Home
Class 10
MATHS
In Delta PQR "seg" PS "is median of" Del...

In `Delta PQR "seg" PS "is median of" Delta PQR. "And" PT bot QR,`
(i) `PR^(2) = PS^(2) + QR xx ST + ( (QR)/2)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
`PR^(2) = PS^(2) + QR xx ST + ((QR)/2)^(2)`
Promotional Banner

Topper's Solved these Questions

  • PYTHAGORAS THEOREM

    CHETAN PUBLICATION|Exercise pratice set - 2.2|2 Videos

  • PYTHAGORAS THEOREM

    CHETAN PUBLICATION|Exercise Practive Set - 2.2|1 Videos

  • PYTHAGORAS THEOREM

    CHETAN PUBLICATION|Exercise pratice set - 2.1|1 Videos

  • MODEL ACTIVITY SHEET

    CHETAN PUBLICATION|Exercise QUESTION|29 Videos

  • QUADRATIC EQUATIONS

    CHETAN PUBLICATION|Exercise ASSIGNMENT|12 Videos

Similar Questions

Explore conceptually related problems

In Delta PQR "seg" PS "is median of" Delta PQR. "And" PT bot QR , (ii) PQ^(2) = PS^(2) - QR xx ST + ((QR)/2)^(2)

Seg Pm is a median of Delta PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

In Delta ABC, "seg" AP "is a median". If BC = 18, AB^(2) + AC^(2) = 260 "find" AP .

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

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

In Delta PQR, M is "the midpoint of" side QR. If PQ = 11, PR = 17 "and" QR = 12 "then find PM" .

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

Delta PQR is an "equilateral triangle", "seg" PM bot side QR, Q -M- R "Prove that" : PQ^(2) = 4QM^(2)