Question-220286

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Question Number 220286 by Tawa11 last updated on 10/May/25

Commented by Tawa11 last updated on 10/May/25

$$\mathrm{perimeter}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{interms}\:\mathrm{of}\:\pi \\ $$

Commented by fantastic last updated on 10/May/25

Commented by fantastic last updated on 10/May/25

Here △ABO∼△ACD ∴((AB)/(AD))=((AO)/(CD))=((BO)/(AC)) ((25)/(2r))=(r/( (√((2r)^2 −(32)^2 )))) r=((5(√(50−2i(√(399)))))/2) ∴perimeter=πr+2r =π(r+((2r)/π)) =π(5(√(50−2i(√(399))))/2+((5(√(50−2i(√(399)))))/π))u

$${Here}\:\bigtriangleup{ABO}\sim\bigtriangleup{ACD} \\ $$$$\therefore\frac{{AB}}{{AD}}=\frac{{AO}}{{CD}}=\frac{{BO}}{{AC}} \\ $$$$\frac{\mathrm{25}}{\mathrm{2}{r}}=\frac{{r}}{\:\sqrt{\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left(\mathrm{32}\right)^{\mathrm{2}} }} \\ $$$${r}=\frac{\mathrm{5}\sqrt{\mathrm{50}−\mathrm{2}{i}\sqrt{\mathrm{399}}}}{\mathrm{2}} \\ $$$$\therefore{perimeter}=\pi{r}+\mathrm{2}{r} \\ $$$$=\pi\left({r}+\frac{\mathrm{2}{r}}{\pi}\right) \\ $$$$=\pi\left(\mathrm{5}\sqrt{\mathrm{50}−\mathrm{2}{i}\sqrt{\mathrm{399}}}/\mathrm{2}+\frac{\mathrm{5}\sqrt{\mathrm{50}−\mathrm{2}{i}\sqrt{\mathrm{399}}}}{\pi}\right){u} \\ $$

Commented by fantastic last updated on 10/May/25

$${i}\:{am}\:{not}\:{sure}\:{if}\:{it}\:{is}\:{right}\:{or}\:{not} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 10/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 10/May/25

((25)/R)=((2R)/(25+7)) ⇒R=(√((25(25+7))/2))=20 P=(2+π)R=20(2+π)

$$\frac{\mathrm{25}}{{R}}=\frac{\mathrm{2}{R}}{\mathrm{25}+\mathrm{7}} \\ $$$$\Rightarrow{R}=\sqrt{\frac{\mathrm{25}\left(\mathrm{25}+\mathrm{7}\right)}{\mathrm{2}}}=\mathrm{20} \\ $$$${P}=\left(\mathrm{2}+\pi\right){R}=\mathrm{20}\left(\mathrm{2}+\pi\right) \\ $$

Commented by fantastic last updated on 10/May/25

sir how can you say that ((AO)/(AD))=((AB)/(AC))

$${sir}\:{how}\:{can}\:{you}\:{say}\:{that}\:\frac{{AO}}{{AD}}=\frac{{AB}}{{AC}} \\ $$

Commented by mr W last updated on 10/May/25

ΔAOB∼ΔACD ⇒((AB)/(AO))=((AD)/(AC)), i.e. ((25)/R)=((2R)/(25+7))

$$\Delta{AOB}\sim\Delta{ACD} \\ $$$$\Rightarrow\frac{{AB}}{{AO}}=\frac{{AD}}{{AC}},\:{i}.{e}.\:\frac{\mathrm{25}}{{R}}=\frac{\mathrm{2}{R}}{\mathrm{25}+\mathrm{7}} \\ $$

Commented by fantastic last updated on 10/May/25