Cones - Math

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The radius of the base of a given cone is three times that of each base of a given cylinder. The heights of the cone and the cylinder are equal.

Which of the following is the greater quantity?

(a) The volume of the cone

(b) The volume of the cylinder

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If we let be the radius of each base of the cylinder, then is the radius of the base of the cone. We can let be their common height.

In the formula for the volume of a cylinder, set and :

$V = \pi $r^{2}$h$

$V = \pi R ^{2}H$

In the formula for the volume of a cone, set and :

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

$V = $\frac{1}{3}$ \pi $(3R)^{2}$H = $\frac{1}{3}$ \pi $(9R^{2}$)H = 3\pi R ^{2}H$

, so $3\pi R ^{2}H > \pi R ^{2}H$ . The cone has the greater volume.

The height of Cone B is three times that of Cone A. The radius of the base of Cone B is one-half the radius of the base of Cone A.

Which is the greater quantity?

(a) The volume of Cone A

(b) The volume of Cone B

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Let be the radius and height of Cone A, respectively. Then the radius and height of Cone B are $$\frac{1}{2}$r$ and , respectively.

(a) The volume of Cone A is $$\frac{1}{3}$ \pi $r^{2}$ h$ .

(b) The volume of Cone B is

$$\frac{1}{3}$ \pi \left ( $\frac{1}{2}$ r\right ) ^{2} \cdot 3 h = $\frac{1}{3}$ \cdot $\frac{1}{4}$ \cdot 3\cdot \pi r ^{2} h= $\frac{1}{4}$ r ^{2} h$ .

Since $$\frac{1}{4}$ r ^{2} h < $\frac{1}{3}$ r ^{2} h$ , the cone in (a) has the greater volume.

The volume of a cone whose height is three times the radius of its base is one cubic yard. Give its radius in inches.

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The volume of a cone with base radius and height is

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

The height is three times this, or . Therefore, the formula becomes

$V =\frac{1}{3}$ \pi $r^{2}$ \cdot 3r$

$V = \pi $r^{3}$$

Set this volume equal to one and solve for :

$\pi $r^{3}$ = 1$

$\pi $r^{3}$ \div \pi = 1 \div \pi$

$r=\sqrt[3]{$\frac{ 1 }{ \pi}$} ={$\frac{ \sqrt[3] {1}$ }{ \sqrt[3]{ \pi}}} ={$\frac{ 1}{ \sqrt[3]{ \pi}$}}={$\frac{1 \cdot \sqrt[3]{ $\pi^{2}$$ }}{ \sqrt[3]{ \pi} \cdot \sqrt[3]{ $\pi^{2}$}} ={$\frac{ \sqrt[3]{ $\pi^{2}$$ }}{ \pi}}$

This is the radius in yards; since the radius in inches is requested, multiply by 36.

$$\frac{ \sqrt[3]{ $\pi^{2}$$ }}{ \pi} \times 36 = $\frac{ 36 \sqrt[3]{ $\pi^{2}$$ }}{ \pi}$

The height of a given cylinder is one half the height of a given cone. The radii of their bases are equal.

Which of the following is the greater quantity?

(a) The volume of the cone

(b) The volume of the cylinder

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Call the radius of the base of the cone and the height of the cone. The cylinder will have bases of radius and height $$\frac{1}{2}$ H$ .

In the formula for the volume of a cylinder, set and $h = $\frac{1}{2}$ H$ :

$V = \pi $r^{2}$h$

$V = \pi R ^{2} \cdot $\frac{1}{2}$H = $\frac{1}{2}$ \pi R ^{2}H$

In the formula for the volume of a cone, set and :

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

$V = $\frac{1}{3}$ \pi $R^{2}$H$

$\frac{1}{2}$ > $\frac{1}{3}$ , so

$$\frac{1}{2}$ \pi R ^{2}H > $\frac{1}{3}$ \pi R ^{2}H$ ,

meaning that the cylinder has the greater volume.

The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?

