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(B) Let the vertices of the quadrilateral be $A(-4,-2), B(-3,-5), C(3,-2),$ and $D(2,3)$. Join $AC$ to form two triangles $	riangle ABC$ and $	riang

Vedclass · Accepted Answer

$28$

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(B) Let the vertices of the quadrilateral be $A(-4,-2), B(-3,-5), C(3,-2),$ and $D(2,3)$. Join $AC$ to form two triangles $	riangle ABC$ and $	riangle ACD$.Area of a triangle $= \frac{1}{2} |x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$Area of $	riangle ABC = \frac{1}{2} |(-4){(-5)-(-2)} + (-3){(-2)-(-2)} + 3{(-2)-(-5)}|$$= \frac{1}{2} |(-4)(-3) + (-3)(0) + 3(3)| = \frac{1}{2} |12 + 0 + 9| = \frac{21}{2} = 10.5$ square units.Area of $	riangle ACD = \frac{1}{2} |(-4){(-2)-(3)} + 3{(3)-(-2)} + 2{(-2)-(-2)}|$$= \frac{1}{2} |(-4)(-5) + 3(5) + 2(0)| = \frac{1}{2} |20 + 15 + 0| = \frac{35}{2} = 17.5$ square units.Area of quadrilateral $ABCD = 	ext{Area of } 	riangle ABC + 	ext{Area of } 	riangle ACD$$= 10.5 + 17.5 = 28$ square units.

Find the area of the quadrilateral whose vertices,taken in order,are $(-4,-2), (-3,-5), (3,-2),$ and $(2,3)$ (in square units).

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