Question

State which statement is/are true of the following figure.

• A. $\frac{1}{2}$ $\times$ area of AFEB = area of $\Delta$ DEF
• B. area of $\Delta$ ABE = area of $\Delta$ DEF
• C. area of $\Delta$ AED = $\frac{1}{2}$ $\times$ area of ABCD
• D. All of the above

Answer 1

The correct option is D

All of the above

In the given figure, ABCD is a parallelogram, AF = FD and BE = EC.

$AF=FD=\frac{AD}{2}$

$BE=EC=\frac{BC}{2}$

$AD=BC$

$\therefore AF=BE$ and we know that $AF||BE$.

Similarly, $FD=EC$ and $FD||EC$

Thus, AFEB and DFEC are parallelograms.

AF = FD, and the parallelograms AFEB and DFEC lie between the same parallels

Thus, ar(AFEB) = ar(DFEC)

Also, ar($\Delta DEF$) = $\frac{1}{2}$ar(DFEC)

$\therefore \frac{1}{2}\times$ ar(AFEB) = ar($\Delta DEF$)

Also, ar($\Delta ABE$) = $\frac{1}{2}$ar(AFEB)

Thus, ar($\Delta ABE$) = ar($\Delta DEF$)

$\Delta AED$ and parallelogram ABCD lie on the same base AD and between the same parallels AD and BC.

Thus, ar($\Delta AED$)=$\frac{1}{2}\times$ ar(ABCD)