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Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant height

LA = 2B

π(r)(l) = 2π(r2)

rl = 2r2

l = 2r

From the diagram, we can see that r2 + h2 = l2. Since h = 9 and l = 2r, some substitution yields

r2 + 92 = (2r)2

r2 + 81 = 4r2

81 = 3r2

27 = r2

B = π(r2) = 27π

LA = 2B = 2(27π) = 54π

SA = B + LA = 81π

A right cone has a radius of 4R and a height of 3R. What is the ratio of the total surface area of the cone to the surface area of just the base?

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We need to find total surface area of the cone and the area of the base.

The area of the base of a cone is equal to the area of a circle. The formula for the area of a circle is given below:

$A=\pi $r^2$$ , where r is the length of the radius.

In the case of this cone, the radius is equal to 4R, so we must replace r with 4R.

To find the total area of the cone, we need the area of the base and the lateral surface area of the cone. The lateral surface area (LA) of a cone is given by the following formula:

$LA=\frac{1}{2}$(2\pi r)(l)$ , where r is the radius and l is the slant height.

We know that r = 4R. What we need now is the slant height, which is the distance from the edge of the base of the cone to the tip.

In order to find the slant height, we need to construct a right triangle with the legs equal to the height and the radius of the cone. The slant height will be the hypotenuse of this triangle. We can use the Pythagorean Theorem to find an expression for l. According to the Pythagorean Theorem, the sum of the squares of the legs (which are 4R and 3R in this case) is equal to the square of the hypotenuse (which is the slant height). According to the Pythagorean Theorem, we can write the following equation:

Let's go back to the formula for the lateral surface area (LA).

$LA=\frac{1}{2}$(2\pi r)(l)$

$LA=\frac{1}{2}$(2\pi(4R))(5R)$

$LA=20\pi $R^2$$

To find the total surface area (TA), we must add the lateral area and the area of the base.

$TA=20\pi $R^2$+16\pi $R^2$=36\pi $R^2$$

The problem requires us to find the ratio of the total surface area to the area of the base. This means we must find the following ratio:

$\frac{36\pi $R^2$}{16\pi $R^2$}$

We can cancel $\pi $R^2$$ , which leaves us with 36/16.

Simplifying 36/16 gives 9/4.

The answer is 9/4.

What is the surface area of a cone with a radius of 6 in and a height of 8 in?

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Find the slant height of the cone using the Pythagorean theorem: _r_2 + _h_2 = _s_2 resulting in 62 + 82 = _s_2 leading to _s_2 = 100 or s = 10 in

SA = πrs + πr_2 = π(6)(10) + π(6)2 = 60_π + 36_π_ = 96_π_ in2

60_π_ in2 is the area of the cone without the base.

36_π_ in2 is the area of the base only.

What is the surface area of a cone with a radius of 4 and a height of 3?

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Here we simply need to remember the formula for the surface area of a cone and plug in our values for the radius and height.

Pi $r^{2}$ + Pi $r$\sqrt{r^{2}$$ + $h^{2}$}= Piast $4^{2}$ + Pi ast $4$\sqrt{4^{2}$$ + $3^{2}$} = 16Pi + 4Pi $\sqrt{25}$ = 16Pi + 20Pi = 36Pi

What is the surface area of the following cone?

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The formula for the surface area of a cone is:

$\pi$ $\pi$ ,

where represents the radius of the cone base and represents the slant height of the cone.

Plugging in our values, we get:

$\pi$ $\pi$

$\pi$ $\pi$

$\pi$

If the surface area of a right angle cone $\small A$ is $\small 48\pi$ , and the distance from the tip of the cone to a point on the edge of the cone's base is $\small 8$ , what is the cone's radius?

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Solving this problem is going to take knowledge of Algebra, Geometry, and the equation for the surface area of a cone: $\small \small A=\pi $r^2$ + \pi r s$ , where $\small r$ is the radius of the cone's base and $\small s$ is the distance from the tip of the cone to a point along the edge of the cone's base. First, let's substitute what we know in this equation:

$\small \small 48\pi=\pi $r^2$ + \pi r (8)=\pi $r^2$ + 8\pi r$

We can divide out $\small \small \pi$ from every term in the equation to obtain:

$\small \small \small $48=r^2$$+8r\Rightarrow0=r^2$+8r-48$

We see this equation has taken the form of a quadratic expression, so to solve for $\small r$ we need to find the zeroes by factoring. We therefore need to find factors of $\small \small -48$ that when added equal $\small \small 8$ . In this case, $\small -4$ and $\small \small 12$ :

$\small \small $r^2$+8r-48=(r+12)(r-4)$

This gives us solutions of $\small \small r=-12$ and $\small r=4$ . Since $\small r$ represents the radius of the cone and the radius must be positive, we know that $\small r=4$ is our only possible answer, and therefore the radius of the cone is $\small 4$ .

