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Question-122055




Question Number 122055 by I want to learn more last updated on 13/Nov/20
Answered by mr W last updated on 13/Nov/20
AB=((2×10π)/4)=5π  PR=10  AP=AO−OP  BR=BO−OR  AP+BR=2×10−(OP+OR)  =20−((26)/2)=7  AB+BR+RP+PA=5π+10+7=17+5π  ⇒(C)
$${AB}=\frac{\mathrm{2}×\mathrm{10}\pi}{\mathrm{4}}=\mathrm{5}\pi \\ $$$${PR}=\mathrm{10} \\ $$$${AP}={AO}−{OP} \\ $$$${BR}={BO}−{OR} \\ $$$${AP}+{BR}=\mathrm{2}×\mathrm{10}−\left({OP}+{OR}\right) \\ $$$$=\mathrm{20}−\frac{\mathrm{26}}{\mathrm{2}}=\mathrm{7} \\ $$$${AB}+{BR}+{RP}+{PA}=\mathrm{5}\pi+\mathrm{10}+\mathrm{7}=\mathrm{17}+\mathrm{5}\pi \\ $$$$\Rightarrow\left({C}\right) \\ $$
Commented by I want to learn more last updated on 13/Nov/20
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by TANMAY PANACEA last updated on 13/Nov/20
sir why PR=10
$${sir}\:{why}\:{PR}=\mathrm{10} \\ $$
Commented by mr W last updated on 13/Nov/20
PR=OQ=radius=10
$${PR}={OQ}={radius}=\mathrm{10} \\ $$

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