Question

If a 2.5 g sample of gold is hammered into an equilateral triangle .28mm thick, how long are the sides?

Answer 1

I think I know what to do, but I would still like some extra help!

Answer 2

Someone please help?

Answer 3

@FaiqRaees

Answer 4

@Redcan do you think you could help me?

Answer 5

\[\large\rm Volume=\frac{base*height*0.28}{2} \]\[\large\rm Volume=mass~in~grams\]

Answer 6

Really? It's that simple? Thank you!

Answer 7

So the total volume would be the lenght of the sides? @FaiqRaees

Answer 8

the base will be length of side

Answer 9

How would I find that?

Answer 10

Um... Okay, let me figure this out.

Answer 11

how would the 2.5g tie in with this?

Answer 12

just find the value of hieght in terms of x

Answer 13

This is a density and volume related problem

Answer 14

no density isn't given

Answer 15

So how could I tie in the 2.5g and the .28mm of thickness with this?

Answer 16

just please find out the value of height, let me take care of the rest

Answer 17

Ok one second

Answer 18

Wait, how exactly would I do that? I don't have any values for x

Answer 19

just in terms of x

Answer 20

\[x^{2}\]

Answer 21

I have to go, but I will post again later. Thanks for the help!

Answer 22

No the height will be
\[\large\rm Hypotenuse^2=Perpendicular^2+Height^2 \]\[\large\rm Height^2=Hypotenuse^2-Perpendicular^2\]\[\large\rm Height^2=x^2-(x/2)^2\]
\[\large\rm Height=\sqrt{x^2-(x/2)^2}\]

Answer 23

got it?

Answer 24

\[\large\rm Volume=\frac{base*height*0.28}{2} \]\[\large\rm 2.5=\frac{x*\sqrt{x^2-(x/2)^2}*0.28}{2} \]\[\large\rm 5=x*\sqrt{x^2-(x/2)^2}*0.28 \]\[\large\rm 17.85=x*\sqrt{x^2-(x/2)^2}\]\[\large\rm 17.85^2=(x*\sqrt{x^2-(x/2)^2})^2\]\[\large\rm 318.9=x^2*(x^2-(x/2)^2)\]\[\large\rm 318.9=x^2*(x^2-x^2/4)\]\[\large\rm 318.9=x^4-x^4/4\]\[\large\rm 318.9=3x^4/4\]\[\large\rm 425.5=x^4\]\[\large\rm 4.54=x\]

Answer 25

length of side = 4.54mm