Difference between revisions of "2010 AMC 10B Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
| − | An alternative way to find the area of the triangle is by using Heron's formula, <math>A=\sqrt{(s)(s-a)(s-b)(s-c)}</math> where <math>s</math> is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is <math>(10+10+12)/2 = 16</math>. Thus the area equals <math>\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48. </math> | + | An alternative way to find the area of the triangle is by using Heron's formula, <math>A=\sqrt{(s)(s-a)(s-b)(s-c)}</math> where <math>s</math> is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is <math>(10+10+12)/2 = 16</math>. Thus the area equals <math>\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48. </math> We know that the width of the rectangle is <math>4</math>, so <math>48/4 = 12</math>, which is the length. The perimeter of the rectangle is <math>2(4+12) = 32</math>. |
(Revised by Flamedragon) | (Revised by Flamedragon) | ||
Revision as of 21:16, 9 February 2014
Contents
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 
.
(Revised by Flamedragon)
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. 
