Find the area of a quadrilateral with vertices (7,-6) , (3,-2) , (8,-4) , and (-4,-5) .
A) 5.5 square units
B) 38 square units
C) 4.9 square units
D) $40$ square units
E) 47 square units

Question

Quizplus · Accepted Answer

Explanation:Using the distance formula, we can find the lengths of the sides of the quadrilateral:AB = sqrt((7-(-4))^2 + (-6-(-5))^2) = sqrt(11^2 + (-1)^2) = sqrt(122)BC = sqrt((3-8)^2 + (-2-(-4))^2) = sqrt((-5)^2 + 2^2) = sqrt(29)CD = sqrt((8-(-4))^2 + (-4-(-5))^2) = sqrt(12^2 + 1^2) = sqrt(145)DA = sqrt((-4-7)^2 + (-5-(-6))^2) = sqrt((-11)^2 + 1^2) = sqrt(122)Using Heron's formula, we can find the area of the quadrilateral:s = (AB + BC + CD + DA) / 2 = (sqrt(122) + sqrt(29) + sqrt(145) + sqrt(122)) / 2 = 2sqrt(122) + 2sqrt(29) + 2sqrt(145) / 2 = sqrt(122) + sqrt(29) + sqrt(145)Area = sqrt(s(s-AB)(s-BC)(s-CD)) = sqrt((sqrt(122) + sqrt(29) + sqrt(145))(sqrt(122) + sqrt(29) + sqrt(145) - sqrt(122))(sqrt(122) + sqrt(29) + sqrt(145) - sqrt(29))(sqrt(122) + sqrt(29) + sqrt(145) - sqrt(145))) = sqrt((sqrt(122) + sqrt(29) + sqrt(145))(sqrt(29) + sqrt(145))(sqrt(122) + sqrt(145))(sqrt(122) + sqrt(29))) = sqrt((sqrt(122) + sqrt(29))^2(sqrt(145) + sqrt(122))^2) = (sqrt(122) + sqrt(29))(sqrt(145) + sqrt(122)) = sqrt(122*145) + sqrt(122*122) + sqrt(29*145) + sqrt(29*122) = sqrt(17690) + sqrt(14884) + sqrt(4205) + sqrt(3546) = 133 + 122 + 65 + 59 = 379Therefore, the area of the quadrilateral is 379 square units.

Find the Area of a Quadrilateral with Vertices (7,-6) 404040 Square Units E) 47 Square Units

Find the Area of a Quadrilateral with Vertices (7,-6) $40$ Square Units
E) 47 Square Units