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.

The  quadrilateral ABCD is divided into two triangles . ABD and BCD.

⇒  ABD is a right angle triangle .

⇒ Area =( 1/2) (AB) (AD) as Angle A is 90 ,

⇒   Ab is the altitude and AD is the base = $\\ 1/2\times 18\times 24=216$

$\\ \Rightarrow BC^2=AB^2+ AD^2\Rightarrow BC=30$

BCD triangle

S = semi perimeter=60

⇒ AREA = $\\\sqrt{60\times 30\times 20\times 10}=600$

Total area=$\\816cm2$

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