Asked in: July 2025 board paper\(\Delta ABC \sim \Delta PBR\), BC = 8 cm, AC = 10 cm, \(\angle B = 90^\circ\), \(\dfrac{BC}{BR} = \dfrac{5}{4}\), then construct \(\Delta PBR\).

Answer: \(\Delta PBR\) with BR = 6.4 cm, \(\angle B = 90^\circ\), PR = 8 cm (and PB = 4.8 cm).

Analysis: In right-angled \(\Delta ABC\), \(AB = \sqrt{10^2 - 8^2} = 6\) cm. \(\dfrac{AB}{PB} = \dfrac{BC}{BR} = \dfrac{AC}{PR} = \dfrac{5}{4}\), so \(BR = \dfrac{4 \times 8}{5} = 6.4\) cm, \(PR = \dfrac{4 \times 10}{5} = 8\) cm, \(PB = \dfrac{4 \times 6}{5} = 4.8\) cm.

Steps of construction:

1. Draw seg BC = 8 cm. At B draw ray BY perpendicular to BC. With C as centre and radius 10 cm, draw an arc cutting ray BY at A. Join AC: this is \(\Delta ABC\).
2. Mark R on seg BC such that BR = 6.4 cm.
3. Through R draw a line parallel to CA, meeting seg BA at P. \(\Delta PBR\) is the required triangle.
4. Check: PB should measure 4.8 cm and PR 8 cm.

[Figure: right angle at B; \(\Delta ABC\) with BC = 8 cm, AC = 10 cm; inside it \(\Delta PBR\) with R on BC, P on BA and PR parallel to AC.]