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

(Solution to Problem 32 — need more precise approximation of pi than 3.14 or 22/7)
 
m (Solution to Problem 32 (Credit to r_fractal))
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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} = 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), the value </math>10040\pi<math> is around </math>31541.59,<math> so </math>\lfloor A \rfloor = \boxed{31541}$.
+
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>.
  
 
==See Also==
 
==See Also==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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Credit to r_fractal
 

Revision as of 20:50, 4 August 2018

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 $\sqrt{100 \pi} = 10\pi$. That means the area of the triangle is $1004 \cdot 10\pi = 10040\pi.$

By using a calculator (or noting that $\pi \approx 3.14159265$), 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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