Answer:
The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). The perimeter of ∆ABC is 30 units, and its area is 2 square units.
Answer:
Perimeter is 32.4 unts
Area is 30.1 square units.
Step-by-step 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,
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]AB=\sqrt{(16-2)^2+(2-8)^2}=\sqrt{196+36}=\sqrt{232}=15.2units[/tex]
[tex]AC=\sqrt{(6-2)^2+(2-8)^2}=\sqrt{16+36}=\sqrt{52}=7.2units[/tex]
[tex]BC=\sqrt{(6-16)^2+(2-2)^2}=\sqrt{100+0}=\sqrt{100}=10units[/tex]
Perimeter=AB+BC+AC=15.2+7.2+10=32.4 units
Now, we find the area of triangle
[tex]semiperimeter=\frac{a+b+c}{2}=\frac{32.4}{2}=16.2 units[/tex]
By Heron's formula
[tex]Area=\sqrt{s(s-a)(s-b)(s-c)}\\\\=\sqrt{16.2(16.2-15.2)(16.2-7.2)(16.2-10)}\\\\=\sqrt{16.2(1)(9)(6.2)}\\\\=\sqrt{903.96}=30.0659\sim30.1units^2[/tex]
Area is 30.1 square units.
