Asked in: March 2026 board paper\(\Delta ABC \sim \Delta ADE\). In \(\Delta ABC\), AB = 6.3 cm, \(\angle CAB = 50^\circ\), AC = 5.6 cm and \(\dfrac{AB}{AD} = \dfrac{7}{5}\). Construct \(\Delta ADE\).

Answer: \(\Delta ADE\) with AD = 4.5 cm, \(\angle A = 50^\circ\), AE = 4 cm.

Analysis: \(\dfrac{AB}{AD} = \dfrac{AC}{AE} = \dfrac{7}{5}\). \(AD = \dfrac{5 \times 6.3}{7} = 4.5\) cm, \(AE = \dfrac{5 \times 5.6}{7} = 4\) cm; \(\angle DAE = \angle CAB = 50^\circ\).

Steps of construction:

1. Draw seg AB = 6.3 cm. At A draw ray AX making 50° with AB. Mark C on AX with AC = 5.6 cm. Join BC to get \(\Delta ABC\).
2. On seg AB mark D such that AD = 4.5 cm, and on seg AC mark E such that AE = 4 cm.
3. Join DE. \(\Delta ADE\) is the required triangle (DE is parallel to BC).

[Figure: \(\Delta ABC\) with \(\angle A = 50^\circ\); points D on AB and E on AC; the smaller \(\Delta ADE\) shares vertex A.]

(By the ray method: divide a ray from A into 7 equal parts, join \(A_7B\), and through \(A_5\) draw a line parallel to \(A_7B\) to get D on AB; then draw DE || BC.)