Question

Triangle ABC has vertices at A(-4, -2), B(1, 7) and C(8, -2)

Answer 1

Answer:

a). Area = 54 square units

b). Perimeter = 33.7 units

Step-by-step explanation:

Vertices of the triangle ABC are A(-4, -2), B(1, 7) and C(8, -2).

(a). Area of the triangle ABC = $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$ (Absolute value)

By substituting the values from the given vertices,

Area = $\frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)]$

        = $\frac{1}{2}[-36+0-72]$

        = $\frac{1}{2}(-108)$

        = (-54) unit²

Therefore, absolute value of the area = 54 square units

(b). Distance between two vertices (a, b) and (c, d)

        d = $\sqrt{(a-c)^{2}+(b-d)^2}$

     AB = $\sqrt{(-4-1)^{2}+(-2-7)^{2}}$

           = $\sqrt{106}$

           = 10.295 units

     BC = $\sqrt{(1-8)^2+(7+2)^2}$

           = $\sqrt{130}$

           = 11.402 units

     AC = $\sqrt{(-4-8)^2+(-2+2)^2}$

           = 12 units

     Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12

                                                                          = 33.697

                                                                          ≈ 33.7 units