Asked in: July 2024 board paper (same type as a 1-mark question with MQ = 9, QP = 4 in March 2025) In the figure, ∠ PQR = 90^° , seg QN ⊥ seg PR, PN = 9, NR = 16, find QN. [Figure: right-angled Δ PQR with the right angle at Q; QN is the perpendicular from Q to the hypotenuse PR.]

Answer: QN = 12 In right-angled Δ PQR , seg QN is the perpendicular to the hypotenuse PR. therefore QN² = PN × NR ...... (theorem of geometric mean) therefore QN² = 9 × 16 = 144 therefore QN = 12 . (For MQ = 9, QP = 4 the same theorem gives NQ² = 36 , NQ = 6.)

Chapter: Pythagoras Theorem· Mathematics Part 2 (Geometry)

Asked in: July 2024 board paper (same type as a 1-mark question with MQ = 9, QP = 4 in March 2025)In the 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…

Asked in: July 2024 board paper (same type as a 1-mark question with MQ = 9, QP = 4 in March 2025)

In the 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; QN is the perpendicular from Q to the hypotenuse PR.]

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Answer: QN = 12

In right-angled \(\Delta PQR\), seg QN is the perpendicular to the hypotenuse PR.

\(\therefore QN^2 = PN \times NR\) ...... (theorem of geometric mean)

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

(For MQ = 9, QP = 4 the same theorem gives \(NQ^2 = 36\), NQ = 6.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Asked in: July 2024 board paper (same type as a 1-mark question with MQ = 9, QP = 4 in March 2025)In the 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…

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