Question

D, E and F are the mid-point of sides BC, CA and AB of $△$ ABC, then area $△$ ABC : area $△$ DEF =

• A. $4:1$
• B. $1:4$
• C. $2:1$
• D. $4:3$

Answer 1

The correct option is A $4:1$

Given: In $\Delta ABC$, $D,E$ and $F$ are midpoints of sides $AB,BC$ and $CA$ respectively.

$BC=EC$

Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.

Therefore,we have:

$DF=d\frac{1}{2}BC\Rightarrow \frac{DF}{BC}=\frac{1}{2}....\left(1\right)$

$\frac{DE}{AC}=\frac{1}{2}....\left(2\right)$ and

$\frac{EF}{AB}=\frac{1}{2}....\left(3\right)$

From (1), (2) and (3) we have

But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar

Therefore, $\Delta ABC\sim \Delta EDF$ [By SSS similarity theorem]

Hence area of $\Delta ABC:$ area of $\Delta DEF=4:1$