CTET 2 Math and Science 07 Jul 2024 Paper

$\mathit{B}\mathit{C}=48\mathit{c}\mathit{m},\mathit{C}\mathit{D}=17\mathit{c}\mathit{m},\mathit{A}\mathit{D}=40\mathit{c}\mathit{m}$
$\angle \mathit{A}=\angle \mathit{B}={90}^{\circ }$
Formula used:
Area of rectangle $=$ Length $\times$ Breadth
Area of right-angled triangle $=1∕2\times$ Base $\times$ Perpendicular

Calculation:
In $△\mathit{D}\mathit{E}\mathit{C}$,
Perpendicular $\left(\mathit{D}\mathit{E}\right)=\sqrt{}\left(\mathit{D}{\mathit{C}}^2-\mathit{E}{\mathit{C}}^2\right)$
$\Rightarrow \mathit{D}\mathit{E}=\sqrt{}\left({17}^2-8^2\right)$
$\Rightarrow \mathit{D}\mathit{E}=\sqrt{}\left(289-64\right)$
$\Rightarrow \mathit{D}\mathit{E}=\sqrt{}225=15\mathit{c}\mathit{m}$
Now,
Area of $△\mathit{D}\mathit{E}\mathit{C}$
$\Rightarrow 1∕2\times \mathit{E}\mathit{C}\times \mathit{D}\mathit{E}$
$\Rightarrow 1∕2\times 8\times 15$
$\Rightarrow 1∕2\times 120=60\mathit{c}{\mathit{m}}^2$
Now, $\mathit{A}\mathit{B}=\mathit{D}\mathit{E}=15\mathit{c}\mathit{m}$
Area of rectangle ABED
$\Rightarrow \mathit{A}\mathit{B}\times \mathit{A}\mathit{D}$
$\Rightarrow 15\times 40=600\mathit{c}{\mathit{m}}^2$
Thus,
Area of quadrilateral ABCD

$\Rightarrow 60+600$
$\Rightarrow 660\mathit{c}{\mathit{m}}^2$