121k views
0 votes
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​)

User Louis XIV
by
3.4k points

2 Answers

1 vote

answer

The pentagon MNPQR has a perimeter of 22 units.

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

p = MN + NP + PQ + QR + RMp=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}}p=

MN

MN

+

NP

NP

+

PQ

PQ

+

QR

QR

+

RM

RM

(1b)

If we know that M(x,y) = (2,4)M(x,y)=(2,4) , N(x,y) = (5,8)N(x,y)=(5,8) , P(x,y) = (8,4)P(x,y)=(8,4) , Q(x,y) = (8,1)Q(x,y)=(8,1) and R(x,y) = (2,1)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=

(5−2)

2

+(8−4)

2

+

(8−5)

2

+(4−8)

2

+

(8−8)

2

+(1−4)

2

+

(2−8)

2

+(1−1)

2

+

(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=

3

2

+4

2

+

3

2

+(−4)

2

+

0

2

+(−3)

2

+

(−6)

2

+0

2

+

0

2

+3

2

p = 22p=22

The pentagon MNPQR has a perimeter of 22 units.

User Hank
by
3.0k points
2 votes

Answer:

The pentagon MNPQR has a perimeter of 22 units.

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.

User Ashutosh Patole
by
3.8k points