We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
259
1
avatar+196 

help

 Apr 26, 2018
 #1
avatar+2340 
+2

The general formula for volume for any pyramid or cone is \(V=\frac{1}{3}Bh\)

 

Let V = volume of the cone or pyramid

Let B = area of the base

Let h = the perpendicular height of the cone or pyramid

 

Since all these figures are cones, we know that the area of the base equals \(V=\frac{1}{3}\pi r^2h\). Ok, let's start solving for the individual volumes. 

 

Cone 1 Cone 2 Cone 3
\(V_1=\frac{1}{3}\pi r^2h;\\ r=11,h=9 \) \(V_2=\frac{1}{3}\pi r^2h;\\ r=6,h=14 \) \(V_3=\frac{1}{3}\pi r^2h;\\ r=14, h=8\)
\(V_1=\frac{1}{3}\pi*11^2*9\) \(V_2=\frac{1}{3}\pi*6^2*14\) \(V_3=\frac{1}{3}\pi*14^2*8\)
\(V_1=3\pi*121\) \(V_2=12\pi*14\) \(V_3=\frac{1}{3}\pi*1568\)
\(V_1=363\pi\text{cm}^3\) \(V_2=168\pi\text{cm}^3\) \(V_3=\frac{1568\pi}{3}\text{cm}^3\approx 500\text{cm}^3\)
     

I estimated for cone 3 because 3 does not divide evenly with 1568. This approximation is good enough for these purposes anyway because we only care about their order of greatness. The otder, therefore, from least to greatest is the following:

 

Cone 2, Cone 1, Cone 3 (option 2)

 Apr 26, 2018

17 Online Users

avatar