question_answer

If D,E,F are the mid-points of sides BC,CA and AB respectively of MBC, then the ratio of the areas of triangles DEF and ABC is:

A) 1:4

B) 1:2

C) 2:3

D) 4:5

Correct Answer: A

Solution :

[a] We know that the line segment joining the mid-points of two sides of the triangle is half the length of the third side.

\[\therefore \,\,\,\,\,\,\,\,DE=\frac{1}{2}AB,\]

\[EF=\frac{1}{2}BC\] and \[DF=\frac{1}{2}AC\]

\[\Rightarrow \,\,\,\,\,\frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}=\frac{1}{2}\]

Thus,    \[\Delta DEF\tilde{\ }\Delta ABC\]

(BY SSS similarity criterion)

\[\therefore \,\,\,\,\,\,\frac{ar(\Delta DEF)}{ar(\Delta ABC)}={{\left( \frac{DF}{AC} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\]