Question

please answer this question​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, ABCD is a rectangle such that CD = 24 cm and BC = 18 cm.

So,

$\rm \: Area_{(ABCD)} = BC \times CD$

$\rm \: = \: 24 \times 18$

$\rm \: = \: 432 \: {cm}^{2}$

$\color{green}\rm\implies \:\boxed{ \rm{ \:Area_{(ABCD)} \: = \: 432 \: {cm}^{2} \: }}$

Now,

BD is a diagonal of rectangle ABCD

We know, Diagonal divides parallelogram in to two triangle of equal areas.

So, it means

$\rm \: Area_{(\triangle\:ABD)} \: = \: \frac{1}{2} \times Area_{(ABCD)} \:$

$\rm \: Area_{(\triangle\:ABD)} \: = \: \frac{1}{2} \times 432 \:$

$\color{green}\rm\implies \:\boxed{ \rm{ \:Area_{(\triangle\:ABD)} \: = \: 216 \: {cm}^{2} \: \: }}$

Further, given that E is the midpoint of BD.

So, it means AE is median of triangle ABD

We know, median divides triangle in to two triangle of equal areas.

So, it means

$\rm \: Area_{(\triangle\:ABE)} = \frac{1}{2} \times Area_{(\triangle\:ABD)}$

$\rm \: Area_{(\triangle\:ABE)} = \frac{1}{2} \times 216$

$\color{green}\rm\implies \:\boxed{ \rm{ \:Area_{(\triangle\:ABE)} \: = \: 108 \: {cm}^{2} \: \: }}$

So, option (a) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$