## anonymous 5 years ago Can someone help me with Surface area of pyramids and cones

1. anonymous

Find the slant height of the regular pyramid or cone

2. anonymous

Do you know what pathagorean therom is?

3. anonymous

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

4. anonymous

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

5. anonymous

so 12^2+15^2=369

6. 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

7. anonymous

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

8. 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.

9. anonymous

Get it?

10. anonymous

so 4(1/2*144*15)

11. anonymous
12. 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

13. anonymous

know im lost

14. anonymous

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

15. anonymous

ya

16. anonymous

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

17. anonymous

ya

18. 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.

19. anonymous

82.486362509205

20. anonymous

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

21. anonymous

* it by 4

22. anonymous

23. anonymous

the base of 144

24. anonymous

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

25. anonymous

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

26. anonymous

so 473.96

27. anonymous

Assuming your algebra is correct, yes..

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