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Question Number 19415 by Tinkutara last updated on 10/Aug/17
PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. What is the area of the region QXROQ bounded by the two semicircular arcs?
$${PS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{of}\:\mathrm{length}\:\mathrm{4}\:\mathrm{and}\:{O} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PS}.\:\mathrm{A}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:{PS}\:\mathrm{as}\:\mathrm{diameter}.\:\mathrm{Let} \\ $$$${X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{this}\:\mathrm{arc}.\:{Q}\:\mathrm{and}\:{R} \\ $$$$\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{arc}\:{PXS}\:\mathrm{such}\:\mathrm{that}\:{QR} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{PS}\:\mathrm{and}\:\mathrm{the}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{drawn}\:\mathrm{with}\:{QR}\:\mathrm{as}\:\mathrm{diameter}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:{PS}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{region}\:{QXROQ}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{semicircular}\:\mathrm{arcs}? \\ $$
Answered by ajfour last updated on 11/Aug/17
Commented by ajfour last updated on 11/Aug/17
A_1 =Area of △QOR=(1/2)×2(√2)×(√2)=2 A_2 =Area of sector OQXR=(1/4)×π×2^2 =π A_3 =Area of semicircle OQR=(1/2)×π×((√2))^2 =π Required area=A_2 −A_1 +A_3 =𝛑−2+𝛑=2𝛑−2 .
$$\mathrm{A}_{\mathrm{1}} =\mathrm{Area}\:\mathrm{of}\:\bigtriangleup\mathrm{QOR}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}}=\mathrm{2} \\ $$$$\mathrm{A}_{\mathrm{2}} =\mathrm{Area}\:\mathrm{of}\:\mathrm{sector}\:\mathrm{OQXR}=\frac{\mathrm{1}}{\mathrm{4}}×\pi×\mathrm{2}^{\mathrm{2}} =\pi \\ $$$$\mathrm{A}_{\mathrm{3}} =\mathrm{Area}\:\mathrm{of}\:\mathrm{semicircle}\:\mathrm{OQR}=\frac{\mathrm{1}}{\mathrm{2}}×\pi×\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\pi \\ $$$$\mathrm{Required}\:\mathrm{area}=\mathrm{A}_{\mathrm{2}} −\mathrm{A}_{\mathrm{1}} +\mathrm{A}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\pi}−\mathrm{2}+\boldsymbol{\pi}=\mathrm{2}\boldsymbol{\pi}−\mathrm{2}\:. \\ $$
Commented by Tinkutara last updated on 11/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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