Find the area of a pentagon having the vertic | vedclass

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(B) Let the vertices of the pentagon be $A(4,3), B(-5,6), C(-7,-2), D(0,-7),$ and $E(3,-6)$.We can divide the pentagon into three triangles: $\Delta A

Vedclass · Accepted Answer

$97$

Vedclass · Answer

(B) Let the vertices of the pentagon be $A(4,3), B(-5,6), C(-7,-2), D(0,-7),$ and $E(3,-6)$.We can divide the pentagon into three triangles: $\Delta ABE, \Delta BCE,$ and $\Delta CDE$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.$1$. Area of $\Delta ABE$ with vertices $A(4,3), B(-5,6), E(3,-6)$:$= \frac{1}{2} |4(6 - (-6)) + (-5)(-6 - 3) + 3(3 - 6)|$$= \frac{1}{2} |4(12) + (-5)(-9) + 3(-3)|$$= \frac{1}{2} |48 + 45 - 9| = \frac{1}{2} |84| = 42$ sq units.$2$. Area of $\Delta BCE$ with vertices $B(-5,6), C(-7,-2), E(3,-6)$:$= \frac{1}{2} |-5(-2 - (-6)) + (-7)(-6 - 6) + 3(6 - (-2))|$$= \frac{1}{2} |-5(4) + (-7)(-12) + 3(8)|$$= \frac{1}{2} |-20 + 84 + 24| = \frac{1}{2} |88| = 44$ sq units.$3$. Area of $\Delta CDE$ with vertices $C(-7,-2), D(0,-7), E(3,-6)$:$= \frac{1}{2} |-7(-7 - (-6)) + 0(-6 - (-2)) + 3(-2 - (-7))|$$= \frac{1}{2} |-7(-1) + 0 + 3(5)|$$= \frac{1}{2} |7 + 15| = \frac{1}{2} |22| = 11$ sq units.Total area of pentagon $ABCDE = 	ext{Area}(\Delta ABE) + 	ext{Area}(\Delta BCE) + 	ext{Area}(\Delta CDE)$$= 42 + 44 + 11 = 97$ sq units.

Find the area of a pentagon having the vertices $(4,3), (-5,6), (-7,-2), (0,-7),$ and $(3,-6)$.

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