Question

In an equilateral triangle ABC, D is the mid-point of AB and E is the mid point of AC. The ar(ΔABC) : ar (ΔADE) =

Answer 1

ΔABC is an equilateral triangle. D and E are mid-points on the sides AB and AC, respectively.



Using mid-point theorem, we have

DE || BC and $\mathrm{DE}=\frac{1}{2}\mathrm{BC}$ .....(1)

In ∆ADE and ∆ABC,

∠DAE = ∠BAC (Common)

∠AED = ∠ACB (Corresponding angles)

∴ ∆ADE $~$ ∆ABC (AA Similarity)

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

$\Rightarrow \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{ADE}\right)}={\left(\frac{2\mathrm{DE}}{\mathrm{DE}}\right)}^2=\frac{4}{1}$ [Using (1)]

Or ar(ΔABC) : ar(ΔADE) = 4 : 1

In an equilateral triangle ABC, D is the mid-point of AB and E is the mid point of AC. The ar(ΔABC) : ar (ΔADE) = 4 : 1.