Q.2 (B)(4) In the above figure, ∠ PQR = 90^° , seg QN ⊥ seg PR, PN = 9, NR = 16, find QN. [Figure: right-angled Δ PQR with the right angle at Q; N lies on hypotenuse PR and seg QN is drawn perpendicular to PR, with PN = 9 and NR = 16.]

In the above figure, \(\angle PQR = 90^\circ\), seg QN \(\perp\) seg PR, PN = 9, NR = 16, find QN.[Figure: right-angled \(\Delta PQR\) with the right angle at Q; N lies on hypotenuse PR and seg QN is drawn perpendicular to PR, with PN = 9 and NR = 16…

Q.2 (B)(4) In the above figure, \(\angle PQR = 90^\circ\), seg QN \(\perp\) seg PR, PN = 9, NR = 16, find QN.

[Figure: right-angled \(\Delta PQR\) with the right angle at Q; N lies on hypotenuse PR and seg QN is drawn perpendicular to PR, with PN = 9 and NR = 16.]

Doubtly Asked 25 Oct 2026 ssc 10th Save WhatsApp Report

1 answer

Best answer

Answer: QN = 12

In a right triangle, the perpendicular from the right angle to the hypotenuse divides it so that \(QN^2 = PN \times NR\) (theorem of geometric mean).

\(QN^2 = 9 \times 16 = 144\), so \(QN = 12\).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?

In the above figure, \(\angle PQR = 90^\circ\), seg QN \(\perp\) seg PR, PN = 9, NR = 16, find QN.[Figure: right-angled \(\Delta PQR\) with the right angle at Q; N lies on hypotenuse PR and seg QN is drawn perpendicular to PR, with PN = 9 and NR = 16…

1 answer

Your answer