Home
Class 10
MATHS
The perpendicular PS on the base QR of D...

The perpendicular PS on the base QR of `Delta PQR` intersects QR at S, such that QS=3 SR. Prove that `2PQ^(2)=2PR^(2)+QR^(2)`.

Text Solution

Verified by Experts

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

Topper's Solved these Questions

  • GEOMETRY

    PREMIERS PUBLISHERS|Exercise SOLUTION TO EXERCISE 4.4|17 Videos
  • GEOMETRY

    PREMIERS PUBLISHERS|Exercise SOLUTION TO EXERCISE 4.5|15 Videos
  • GEOMETRY

    PREMIERS PUBLISHERS|Exercise SOLUTION TO EXERCISE 4.2|21 Videos
  • CO-ORDINATE GEOMETRY

    PREMIERS PUBLISHERS|Exercise Other Important Objective|25 Videos
  • MENSURATION

    PREMIERS PUBLISHERS|Exercise OTHER IMPORTANT OBJECTIVE TYPE QUESTIONS|16 Videos

Similar Questions

Explore conceptually related problems

The perpendicular from A on side BC at a Delta ABC intersects BC at D such that DB=3 CD. Prove that 2 AB^(2)=2AC^(2)+BC^(2)

In the adjoining figure , Delta PQR is an equilateral triangle. QR = RN . Prove that PN^(2) =3PR^(2)

If a point Q lies between two points P and R such that PQ = QR, prove that PQ = 1/2 PR.

in the adjoining figure, Delta PQR is an equilateral triangle. Point S is on seg QR such that QS = 1/3 QR , Prove that 9 PS^(2) = 7 PQ^(2)

In the adjoining figure, angle PQR = 90^(@)T is the midpoint of side QR. Prove that PR^(2) = 4PT^(2) - 3PQ^(2)

If p_(2),p_(2),p_(3) are the perpendiculars from the vertices of a triangle to the opposite sides, then prove that p_(1)p_(2)p_(3)=(a^(2)b^(2)c^(2))/(8R^(3))

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

In the adjoining figure, M is the midpoint of QR. anglePRQ = 90^(@) prove that PQ^(2) = 4 PM^(2) - 3 PR^(2)

PQR is a triangle right angled at P and M is a point on QR such that PM bot QR . Show that PM^(2) = QM .MR.

Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT xx TR= ST xx TQ .