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Since ∠ = ° A 90 ∴ By Pythagoras theorem,
In ∆ ABD, BD² = AB² + AD² BD=√ AB ²+AD²=√18²+24² = √ 324+ 576 =√900 = 30 Area of Δ ABD=1/2 (18) (24) =(18) × (12) = 216 m² In ∆ BCD, sides are 30, 40, 50 ⇒ By Pythagoras theorem ∠CBD =90 ° ∴ DC² = BD² + BC ² , ∴ (50)²= (30)²+ (40)² Thus, area of Δ BCD =1/2(40) (30) 600 m ² Hence, area of quadrilateral ABCD = Area of ∆ ABD + area of
∆ BCD = 216 + 600 = 816 m²

See, draw diagonal BD which divide the quad ABCD into two triangle, ΔABD and ΔBDC given,∠A =90° so area of ΔADB = 1/2 ×base ×height =1/2×18×24 =216 m² now using pythagoras' formula, as ΔABD is right anged triangle, BD= √((24)²+ (18)²) =√900 = 30 m since , BD is 30 m then ΔDBC become right angled triangled, ∠B= 90° so, area ΔBDC = 1/2×base ×height =1/2×30×40= 600 m² area of quad ABCD = area ΔABD + area of ΔBDC = 216 +600= 816 m²