Question

In the given figure, if $ABCD$ is a parallelogram and $E$ is the mid-point of $BC$, then area $\left(△DEC\right)=k$ area $\left(ABCD\right)$. Find $k$

Answer 1

Given ABCD is parallelogram.

Let the diagonal BD divides parallelogram ABCD into $2$ triangles.

i.e., $\Delta$ ABD; $\Delta$ BCD

Area of ABD$+$ Area of BCD

$=$Area of ABCD

$\because$ ABD$=$BCD

$\therefore 2$(Area of BCD)$=$ Area of ABCD ……….$\left(1\right)$

'E' is M.P of BC, so DE divides $\Delta$ BCD into $2$ equal triangles

i.e., $\Delta$BED & $\Delta$ DEC

Area of $\Delta$BCD$=$ Area of $\Delta$BED$+$ Area of $\Delta$ DEC.

Area of $\Delta$BCD$=2$(Area of $\Delta$DEC)

By $\left(1\right)$

$2[2$(Area of DEC)$]=$ Area of parallelogram ABCD

$4$(Area of DEC)$=$Area of ABCD

$\therefore$ Area of $\Delta$DEC$=\frac{1}{4}$(Area of parallelogram ABCD).