Pyramids and cones?
Could I get any help with these two geometry-related questions? They're about solid shapes.
- MathmomLv 73 years agoFavorite Answer
Pyramids (KABCD and KEFGH) are simlar.
Ratio of volume of larger pyramid to volume of smaller pyramid
= (V₁ + V₂) / V₁
= (V₁ + V₁) / V₁
Ratio of volumes = (Ratio of sides)^3
Assuming p is base edge of larger pyramid, then
p/q = ∛2
H = height of cone TCD
h = height of cone TAB
Cones are similar
Volume of cone TAB = v
Volume of frustum ABCD = 26v
Ratio of volumes
= volume of larger cone / volume of smaller cone
= (v + 26v) / v
Ratio of heights = ∛27
H/h = 3
H = 3h
Height of frustum = H−h = 3h−h = 2h
Let a = area of base AB
Volume of cone TAB = 1/3 * a * h = ah/3
Volume of frustum ABCD = 26ah/3
Cylinder has base area = a and height = 2h
Volume of cylinder = a * 2h = 2ah
Ratio between volume of frustum and cylinder
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Note: I've tried last problem different ways, but I always end up with 13/3, which is not one of the choices listed.
Since larger cone has volume 27 times larger than smaller cone, it must have height 3 times larger. So cone TAB has height = 1/3 of cone TCD, and frustum ABCD has height = 2/3 of cone TCD. Therefore, frustum has height 2 times that of smaller cone TAB
If v = volume of cone TAB, then 26v = volume of frustum (given)
Also, volume of cone = 1/3 volume of cylinder of same height and radius.
So any cylinder with same height and base as cone TAB h volume = 3v, and cylinder with same base but twice the height has volume = 6v
26v/(6v) = 13/3
- Anonymous3 years ago