Question

A park, in the shape of a quadrilateral ABCD, has Z C = 90°. AB = 9 m, BC = 12m
CD=5 m and AD = 8 m. How much area does it occupy?​

Answer 1

Step-by-step explanation:

Connect D. to B to get hypotenuse for right angled triangle BCD. DB = Sqrt (CB^2+CD^2) = Sqrt (169)=13m

NOW area of ∆BCD = 1/2*12*5=30sq.m

For ∆ABD, we now have all three sides. So, using formula sqrt(s*(s-DB)*(s-AD)*(s-AB) where s= (DB+AD+AB)/2 = 15m

Area ∆ABD = sqrt (15*2*7*6)= 35.496 sq m

So now area of quadrilateral ABCD = ∆BCD + ∆ABD= 65.496 sq.m