While preparing a PowerPoint presentation, ∆ ABC is enlarged along the side BC to ∆ AB'C', as shown in the diagram, such that BC ∶ B'C' is 3 ∶ 5. Find :

(a) AB ∶ BB'

(b) length AB, if BB' = 4 cm.

(c) Is ∆ ABC ~ ∆ AB'C' ? Justify your answer.

(d) ar (∆ ABC) : ar (quad. BB'C'C).

Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ ∆ ABC and ∆ AB'C' are similar triangles.

(a) We know that,

Ratio of corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{AB'} = \dfrac{BC}{B'C'} \\[1em] \Rightarrow \dfrac{AB}{AB'} = \dfrac{3}{5}$

Let AB = 3x and AB' = 5x

From figure,

⇒ AB' = AB + BB'

⇒ 5x = 3x + BB'

⇒ BB' = 5x - 3x = 2x.

⇒ AB : BB' = 3x : 2x = 3 : 2.

Hence, AB : BB' = 3 : 2.

(b) As,

$\Rightarrow \dfrac{AB}{BB'} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{AB}{4} = \dfrac{3}{2} \\[1em] \Rightarrow AB = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow AB = 6 \text{ cm}.$

Hence, AB = 6 cm.

(c) Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ BC || B'C'

In ∆ ABC and ∆ AB'C',

⇒ ∠BAC = ∠B'AC' (Common angle)

⇒ ∠ABC = ∠AB'C' (Corresponding angle are equal)

∴ ∆ ABC ~ ∆ AB'C' (By A.A. axiom).

Hence, proved that ∆ ABC ~ ∆ AB'C'.

(d) We know that,

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

$\therefore \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{AB}{AB'}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\dfrac{3}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \dfrac{9}{25}.$

Let area of ∆ ABC = 9a and area of ∆ AB'C' = 25a.

Area of quadrilateral BB'C'C = Area of ∆ AB'C' - Area of ∆ ABC = 25a - 9a = 16a.

∴ Area of ∆ ABC : Area of quadrilateral BB'C'C = 9a : 16a = 9 : 16.

Hence, area of ∆ ABC : area of quadrilateral BB'C'C = 9 : 16.