Difference between revisions of "2010 AMC 10B Problems/Problem 7"
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==Solution 3== | ==Solution 3== | ||
| − | Note that a triangle with side lengths <math>10,10,12</math> is essentially <math>2</math> | + | Note that a triangle with side lengths <math>10,10,12</math> is essentially <math>2</math> “6,8,10” right triangles stuck together. Hence, the height is <math>8</math>, and our area is <math>48</math>. |
So, the length of the rectangle is <math>\frac{48}{4}=12</math>, and our perimeter <math>P=2(4+12)=\boxed{\textbf{(D)}\ 32}</math> | So, the length of the rectangle is <math>\frac{48}{4}=12</math>, and our perimeter <math>P=2(4+12)=\boxed{\textbf{(D)}\ 32}</math> | ||
Revision as of 00:59, 11 February 2018
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 1
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:
Solution 2
An alternative way to find the area of the triangle is by using Heron's formula, 
where 
is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is 
. Thus the area equals 
We know that the width of the rectangle is , so 
, which is the length. The perimeter of the rectangle is 
.
(Solution by Flamedragon)
Solution 3
Note that a triangle with side lengths 
is essentially 
“6,8,10” right triangles stuck together. Hence, the height is 
, and our area is .
So, the length of the rectangle is 
, and our perimeter 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
