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== | ||
| + | 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 | ||
| + | area of the triangle. What is the perimeter of this rectangle? | ||
| + | |||
| + | <math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36</math> | ||
| + | |||
| + | ==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. | + | 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 | + | 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{\ | + | <math> \boxed{\textbf{(D)}\ 32} </math> |
| + | |||
| + | ==See Also== | ||
| + | {{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 
, , and 
. A rectangle has width 
and area equal to the
area of the triangle. What is the perimeter of this rectangle?
Solution
The triangle is isosceles. The height of the triangle is therefore given by 
Now, the area of the triangle is 
We have that the area of the rectangle is the same as the area of the triangle, namely 
. We also have the width of the rectangle: .
The length of the rectangle therefore is:
The perimeter of the rectangle then becomes:
The answer is:
See Also
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
