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Question

Two circles intersect in A and B. Quadrilaterals PCBA and ABDE are inscribed in these circles such that PAE and CBD are line segments. If P=95o and C=40o. Also, AED=z. Then the value of z is:
244576_96618f81570844ceaaeca5b0b0dde134.png

A
65o
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B
105o
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C
95o
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D
85o
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Solution

The correct option is D 85o
GivenTwocirclesintersectatA&B.ThequadrilateralPCBAisinscribedinonecircleandthequadrilateralEDBAisinscribedinanothercirclesuchthatthepointsP,A&BarecollinearandC,B&Darecollinear.AED=z,APC=95o&PCB=40o.Tofindoutz=?solutionAPC+ABC=180o(sincethesumoftheoppositeanglesofacyclicquadrilateralis180o).ABC=180oAPC=180o95o=85o.AgainABC+ABD=180o(linearpair).ABD=180oABC=180o85o=95o.NowABD+AED=180o(sincethesumoftheoppositeanglesofacyclicquadrilateralis180o).AED=z=180oABD=180o95o=85o.Hence,optionDiscorrect.

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