# Frustum of cone question and enlargement of triangle length. pic included - gcse math

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• May 30th 2011, 02:23 PM
lFrenzied
Frustum of cone question and enlargement of triangle length. pic included - gcse math
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At a complete loss. Please do not simply tell me the answer, but give me a step-by-step process of how you got to that answer.
• May 30th 2011, 06:12 PM
Wilmer
That "triangle" is NOT a triangle!
Agree with y = 2.5; that leaves a triangle dimensions 3.5-7-11 which is impossible...OK?
• May 30th 2011, 07:15 PM
Wilmer
On the cone problem.
I assume you know the volume of a cone = pi r^2 h / 3

volume larger cone = pi r^2 2h / 3 = 360 ; simplify to get: p1 r^2 = 540 [1]

volume smaller cone = pi (r/2)^2 h ; simplify, substitute [1] to get volume = 45

cup volume is difference...OK?
• May 30th 2011, 07:17 PM
Soroban
Hello, lFrenzied!

Quote:

Frustum of cone

The a diagram shows a paper cup of height $h.$
It is part of a cone of height $2h$ and volume $360\text{ cm}^2.$

Find the volume of the cup.

Code:

      :      :  2r  :       *-------+-------* -       *      :      *  :         *    h:    *  :         *    : r  *    :           *---+---*    2h           *  :  *      :             *h: *      :             *:*        :               *        -

$\text{The volume of the entire cone is: }\:V_1 \:=\:\tfrac{\pi}{3}(2r)^2(2h) \:=\:\tfrac{8\pi}{3}r^2h\text{ cm}^3$

$\text{The volume of the small cone is: }\:V_2 \:=\:\tfrac{\pi}{3}(r^2)(h) \:=\:\tfrac{\pi}{3}r^2h\text{ cm}^3$

$\text{The volume of the cup is: }\:V \:=\:V_1 - V_2 \:=\:\tfrac{8\pi}{3}r^2h - \tfrac{\pi}{3}r^2h \:=\:\tfrac{7\pi}{3}r^2h\text{ cm}^3$

$\text{We see that the cup has }\tfrac{7}{8}\text{ the volume of the entire cone.}$

$\text{Therefore: }\;V \;=\;\tfrac{7}{8}(360) \:=\:315\text{ cm}^3.$

• May 31st 2011, 04:40 AM
bjhopper
Another way
V of smaller cone / V of larger cone= (1/2 r )^2*h / r^2*2h = 1/8 Cup = 7/8 of 360