Dear Student In △ABC by Pythagoras theoremAC2=AB2+BC2152-92=BC2BC=225-81=144=12Area of quadrilateral= Area of △ABC+BCD72=12×BC×9+12×BC×DE14412=9+DE12-9=DEDE=3 cm We hope that this answer solves your query Regards

Pls solve sum no. 159

Dear Student.

I n △ A B C b y P y t h a g o r a s t h e o r e m A C^{2} = A B^{2} + B C^{2} 15^{2} - 9^{2} = B C^{2} B C = \sqrt{225 - 81} = \sqrt{144} = 12 A r e a o f q u a d r i l a t e r a l = A r e a o f △ A B C + B C D 72 = \frac{1}{2} \times B C \times 9 + \frac{1}{2} \times B C \times D E \frac{144}{12} = 9 + D E 12 - 9 = D E D E = 3 c m

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