Question

In $\Delta$ABC, if $\Delta$DFE is formed by joining the midpoints of the sides, the area of $\Delta$DFE is times the area of $\Delta$ ABC.

• A. 2
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. 4

Answer 1

The correct option is C $\frac{1}{4}$


When a triangle is formed by joining the midpoints of the sides of the $\Delta$ ABC,
AD = DB, BF = FC and AE = EC.
So we can observe that D divides AB in the ratio 1:1.
Similarly, E and F also divide AC and BC in the ratio 1:1.
Hence, $DE\parallel BC,DF\parallel ACandEF\parallel AB$.
So $\angle ADE=\angle B,\angle AED=\angle C,\angle BDF=\angle A,\angle BFD=\angle C,\angle CFE=\angle CBDand\angle CEF=\angle CAB.$
Therefore, $\Delta ADE\sim \Delta DBF\sim \Delta FED\sim \Delta EFC.$
The scale factor of these triangles is $\frac{AD}{AB}=\frac{1}{2}$ with respect to $\Delta$ ABC.
Hence the areas of these small triangles is
${\left(\frac{1}{2}\right)}^{2}Area\left(\Delta ABC\right)=\frac{1}{4}Area\left(\Delta ABC\right)$