Find the area of the triangle with the given vertices. $A ( 8 , - 3 , - 3 ) , B ( 8 , - 8 , - 9 ) , C ( - 2,6 , - 6 )$
A) $\sqrt { 2715.25 }$
B) $2 \sqrt { 2715.25 }$
C) $\frac { \sqrt { 2715.25 } } { 2 }$
D)$4 \sqrt { 2715.25 }$
E) $\frac { \sqrt { 2715.25 } } { 4 }$

Question

Find the area of the triangle with the given vertices. $A ( 8 , - 3 , - 3 ) , B ( 8 , - 8 , - 9 ) , C ( - 2,6 , - 6 )$
A) $\sqrt { 2715.25 }$ 
B) $2 \sqrt { 2715.25 }$
C) $\frac { \sqrt { 2715.25 } } { 2 }$
D)$4 \sqrt { 2715.25 }$
E) $\frac { \sqrt { 2715.25 } } { 4 }$

Quizplus · Accepted Answer

Explanation:The area of the triangle with vertices $A ( 8 , - 3 , - 3 ) , B ( 8 , - 8 , - 9 ) , C ( - 2,6 , - 6 )$ is given by the formula:$$Area = \frac{1}{2} \sqrt{(x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2))^2 + (x_1(z_2-z_3) + x_2(z_3-z_1) + x_3(z_1-z_2))^2 + (y_1(z_2-z_3) + y_2(z_3-z_1) + y_3(z_1-z_2))^2}$$where $x_1, y_1, z_1$ are the coordinates of point A, $x_2, y_2, z_2$ are the coordinates of point B, and $x_3, y_3, z_3$ are the coordinates of point C.Plugging in the values of the coordinates, we get:$$Area = \frac{1}{2} \sqrt{(8(-8-(-6)) + 8(6-(-3)) + (-2)(-3-(-8)))^2 + (8(-9-(-6)) + 8(-6-(-3)) + (-2)(-3-(-9)))^2 + (-3(-9-(-6)) + (-8)(6-(-3)) + 6(-3-(-8)))^2}$$$$Area = \frac{1}{2} \sqrt{(8(-2) + 8(9) + (-2)(-5))^2 + (8(-3) + 8(3) + (-2)(6))^2 + (-3(-3) + (-8)(9) + 6(5))^2}$$$$Area = \frac{1}{2} \sqrt{(-16 + 72 + 10)^2 + (0 + 0 + (-12))^2 + (9 + (-72) + 30)^2}$$$$Area = \frac{1}{2} \sqrt{(66)^2 + (-12)^2 + (-33)^2}$$$$Area = \frac{1}{2} \sqrt{4356 + 144 + 1089}$$$$Area = \frac{1}{2} \sqrt{5589}$$$$Area = \frac{\sqrt{5589}}{2}$$$$Area = \sqrt { 2715.25 }$$Therefore, the area of the triangle is $\sqrt { 2715.25 }$.

Find the Area of the Triangle with the Given Vertices A(8,−3,−3),B(8,−8,−9),C(−2,6,−6)A ( 8 , - 3 , - 3 ) , B ( 8 , - 8 , - 9 ) , C ( - 2,6 , - 6 )A(8,−3,−3),B(8,−8,−9),C(−2,6,−6)

Find the Area of the Triangle with the Given Vertices $A ( 8 , - 3 , - 3 ) , B ( 8 , - 8 , - 9 ) , C ( - 2,6 , - 6 )$