Question

Can someone help me with Surface area of pyramids and cones

Answer 1

Find the slant height of the regular pyramid or cone

Answer 2

Do you know what pathagorean therom is?

Answer 3

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

Answer 4

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

Answer 5

so 12^2+15^2=369

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

Get it?

Answer 10

so 4(1/2*144*15)

Answer 11

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

Answer 12

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

Answer 13

know im lost

Answer 14

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

Answer 15

ya

Answer 16

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

Answer 17

ya

Answer 18

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.

Answer 19

82.486362509205

Answer 20

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

Answer 21

* it by 4

Answer 22

right, and then add what?

Answer 23

the base of 144

Answer 24

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

Answer 25

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

Answer 26

so 473.96

Answer 27

Assuming your algebra is correct, yes..