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

You can use the fact that perimeter of a triangle is sum of its sides' lengths.

The perimeter of the specified triangle is 22.81 units

The area of the specified triangle is $24 \: \rm unit^2$

How to find the perimeter of a figure?

To find the perimeter of a figure, get its sides' lengths and add them. This is the perimeter of that given figure.

How to find the perimeter of the given figure using coordinates?

We can find the distance between two points and that will represent lengths of sides of triangle.

Suppose we have A(x,y) and B(a,b). The distance between A and B is the length of line segment AB and its obtained by

$|AB| = \sqrt{(x-b)^2 + (y-b)^2}$

Thus, using above distance formula, as we have

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

$|AB| = \sqrt{(-2-6)^2 + (2-2)^2} = 8\: \rm units\\\\|AC| = \sqrt{(6-0)^2 + (2-8)^2} = 6\sqrt{2} \approx 8.485\: \rm units\\\\|BC| = \sqrt{(-2-0)^2 + (2-8)^2} = 2\sqrt{10} \approx 6.325\: \rm units$

Thus, the perimeter of the given triangle is

$|AB| + |BC| + |AC| \approx 22.81 \: \rm units$

The formula for area of triangle from its vertices' coordinates is

$Area = (1/2) \times | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

Since we have

$(x_1, y_1) = (-2,2)\\(x_2, y_2) = (6, 2)\\(x_3, y_3) = (0, 8)$

$Area = (1/2) \times | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |\\Area = (1/2) \times | -2(2 - 8) + 6(8-2) + 8(2-2) | = (1/2) \times 48 = 24 \rm \: unit^2$

Thus,

The perimeter of the specified triangle is 22.81 units

The area of the specified triangle is $24 \: \rm unit^2$

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