Difference between revisions of "2008 iTest Problems/Problem 32"
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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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Revision as of 19:55, 4 August 2018
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 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100 | ||
