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Q.11. In the given figure, two circles touch each other internally at point P, where Q is the centre of larger circle. If RS = 17 cm, TU = 9 cm and TV RP, then find the area of shaded region.


Hi, 



Let the center of smaller circle is Oand smaller and larger circle has radii r and R cmJoin PU, OU, USUQ= QT-9 = R-9 QS= R-17 PSis the diameter of smaller circle angle PUS= 90 degreeangle in the semicircle Also PRTV or OQTVIn right angle triangle PUS, UQis perpendicular to hypotenuse PSand we know that UQ2= PQ×QS since triangle PQU and UQS will be similar R-92= RR-17R2+81 -18R = R2-17RR = 81 so  QS= R-17 = 64 cm and UQ= R-9 = 72 cm Now OQ= OS- QS = r-64 so in right triangle OQU OU2= OQ2+QU2r2= r-642+722r2= r2+4096-128r+5184r = 72.5 cm so area of shaded region = πR2-r2 = 3.14×812-72.52=4096.951 cm2

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