Let area of square PQRS =Acm2 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 ar(Δ‌AMP)=acm2 ar (ΔADN) = 5a ⇒‌‌
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×10×20=5a⇒a=20cm2 Now area of square ABCD =16a + area of sq. PQRS ⇒20×20=16(20)+A2 ⇒A2=400−320=80cm2