Difference between revisions of "2010 AMC 10B Problems/Problem 7"
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<math> \boxed{\textbf{(D)}\ 32} </math> | <math> \boxed{\textbf{(D)}\ 32} </math> | ||
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+ | 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> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2010|ab=B|num-b=6|num-a=8}} |
Revision as of 22:59, 3 February 2012
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:
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
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 |