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allsm [11]
3 years ago
8

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​)

Mathematics
1 answer:
Gekata [30.6K]3 years ago
4 0

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.

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