The formula for the surface area of a cone is:

$\displaystyle A =$$\displaystyle ($$\displaystyle r$$\displaystyle )$$\displaystyle ($$\displaystyle r$$\displaystyle )$$\displaystyle ($$\displaystyle \pi$$\displaystyle )$$\displaystyle +$$\displaystyle ($$\displaystyle r$$\displaystyle )$$\displaystyle ($$\displaystyle l$$\displaystyle )$$\displaystyle ($$\displaystyle \pi$$\displaystyle )$,

where $\displaystyle r$ represents the radius of the cone base and $\displaystyle l$ represents the slant height of the cone.

Plugging in our values, we get:

$\displaystyle A =$$\displaystyle ($$\displaystyle 2$$\displaystyle )$$\displaystyle ($$\displaystyle 2$$\displaystyle )$$\displaystyle ($$\displaystyle \pi$$\displaystyle )$$\displaystyle +$$\displaystyle ($$\displaystyle 2$$\displaystyle )$$\displaystyle ($$\displaystyle 6$$\displaystyle )$$\displaystyle ($$\displaystyle \pi$$\displaystyle )$

$\displaystyle A =$$\displaystyle 4$$\displaystyle \pi$$\displaystyle +$$\displaystyle 12$$\displaystyle \pi$

$\displaystyle A =$$\displaystyle 16$$\displaystyle \pi$$\displaystyle m^2$

To figure out $\displaystyle \small h$, we must use the equation for the surface area of a cone, $\displaystyle \small A=\pi r^2+\pi rs$, where $\displaystyle \small r$ is the radius of the base of the cone and $\displaystyle \small s$ 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 $\displaystyle \small s$ by plugging what we know into the equation:

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

This equation can be reduced to:

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

$\displaystyle \small 9=s$

For a normal right angle cone, $\displaystyle \small s$ 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 $\displaystyle \small h$ using the Pythagorean theorem:

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

so

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

Our $\displaystyle \small h$ is therefore:

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

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

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 $\displaystyle 2\pi r$ (where r is the radius of the circle on the bottom). The circumference of the bottom is also $\displaystyle 30\%$ of the circumfrence of the original larger circle, which is $\displaystyle .3C=.3(2\pi R)=.3(10\pi)=3\pi$ (where R is the radius of the original, larger circle)

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

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

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

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

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

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

The height of the cone is $\displaystyle 4$, so the Pythagorean Theorem will give the slant height, $\displaystyle l=\sqrt{3^2+4^2}=5$

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

Solving this problem is going to take knowledge of Algebra, Geometry, and the equation for the surface area of a cone: $\displaystyle \small \small A=\pi r^2 + \pi r s$, where $\displaystyle \small r$ is the radius of the cone's base and $\displaystyle \small s$ 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:

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

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

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

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

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

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

To solve this problem, we will need to use the formula for finding the surface area of a cone, $\displaystyle \small A=\pi r^2+\pi rs$, where $\displaystyle \small s$ 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:

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

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

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

To find out the radius, we must use our knowledge of the formula for the surface area of a cone: $\displaystyle \small A= \pi r^2 + \pi rs$, where $\displaystyle \small r$ is the radius of the cone and $\displaystyle \small s$ 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:

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

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

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

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

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

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

The Surface Area of a cone is:

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

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.

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

The surface area of a cone is:

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

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

$\displaystyle A = \pi r^2$

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

$\displaystyle 4=r^2$

$\displaystyle 2=r$

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

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

The Surface Area of a cone is:

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

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.

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

$\displaystyle (3)^2 + (10)^2 = s^2$

$\displaystyle 109= s^2$

$\displaystyle \sqrt{109} = s$

Substitute slant height and radius into the Surface Area equation.

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