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Difference between revisions of "2008 iTest Problems/Problem 32"

m (Solution to Problem 32 (Credit to r_fractal))
(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>
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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^2} = 10\pi</math>.  That means the area of the triangle is <math>1004 \cdot 10\pi = 10040\pi.</math>
  
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By using a calculator (or noting that <math>\pi \approx 3.14159265</math>), the value <math>10040\pi</math> is around <math>31541.59,</math> so <math>\lfloor A \rfloor = \boxed{31541}</math>.
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By noting that <math>\pi \approx 3.14159265</math> (or by using a calculator), the value <math>10040\pi</math> is around <math>31541.59,</math> so <math>\lfloor A \rfloor = \boxed{31541}</math>.
  
 
==See Also==
 
==See Also==
 
{{2008 iTest box|num-b=31|num-a=33}}
 
{{2008 iTest box|num-b=31|num-a=33}}
  
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[[Category:Intermediate Geometry Problems]]
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[[Category:Introductory Geometry Problems]]

Latest revision as of 12:30, 28 February 2020

Problem

A right triangle has perimeter $2008$, and the area of a circle inscribed in the triangle is $100\pi^3$. Let A be the area of the triangle. Compute $\lfloor A\rfloor$.

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 $A = sr.$


The semiperimeter of the triangle is $\tfrac{2008}{2} = 1004.$ The radius of the incircle is $\sqrt{100 \pi^2} = 10\pi$. That means the area of the triangle is $1004 \cdot 10\pi = 10040\pi.$


By noting that $\pi \approx 3.14159265$ (or by using a calculator), the value $10040\pi$ is around $31541.59,$ so $\lfloor A \rfloor = \boxed{31541}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 31 		Followed by:
Problem 33
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