CBSE2026Class Class 10 · Mathematics3 Marks · Short✅ Verified
(B) [OR] In the given figure, if a circle touches the side QR of $\triangle PQR$ at S, and touches the extended sides PQ and PR at M and N respectively, then prove that: $PM = \frac{1}{2}(PQ+QR+PR)$
✅ Answer & Solution
Tangent segments drawn from an external point to a circle are equal in length. From P: $PM = PN$ ...(i). From Q: $QM = QS$ ...(ii). From R: $RN = RS$ ...(iii). Perimeter of $\triangle PQR = PQ+QR+PR = PQ+(QS+SR)+PR$. Using (ii) and (iii): $= PQ+QM+RN+PR$. Now, $PM = PQ+QM$ and $PN = PR+RN$, so $PM+PN = PQ+QM+PR+RN =$ Perimeter. Using (i), $PM=PN$: Perimeter $=2PM$. $\therefore PM = \frac{1}{2}(PQ+QR+PR)$ (Proved)
