Difference between revisions of "2008 iTest Problems/Problem 32"
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Rockmanex3 (talk | contribs) (Problem is more of an introductory geo problem.) |
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− | The semiperimeter of the triangle is <math>\tfrac{2008}{2} = 1004.</math> The radius of the incircle is <math>\sqrt{100 \pi} = 10\pi</math>. That means the area of the triangle is <math>1004 \cdot 10\pi = 10040\pi.</math> | + | The semiperimeter of the triangle is <math>\tfrac{2008}{2} = 1004.</math> The radius of the incircle is <math>\sqrt{100 \pi^2} = 10\pi</math>. That means the area of the triangle is <math>1004 \cdot 10\pi = 10040\pi.</math> |
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{{2008 iTest box|num-b=31|num-a=33}} | {{2008 iTest box|num-b=31|num-a=33}} | ||
− | [[Category: | + | [[Category:Introductory Geometry Problems]] |
Latest revision as of 13:30, 28 February 2020
Problem
A right triangle has perimeter , and the area of a circle inscribed in the triangle is
. Let A be the area of the triangle. Compute
.
Solution
We know the perimeter and area of incircle, so we can find the semiperimeter of the triangle and radius of the incircle and plug the values into the area formula
The semiperimeter of the triangle is The radius of the incircle is
. That means the area of the triangle is
By noting that (or by using a calculator), the value
is around
so
.
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 31 |
Followed by: Problem 33 | |
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