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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

You told me that you had seen a problem similar to this one, but I haven't seen it anywhere. I looked on the keyword search page and typed in keywords, but I couldn't find it anywhere. Maybe you were thinking of a different problem. Anyway, I have tried this problem several times and I am still confused on it. If you could help, that would be great.

Find the surface area and volume of the pool shown below when the sides, the twelve "bumpers" making up the perimeter of the pool, are 5 ft each and the depth of the pool is 6 ft.

Use your logic and the formulas you have learned so far for the area of polygons, surface area, and volume to calculate the surface area of the inside of the pool in the picture (the pool liner) and the volume of the pool if it was filled all the way to the top. Use the shape of the pool and include formulas that you have learned in class. (You cannot add or take out water to find the volume.)

- This 12 sided pool is not a regular polygon because the inner angles are not all equal. You will need to break the base of the pool down into 2 regular polygons.

- Show your work step-by-step just as you have done for #1-4 above. You may need to include some written explanations for what you are doing in each step. Show all of the work to find all of the areas necessary for the surface area and all of the work to find the volume.

Here is a link to the picture I was given: https://imgur.com/k7jAdCG

Hope you can help. Thank you,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

Thanks for the picture. That's a big help. I've provided a link in one of your other posts but now I've seen the picture I don't think what I've said there is correct. I had assumed that the ends are each half a regular octagon but the diagram there has the distance across the pool as about 13, whereas, if the middle part is a square it should be 10 (=2 x 5). Does the original wording of the question make it clear which is correct. I'll make a new diagram based on a square middle section but I think it won't then give regular ends so the question may not be do-able.

I've read WyoCowboy's post again. It doesn't say the middle is a square ... that poster has just assumed it is. But both their diagram and your picture show a rectangular middle, width AJ in my diagram and length 10 as it's made form two 5 sections. SO let's assume the correct version is octagonal ends and rectangle in the middle. So my help to WyoCowboy will still work. I'll add a post there too.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

I tried figuring it out but I am very lost. I am going to send you the original work I had and maybe we can work from there?

I split the pool into 2 polygons with 6 sides each.

360 / 6 = 60

180 - 60 / 2 = 60

Tan (60) = opp. / adj.

Tan (60) = h (s/2)

h= tan 60 * 2.5 (half of the side)

1.732 * 2.5

= 4.330

(½) * base * height

(½) * 5 * 4.330

= 10.825 (6)

=64.95 (the area of one polygon)

64.95 (2)

= 129.9

130 ft^2 is the surface area of the pool.

129.9 (6)

= 779.4

180 ft^2 is the volume of the pool.

When I sent this, my teacher said, "The two shapes must be REGULAR polygons. Two 6-sided figures will not have equal sides and equal angles. Take another look at the pool."

Please let me know if you have any questions and what I can fix.

Thank you,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

I'm not surprised you are having difficulty. I think the person who set the question made a mistake and until we get that person to change the question you won't be able to do it. I'll give my reasoning and send a copy of this post to WyoCowboy who is also struggling to complete this.

Here's the picture you sent me:

and here's the one WyoCowboy sent. I've added some to the picture to show the pool split into two shapes.

from this question you wrote:

This 12 sided pool is not a regular polygon because the inner angles are not all equal. You will need to break the base of the pool down into 2 regular polygons.

As you can see the two ends do not make a hexagon. They make an octagon and it certainly looks regular so no problem so far.

But what's left is a rectangle, not a square! So it's not a regular shape. Here's my accurate construction of one end. Each square of my grid represents one foot. You can clearly see that the diagonal of the octagon, which would be one of the middle shape measurements is not 10 feet. It's closer to 13.

Please raise this with your teacher and ask them for clarification. You may copy from this post if it helps but I would recommend that you do not tell them it's from MIF.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

Okay, I emailed my teacher. I will let you know what she says.

Thank you for all the help,

Kayla

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

My teacher said, "Yes, the two shapes are an octagon and a rectangle. Here is an image with the triangles of the octagon marked to help with your work."

I will provide you with the picture that she attached. https://imgur.com/yjoAILm

Hope this helps,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

Ok, we have something we can work with. The triangle AFK (see an earlier post) is the key to this problem. It is isosceles so you'll have to split it in half to make a right angled triangle. The internal angles of a regular octagon are each 135 so angle FAK is 67.5. With that you can work out the height of the triangle, then its area and then the area of the whole octagon.

For the rectangle the length is AJ and the width is 10. AJ is twice AK which you can get with more trig in the half triangle already used.

Have a try and post back when you're ready.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,186

bob bundy wrote:

but I would recommend that you do not tell them it's from MIF.

Why? Can they even sue this site?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

Okay so do you use 67.5 as the tangent?

