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3 years ago

What is the approximate volume of a cone with a height of 9 ft and radius of 3 ft?

Mathematics

2 answers:

Nesterboy [21]3 years ago

7 0

Answer: The answer is 84 ft³.

Step-by-step explanation: Given that a cone has a height of 9 ft and a radius of 3 feet. We are to find the approximate volume of the cone.

Also, it is given to assume that

$\pi=3.14.$

We know that the volume of a cone with height 'h' and radius 'r' is given by

$V=\dfrac{1}{3}\pi r^2h.$

Here, r = 3 ft and h=9 ft.

Therefore, the volume of the cone will be

$V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times 3.14\times 3^2\times 9=\dfrac{254.34}{3}=84.78\sim 84~\textup{ft}^3.$

Thus, the required volume will be 84 ft³.

mr Goodwill [35]3 years ago

4 0

Cone Volume = (π • r² • h) ÷ 3

Cone Volume = (3.14 * 3^2 * 9) / 3

Cone Volume = (3.14 * 27)

Cone Volume = 84.82 cubic feet

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$A=\frac{3\sqrt{3}}{2} a^2$

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$a$ is the side length of the regular hexagon

Also, it's Perimeter is given by:

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Thus, all that we need to do is to find the side length of any one of the sides and to do that let us have a look at at the data of vertices points given and find out which points are definitely adjacent to each other and are also easy to calculate.

A quick search will yield that D(8, 0) and E(4, 0) are definitely adjacent to each other.

Please check the attached file here for a better understanding of the diagram of the original regular hexagon. Points D and E indeed are adjacent to each other.

Let us now find the distance between the points D and E using the distance formula which is as:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Where $d$ is the distance.

$(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of points D and E respectively. (please note that interchanging the values of the coordinates will not alter the distance $d$ )