Question

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit? 20.6 units 22.7 units 25.6 units 27.6 units

Answer 1

we know that

the perimeter of a polygon is the sum of the length sides

in this problem we have a triangle

so

the polygon has three sides

Let

$A(-5,4)\\B(1,4)\\C(3,-4)$

the perimeter is equal to

$P=AB+BC+AC$

The formula to calculate the distance between two points is equal to

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

Step 1

Find the distance AB

$A(-5,4)\\B(1,4)$

substitutes the values in the formula

$d=\sqrt{(4-4)^{2}+(1+5)^{2}}$

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

$dAB=6\ units$

Step 2

Find the distance BC

$B(1,4)\\C(3,-4)$

substitutes the values in the formula

$d=\sqrt{(-4-4)^{2}+(3-1)^{2}}$

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

$d=\sqrt{68}$

$dBC=8.25\ units$

Step 3

Find the distance AC

$A(-5,4)\\C(3,-4)$

substitutes the values in the formula

$d=\sqrt{(-4-4)^{2}+(3+5)^{2}}$

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

$d=\sqrt{128}$

$dAC=11.31\ units$

Step 4

Find the perimeter

the perimeter is equal to

$P=AB+BC+AC$

substitutes the values

$P=6+8.25+11.31=25.56\ units=25.6\ units$

therefore

the answer is

$25.6\ units$

Answer 2

11.3 plus 6 plus 8.2 = 25.5
 
The perimeter is 25.5