Question

He 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

Answer:

Note :

The perimeter is :  AB + BC + CA

AB =√((XB - XA)² + (YB - YA)²) = √((16-2)²+(2-8)²) = √(12²+6²) = √(144+36)=√180

AB = √(36×5) = 6√5 units

same method for :  BC ; CA   ..........

Answer 2

Answer:

Perimeter = 32.44 unit

Area of triangle = 30 unit²

Step-by-step explanation:

By distance formula distance between (a,b) and (c,d) is $\sqrt{(c-a)^2+(d-b)^2}$

So, we have

      $AB=\sqrt{(16-2)^2+(2-8)^2}=15.23\\\\BC=\sqrt{(16-6)^2+(2-2)^2}=10\\\\AC=\sqrt{(2-6)^2+(8-2)^2}=7.21$

Perimeter = 15.23 + 10 + 7.21 = 32.44 unit

We have $s=\frac{15.23+10+7.21}{2}=16.22$

Area of triangle$=\sqrt{s(s-AB)(s-BC)(s-AC)}=\sqrt{16.22(16.22-15.23)(16.22-10)(16.22-7.21)}=30$

Area of triangle = 30 unit²