A line PQ is drawn parallel to the base BC of ΔABC which meets sides AB and AC at points P and Q respectively. If AP = $\dfrac{1}{3}$PB; find the value of :

(i) $\dfrac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}}$

(ii) $\dfrac{\text{Area of ΔAPQ}}{\text{Area of trapezium PBCQ}}$

Given, AP = $\dfrac{1}{3}$PB

So, $\dfrac{AP}{PB} = \dfrac{1}{3}$

Let AP = x and PB = 3x.

AB = AP + PB = x + 3x = 4x.

$\therefore \dfrac{AP}{AB} = \dfrac{x}{4x} = \dfrac{1}{4}$.

In ∆APQ and ∆ABC,

∠APQ = ∠ABC and ∠AQP = ∠ACB [Corresponding angles are equal]

Hence, ∆APQ ~ ∆ABC by AA criterion for similarity

(i) We know that,

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

$\dfrac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}} = \dfrac{\text{AB}^2}{\text{AP}^2} \\[1em] = \dfrac{4^2}{1^2} \\[1em] = \dfrac{16}{1} \\[1em] = 16 : 1$

Hence, $\dfrac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}}$ = 16 : 1.

(ii) From figure,

Area of Trapezium PBCQ = Area of ΔABC – Area of ΔAPQ

From part (i) above we get,

$\dfrac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}} = \dfrac{16}{1}$

Let Area of ΔABC = 16a and Area of ΔAPQ = a

Area of trapezium PBCQ = 16a - a = 15a.

$\dfrac{\text{Area of ΔAPQ}}{\text{Area of trap. PBCQ}} = \dfrac{a}{15a} = \dfrac{1}{15}$ = 1 : 15.

Hence, $\dfrac{\text{Area of ΔAPQ}}{\text{Area of trap. PBCQ}}$ = 1 : 15.