Asked in: July 2024 and June 2026 board papers (same type with LM = 6, MN = 10, TN = 8 in July 2025) In Δ PQR, seg RS bisects ∠ PRQ. If PR = 15, RQ = 20, PS = 12, then find SQ. [Figure: Δ PQR with point S on side PQ; seg RS bisects ∠ R .]

Answer: SQ = 16 In Δ PQR, seg RS bisects ∠ PRQ. therefore PR/RQ = PS/SQ ...... (theorem of angle bisector of a triangle) therefore 15/20 = 12/SQ therefore SQ = (12 × 20)/15 = 16.

Chapter: Similarity· Mathematics Part 2 (Geometry)

Asked in: July 2024 and June 2026 board papers (same type with LM = 6, MN = 10, TN = 8 in July 2025)In \(\Delta PQR\), seg RS bisects \(\angle PRQ\). If PR = 15, RQ = 20, PS = 12, then find SQ.[Figure: \(\Delta PQR\) with point S on side PQ; seg RS b…

Asked in: July 2024 and June 2026 board papers (same type with LM = 6, MN = 10, TN = 8 in July 2025)

In \(\Delta PQR\), seg RS bisects \(\angle PRQ\). If PR = 15, RQ = 20, PS = 12, then find SQ.

[Figure: \(\Delta PQR\) with point S on side PQ; seg RS bisects \(\angle R\).]

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Answer: SQ = 16

In \(\Delta PQR\), seg RS bisects \(\angle PRQ\).

\(\therefore \dfrac{PR}{RQ} = \dfrac{PS}{SQ}\) ...... (theorem of angle bisector of a triangle)

\(\therefore \dfrac{15}{20} = \dfrac{12}{SQ}\)

\(\therefore SQ = \dfrac{12 \times 20}{15} = 16\).

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Asked in: July 2024 and June 2026 board papers (same type with LM = 6, MN = 10, TN = 8 in July 2025)In \(\Delta PQR\), seg RS bisects \(\angle PRQ\). If PR = 15, RQ = 20, PS = 12, then find SQ.[Figure: \(\Delta PQR\) with point S on side PQ; seg RS b…

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