Use only Euclidean geometry to solve this problem.
There exists a right isosceles triangle with an incircle. What percentage of the area of the triangle does the area of the circle take up?
Answer as a percent.
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Without loss of generality: let the shorter sides of the triangle be 1 and the hypotenuse 2 such that: a r e a o f t r i a n g l e = A T = 2 1 . Now the radius of the incircle can be given by: r = a + b + c 2 A T = 2 + 2 1 = 2 2 − 2 Let I A be the area of the incircle such that ⇒ I A = π × ( 2 2 − 2 ) 2 ≈ 0 . 2 6 9 5 = x ∴ p e r c e n t a g e o f a r e a = 1 / 2 x × 1 0 0 ≈ 5 3 . 9
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A right isosceles triangle is a 4 5 ∘ - 9 0 ∘ - 4 5 ∘ triangle (see figure above). Let the equal sides be of length a units, therefore, the hypothenuse b = 2 a . Let the radius for the incircle be r .
The area of △ A B C , A = 2 a 2 = 2 ( a + a + b ) r
⇒ a 2 = ( 2 a + 2 a ) r ⇒ r = ( 2 + 2 ) a
Now the area of incircle A ∘ = π r 2
Therefore, the ratio,
x = A A ∘ = 2 a 2 π r 2 = ( 2 + 2 ) 2 2 π = 6 + 4 2 2 π
= 3 + 2 2 π = 0 . 5 3 9 0 1 2 0 8 4 = 5 3 . 9 %