Let Δ ABC ∼ Δ QPR and \(\rm \frac{area \ of \ \Delta ABC}{area \ of \ \Delta PQR}=\frac{64}{25}\) If QP = 12 cm, PR = 10 cm and AC = 15 cm, then the length of AB is:

Question

Question

This question was previously asked in

AAI Junior Assistant (Fire Service) Official Paper (Held On: 15 Nov 2022 Shift 1)

Answer 1

Given:

Δ ABC ∼ Δ QPR

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR}=\frac{64}{25}\)

Formula used:

In similar triangles, the ratio of the areas is equal to the square of the ratio of the corresponding sides.

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR} = (\frac{corresponding \ side \ of \ Δ ABC}{corresponding \ side \ of \ Δ PQR})^2\)

Calculation:

5-5-2025 IMG-1083 Prakhar -2

\(\frac{64}{25} = (\frac{AB}{QP})^2\)

⇒ \(\frac{64}{25} = (\frac{AB}{12})^2\)

⇒ \(\frac{64}{25} = \frac{AB^2}{144}\)

⇒ \(\frac{64}{25} × 144 = AB^2\)

⇒ \(\frac{64 × 144}{25} = AB^2\)

⇒ \(\frac{9216}{25} = AB^2\)

⇒ 368.64 = \(AB^2\)

⇒ AB = \(\sqrt{368.64}\)

⇒ AB = 19.2 cm

∴ The correct answer is option (4).