# A field is in a shape of a quadrilateral ABCD, in which AB= 18m, AD= 24m, BC= 40m, DC= 50m and angle A= 90 degree. Find the area of the field.

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∴ 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

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