Δ ABC ~ Δ ADE,In Δ ABC, AB = 6.3 cm, ∠CAB = 50º, AC = 5.6 cm and \(\dfrac{AB}{AD} = \dfrac{7}{5}\).Construct Δ ADE.

Answer: AD = 4.5 cm, AE = 4 cm, ∠DAE = 50º; construct Δ ADE with these measurements.

Since Δ ABC ~ Δ ADE and \(\dfrac{AB}{AD} = \dfrac{AC}{AE} = \dfrac{7}{5}\):

AD = \(\dfrac{5}{7} \times 6.3\) = 4.5 cm, AE = \(\dfrac{5}{7} \times 5.6\) = 4 cm, and ∠DAE = ∠CAB = 50º (corresponding angles of similar triangles).

Construction:

1. Draw seg AD = 4.5 cm.
2. At A, draw a ray making an angle of 50º with seg AD.
3. With A as centre and radius 4 cm, draw an arc cutting this ray at E.
4. Join DE. Δ ADE is the required triangle, similar to Δ ABC with \(\dfrac{AB}{AD} = \dfrac{7}{5}\).