Question Number 187301 by Rupesh123 last updated on 15/Feb/23 | ||
Answered by HeferH last updated on 16/Feb/23 | ||
$$\:{chord}\:{theorem}: \\ $$$$\mathrm{5}^{\mathrm{2}} \:=\:{x}\left(\mathrm{5}\:+\:\mathrm{3}\right) \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:=\:{x} \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:+\:\mathrm{8}\:=\:\mathrm{2}{r} \\ $$$$\:\frac{\mathrm{225}}{\mathrm{16}}\:=\:{r} \\ $$$$\:{A}\:\neq\:\:\frac{\mathrm{225}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{2}} }\pi\:\:=\:\frac{\mathrm{11025}\pi}{\mathrm{256}}\:\left({i}\:{assumed}\right. \\ $$$$\left.\:{that}\:{the}\:{line}\:{was}\:{diameter}\:{which}\:{is}\:{false}\right) \\ $$$$\: \\ $$ | ||
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 | ||
$${Yes}\:{sir},\:{you}\:{are}\:{right},\:{im}\:{going}\:{to}\:{correct}\:{it}\: \\ $$ | ||
Commented by HeferH last updated on 15/Feb/23 | ||
Commented by HeferH last updated on 15/Feb/23 | ||
$${b}\left(\mathrm{5}\:+\:\mathrm{3}\right)\:=\:\mathrm{5}{a} \\ $$$$\mathrm{3}\left({b}\:+\:\mathrm{5}\right)\:=\:\mathrm{5}{a} \\ $$$$\:\mathrm{8}{b}\:=\:\mathrm{3}{b}\:+\:\mathrm{15} \\ $$$$\:{b}\:=\:\mathrm{3} \\ $$$$\:\Rightarrow\:{a}\:=\:\frac{\mathrm{24}}{\mathrm{5}} \\ $$$$\:{r}^{\mathrm{2}} \:=\:\frac{\mathrm{25}\:+\:\left(\mathrm{5}\:+\:{a}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{\mathrm{25}\:+\:\frac{\mathrm{2401}}{\mathrm{25}}}{\mathrm{4}}\:=\:\frac{\mathrm{625}+\mathrm{2401}}{\mathrm{100}} \\ $$$$\:=\:\frac{\mathrm{3026}}{\mathrm{100}}\:=\:\frac{\mathrm{1513}}{\mathrm{50}}\: \\ $$$$\:{A}\:=\:\frac{\mathrm{1513}\pi}{\mathrm{50}} \\ $$$$\: \\ $$ | ||
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 | ||
$${AB}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{34}} \\ $$$${BC}=\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}+\mathrm{3}\right)^{\mathrm{2}} }=\sqrt{\mathrm{89}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}} \\ $$$$\mathrm{2}{R}=\frac{{BC}}{\mathrm{sin}\:\theta}=\frac{\sqrt{\mathrm{89}}×\sqrt{\mathrm{34}}}{\mathrm{5}}=\frac{\sqrt{\mathrm{3026}}}{\mathrm{5}} \\ $$$${R}=\frac{\sqrt{\mathrm{3026}}}{\mathrm{10}} \\ $$$${area}\:{of}\:{circle}\:=\pi{R}^{\mathrm{2}} =\frac{\mathrm{3026}\pi}{\mathrm{100}}=\frac{\mathrm{1513}\pi}{\mathrm{50}} \\ $$ | ||
Commented by Rupesh123 last updated on 15/Feb/23 | ||
Excellent | ||