ABCD is a rectangle where side BC is twice side AB. If △ACQ ~ △BAP, find area of △BAP : area of △ACQ.

Given,

ABCD is a rectangle where side BC is twice side AB.

⇒ BC = 2AB

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (2AB)²

⇒ AC² = AB² + 4AB²

⇒ AC² = 5AB²

⇒ AC = $\sqrt{5}$ AB.

We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\therefore \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{AC^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{(\sqrt{5}BA)^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{5BA^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{1}{5} \\[1em]$

Area of △BAP : Area of △ACQ = 1 : 5.

Hence, Area of △BAP : Area of △ACQ = 1 : 5.