First, find the lateral surface area of the cone.

\(\displaystyle \text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})\)

Plug in the given slant height and radius.

\(\displaystyle \text{Lateral Area}=(18)(12)\pi=216\pi\)

Next, find the surface area of the cylinder:

\(\displaystyle \text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2\)

The surface area of the cylinder is the sum of the lateral surface area of the cylinder and its two bases. However, since one base of the cylinder is covered up by the cone, we will need to subtract that area out of the total surface area of the cylinder.

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2\)

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(12)(15)+\pi(12)^2=504\pi\)

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

\(\displaystyle \text{Surface Area of Figure}=216\pi+504\pi=720\pi=2261.95\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

First, find the lateral surface area of the cone.

\(\displaystyle \text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})\)

Plug in the given slant height and radius.

\(\displaystyle \text{Lateral Area}=(17)(10)\pi=170\pi\)

Next, find the surface area of the cylinder:

\(\displaystyle \text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2\)

The surface area of the cylinder is the sum of the lateral surface area of the cylinder and its two bases. However, since one base of the cylinder is covered up by the cone, we will need to subtract that area out of the total surface area of the cylinder.

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2\)

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(10)(15)+(10)^2\pi=400\pi\)

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

\(\displaystyle \text{Surface Area of Figure}=170\pi+400\pi=570\pi=1790.71\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

First, find the lateral surface area of the cone.

\(\displaystyle \text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})\)

Plug in the given slant height and radius.

\(\displaystyle \text{Lateral Area}=(25\sqrt3)(22)\pi=(550\sqrt3)\pi\)

Next, find the surface area of the cylinder:

\(\displaystyle \text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2\)

The surface area of the cylinder is the sum of the lateral surface area of the cylinder and its two bases. However, since one base of the cylinder is covered up by the cone, we will need to subtract that area out of the total surface area of the cylinder.

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2\)

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(22)(20)+\pi(22)^2=1364\pi\)

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

\(\displaystyle \text{Surface Area of Figure}=(550\sqrt3)\pi+1364\pi=7277.90\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

First, find the lateral surface area of the cone.

\(\displaystyle \text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})\)

Plug in the given slant height and radius.

\(\displaystyle \text{Lateral Area}=(16)(12)\pi=192\pi\)

Next, find the surface area of the cylinder:

\(\displaystyle \text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2\)

The surface area of the cylinder is the sum of the lateral surface area of the cylinder and its two bases. However, since one base of the cylinder is covered up by the cone, we will need to subtract that area out of the total surface area of the cylinder.

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2\)

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

\(\displaystyle \text{Surface Area of Cylindrical Portion}=2\pi(12)(13)+\pi(12)^2=456\pi\)

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

\(\displaystyle \text{Surface Area of Figure}=192\pi+456\pi=648\pi=2035.75\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

The surface area of a cone can be calculated using the formula

\(\displaystyle S = \pi r (r+l)\)

where \(\displaystyle r\) is the radius of the base and \(\displaystyle l\) is the slant height. The slant height is known to be 20.

The radius of the circular base can be found by dividing its circumference 50 by \(\displaystyle 2 \pi\):

\(\displaystyle r = \frac{50}{2 \pi} \approx \frac{50}{2 \cdot 3.1416} \approx 7.958\)

Set \(\displaystyle r = 7.958\) and \(\displaystyle l = 20\), and evaluate:

\(\displaystyle S = \pi r (r+l)\)

\(\displaystyle S \approx 3.1416 \cdot 7.958 \cdot (7.958 + 20)\)

\(\displaystyle \approx 3.1416 \cdot 7.958 \cdot 27.958\)

\(\displaystyle \approx 699\)

The area \(\displaystyle A\) and the radius \(\displaystyle r\) of the circular base are related by the formula

\(\displaystyle \pi r^{2}= A\)

The area of the circle is 50, so set \(\displaystyle A = 50\), then solve for \(\displaystyle r\):

\(\displaystyle \pi r^{2}= 50\)

Divide by \(\displaystyle \pi\):

\(\displaystyle \frac{\pi r^{2}}{\pi}= \frac{50}{\pi}\)

\(\displaystyle r^{2} \approx \frac{50}{3.1416}\)

\(\displaystyle r^{2} \approx 15.9155\)

Take the square root of both sides:

\(\displaystyle r \approx \sqrt{15.9155}\)

\(\displaystyle r \approx 3.9894\)

The surface area of the cone is the sum of the areas of the base and the lateral area. The lateral area of the cone can be found from the radius \(\displaystyle r\) and the slant height \(\displaystyle l\) using the formula

\(\displaystyle L = \pi r l\)

Set \(\displaystyle r \approx 3.9894\) and \(\displaystyle l = 20\):

\(\displaystyle L \approx 3.1416 \cdot 3.9894 \cdot 20\)

\(\displaystyle L \approx 251\)

Add the area of the base, 50:

\(\displaystyle S \approx 251 + 50 = 301\)

The area \(\displaystyle A\) and the radius \(\displaystyle r\) of the circular base are related by the formula

\(\displaystyle \pi r^{2}= A\)

The area of the circle is 50, so set \(\displaystyle A = 50\), then solve for \(\displaystyle r\):

\(\displaystyle \pi r^{2}= 50\)

Divide by \(\displaystyle \pi\):

\(\displaystyle \frac{\pi r^{2}}{\pi}= \frac{50}{\pi}\)

\(\displaystyle r^{2} \approx \frac{50}{3.1416}\)

\(\displaystyle r^{2} \approx 15.9155\)

Take the square root of both sides:

\(\displaystyle r \approx \sqrt{15.9155}\)

\(\displaystyle r \approx 3.9894\)

Now examine the figure below.

