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 = [tex]\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})][/tex] (Absolute value)
By substituting the values from the given vertices,
Area = [tex]\frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)][/tex]
= [tex]\frac{1}{2}[-36+0-72][/tex]
= [tex]\frac{1}{2}(-108)[/tex]
= (-54) unit²
Therefore, absolute value of the area = 54 square units
(b). Distance between two vertices (a, b) and (c, d)
d = [tex]\sqrt{(a-c)^{2}+(b-d)^2}[/tex]
AB = [tex]\sqrt{(-4-1)^{2}+(-2-7)^{2}}[/tex]
= [tex]\sqrt{106}[/tex]
= 10.295 units
BC = [tex]\sqrt{(1-8)^2+(7+2)^2}[/tex]
= [tex]\sqrt{130}[/tex]
= 11.402 units
AC = [tex]\sqrt{(-4-8)^2+(-2+2)^2}[/tex]
= 12 units
Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12
= 33.697
≈ 33.7 units
