## anonymous 5 years ago I need some help.. I've already done these problems and i got them back and they need fixed.. will someone please help me

1. anonymous

find the surface area of the solids. the pyramids are regular and the prisms, cylinders, and cones are right

2. anonymous

For the first one, you can find the surface area of the cylinder part ( the part that looks like a can) by using 2(pi)r * height

3. anonymous

Then add the area of the bottom circle (pi)r^2

4. anonymous

and then add SA of the cone, pi*r*s s is the slant height

5. anonymous

thats what i did and i got 211.95cmin^2

6. anonymous

ok, let me try and work it out with the numbers, just a sec

7. anonymous

Do you know the right answer?

8. anonymous

no

9. anonymous

I got 203.47cm^2

10. anonymous

wekkm 203.58 cm^2 when I use the pi button

11. anonymous

oops, I mean "well"

12. anonymous

for the cylinder, it's 30.6pi, the circle base is 9pi, and the cone is 25.2pi

13. anonymous

so it would be 30.6pi+9pi+25.2pi

14. anonymous

yes, and with that I got 64.8pi

15. anonymous

For the 2nd picture, there are 6 sides that are rectangles, so that would be 6(base*height). I can't tell in the picture, is the height of each triangle 5, or what does the 5 go to?

16. anonymous

to the slant eight of the triangle

17. anonymous

height*

18. anonymous

6 there are 6 triangles, each with a base of 3 and slant height of 5.....that would be 6 * 1/2 (base * height)

19. anonymous

So, 6(3*6) +6(1/2)(3*5)

20. anonymous

and then would you have to add the area of the base?

21. anonymous

855+64.8pi

22. anonymous

where did you get 855?

23. anonymous

6(3*6) +6(1/2)(3*5)

24. anonymous

6(3*12) +6(1/2)(3*5) that would be a 12 in the first ( )

25. anonymous

153

26. anonymous

64.8pi went to the first drawing...the one above is for the 2nd picture

27. anonymous

I get 261 in^2

28. anonymous

i thought the first pic was 203.58cm^2

29. anonymous

yes, that's what I got....it was 64.8pi, which = 203.58cm^2

30. anonymous

ok

31. anonymous

for the 2nd drawing I get 261in^2 + area of that hexagon base

32. anonymous

261 in^2 so this is the second pic surface area?

33. anonymous

yes, but I dont' know how to find area of the base

34. anonymous

will u help me with some other ones that ive already done i just got them wrong

35. anonymous

i will try

36. anonymous

okay.. for this one i needed to find the value of x.. and i got 3

37. anonymous

v=1/3 pi r^2 h

38. anonymous

I get 3 also

39. anonymous

so redoing it i get $56.5=1/3\times \pi \times x ^{2}\times6$

40. anonymous

3 meters

41. anonymous

$56.5=6.28x ^{2}$

42. anonymous

yes, i get that too

43. anonymous

Did it count off because you didn't put meters?

44. anonymous

maybe

45. anonymous

oh, wait, is x the diameter? We only found the radius.

46. anonymous

so x = 6 meters

47. anonymous

56.5/6.28=8.996

48. anonymous

yeah, then sq rt of that gives the radius of that circle as 3 meters

49. anonymous

but in the drawing, the x represents the diameter, which would be twice the radius, or 6 meters

50. anonymous

I've got to go....good luck!

51. anonymous

thank u

52. anonymous

you're welcome