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In \(\Delta ABC\) and \(\Delta PQR\), in a one to one correspondence of vertices, \(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\), then ....................... statement is true.[Figure: a small triangle ABC with A at the top left, B below it an…

Q.1 (A)(1) In \(\Delta ABC\) and \(\Delta PQR\), in a one to one correspondence of vertices, \(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\), then ....................... statement is true.

[Figure: a small triangle ABC with A at the top left, B below it and C to the right of B; a larger similarly shaped triangle with Q at the top, R below Q and P to the right of R, so that A corresponds to Q, B to R and C to P.]

A. \(\Delta PQR \sim \Delta ABC\)

B. \(\Delta PQR \sim \Delta CAB\)

C. \(\Delta CBA \sim \Delta PQR\)

D. \(\Delta BCA \sim \Delta PQR\)

1 answer

Best answer

Answer: B. \(\Delta PQR \sim \Delta CAB\)

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

The given ratios pair the sides as AB with QR, BC with PR and CA with PQ, i.e. A ↔ Q, B ↔ R, C ↔ P.

Writing the vertices in the same order, \(\Delta CAB \sim \Delta PQR\), which is the same statement as \(\Delta PQR \sim \Delta CAB\): \(\dfrac{PQ}{CA} = \dfrac{QR}{AB} = \dfrac{PR}{BC}\).

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