The given diagram shows two isosceles triangles which are similar. In the given diagram, PQ and BC are not parallel; PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.

Calculate :

(i) the length of AP,

(ii) the ratio of the areas of triangle APQ and triangle ABC.

(i) Given,

△APQ ~ △ABC.

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{AQ}{AC} = \dfrac{AP}{AB} \\[1em] \Rightarrow \dfrac{AQ}{AP + PC} = \dfrac{AP}{AQ + QB} \\[1em] \Rightarrow \dfrac{3}{AP + 4} = \dfrac{AP}{3 + 12} \\[1em] \Rightarrow \dfrac{3}{AP + 4} = \dfrac{AP}{15} \\[1em] \Rightarrow AP(AP + 4) = 45 \\[1em] \Rightarrow AP^2 + 4AP - 45 = 0 \\[1em] \Rightarrow AP^2 + 9AP - 5AP - 45 = 0 \\[1em] \Rightarrow AP(AP + 9) - 5(AP + 9) = 0 \\[1em] \Rightarrow (AP - 5)(AP + 9) = 0 \\[1em] \Rightarrow AP = 5 \text{ or } -9.$

Since, length cannot be negative.

Hence, AP = 5 units.

(ii) We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\Rightarrow \dfrac{\text{Area of △APQ}}{\text{Area of △ABC}} = \dfrac{PQ^2}{BC^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APQ}}{\text{Area of △ABC}} = \dfrac{AP^2}{BC^2} \space [\because \text{PQ = AP}] \\[1em] \Rightarrow \dfrac{\text{Area of △APQ}}{\text{Area of △ABC}} = \dfrac{5^2}{15^2} = \dfrac{25}{225} = \dfrac{1}{9}.$

Hence, the ratio of the areas of △APQ and △ABC = 1 : 9.