What Are the Solutions to These Various Geometry Problems?

[blahblah]

can anyone help me with any of these answers

3. Given a sphere with diameter 12 cm, find the surface area of the smallest cylinder containing the sphere.

4. A rectangular box has sides of length 4 cm, 10 cm and 12 cm. What are the lengths of each of the four diagonals of the box?

5. A room measures 4 m by 7 m and the ceiling is 3 m high. A litre of paint covers 20 square metres. How many litres of paint will it take to paint all but the floor of the room?

6. Find the dimensions of the smallest possible cubical box to hold a sphere of radius r. What percent of the volume will be air (percent nearest)?

7. The interior angles of an n-gon have an average measure of 175 degrees. Calculate n.

 

[MarkFL]

Re: assignment help pleasse

Let's do these one at a time...it's easier to keep track of what we're doing that way.

3.) What must the radius and height of the cylinder be?

Note to others: I have to run for a few hours, so feel free to jump in. ;)

 

[blahblah]

Re: assignment help pleasse

radius 6? and height I am not sure

 

[Jameson]

Re: assignment help pleasse

I picture this with the cylinder sitting upright like a soda can. You are correct that to have a sphere inside of it this cylinder needs to be at least 12cm long, or have a radius of 6cm as you said. Now think about the height of the cylinder. What is the height of the sphere?

 

[blahblah]

Re: assignment help pleasse

I picture this with the cylinder sitting upright like a soda can. You are correct that to have a sphere inside of it this cylinder needs to be at least 12cm long, or have a radius of 6cm as you said. Now think about the height of the cylinder. What is the height of the sphere?

6? I am not sure

 

[Jameson]

Re: assignment help pleasse

6? I am not sure

Try drawing a picture. Here's how I described it.

Now from the bottom of the sphere to the top of the sphere is the distance we're looking for. What do you think that is? I can't tell if you're just guessing or if you have a reason behind what you're saying so can you try to explain what you're thinking? :)

 

[schu0160]

I could also use help with this...? I given the sphere has a diameter of 12cm does that mean the radius is 6cm? And how do I find the SA of the smallest cylinder to contain the sphere?

 

[earboth]

I could also use help with this...? I given the sphere has a diameter of 12cm does that mean the radius is 6cm? And how do I find the SA of the smallest cylinder to contain the sphere?

Welcome on board!

1. If you have a new question please start a new thread.

2. According to the drawing in Jameson's post the height of the smallest cylinder must be as long as the diameter of the sphere.

You now know the radius and the height of the cylinder.

The surface area consists of 2 circles and one curved surface which is actually a rectangle:

$$\displaystyle{A_{surface} = \underbrace{2 \cdot \pi \cdot r^2}_{\text{area of 2 circles}} + \underbrace{2 \pi r \cdot h}_{\text{area of curved surface}}}$$

Replace the variables by the known values.

 

[MarkFL]

Although we normally ask that no more than 2 questions be asked per topic, and that effort be shown when posting questions, enough time has gone by that I am going to provide full solutions to the questions for the benefit of our readers.

3. Given a sphere with diameter 12 cm, find the surface area of the smallest cylinder containing the sphere.

The smallest cylinder that can contain a sphere of radius $r$ will have radius $r$ and height $h=2r$. And so the surface area $S$ of this cylinder is:

\(\displaystyle S=2\pi r^2+2\pi r(2r)=2\pi r^2\left(1+2 \right)=6\pi r^2\)

4. A rectangular box has sides of length 4 cm, 10 cm and 12 cm. What are the lengths of each of the four diagonals of the box?

