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Question Number 187301 by Rupesh123 last updated on 15/Feb/23

Answered by HeferH last updated on 16/Feb/23

$$chordtheorem:$$$$5^2=x(5+3)$$$$\frac{25}{8}=x$$$$\frac{25}{8}+8=2r$$$$\frac{225}{16}=r$$$$A\ne \frac{{225}^2}{{16}^2}\pi =\frac{11025\pi }{256}(iassumed$$$$thatthelinewasdiameterwhichisfalse)$$$$$$

Commented by Rupesh123 last updated on 15/Feb/23

Good!

Commented by Rupesh123 last updated on 15/Feb/23

Please note that it doesn't say the bottom of the square is coincident with the diameter

Commented by HeferH last updated on 15/Feb/23

$$Yessir,youareright,imgoingtocorrectit$$

Commented by HeferH last updated on 15/Feb/23

Commented by HeferH last updated on 15/Feb/23

$$b(5+3)=5a$$$$3(b+5)=5a$$$$8b=3b+15$$$$b=3$$$$\Rightarrow a=\frac{24}{5}$$$$r^2=\frac{25+{(5+a)}^2}{4}=\frac{25+\frac{2401}{25}}{4}=\frac{625+2401}{100}$$$$=\frac{3026}{100}=\frac{1513}{50}$$$$A=\frac{1513\pi }{50}$$$$$$

Commented by Rupesh123 last updated on 16/Feb/23

Excellent

Answered by mr W last updated on 15/Feb/23

Commented by mr W last updated on 15/Feb/23

$$AB=\sqrt{3^2+5^2}=\sqrt{34}$$$$BC=\sqrt{5^2+{(5+3)}^2}=\sqrt{89}$$$$\sin \theta =\frac{5}{\sqrt{34}}$$$$2R=\frac{BC}{\sin \theta }=\frac{\sqrt{89}\times \sqrt{34}}{5}=\frac{\sqrt{3026}}{5}$$$$R=\frac{\sqrt{3026}}{10}$$$$areaofcircle=\pi R^2=\frac{3026\pi }{100}=\frac{1513\pi }{50}$$

Commented by Rupesh123 last updated on 15/Feb/23

Excellent

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