sqrt{64} = 8$

Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$

We have that the area of the rectangle is the same as the area of the triangle, namely $48$. We also have the width of the rectangle: $4$.

The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$

The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$

The answer is:

$\boxed{\textbf{(D)}\ 32}$

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 AMC 10 Problems and Solutions