Question

In MBC, DE is drawn parallel to BC. If AD: DB=2:3, DE =6cm and AE =3.6cm, find BC and AC. 

Answer 1

Given : DE = 6 cm , AE = 3.6 cm , ${\displaystyle \frac{\text{AD}}{\text{DB}}=\frac{2}{3}}$ , DE || BC

To find : BC  and AC 

Sol : In Δ ABC , DE || BC

∴ By BPT ${\displaystyle \frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}}$

${\displaystyle \frac{2}{3}=\frac{3.6}{\text{x}}}$

x = ${\displaystyle \frac{3.6\times 2}{2}}$

= 1.8 × 3

x = 5.4 = EC

∴ AC = 3.6 + 5.4 = 9 cm

AC = 9 cm

In ΔADE and Δ ABC

∠ ADE = ∠ ABC

Similarly ∠AED = ∠ ACB (corresponding angles)

∴ ΔADE ∼ Δ ABC    (AA corollary)

${\displaystyle \frac{\text{AE}}{\text{AC}}=\frac{\text{DE}}{\text{BC}}}$ (similar sides of angles)

${\displaystyle \frac{3.6}{9}=\frac{6}{\text{y}}}$

y = ${\displaystyle \frac{9\times 6}{3.6}}$

y = 15

BC = 15 cm