Answer:
Volume of the cone in ascending order.
[tex]V_{2}=270\pi\ units^{3}<V_{3}=300\pi\ units^{3}<V_{4}=363\pi\ units^{3}<V_{1}=400\pi\ units^{3}[/tex]
cone with DIAMETER of 18 & height of 10
cone with RADIUS of 10 & height of 9
cone with RADIUS of 11 & height of 9
cone with DIAMETER of 20 & height of 12
Step-by-step explanation:
Let [tex]V_{2}. V_{3}. and\ V_{4}.[/tex] be the volume of the cone.
Let d, r and h be the diameter, radius and height of the cone.
Given:
[tex]d_{1} = 20\ and\ h_{1}=12[/tex]
[tex]d_{2} = 18\ and\ h_{2}=10[/tex]
[tex]r_{3} = 10\ and\ h_{3}=9[/tex]
[tex]r_{4} = 11\ and\ h_{14}=9[/tex]
Arrange the cones in order from lease volume to greatest volume.
Solution:
The volume of the cone is given below.
[tex]V=\pi r^{2} \frac{h}{3}[/tex]----------------(1)
where: r is radius of the base of cone.
and h is height of the cone.
The volume of the cone for [tex]d_{1} = 20\ and\ h_{1}=12[/tex]
[tex]r_{1} = \frac{d_{1}}{2}[/tex]
[tex]r_{1} = \frac{20}{2}=10\ units[/tex]
[tex]V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}[/tex]
[tex]V_{1}=\pi (10)^{2} \frac{12}{3}[/tex]
[tex]V_{1}=\pi\times 100\times 4[/tex]
[tex]V_{1}=400\pi\ units^{3}[/tex]
Similarly, for volume of the cone for [tex]d_{2} = 18\ and\ h_{2}=10[/tex]
[tex]r_{2} = \frac{d_{2}}{2}[/tex]
[tex]r_{2} = \frac{18}{2}=9\ units[/tex]
[tex]V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}[/tex]
[tex]V_{2}=\pi (9)^{2} \frac{10}{3}[/tex]
[tex]V_{2}=\pi\times 81\times \frac{10}{3}[/tex]
[tex]V_{2}=\pi\times 27\times 10[/tex]
[tex]V_{2}=270\pi\ units^{3}[/tex]
Similarly, for volume of the cone for [tex]r_{3} = 10\ and\ h_{3}=9[/tex]
[tex]V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}[/tex]
[tex]V_{3}=\pi (10)^{2} \frac{9}{3}[/tex]
[tex]V_{3}=\pi\times 100\times 3[/tex]
[tex]V_{3}=\pi\times 300[/tex]
[tex]V_{3}=300\pi\ units^{3}[/tex]
Similarly, for volume of the cone for [tex]r_{4} = 11\ and\ h_{4}=9[/tex]
[tex]V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}[/tex]
[tex]V_{4}=\pi (11)^{2} \frac{9}{3}[/tex]
[tex]V_{4}=\pi\times 121\times 3[/tex]
[tex]V_{4}=\pi\times 363[/tex]
[tex]V_{4}=363\pi\ units^{3}[/tex]
So, the volume of the cone in ascending order.
[tex]V_{2}=270\pi\ units^{3}<V_{3}=300\pi\ units^{3}<V_{4}=363\pi\ units^{3}<V_{1}=400\pi\ units^{3}[/tex]