For a right circular cone , the radius is and the height of the cone is . What is the surface area of the cone in terms of $\pi$ ?

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To solve this problem, we will need to use the formula for finding the surface area of a cone, $\small A=\pi $r^2$+\pi rs$ , where $\small s$ is the length of the diagonal from the circle edge of the cone to the top. Since we are not given s, we must find it by using Pythagorean's Theorem:

$\small $s=$\sqrt{r^2$$+h^2$$}$=$\sqrt{10^2$$+3^2$}$=$\sqrt{109}$$ .

$\small 109$ is a prime number, so we cannot factor the radical any further. Therefore, our equation for our surface area of $\small C$ becomes:

$\small $A=\pi(10^2$)+\pi(10)$\sqrt{109}$=100\pi + 10\pi$\sqrt{109}$$ , which is our final answer.

If the surface area of cone $\small D$ is $\small 55\pi$ , and the distance $\small s$ between the cone's tip and a point on the cone's circular base is $\small 6$ , what is the radius $\small r$ of the cone?

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To find out the radius, we must use our knowledge of the formula for the surface area of a cone: $\small A= \pi $r^2$ + \pi rs$ , where $\small r$ is the radius of the cone and $\small s$ is the distance from the tip of the cone to any point along the circumference of the cone's base. We can plug in what we already know into the above equation:

$\small 55\pi=\pi $r^2$+6\pi r$

We can divide out $\small \pi$ from each term to obtain:

$\small $55=r^2$$+6r\Rightarrow0=r^2$+6r-55$

We now can recognize that the above is a quadratic expression, so to solve for $\small r$ we can find the zeroes of the equation by factoring. We need two numbers which will multiply to $\small -55$ but will add to $\small 6$ (in this case $\small 11$ and $\small -5$ ). Therefore, we can factor the above to the following:

$\small $r^2$+6r-55=(r+11)(r-5)$ .

Our two solutions are therefore $\small r=-11$ and $\small r=5$ . Since $\small r$ represents the radius of the base of the cone, it must be positive, and that leaves $\small 5$ as our one and only answer.

The surface area of cone $\small E$ is $\small 70\pi$ . If the radius of the base of the cone is , what is the height of the cone?

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To figure out $\small h$ , we must use the equation for the surface area of a cone, $\small A=\pi $r^2$+\pi rs$ , where $\small r$ is the radius of the base of the cone and $\small s$ is the length of the diagonal from the tip of the cone to any point on the base's circumference. We therefore first need to solve for $\small s$ by plugging what we know into the equation:

$\small $70\pi=(5^2$)\pi +5s\pi=25\pi +5s\pi$

This equation can be reduced to:

$\small 45\pi=5s\pi$

$\small 9=s$

For a normal right angle cone, $\small s$ represents the line from the tip of the cone running along the outside of the cone to a point on the base's circumference. This line represents the hypotenuse of the right triangle formed by the radius and height of the cone. We can therefore solve for $\small h$ using the Pythagorean theorem:

$\small $s^2$ = $r^2$$+h^2$$

so

$\small $h=$\sqrt{s^2$$-r^2$}$$

Our $\small h$ is therefore:

$\small $h=$\sqrt{9^2$$-5^2$}$=$\sqrt{81-25}$=$\sqrt{56}$=2$\sqrt{14}$$

The height of cone $\small E$ is therefore $\small 2$\sqrt{14}$$

Use the following formula to answer the question.

$SA=\pi $r^2$+\pi rl$

The slant height of a right circular cone is . The radius is , and the height is . Determine the surface area of the cone.

