Question

ABCD is a parallelogram and X is the mid-point of AB. If $Area\left(\square AXCD\right)=$ $24$ ${cm}^2$, Find $Area\left(\Delta ABC\right)$.

• A. $16$ $c{m}^2$
• B. $26$ $c{m}^2$
• C. $12$ $c{m}^2$
• D. $24$ $c{m}^2$

Answer 1

The correct option is A $16$ $c{m}^2$

We know that the area of a triangle is half of that of a parallelogram if they are on the same base and between the same parallels.

In case $△XCB$, base is half.

$\therefore$ Area of $△XCB=\frac{1}{2}\times \frac{1}{2}\times ar\left(ABC\right)$

$=\frac{1}{4}\times ar\left(ABC\right)$

Now, $ar\left(AXCD\right)=ar\left(ABCD\right)-ar\left(XCB\right)$

$=ar\left(ABCD\right)-\frac{1}{4}\times ar\left(ABCD\right)$

$=\frac{3}{4}\times ar\left(ABCD\right)$

$\Rightarrow$ $ar\left(AXCD\right)=\frac{3}{4}\times ar\left(ABCD\right)$

$\Rightarrow$ $24=\frac{3}{4}\times ar\left(ABCD\right)$

$\Rightarrow$ $ar\left(ABCD\right)=32c{m}^2$

We know that a diagonal divides a parallelogram into two triangles of equal area.

$\Rightarrow$ $ar\left(ABC\right)=\frac{1}{2}\times ar\left(ABCD\right)$

$=\frac{1}{2}\times 32$

$=16c{m}^2$