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Sorry, could not draw the figure here. Steps to draw the figure-Draw IIgm PQRS with PS and QR as base. Join PR and QS. Now the solution is as follows:

Given: In thr figure, PQRS is a parallelogram, Diagonal PR and QS meets at point E. ar(ΔSEP)+ar(ΔQER)=12cm² To find: area of parallelogram PQRS Construction: Draw perpedicular AB passing through E Solution: In parallelogram PQRS SP=QR and SPIIQR ..(1) ar(ΔSEP)+ar(ΔQER)=12cm² 1/2×PS×AE+1/2×BE×QR=12cm² (area of Δ=base×height, AE is now height of ΔSEP and BE of ΔQER) 1/2×QR×AE+1/2×BE×QR=12cm² (from (1) ) 1/2QR(AE+BE)=12cm² QR×AB=24cm² QR and AB are the base and height of IIgm PQRS Therefore, area of IIgm PQRS= 24cm²