Find the area of the quadrilateral ABCD in which AB = 9 m, BC = 40 m, `angle ABC = 90^(@)`, CD = 15 m and AD = 28 m.

Explanation

Area of right `Delta`ABC
`=((1)/(2)xx "base" xx "height")`
`=((1)/(2) xx 40 xx 9)m^(2) = 180 m^(2)`.
Also, from right `Delta`ABC, we get
`AC^(2) = AB^(2) + BC^(2)`
`=[(9)^(2)+(40)^(2)]m^(2)`
`=(81 + 1600)m^(2) = 1681 m^(2)`
`rArr" "AC = sqrt(1681)m = 41 m`
In `Delta`ACD, we have AC = 41 m, CD = 15 m and AD = 28 m.
Let a = 41 m, b = 15 m and c = 28 m. Then,
`s=(1)/(2)(41 + 15 + 28)m=((1)/(2) xx 84)m = 42m`.
`therefore" "(s-a)=(42-41)m = 1m, (s-b)=(42-15)m = 27 m and (s-c) = (42-28)m = 14 m`.
`therefore" ""area of "Delta ACD = sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(42 xx 1 xx 27 xx 14)m^(2)`
`= (14 xx 9)m^(2) = 126 m^(2)`.
Area of the quad. ABCD = area(`Delta`ABC) + area(`Delta`ACD)
`=(180 + 126)m^(2) = 306 m^(2)`.