In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If 'x' be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of 'x'.

Given,

ΔABC ~ ΔADE

Given,

AE : EC = 4 : 7

Let AE = 4y and EC = 7y.

So, AC = 4y + 7y = 11y.

From figure,

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{AE}{AC} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{4y}{11y} = \dfrac{6.6}{BC} \\[1em] \Rightarrow \dfrac{4}{11} = \dfrac{6.6}{BC} \\[1em] \Rightarrow BC = \dfrac{11 \times 6.6}{4} = 18.15 \text{ cm}$

As ΔABC ~ ΔADE, we have :

∠ABC = ∠ADE and ∠ACB = ∠AED

So, DE || BC as ∠ADE and ∠ABC are corresponding angles.

$\therefore \dfrac{AB}{AD} = \dfrac{AC}{AE} = \dfrac{11y}{4y} = \dfrac{11}{4}$

Let perpendicular from A to DE meet DE at point P. Then,

AP = x

Let perpendicular from A to BC meet BC at point Q.

In ∆ADP and ∆ABQ,

∠ADP = ∠ABQ [Corresponding angles are equal.]

∠APD = ∠AQB [Both = 90°]

∴ ∆ADP ~ ∆ABQ [By AA]

$\Rightarrow \dfrac{AD}{AB} = \dfrac{AP}{AQ} \\[1em] \Rightarrow \dfrac{4}{11} = \dfrac{x}{AQ} \\[1em] \Rightarrow AQ = \dfrac{11}{4}x.$

Hence, BC = 18.15 cm and AQ = $\dfrac{11}{4}x.$