Question

In PQR, Q = 90 and QM is perpendicular to PR. Prove that: (i) PQ2 = PM  PR (ii) QR2 = PR  MR (iii) PQ2 + QR2 = PR2

Answer 1

Answer:

Solution In ∆Pqr, ∠Q = 90° and Qm is Perpendicular to Pr. Prove That: (I) Pq2 = Pm × Pr (Ii)Qr2 = Pr  Mr (Iii) Pq2 + Qr2 = Pr2 Concept: Axioms of Similarity of Triangles.

In PQR, Q = 90 and QM is perpendicular to PR. Prove that: (i) PQ2 = PM PR (ii) QR2 = PR ...

4 days ago · In PQR, Q = 90 and QM is perpendicular to PR. Prove that: (i) PQ2 = PM PR (ii) QR2 = PR MR (iii) PQ2 + QR2 = PR2. Log in to add ...

In triangle PQR , QM perpendicular to PR and PR2 - PQ2 = QR2 . prove that QM2 = PM×MR ...

19-Feb-2018 ·

So, ∆ PQR is a right angled triangle at Q. In ∆ QMR & ∆PMQ ∠QMR = ∠PMQ [ Each 90°] ∠MQR = ∠QPM [