NTSE Stage 2 SAT 2019 Exam Question Paper

Let area of square PQRS $=\mathit{A}\mathit{c}{\mathit{m}}^2$
BODM is a parallelogram
M is the mid-point of AD
∴P is also the mid-point of AS (By mid-point theorem)
AP=PS=k[Let]
Draw LT⊥ AP
Δ AMP ≅ Δ LAT
(By RHS congruence criterion)
⇒MP=AT=x
Join LP
It is clear that
ar Δ ALT = ar Δ AMP = ar Δ LAT = ar Δ LPQ=3ar (trap APQL)
Let $\mathit{a}\mathit{r}\left(\mathit{\Delta }\mathrm{AMP}\right)=\mathit{a}\mathit{c}{\mathit{m}}^2$
ar (ΔADN) = 5a
$\Rightarrow \frac{1}{2}\times 10\times 20=5\mathit{a}\Rightarrow \mathit{a}=20{\mathit{c}\mathit{m}}^2$
Now area of square ABCD =16a + area of sq. PQRS
$\Rightarrow 20\times 20=16\left(20\right)+{\mathit{A}}^2$
$\Rightarrow {\mathit{A}}^2=400-320=80{\mathit{c}\mathit{m}}^2$