Question

ABCD is a parallelogram and X is mid-point of AB. If $\mathrm{ar}\left(\mathrm{AXCD}\right)=24{\mathrm{cm}}^2$, then $\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=24{\mathrm{cm}}^2$. Write if it is true or false and justify your answer.

• A. True
• B. False

Answer 1

The correct option is B False

Step $1:$ Drawing the diagram:

ABCD is a parallelogram and X is mid-point of AB.

From C, draw a height h at side AB, such that it meets side AB at O.

Step $2:$ Verifying the $\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=24{\mathrm{cm}}^2$ or not:

As, $\mathrm{ar}\left(\mathrm{AXCD}\right)=24{\mathrm{cm}}^2$

X is the midpoint of AB, therefore,

$\mathrm{AX}=\mathrm{XB}=\frac{1}{2}\mathrm{AB}$,

Area of parallelogram $=\mathrm{BASE}\times \mathrm{HEIGHT}$,

$\mathrm{area}\left(\mathrm{ABCD}\right)=\mathrm{AB}\times \mathrm{h}$…………..$\left(1\right)$

Area of Triangle $=\frac{1}{2}\times \mathrm{B}\times \mathrm{H}$,

$\begin{array}{rcl}\mathrm{area}\mathrm{of}\mathrm{\Delta XBC}& =& \frac{1}{2}\times \mathrm{XB}\times \mathrm{h}\\ & =& \frac{1}{2}\times \frac{\mathrm{AB}}{2}\times \mathrm{h}\\ & =& \frac{\mathrm{area}\left(\mathrm{ABCD}\right)}{4}(\mathrm{from}(1))\end{array}$

$\begin{array}{rcl}\therefore \mathrm{ar}\left(\mathrm{AXCD}\right)& =& \mathrm{ar}\left(\mathrm{ABCD}\right)-\mathrm{ar}(∆\mathrm{XBC})\\ & =& \mathrm{ar}\left(\mathrm{ABCD}\right)-\frac{\mathrm{ar}\left(\mathrm{ABCD}\right)}{4}\\ & =& \frac{3\times \mathrm{ar}\left(\mathrm{ABCD}\right)}{4}\end{array}$

But, $\mathrm{ar}\left(\mathrm{AXCD}\right)=24{\mathrm{cm}}^2$,

$\begin{array}{rcl}\therefore \frac{3\times \mathrm{ar}\left(\mathrm{ABCD}\right)}{4}& =& 24{\mathrm{cm}}^2\\ \mathrm{ar}\left(\mathrm{ABCD}\right)& =& \frac{24\times 4}{3}\\ \mathrm{ar}\left(\mathrm{ABCD}\right)& =& 32{\mathrm{cm}}^2\end{array}$

As we know that diagonal divides parallelogram in two equal areas

$\begin{array}{rcl}\therefore \mathrm{ar}\left(\mathrm{\Delta ABC}\right)& =& \frac{1}{2}\times \mathrm{ar}\left(\mathrm{ABCD}\right)\\ & =& \frac{1}{2}\times 32{\mathrm{cm}}^2\\ & =& 16{\mathrm{cm}}^2\end{array}$

Hence, the statement is false.