Question

In Fig. 7.255, DE || BC, AD = 1 cm, BD = 2 cm. What is the ratio of the area (ΔABC) to the areas (ΔADE)?

Answer 1

In ∆ABC and ∆ADE,

∠B = ∠ADE (Corresponding angles)

∠A = ∠A (Common)

∴ ∆ABC $~$ ∆ADE (AA Similarity)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{AD}}\right)}^2$ (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{3\mathrm{cm}}{1\mathrm{cm}}\right)}^2$ (AB = AD + BD = 1 cm + 2 cm = 3 cm)

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}=\frac{9}{1}$

∴ ar(ΔABC) : (ΔADE) = 9 : 1

Thus, the ratio of the area(ΔABC) to the area(ΔADE) is 9 : 1.