Question

What is the perimeter of △LMN?

Answer 1

we know that

The distance 's formula between two points is equal to

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

Step $1$

Find the distance MN

$M(-2,1)\\N(-1,4)$

Substitute in the distance's formula

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

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

$dMN=\sqrt{10}\ units$

Step $2$

Find the distance NL

$N(-1,4)\\L(2,4)$

Substitute in the distance's formula

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

$d=\sqrt{(0)^{2}+(3)^{2}}$

$dNL=3\ units$

Step $3$

Find the distance LM

$L(2,4)\\M(-2,1)$

Substitute in the distance's formula

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

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

$dLM=5\ units$

Step $4$

Find the perimeter of the triangle LMN

we know that

The perimeter of a triangle is equal to the sum of the three length sides

In this problem

$Perimeter=MN+NL+LM$

substitute the values in the formula

$Perimeter=(\sqrt{10}+3+5)\ units$

$Perimeter=(8+\sqrt{10})\ units$

therefore

the answer is the option D

the perimeter of the triangle LMN is equal to $(8+\sqrt{10})\ units$

Answer 2

I  solved this via the Pythagorean theorem. Side LM is 5 and side NM is 3.16. Side NL is also 3. 3+5+3.16=11.16. The decimal form of 8+sqrt(10) is 11.16. D is the answe