If A (-5, 7),B (-4, -5),C (-1, -6),and D (4, | vedclass

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(C) To find the area of the quadrilateral $ABCD$,we can divide it into two triangles by drawing a diagonal $BD$.The area of a triangle with vertices $

Vedclass · Accepted Answer

$72$

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(C) To find the area of the quadrilateral $ABCD$,we can divide it into two triangles by drawing a diagonal $BD$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.$1$. Area of $\Delta ABD$ with vertices $A(-5, 7)$,$B(-4, -5)$,and $D(4, 5)$:$	ext{Area} = \frac{1}{2} |-5(-5 - 5) + (-4)(5 - 7) + 4(7 - (-5))|$$= \frac{1}{2} |-5(-10) - 4(-2) + 4(12)|$$= \frac{1}{2} |50 + 8 + 48| = \frac{106}{2} = 53$ square units.$2$. Area of $\Delta BCD$ with vertices $B(-4, -5)$,$C(-1, -6)$,and $D(4, 5)$:$	ext{Area} = \frac{1}{2} |-4(-6 - 5) + (-1)(5 - (-5)) + 4(-5 - (-6))|$$= \frac{1}{2} |-4(-11) - 1(10) + 4(1)|$$= \frac{1}{2} |44 - 10 + 4| = \frac{38}{2} = 19$ square units.Total area of quadrilateral $ABCD = 	ext{Area}(\Delta ABD) + 	ext{Area}(\Delta BCD) = 53 + 19 = 72$ square units.

If $A (-5, 7)$,$B (-4, -5)$,$C (-1, -6)$,and $D (4, 5)$ are the vertices of a quadrilateral,find the area of the quadrilateral $ABCD$ (in square units).

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