Answer:
Perimeter = 32.44 units
Area = 30 square units
Step-by-step explanation:
Given
Vertices
A(2,8), B(16,2) and C(6,2)
WE have to determine the lengths of all sides before finding the perimeter and area.
The formula of modulus is:
[tex]d = \sqrt{(x_{2}- x_{1})^{2} +(y_{2}-y_{1})^{2}}\\AB=\sqrt{(16-2)^{2} +(2-8)^{2}}\\=\sqrt{(14)^{2} +(-6)^{2}}\\=\sqrt{196+36}\\ =\sqrt{232}\\=15.23\\\\BC=\sqrt{(6-16)^{2} +(2-2)^{2}}\\=\sqrt{(-10)^{2} +(0)^{2}}\\=\sqrt{100+0}\\ =\sqrt{100}\\=10\\\\AC=\sqrt{(6-2)^{2} +(2-8)^{2}}\\=\sqrt{(4)^{2} +(-6)^{2}}\\=\sqrt{16+36}\\ =\sqrt{52}\\=7.21\\\\[/tex]
So the perimeter is:
[tex]Perimeter=AB+BC+AC\\=15.23+10+7.21\\=32.44\ units[/tex]
Using hero's formula,
[tex]s=\frac{perimeter}{2}\\s=\frac{32.44}{2}\\ s=16.22\\Area=\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{16.22(16.22-15.23)(16.22-10)(16.22-7.21)}\\=\sqrt{(16.22)(0.99)(6.22)(9.01)}\\=\sqrt{899.91}\\=29.99\ square\ units[/tex]
Rounding off will give us 30 square units ..
