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# I'm stuck? And am I solving it right? :/

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1. So I drew the diagram below to find the surface area of the prism.
which is
S.A: 2(l x 2x) +2(1/2 x 2x x h) +l x 2x
= 6xl+ 2xh

To get h
I calculated the volume which is:
Area of triangle x height

Volume: 1/2 x 2x x h
=xh

so I guess I should make h the subject then substitute it into the surface area equation??
Am I right?

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2. What paper is this from?

So I drew the diagram below to find the surface area of the prism.
which is
S.A: 2(l x 2x) +2(1/2 x 2x x h) +l x 2x
= 6xl+ 2xh

To get h
I calculated the volume which is:
Area of triangle x height

Volume: 1/2 x 2xx h
=xh

so I guess I should make h the subject then substitute it into the surface area equation??
Am I right?

Cut the triangle in half, so the hypotenuse is 2x, the base is x so the height is...
4. (Original post by zetamcfc)
Cut the triangle in half, so the hypotenuse is 2x, the base is x so the height is...
I see, so the height is x therefore the S.A:
2(l x 2x) +2(1/2 x 2xx x) + l x 2x
= 6xl+ 2x2

Now what do I next?
I see, so the height is x therefore the S.A:
2(l x 2x) +2(1/2 x 2xx x) + l x 2x
= 6xl+ 2x2

Now what do I next?
Nope, think Pythagoras, hypotenuse is 2x and the base is x and the height h, so,

(2x)^2 - x^2 = h^2 so what is h.
6. (Original post by zetamcfc)
Nope, think Pythagoras, hypotenuse is 2x and the base is x and the height h, so,

(2x)^2 - x^2 = h^2 so what is h.
Oh yeaah... so
h2= 4x2 - x2
h2= 3x2
h= 3x

Thank you!!

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