# Pyramids and cones?

Could I get any help with these two geometry-related questions? They're about solid shapes.

### 2 Answers

- MathmomLv 73 years agoFavorite Answer
25.

Pyramids (KABCD and KEFGH) are simlar.

Ratio of volume of larger pyramid to volume of smaller pyramid

= (V₁ + V₂) / V₁

= (V₁ + V₁) / V₁

= 2

Ratio of volumes = (Ratio of sides)^3

Assuming p is base edge of larger pyramid, then

p/q = ∛2

——————————————————————————————

26.

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

= 27

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

= (26ah/3)/(2ah)

= 13/3

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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
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I wanted to give you five stars, so sorry for the inconvenience. It was a very thorough answer, thank you a lot :D