you can solve for the volume:V = Bh/3V = 3(4π) / 3V = 12π/3V = 4πV ≈ 13 m^3-Look at diagram for #12. It shows a nice right triangle, which hasbase = 1/2 base length of pyramid = 10/2 = 5height = height of pyramid ----> unknownhypotenuse = slant height of pyramid = 12So you can use Pythagorean theory to find height of Pyramidh² = 12² − 5² = 119 -----> h = √119V = 1/3 * base area * heightV = 1/3 * (10*10) + √119V = 100√119/3 ≈ 363.623737155Do the same with #13 and #14------------------------------Volume of cone:V = 1/3 π r² hIt doesnt matter if top of cone is straight above centre of base, or cone is slanted and tip of cone is off-centre. So long as you use vertical height#17V = 1/3 π (2)² * 4 = 16π/3Do the same with #18 and #19-If Ive done this correctly,......
A = πr^2
A = π(2^2)
A = 4π
Now that you know the area of the base and the height, you can solve for the volume:
V = Bh/3
V = 5.5(4π) / 3
V = 22π/3
V ≈ 23 in^3
19.) You have the radius and the height. The area of the base is:
A = πr^2
A = π(2^2)
A = 4π
Now that you know the area of the base and the height, you can solve for the volume:
V = Bh/3
V = 3(4π) / 3
V = 12π/3
V = 4π
V ≈ 13 m^3
Look at diagram for #12. It shows a nice right triangle, which has
base = 1/2 base length of pyramid = 10/2 = 5
height = height of pyramid ----> unknown
hypotenuse = slant height of pyramid = 12
So you can use Pythagorean theory to find height of Pyramid
h² = 12² − 5² = 119 -----> h = √119
V = 1/3 * base area * height
V = 1/3 * (10*10) + √119
V = 100√119/3 ≈ 363.623737155
Do the same with #13 and #14
------------------------------
Volume of cone:
V = 1/3 π r² h
It doesn't matter if top of cone is straight above centre of base, or cone is slanted and tip of cone is off-centre. So long as you use vertical height
#17
V = 1/3 π (2)² * 4 = 16π/3
Do the same with #18 and #19
If I've done this correctly, then here are the answers for 12~14. I'll do the cone ones in a minute:
12. 1091 cubic meters
13. 11144 cubic meters
14. 1689 cubic meters
17. 17 cubic feet
18. 23 cubic inches
19. 13 cubic meters
hope that helps
Try to use Wolfram Alpha to answer it (Link in the source).
