Draw a quadrilateral in the Cartesian plane,w | vedclass

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(A) Let $ABCD$ be the given quadrilateral with vertices $A(-4, 5), B(0, 7), C(5, -5),$ and $D(-4, -2).$To find the area of quadrilateral $ABCD$,we div

Vedclass · Accepted Answer

$121/2 	ext{ unit}^2$

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(A) Let $ABCD$ be the given quadrilateral with vertices $A(-4, 5), B(0, 7), C(5, -5),$ and $D(-4, -2).$To find the area of quadrilateral $ABCD$,we divide it into two triangles by drawing the diagonal $AC$.Area $(ABCD) = 	ext{Area}(\Delta ABC) + 	ext{Area}(\Delta ACD).$The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|.$For $\Delta ABC$ with vertices $A(-4, 5), B(0, 7), C(5, -5):$Area $(\Delta ABC) = \frac{1}{2} |-4(7 - (-5)) + 0(-5 - 5) + 5(5 - 7)|$$= \frac{1}{2} |-4(12) + 0(-10) + 5(-2)|$$= \frac{1}{2} |-48 - 10| = \frac{1}{2} |-58| = 29 	ext{ unit}^2.$For $\Delta ACD$ with vertices $A(-4, 5), C(5, -5), D(-4, -2):$Area $(\Delta ACD) = \frac{1}{2} |-4(-5 - (-2)) + 5(-2 - 5) + (-4)(5 - (-5))|$$= \frac{1}{2} |-4(-3) + 5(-7) - 4(10)|$$= \frac{1}{2} |12 - 35 - 40| = \frac{1}{2} |-63| = 31.5 	ext{ unit}^2.$Total Area $(ABCD) = 29 + 31.5 = 60.5 = \frac{121}{2} 	ext{ unit}^2.$

Draw a quadrilateral in the Cartesian plane,whose vertices are $(-4, 5), (0, 7), (5, -5),$ and $(-4, -2).$ Also,find its area.

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