\(\Delta ABC \sim \Delta PBR\), BC = 8 cm, AC = 10 cm, ∠B = 90º, \(\dfrac{BC}{BR} = \dfrac{5}{4}\), then construct Δ PBR.

Answer: BR = 6.4 cm, PB = 4.8 cm, PR = 8 cm; construct Δ PBR from these three sides.

In right Δ ABC, AB = \(\sqrt{AC^2 - BC^2} = \sqrt{100 - 64} = \sqrt{36}\) = 6 cm.

Since Δ ABC ~ Δ PBR, \(\dfrac{AB}{PB} = \dfrac{BC}{BR} = \dfrac{AC}{PR} = \dfrac{5}{4}\).

So BR = \(\dfrac{4}{5} \times 8\) = 6.4 cm, PB = \(\dfrac{4}{5} \times 6\) = 4.8 cm, PR = \(\dfrac{4}{5} \times 10\) = 8 cm.

Construction:

1. Draw BR = 6.4 cm.
2. At B draw a ray making 90º with BR (since ∠B = 90º).
3. With B as centre and radius 4.8 cm, cut the ray at P; join PR.
4. Δ PBR is the required triangle (check: PR = 8 cm, and 4.8² + 6.4² = 23.04 + 40.96 = 64 = 8²).