Question

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

A) Determine the exact value of the area of Triangle ABC.

B) Determine the perimeter of Triangle ABC to the nearest tenth.

Answer 1

a). Area = 54 square units

b). Perimeter = 33.7 units

Explanation:

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

(a). Area of the triangle ABC =
`(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 =
`(1)/(2)[(-4)(7+2)+(1)(-2+2)+8(-2-7)]`

=
`(1)/(2)[-36+0-72]`

=
`(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)}`

=
`√(106)`

= 10.295 units

BC =
`√((1-8)^2+(7+2)^2)`

=
`√(130)`

= 11.402 units

AC =
`√((-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