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Elden [556K]
3 years ago
15

Arrange the cones in order from lease volume to greatest volume

Mathematics
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

Volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

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 V_{2}. V_{3}. and\  V_{4}. be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

d_{1} = 20\ and\ h_{1}=12

d_{2} = 18\ and\ h_{2}=10

r_{3} = 10\ and\ h_{3}=9

r_{4} = 11\ and\ h_{14}=9

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

V=\pi r^{2} \frac{h}{3}----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for d_{1} = 20\ and\ h_{1}=12

r_{1} = \frac{d_{1}}{2}

r_{1} = \frac{20}{2}=10\ units

V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}

V_{1}=\pi (10)^{2} \frac{12}{3}

V_{1}=\pi\times 100\times 4

V_{1}=400\pi\ units^{3}

Similarly, for volume of the cone for d_{2} = 18\ and\ h_{2}=10

r_{2} = \frac{d_{2}}{2}

r_{2} = \frac{18}{2}=9\ units

V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}

V_{2}=\pi (9)^{2} \frac{10}{3}

V_{2}=\pi\times 81\times \frac{10}{3}

V_{2}=\pi\times 27\times 10

V_{2}=270\pi\ units^{3}

Similarly, for volume of the cone for r_{3} = 10\ and\ h_{3}=9

V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}

V_{3}=\pi (10)^{2} \frac{9}{3}

V_{3}=\pi\times 100\times 3

V_{3}=\pi\times 300

V_{3}=300\pi\ units^{3}

Similarly, for volume of the cone for r_{4} = 11\ and\ h_{4}=9

V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}

V_{4}=\pi (11)^{2} \frac{9}{3}

V_{4}=\pi\times 121\times 3

V_{4}=\pi\times 363

V_{4}=363\pi\ units^{3}

So, the volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

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