can you find it....-3

It's a simple problem first take $BC =x$ , then $AT =\dfrac x2$ . And we know that $\boxed{\text{Area of trapezium}=\dfrac{\text{Sum of parallel sides} \times \text{ Height}}{2}}$

Take height as $y$ and area is given 45 $\text{ cm}^2$ . Therefore $xy=60$ . And half the area is triangle ACD, hence its area is $60/2=30.\square$

$\triangle ABC+\triangle ACT=45$

Since $T$ is the midpoint of $AD$ , $\triangle ADC=2\times \triangle ATC$

But $\triangle ABC=\triangle ADC$ . It follows that $\triangle ABC=2\times \triangle ATC$

$\large\therefore$ $2\times \triangle ATC+\triangle ATC=45$ . It follows that $3\times \triangle ATC=45$ . From here $\triangle ATC=15$ .

Finally,

$\triangle ADC=2\times 15=$ $\boxed{\color{#3D99F6}30}$ $\boxed{\text{answer}}$