Find the area of a quadrilateral ABCD whose vertices are A(-4, 8), B(-3, -4), C(0, -5) and D(5, 6).

Explanation

Area of quad. ABCD = `ar(Delta ABC) + ar(Delta ACD).`

For `Delta ABC`, we have
`(x_(1) =-4, y_(1) = 8), (x_(2) = -3, y_(2) = -4)"and" (x_(3) =0, y_(3) = -5)`
`therefore ar(DeltaABC) = (1)/(2)|x_(1)(y_(2)-y_(3)) + x_(2)(y_(3) -y_(1)) + x_(3)(y_(1) - y_(2))|`
`=(1)/(2)|(-4) *(-4+5) + (-3) * (-5-8) + 0 *(8+4)|`
`= (1)/(2)|(-4) xx 1 + (-3) xx (-13) +0|`
`=(1)/(2)|-4+39+0| =(35)/(2)` sq units.
For `Delta ACD`, we have
`(x_(1) = -4, y_(1) = 8), (x_(2) =0, y_(2) =-5) "and"(x_(3)= 5, y_(3) = 6).`
`therefore ar(DeltaACD) = (1)/(2)|x_(1)(y_(2) -y_(3)) + x_(2)(y_(3)-y_(1)) + x_(3)(y_(1) -y_(2))|`
` = (1)/(2) |(-4) *(-5-6) + 0 *(6-8) + 5 * (8+5)|`
` = (1)/(2)|(-4) xx (-11) +0 xx (-2) +5 xx 13|`
` = (1)/(2)| 44 + 0 +65| = (109)/(2)` sq unit.
`therefore ar("quad". ABCD) = ar (Delta ABC) + ar(Delta ACD)`
`=((35)/(2) + (109)/(2)) "sq units" = (144)/(2)` sq units
= 72 sq units.
Hence, the area of the given quadrilateral ABCD is 72 sq units.