Question

In $\Delta ABC$, D, E and F are the mid points of the sides BC, CA and AB respectively. The area of $\Delta ABC$ is $24c{m}^2$, then the area of $\Delta DEF$ is

• A. $7c{m}^2$
• B. $6c{m}^2$
• C. $18c{m}^2$
• D. $16c{m}^2$

Answer 1

The correct option is B $6c{m}^2$

In $△ABC$,

D and E are midpoints of sides $BC$ and $AC$ respectively.

By midpoint theorem,

$DE=\frac{1}{2}AB$ and $DE\parallel AB$ ...(1)

In $△CAB$ and $△CED$,

$\angle C$ is the common angle.

$\angle CAB=\angle CED$ ....alternate angles since $DE\parallel AB$

$\therefore △CAB\sim △CED$ ...AA test of similarity

$\therefore \frac{CA}{CE}=\frac{AB}{DE}=\frac{BC}{DC}=\frac{2}{1}$ ...(2) ....C.S.S.T and from (1)

Similarly, we can prove $△ABC\sim △AFE\sim △FBD$ ...AA test of similarity

Hence, $\frac{EF}{BC}=\frac{DE}{AB}=\frac{DF}{AC}=\frac{1}{2}$ ....C.S.S.T

$\therefore △ABC\sim △DEF$ ...SSS test of similarity

By theorem on ratio of areas of similar triangles, we get

$\frac{A\left(△DEF\right)}{A\left(△ABC\right)}={\left(\frac{DE}{AB}\right)}^2$

$A\left(△DEF\right)=\frac{1}{4}\times 24=6c{m}^2$.