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Mathematics 6 Online
OpenStudy (anonymous):

What is the length of the diagonal of the square shown below?

OpenStudy (anonymous):

OpenStudy (anonymous):

8

OpenStudy (anonymous):

@hackYou thats not an answer...

OpenStudy (anonymous):

That's the diagonal of the rectangle shown.

OpenStudy (anonymous):

@hackYou my bad i put the wrong directions

OpenStudy (anonymous):

@hackYou What is the length of side s of the square shown below?

OpenStudy (anonymous):

OpenStudy (mertsj):

To find the leg pf a 45-45-90 triangle, divide the hypotenuse by sqrt2

OpenStudy (anonymous):

@Mertsj thanks mate!

OpenStudy (anonymous):

@Mertsj I divided 8 by 2 and I got 4, but the correct answer is 4Root2. Isnt 8 the hyptoenuse?

OpenStudy (mertsj):

Did you divide by 2 or sqrt2 as I said?

OpenStudy (anonymous):

@Mertsj I divided 8 by 2 and clicked 4, and it was wrong -_-

OpenStudy (mertsj):

Do you know that these two numbers are NOT the same? \[2, \sqrt{2}\]

OpenStudy (mertsj):

It is a simple yes or no question.

OpenStudy (anonymous):

@mertsj yes i know those 2 numbers are not the same...I know what i did wrong

OpenStudy (anonymous):

@Mertsj so would this answer be 14√2?

OpenStudy (anonymous):

OpenStudy (mertsj):

\[\frac{8}{\sqrt{2}}=\frac{8}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{8\sqrt{2}}{2}=4\sqrt{2}\]

OpenStudy (anonymous):

@Mertsj ok so what formula did you use to find that?

OpenStudy (mertsj):

Well not you have posted a completely different problem. If you divide the hypotenuse by sqrt2 to get the leg, what should you do to the leg to get the hypotenuse?

OpenStudy (anonymous):

@Mertsj I dont know dude, honestly, I am so lost in class, but I assume you divide it by sqrert2

OpenStudy (mertsj):

|dw:1393206410614:dw|

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