Question

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

Answer 1

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

Find the distance AB

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

Find the distance BC

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

Find the distance AC

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

Find the distance DC

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

Find the perimeter of the triangle

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

Find the area of the triangle

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

Answer 2

Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.

The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.

Given A(2,8),B(16,2),C(6,2)
we calculate 
AB=sqrt((16-2)^2+(2-8)^2)=sqrt(14^2+6^2)=sqrt(232);
BC=sqrt((6-16)^2+(2-2)^2)=sqrt(10^2+0)=10
CA=sqrt((2-6)^2+(8-2)^2)=sqrt(4^2+6^2)=sqrt(16+36)=sqrt(52)

Perimeter=AB+BC+CA=32.443 units

For the area, we note that BC is horizontal (parallel to the x-axis), so 
area = (1/2)bast * height
=(1/2)10*(ya-yb)
=(1/2)10*(8-2)
=(1/2)10*6
=30 unit^2