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3 years ago

The length of a rectangle is triple the width. If the perimeter is 104 cm, find the length of the rectangle.

Mathematics

1 answer:

kirill115 [55]3 years ago

5 0

Answer:

Step-by-step explanation:

perimeter of rectangle = 2 ( l+ b )

let width = w length = 3w

according to formula

104 = 2 ( 3w + w )

104 = 2 ( 4 w )

104 = 8 w

104 /8 =w

13 =w

l = 3w

l = 3 ( 13) l = 39 hoping this will be helpful, have amazing day

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Answer:

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Step-by-step explanation:

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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)

Gekata [30.6K]

Answer:

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Step-by-step explanation:

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$p = MN + NP + PQ + QR + RM$ (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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3 years ago