The slant height can be calculated from the height \(\displaystyle h\) and the radius \(\displaystyle r\) by way of the Pythagorean Theorem:

\(\displaystyle l = \sqrt{h^{2}+r^{2}}\)

\(\displaystyle l = \sqrt{20^{2}+ 3.9894^{2}}\)

\(\displaystyle \approx \sqrt{400 + 15.9153 }\)

\(\displaystyle \approx \sqrt{4 15.9153 }\)

\(\displaystyle \approx 20.3940\)

The surface area of the cone is the sum of the areas of the base and the lateral area. The lateral area of the cone can be found from the radius \(\displaystyle r\) and the slant height \(\displaystyle l\) using the formula

\(\displaystyle L = \pi r l\)

Set \(\displaystyle r \approx 3.9894\) and \(\displaystyle l = 20.3940\):

\(\displaystyle L \approx 3.1416 \cdot 3.9894 \cdot 20.3940\)

\(\displaystyle L \approx 256\)

Add the area of the base, 50, to this lateral area to obtain the surface area:

\(\displaystyle S \approx 256 + 50 = 306\).

The surface area of a cone can be calculated using the formula

\(\displaystyle S = \pi r (r+l)\)

where \(\displaystyle r\) is the radius of the base and \(\displaystyle l\) is the slant height. 

The radius of the circular base can be found by dividing its circumference 50 by \(\displaystyle 2 \pi\):

\(\displaystyle r = \frac{50}{2 \pi} \approx \frac{50}{2 \cdot 3.1416} \approx 7.958\)

Examine the figure below. 

The slant height can be calculated from the height \(\displaystyle h\) and the radius \(\displaystyle r\) by way of the Pythagorean Theorem:

\(\displaystyle l = \sqrt{h^{2}+r^{2}}\)

\(\displaystyle l = \sqrt{20^{2}+ 7.958^{2}}\)

\(\displaystyle \approx \sqrt{400 + 63.3298}\)

\(\displaystyle \approx \sqrt{4 63.3298}\)

\(\displaystyle \approx 21.5251\)

Set \(\displaystyle r \approx 7.958\) and \(\displaystyle l \approx 21.5251\) in the surface area formula:

\(\displaystyle S = \pi r (r+l)\)

\(\displaystyle S \approx 3.1416 \cdot 7.958 \cdot ( 7.958+21.5251 )\)

\(\displaystyle \approx 3.1416 \cdot 7.958 \cdot 29.4831\)

\(\displaystyle \approx 737\)

The formula for the volume of a tetrahedron is:

 \(\displaystyle \small V=\frac{a^3}{6\sqrt{2}}\).  

When \(\displaystyle \small V= \frac{4}{3\sqrt{2}}\) we have \(\displaystyle \small \frac{4}{3\sqrt{2}}=\frac{a^3}{6\sqrt{2}}\) .  

Multiplying the left side by \(\displaystyle \small \frac{2}{2}\) gives us,

 \(\displaystyle \small \frac{8}{6\sqrt{2}}=\frac{a^3}{6\sqrt{2}}\) , or \(\displaystyle \small 8=a^3\). 

Finally taking the third root of both sides yields \(\displaystyle \small 2=a\)

A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to:

 \(\displaystyle \small \left(\frac{1}{4}\right)\left(16\sqrt{3}\right) ft^2\) , or \(\displaystyle \small 4\sqrt{3} ft^2\)

Since the surface of one face is an equilateral triangle, and we know that,

 \(\displaystyle \small Area=\left(\frac{1}{2}\right)b\times h\) , the problem can be expressed as:

\(\displaystyle \small 4\sqrt{3}=\left(\frac{1}{2}\right)b\times h\)

In an equilateral triangle, the height \(\displaystyle \small h\), is equal to \(\displaystyle \small \frac{1}{2}(b)\sqrt{3}\) so we can substitute for \(\displaystyle \small h\) like so:

\(\displaystyle \small 4\sqrt{3}=\left(\frac{1}{2}\right)b \times \left(\frac{1}{2}\right)b\sqrt{3}\)

Solving for \(\displaystyle \small b\) gives us the length of one edge.

\(\displaystyle \small 4\sqrt{3}=\frac{b^2\sqrt{3}}{4}\)

\(\displaystyle \small (4)(4\sqrt{3})=(4)\left(\frac{b^2\sqrt{3}}{4}\right)\)

\(\displaystyle \small 16\sqrt{3}=b^2\sqrt{3}\)

\(\displaystyle \small 16=b^2\)

\(\displaystyle \small b=\pm 4\) 

However, we know that the edge of the tetrahedron is a positive number so \(\displaystyle \small b=4\).

Since the base \(\displaystyle \small b\) is the same as one edge of the tetrahedron, and a tetrahedron has six edges we multiply \(\displaystyle \small 6 \times 4\) to arrive at \(\displaystyle \small 24 ft^2\)