Solve the foloowing questions : `DeltaPQR,"seg " RS" is the bisector of " anglePRQ. PS=8, SQ=6, PR=20, "then find " QR`

In DeltaPQR seg RS is the bisector of anglePQR,PS=8, SQ=6,PR=20" then " QR = …………….

In DeltaPQR seg RS is bisector of anglePRQ.PS=6, SQ=8,PR=15 . Find QR.

Solve the any one of following sub questions: In Delta PQR, "seg" PM "is a median" PM = 10 and PQ^(2) + PR^(2) = 328 "then find" QR

Solve the following sub questions: in Delta PQR, angle PQR = 90^(@) "seg" QS bot "hypotenuse" PR, PS = 16 , RS = 9 "find" QS

In triangle PQR , RS is the bisector of angle R .if PQ = 6cm , QR = 8cm, RP = 4cm then PS is equal to ……………

In Delta PQR, Point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13 then find QR.

In the adjoining figure, if PQ = 6, QR = 10, PS = 8, then find TS.

Solve the following questions : DeltaABC~DeltaPQRA(DeltaABC):A(DeltaPQR)=9:16 "Find " BC: QR .

Solve the following sub question : in the adjoining figure, seg PS bat side QR. If PQ = a, PR = b QS = c and Rs = d then complete the following activity to prove that (a + b) (a - b) = (c + d) (c - d) Proof : In Delta PSQ angle PSQ = 90 ^(@) square^(2) = PS^(2) + square^(2) PS^(2) = square^(2) - square^(2) In Delta PSR, angle PSR = 90^(@) square^(2) = PS^(2) + square^(2) PS^(2) = square^(2) - square^(2) = square^(2) - square^(2) a^(2) - c^(2) = b^(2) - d^(2) a^(2) - b^(2) = C^(2) - d^(2) square xx square = square xx square

In the adjoining figure, point Q is the point of contact of tangent and circle . If PQ = 12, PR = 8, then find Ps and RS .