Question

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar $\left(△ADF\right)=ar\left(△ECF\right)$
(ii) IF the area of $△DFB=3c{m}^3$ , Find the area of $|{|}^{gm}ABCD.$

Answer 1

Given : In $|{|}^{gm}ABCD.$, BC is produced to E such that CE = BC
AE intersects CD at F
To Prove:
(i) ar $\left(△ADF\right)=ar\left(△ECF\right)$
(ii) IF ar $\left(△DFB\right)=3c{m}^2,$ Find the area of $\left(||gmABCD\right)$

Proof:
(i) In $△ADFand△ECFAD=CE\angle AFD=\angle CFE\angle ADF=\angle FCE\therefore △ADF\cong △ECF\therefore ar\left(△ADF\right)=ar\left(△CEF\right)\left(ii\right)\therefore AF=CF$
and $AF=EF$
$\because$ BF is the median of $△BCD\therefore ar\left(△BFD\right)=ar\left(△BFC\right)=\frac{1}{2}ar\left(△BCD\right)=\frac{1}{2}\left[\frac{1}{2}ar\right(|{|}^{gm}ABCD\left)\right]=\frac{1}{4}ar\left(|{|}^{gm}ABCD\right)\therefore Butar\left(△BFD\right)=3c{m}^2$

$\therefore$ Area of $|{|}^{gm}ABCD=4\times ar\left(△BFD\right)=4\times 3=12c{m}^2$