3 years ago

Find the perimeter of a square with a side length of 4 minus 2 square root of 6 inches.

1 answer:

8 0

Perimeter of a square = 4s
Required perimeter = $4(4-2\sqrt{6})=16-8\sqrt{16}$

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tigry1 [53]

20 divided by 40=2
20x40=800
800 divided by 40= 20

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ankoles [38]

<h2>Answer:</h2>

$\boxed{\overline{MN}=37.96}$

<h2>Step-by-step explanation:</h2>

For a better understanding of this problem, see the figure below. Our goal is to find $\overline{MN}$ . Since:

$\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm$

and $\overline{MN}$ is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:

$\overline{RN}=\overline{QN}$

By Pythagorean theorem, for triangle NQP:

$\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}$

Applying Pythagorean theorem again, but for triangle MQN:

$\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}$

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velikii [3]

Search I think google have the have answer most of time

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sattari [20]

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lorasvet [3.4K]

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