In the given figure, PR, RT and PT are tangents to the circle at Q, S and U respectively.
If PR = (RT + 3) cm, PR = (PT + 1) cm and perimeter of ∆PRT is 26 cm, then QR + RT equal to:
(A) 9 cm                 (B) 7 cm                (C) 13 cm                 (D) 11 cm

Hi, perimeter of triangle PRT = 26 cm PR+RT+PT= 26 PR+PR-3+PR-1 = 263PR= 30 PR = 10 cm so RT = 7 cm and PT= 9 cm now we know that tangents from external point to circle are equalso lets say RS= SQ= x PQ= PU=  y TU= TS= z x+y = RP= 10 .....1y+z= PT= 9 ......2z+x= RT= 7 .....3add all 2x+y+z= 26 x+y+z = 13 ......4subtracting equation 1,2,3 one by one from equation 4 z =3, x = 4,y = 6 so QR+RT = x+7 = 4+7 = 11 cm

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