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Mathematics 20 Online
Subshilava:

Please help me with this question posted below

Subshilava:

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toga:

@midnight97

toga:

@snowie

snowie:

@oliver69

toga:

@axie

Luigi0210:

Using this as a reference would be very helpful.

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Luigi0210:

The first one for example, it is a 45-45-90 triangle, meaning that the sides would be the same. The hypotenuse would be multiplied by \(\sqrt{2} \) however

Luigi0210:

Soo the first one, to find x, it would be \(\large (8\sqrt{2})*\sqrt{2} =(8*2)=?\)

Subshilava:

16?

Luigi0210:

yess

Luigi0210:

And for the 30-60, they gave you the side as 6, to find the hypotenuse, just multiply that by 2 So, \(\large 2*(6) \)

Subshilava:

12?

Luigi0210:

yeess

Luigi0210:

You just gotta follow the measurements for 30-60-90 and 45-45-90 triangles

Subshilava:

@luigi0210 wrote:
You just gotta follow the measurements for 30-60-90 and 45-45-90 triangles
1 Ok so how would I set up the one squared I know the other one 60*16

Subshilava:

Well it wouldn’t be 60*16

Luigi0210:

@subshilava wrote:
Well it wouldn’t be 60*16
You're right, it wouldn't be. You're given the hypotenuse instead, so you have to do a little backwards work. You have it as \(\color{red}{7\sqrt{2}}\) If you look at the reference, the general measurement is \(\color{yellow}{x\sqrt{2}} \) To get just "x" you would have to divide: \[\Large \frac{x\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{\sqrt2}\] Giving you: \[\Large x=7\] So your x in this case is 7

Subshilava:

@luigi0210 wrote:
@subshilava wrote:
Well it wouldn’t be 60*16
You're right, it wouldn't be. You're given the hypotenuse instead, so you have to do a little backwards work. You have it as \(\color{red}{7\sqrt{2}}\) If you look at the reference, the general measurement is \(\color{yellow}{x\sqrt{2}} \) To get just "x" you would have to divide: \[\Large \frac{x\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{\sqrt2}\] Giving you: \[\Large x=7\] So your x in this case is 7
So why do they add other numbers of the angle? Instead of just the one you need?

Luigi0210:

The angles are just so you know which triangle you are dealing with, different triangles have different measurements

Subshilava:

So for the last one 8 squared 3

Subshilava:

@luigi0210 wrote:
You actually got the right answer, but for the wrong side. If you lined it up with the referenced, you would see you just had to divide by 2 to get x. So it was just 8
Oh okay tysm for helping me

Luigi0210:

Of course, good luck with mathh

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