Find the length of the diagonal of a cube of side 5m

Mathematics

2 answers:

Law Incorporation [45]2 years ago

7 0

Let the length of diagonal be <em>d</em>

Given, Side = 5m

Using Pythagoras theorum,

$\sqrt{ 3} (side) = d \\ \sqrt{3}(5) = d \\ d = 5\sqrt{3} m = 8.66 \: m$

<em>Hope </em><em>this</em><em> </em><em>helps </em><em>:</em><em>)</em>

Darina [25.2K]2 years ago

5 0

<h3>SOLUTION:</h3>

diagonal-

= root 3a

= root 3·5

≈ 8.66025

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4 years ago

Please help me <br> Show your work <br> 10 points

Svet_ta [14]

<h2>Answer</h2>

After the dilation $\frac{5}{3}$ around the center of dilation (2, -2), our triangle will have coordinates:

$R'=(2,3)$

$S'=(2,-2)$

$T'=(-3,-2)$

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

$(x,y)$ → $(x-2, y+2)$

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor $\frac{5}{3}$ . Therefore our second partial rule will be:

$(x,y)$ → $\frac{5}{3} (x-2,y+2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )$

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)$