Question

Find the perimeter of the pentagon MNPQR with vertices ​M(2​, 4​), ​N(5​, 8​), ​P(​8, 4​), ​Q(8​, 1​), and ​R(2​, 1​)

Answer 1

Answer:

The pentagon MNPQR has a perimeter of 22 units.

Step-by-step explanation:

Geometrically speaking, the perimeter of the pentagon is the sum of the lengths of each side, that is:

$p = MN + NP + PQ + QR + RM$ (1)

$p = \sqrt{\overrightarrow{MN}\,\bullet \, \overrightarrow{MN}} + \sqrt{\overrightarrow{NP}\,\bullet \, \overrightarrow{NP}} + \sqrt{\overrightarrow{PQ}\,\bullet \, \overrightarrow{PQ}} + \sqrt{\overrightarrow{QR}\,\bullet \, \overrightarrow{QR}} + \sqrt{\overrightarrow{RM}\,\bullet \, \overrightarrow{RM}}$ (1b)

If we know that $M(x,y) = (2,4)$, $N(x,y) = (5,8)$, $P(x,y) = (8,4)$, $Q(x,y) = (8,1)$ and $R(x,y) = (2,1)$, then the perimeter of the pentagon MNPQR is:

$p =\sqrt{(5-2)^{2}+(8-4)^{2}} + \sqrt{(8-5)^{2}+(4-8)^{2}}+\sqrt{(8-8)^{2}+(1-4)^{2}}+\sqrt{(2-8)^{2}+(1-1)^{2}}+\sqrt{(2-2)^{2}+(4-1)^{2}}$$p = \sqrt{3^{2}+4^{2}} + \sqrt{3^{2}+(-4)^{2}}+\sqrt{0^{2}+(-3)^{2}}+\sqrt{(-6)^{2}+0^{2}}+\sqrt{0^{2}+3^{2}}$

$p = 22$

The pentagon MNPQR has a perimeter of 22 units.