- anonymous

Can someone help me with Surface area of pyramids and cones

- jamiebookeater

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- anonymous

Find the slant height of the regular pyramid or cone

##### 3 Attachments

- anonymous

Do you know what pathagorean therom is?

- anonymous

yeah isnt it a^2+b^2=c^2

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## More answers

- anonymous

Ok so find the length of the hypotenuse of the pyramid (for the first one)

- anonymous

so 12^2+15^2=369

- anonymous

369=c^2 you still need to get rid of that exponent.. so you get \[3\sqrt{41}\]. Now that you have all the lengths apply the forumla for area of a triangle. Area = 1/2 * base * height

- anonymous

Because you have 4 triangles you could change your forumla to reflect that A=4(1/2*b*h)

- anonymous

Then don't forget the add the area of the base of the pyramid itself... width * height. Add them both up, and that's surface area.

- anonymous

Get it?

- anonymous

so 4(1/2*144*15)

- anonymous

http://www.ajdesigner.com/phptriangle/isosceles_triangle_area_k.php

- anonymous

Take the area of one of the isosceles triangles, multiply it by 4 and add it to the area of the base.
Forget that 1/2 B*H I messed up

- anonymous

know im lost

- anonymous

Do you see the forumla for the area of an isocoles triangle I sent you?

- anonymous

ya

- anonymous

Ok so do you see how each side of the pyramid is an isocles triangle?

- anonymous

ya

- anonymous

Ok so on the forumla you know that B=12, and C and A are the same and using pythagorean theorm we know it is \[3\sqrt{41}\]
So plug all those into the forumla and you will have the area of ONE side of the pyramid.

- anonymous

82.486362509205

- anonymous

I'm going to take your word on that... lol. Now what do you think you do?

- anonymous

* it by 4

- anonymous

right, and then add what?

- anonymous

the base of 144

- anonymous

There you go, that's the surface of a pyramid.

- anonymous

As long as your algebra was right when putting those numbers in.

- anonymous

so 473.96

- anonymous

Assuming your algebra is correct, yes..

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