Question

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). The perimeter of ∆ABC is ______units, and its area is ______ square units.

Answer 1

Perimeter is 32.4 unts

Area is 30.1 square units.

Explanation:

Given the vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). we have to find the area and perimeter of triangle ABC.

By distance formula,

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

`AB=√((16-2)^2+(2-8)^2)=√(196+36)=√(232)=15.2units`

`AC=√((6-2)^2+(2-8)^2)=√(16+36)=√(52)=7.2units`

`BC=√((6-16)^2+(2-2)^2)=√(100+0)=√(100)=10units`

Perimeter=AB+BC+AC=15.2+7.2+10=32.4 units

Now, we find the area of triangle

`semiperimeter=(a+b+c)/(2)=(32.4)/(2)=16.2 units`

By Heron's formula

`Area=√(s(s-a)(s-b)(s-c))\\\\=√(16.2(16.2-15.2)(16.2-7.2)(16.2-10))\\\\=√(16.2(1)(9)(6.2))\\\\=√(903.96)=30.0659\sim30.1units^2`

Area is 30.1 square units.