A circle is inscribed in a right isosceles triangle whose legs are 8 inches long. How many square inches are in the area of the shaded portion? Express your answer as a decimal rounded to the nearest hundredth.

HyperBear Jan 12, 2020

#2**0 **

Here is the link : https://latex.artofproblemsolving.com/2/e/c/2ec388968eb94bbd38898fefa6dcdf9224e84ebf.png

HyperBear Jan 13, 2020

#5**+2 **

A circle is inscribed in a right isosceles triangle whose legs are 8 inches long. How many square inches are in the area of the shaded portion? Express your answer as a decimal rounded to the nearest hundredth.

Think of it as a co-ordinate geometry question

the vertices are (0,0) (0,8) (8,0)

Let (r,r) be the centre of the circle

x=0, y=0 and y=-x+8 are all tangents to the circle

Points on the circumference are (r,0), (0,r) (4,4)

Distance from the centre to (0,0) = \(\sqrt{2r^2}=\sqrt2\;r\)

Distance from origin to (4,4) =\( \sqrt{32}=4\sqrt2\)

Distance from origin to (4,4) in terms of raius r is \( \sqrt2 \;r+r=r(\sqrt2+1)\)

So

\(r(\sqrt2+1)=4\sqrt2\\ r=\frac{4\sqrt2}{\sqrt2+1}=\frac{4\sqrt2(\sqrt2-1)}{1}=8-4\sqrt2=4(2-\sqrt2)\)

Area of circle = \(\pi( 4(2-\sqrt2))^2=16\pi(4+2-4\sqrt2)=32\pi(3-2\sqrt2)\)

Area of triangle = 0.5*8*8 = 32

\(\text{Shaded area}=32-32\pi(3-2\sqrt2)\\ \text{Shaded area}=32(1-3\pi+2\pi\sqrt2)\\ \text{Shaded area}\approx 14.75\; square\; inches\)

You need to check my working.

Melody Jan 14, 2020