There are 3 pairs of rectangular faces of the box, and to find the diagonal $d$ of a rectangle of base $b$ and height $h$, we may use the Pythagorean theorem as follows:

\(\displaystyle b^2+h^2=d^2\,\therefore\,d=\sqrt{b^2+h^2}\)

Now, for the diagonal of the entire box, consider the following diagram:

View attachment 1383

The box has the dimensions $L$, $W$ and $H$. As we can see the diagonal $d$ of the entire box is found from:

\(\displaystyle b^2+H^2=d^2\)

But, we see that:

\(\displaystyle b^2=L^2+W^2\)

Hence:

\(\displaystyle d=\sqrt{L^2+W^2+H^2}\)

And so the four diagonals of the box are:

\(\displaystyle d_1=\sqrt{L^2+W^2}\)

\(\displaystyle d_2=\sqrt{L^2+H^2}\)

\(\displaystyle d_3=\sqrt{W^2+H^2}\)

\(\displaystyle d_4=\sqrt{L^2+W^2+H^2}\)

Letting $L=4\text{ cm},\,W=10\text{ cm},\,H=12\text{ cm}$ we have:

\(\displaystyle d_1=\sqrt{4^2+10^2}=2\sqrt{29}\text{ cm}\)

\(\displaystyle d_2=\sqrt{4^2+12^2}=4\sqrt{10}\text{ cm}\)

\(\displaystyle d_3=\sqrt{10^2+12^2}=2\sqrt{61}\text{ cm}\)

\(\displaystyle d_4=\sqrt{4^2+10^2+12^2}=2\sqrt{65}\text{ cm}\)

5. A room measures 4 m by 7 m and the ceiling is 3 m high. A litre of paint covers 20 square metres. How many litres of paint will it take to paint all but the floor of the room?

We have two walls that are 4 m by 3 m, two walls that are 7 m by 3 m and the ceiling which is 4 m by 7 m, and so the total surface $S$ in square meters to be painted is:

\(\displaystyle S=2(4\cdot3+7\cdot3)+4\cdot7=94\)

To find the amount of paint $P$ in liters, we find:

\(\displaystyle P=\frac{1}{20}\frac{\text{l}}{\text{m}^2}\cdot94 \text{ m}^2=\frac{47}{10}\text{ l}=4.7\text{ l}\)

6. Find the dimensions of the smallest possible cubical box to hold a sphere of radius r. What percent of the volume will be air (percent nearest)?

The side lengths of the cubical box will have to be equal to twice the radius of the sphere, and so its volume is:

\(\displaystyle V_{\text{cube}}=(2r)^3=8r^3\)

The volume of the sphere is:

\(\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi r^3\)

And so the portion of the box that is air is the volume of the box minus the volume of the sphere all divide by the volume of the box:

\(\displaystyle V_{\text{air}}=\frac{V_{\text{cube}}-V_{\text{sphere}}}{V_{\text{cube}}}=1-\frac{\frac{4}{3}\pi r^3}{8r^3}=1-\frac{\pi}{6}\approx48\%\)

7. The interior angles of an n-gon have an average measure of 175 degrees. Calculate n.

If we pick some central point within the polygon and draw radial line segments from this point to each of the $n$ vertices, then we have $n$ triangles the sum of whose interior angles is $n\cdot180^{\circ}$. If we then subtract the sum of the angles surrounding the central point, we then find that the sum $S$ of the interior angles of the $n$-gon is:

\(\displaystyle S=n\cdot180^{\circ}-360^{\circ}=(n-2)180^{\circ}\)

Now, if the interior angles of an $n$-gon have an average measure of $175^{\circ}$, then we know:

\(\displaystyle \frac{S}{n}=175^{\circ}\implies S=n\cdot175^{\circ}\)

And so, to find $n$, we may equate the two expressions for $S$:

\(\displaystyle (n-2)180^{\circ}=n\cdot175^{\circ}\)

Divide through by $5^{\circ}$:

\(\displaystyle 36(n-2)=35n\)

Distribute on the left:

\(\displaystyle 36n-72=35n\)

Add $72-35n$ to both sides:

\(\displaystyle n=72\)

Thus, the $n$-gon has 72 sides.