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Notice that the height of the cone is not needed to answer this question and is simply extraneous information. We are told that the radius is , and the slant height is .

First plug these numbers into the equation provided.

$SA=\pi $r^2$+\pi rl=\pi $(2)^2$+\pi (2)(6)=4\pi $cm^2$+12\pi $cm^2$$

Then simplify by combining like terms.

$4\pi $cm^2$+12\pi $cm^2$=16\pi $cm^2$$

A circle of radius five is cut into two pieces, and . The larger section is thrown away. The smaller section is curled until the two straight edges meet, and a bottom is made for the cone.

What is the area of the bottom?

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When the smaller portion of the circle is curled in, it will make the top of a cone. The circumfrence of the circle on the bottom is $2\pi r$ (where r is the radius of the circle on the bottom). The circumference of the bottom is also of the circumfrence of the original larger circle, which is $.3C=.3(2\pi R)=.3(10\pi)=3\pi$ (where R is the radius of the original, larger circle)

Therefore we use the circumference formula to solve for our new r:

$2\pi r=3\pi, r=\frac{3}{2}$$

Substituting this value into the area formula, the area of the small circle becomes:

$A=\pi $r^2$=\frac{9\pi}{4}$$

A cone has a bottom area of $9\pi$ and a height of , what is the surface area of the cone?

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The area of the bottom of the cone yields the radius,

$A=\pi $r^2$, r=$\sqrt{\frac{A}{\pi}$}=$\sqrt{\frac{9\pi}{\pi}$}=3$

The height of the cone is , so the Pythagorean Theorem will give the slant height,

The area of the side of the cone is $A=\pi rl=\pi \cdot 3 \cdot 5=15\pi$ and adding that to the $9\pi$ given as the area of the circle, the surface area comes to $24\pi$

Find the surface area of a cone with a base diameter of and a slant height of .

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The Surface Area of a cone is:

$SA=\pi rs +\pi $r^2$$

Given the base diameter is 6, the radius will be 3. The given slant height is 10.

Substitute the radius and slant height into the equation to find surface area.

$SA=\pi (3)(10) +\pi $(3)^2$ = 30\pi + 9\pi = 39\pi$

Find the surface area of a cone with a base area of $4\pi$ and a slant height of .

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The surface area of a cone is:

$SA= \pi r s + \pi $r^2$$

Given the base area is $4\pi$ , the base of the cone resembles a circle. Using the base area, it is necessary to find the radius.

$A = \pi $r^2$$

$4\pi = \pi $r^2$$

Since radius of the base is 2, and slant height is 6, substitute these into the surface area equation.

$SA= \pi (2)(6) + \pi $(2)^2$=12\pi+4\pi=16\pi$

Find the surface area of a cone with a base diameter of and a height of .

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The Surface Area of a cone is:

$SA=\pi rs + \pi $r^2$$

Given the base diameter is 6, the radius of the base is 3. The height is 10. We will substitute these values to find the slant height by using the Pythagorean Theorem.

$$\sqrt{109}$ = s$

Substitute slant height and radius into the Surface Area equation.

$SA=\pi (3)($\sqrt{109}$) + \pi $(3)^2$ = 3\pi $\sqrt{109}$ + 9\pi$

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

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The formula for the surface area of a cone with base of radius and slant height is

$A = \pi $r^{2}$ + \pi r L$ .

The diameter of the base is $$\frac{1}{5}$L$ ; the radius is half this, so

$r = $\frac{1}{2}$ \cdot $\frac{1}{5}$L= $\frac{1}{10}$L$

Substitute in the surface area formula:

$A = \pi \left ( $\frac{1}{10}$L \right ) ^{2} + \pi \left ( $\frac{1}{10}$L \right ) L$

$A = $\frac{1}{100}$ \pi $L^{2}$ + $\frac{1}{10}$ \pi $L^{2}$$

$A = $\frac{1}{100}$ \pi $L^{2}$ + $\frac{10}{100}$ \pi $L^{2}$$

$A = $\frac{11}{100}$ \pi $L^{2}$ = $\frac{11\pi $L^{2}$$}{100}$