Question

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

Answer 1

$ar\left(AXCD\right)=24c{m}^2$

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

area $\left(ABCD\right)=AB\times h$ (h = height of ABCD)

area of $\Delta XBC=\frac{1}{2}\times XB\times h$

$=\frac{1}{2}\times \frac{AB}{2}\times h$

area of $\left(\Delta XBC\right)=\frac{area\left(ABCD\right)}{4}$

area $\left(AXCD\right)=ar\left(ABCD\right)$ $-ar\Delta XBC$

$=ar\left(ABCD\right)-\frac{ar\left(ABCD\right)}{4}$

$ar\left(AXCD\right)=\frac{3ar\left(ABCD\right)}{4}=24c{m}^2$ (Given)

$ar\left(ABCD\right)=4\times \frac{24}{3}=32c{m}^2$

$ar\left(\Delta ABC\right)=\frac{1}{2}ar\left(ABCD\right)=\frac{1}{2}\left(32\right)=16c{m}^2$

[diagonal divides parallelogram to two equal areas].

$\therefore ar\left(\Delta ABC\right)=16c{m}^2$

So the answer is false.