Select the appropriate alternative. In triangle ABC and triangle PQR , in a one to one correspondence AB/QR = BC/PR = CA/PQ then (A) triangle PQR sim triangle ABC (B) triangle PQR sim triangle CAB (C) triangle CBA sim triangle PQR (D) triangle BCA sim triangle PQR

In ΔABC and ΔPQR, in a one to one correspondence AB/QR = BC/PR = CA/PQ then (A) ΔPQR ~ ΔABC (B) ΔPQR ~ ΔCAB (C) ΔCBA ~ ΔPQR (D) ΔBCA ~ ΔPQR

Select the appropriate alternative.

In \(\triangle ABC\) and \(\triangle PQR\), in a one to one correspondence \(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\) then

(A) \(\triangle PQR \sim \triangle ABC\) (B) \(\triangle PQR \sim \triangle CAB\) (C) \(\triangle CBA \sim \triangle PQR\) (D) \(\triangle BCA \sim \triangle PQR\)

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: (B) \(\triangle PQR \sim \triangle CAB\)

Read the correspondence from the equal ratios: side AB ↔ side QR, BC ↔ PR, CA ↔ PQ.

The vertex common to AB and CA is A; the vertex common to QR and PQ is Q, so A ↔ Q. Similarly B (common to AB, BC) ↔ R (common to QR, PR), and C ↔ P.

So \(\triangle ABC \sim \triangle QRP\). Writing P first: P ↔ C, Q ↔ A, R ↔ B, i.e. \(\triangle PQR \sim \triangle CAB\).

The other options pair the vertices wrongly (for example, (A) pairs P with A).

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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In ΔABC and ΔPQR, in a one to one correspondence AB/QR = BC/PR = CA/PQ then (A) ΔPQR ~ ΔABC (B) ΔPQR ~ ΔCAB (C) ΔCBA ~ ΔPQR (D) ΔBCA ~ ΔPQR

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