Question

lf D, E, F are the midpoints of the sides BC, CA, AB respectively of $\Delta$ABC, then the ratio area $\Delta$DEF : area $\Delta$ABC is equal to

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

Answer 1

The correct option is D 1 : 4

Given,

$D,E,F$ are the mid point of the sides $BC,CA,AB$, respectively of $△ABC$.

We know that if two triangles are similar, the ratio of their area is always equal to the square of the ration of their corresponding side.

$\therefore \frac{Areaof△DEF}{Areaof△ABC}=\frac{D{E}^2}{A{C}^2}$

Since, $DECF$ is a parallelogram, opposite sides are equal, i.e., $DE=FC$.

Therefore,

$\frac{Areaof△DEF}{Areaof△ABC}=\frac{F{C}^2}{A{C}^2}$

$\frac{Areaof△DEF}{Areaof△ABC}=\frac{(AC/2{)}^2}{A{C}^2}$

$\frac{Areaof△DEF}{Areaof△ABC}=\frac{A{C}^2}{4A{C}^2}$

$\frac{Areaof△DEF}{Areaof△ABC}=\frac{1}{4}$

Hence, this is the answer.