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Question Number 203009 by cortano12 last updated on 07/Jan/24

Answered by som(math1967) last updated on 07/Jan/24

$$Perimeter=80$$$$AD=AG+GD=8+8=16unit$$$$AB=\frac{80}{2}-16=24unit$$$$FB=24-8=16$$$$FQ=16-R$$$$OP=8+R$$$${(8+R)}^2={(8-R)}^2+{(16-R)}^2$$$$32R=256-32R+R^2$$$$R^2-64R+256=0$$$$R=\frac{64-\sqrt{{64}^2-4\times 256}}{2}$$$$=\frac{64-\sqrt{(64+32)(64-32)}}{2}$$$$=\frac{64-\sqrt{96\times 32}}{2}$$$$=\frac{64-32\sqrt{3}}{2}$$$$=16(2-\sqrt{3})$$$$=4.28unit\left(approx\right)$$

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