360 / 8 = 45

180 - 45 / 2 = 67.5

h = tan (67.5) * s/2

h = 2.414 * s/2

I am not sure if this is correct because I am a little confused on what the next step is. Also when I first started with this problem, I assumed that 5 was the length of the side because that is what the problem states. (The twelve "bumpers" making up the perimeter of the pool are 5 ft each.) So, I thought that s/2 would be 2.5. My teacher then stated, "The two shapes must be REGULAR polygons. Two 6-sided figures will not have equal sides and equal angles." I am confused about this because since they are regular polygons, wouldn't the sides and angles be equal?

I'm sorry if I am doing this incorrect, I am just very confused on how to complete this problem.

Thank you for everything,

Kayla

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,917

Monox D. I-Fly wrote:

bob bundy wrote:but I would recommend that you do not tell them it's from MIF.

Why? Can they even sue this site?

I presume that the teacher would not be amenable to the copying of work from outside sources.

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**Libera****Member**- Registered: 2018-03-07
- Posts: 16

kayla1dance wrote:

Hello,

Okay so do you use 67.5 as the tangent?

360 / 8 = 45

180 - 45 / 2 = 67.5h = tan (67.5) * s/2

h = 2.414 * s/2I am not sure if this is correct because I am a little confused on what the next step is. Also when I first started with this problem, I assumed that 5 was the length of the side because that is what the problem states. (The twelve "bumpers" making up the perimeter of the pool are 5 ft each.) So, I thought that s/2 would be 2.5. My teacher then stated, "The two shapes must be REGULAR polygons. Two 6-sided figures will not have equal sides and equal angles." I am confused about this because since they are regular polygons, wouldn't the sides and angles be equal?

I'm sorry if I am doing this incorrect, I am just very confused on how to complete this problem.

Thank you for everything,

Kayla

Look at the answer 2018-03-09 20:47:23 by bob bundy: "here's the picture WyoCowboy sent. I've added some to the picture to show the pool split into two shapes: the two ends make an octagon, what's left is a rectangle." And your teacher confirmed "Yes, the two shapes are an octagon and a rectangle" (Yesterday 06:55:13).

In your last message (Today 07:07:27) you wrote "I thought that s/2 would be 2.5" and I think you are right.

In your first work (2018-03-09 07:32:38) you had referred to exagons, but you have already rectified: the note of your teacher is outdated by now, in my opinion.

Go on, keeping the picture close.

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

Okay so if s/2 is 2.5, then

h = 2.414 * s/2

h = 2.414 * 2.5

= 6.035

So, the height is 6.035 ft.

I took 6.035 * 5 (for the length of the bumpers)

h = 30.175 (this is the area of one polygon)

30.175 (2) = 60.35 (The area of the whole octagon)

Would AK = 6 1/2 ? You said that AJ is twice of AK, so would that make AJ = 13 ? Then taking 10(13), then making the area of the whole rectangle = 130 ?

130 + 60.35 = 190.35. Is this the surface area of the whole pool? I must also find the volume, but I am unsure of how to complete that.

I probably did this incorrectly, but I just wanted to give it a shot.

Sorry about the confusion,

Kayla

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**Libera****Member**- Registered: 2018-03-07
- Posts: 16

kayla1dance wrote:

I took 6.035 * 5 (for the length of the bumpers)

h = 30.175 (this is the area of one polygon)

Kayla

Which polygon do you calculate the area of? AFGHJ? Why do you calculate it so?

Refer to http://www.mathsisfun.com/geometry/area.html .

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**Libera****Member**- Registered: 2018-03-07
- Posts: 16

kayla1dance wrote:

Would AK = 6 1/2 ?

Kayla

How have you got this value?

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**Libera****Member**- Registered: 2018-03-07
- Posts: 16

kayla1dance wrote:

Is this the surface area of the whole pool?

Kayla

I agree on these last steps.

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**Libera****Member**- Registered: 2018-03-07
- Posts: 16

kayla1dance wrote:

I must also find the volume, but I am unsure of how to complete that.

Kayla

http://www.mathsisfun.com/geometry/prisms.html

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

Apologies for the slow response. I've been helping to organise a dance show.

Okay so if s/2 is 2.5, then

h = 2.414 * s/2

h = 2.414 * 2.5

= 6.035

So, the height is 6.035 ft.

Yes, that's good for the height of one triangle (angles 67.5 - 67.5 - 90)

I took 6.035 * 5 (for the length of the bumpers)

h = 30.175 (this is the area of one polygon)30.175 (2) = 60.35 (The area of the whole octagon)

Whoops! Not quite! Area of a triangle is half base x height = 0.5 x 5 x 6.035

Then multiply by 8 as the octagon is made up of 8 of these triangles.

Would AK = 6 1/2 ? You said that AJ is twice of AK, so would that make AJ = 13 ? Then taking 10(13), then making the area of the whole rectangle = 130 ?

The length on my diagram does look like 6.5 but that's not the exact value if you use trig to calculate it.

In the right angled triangle with angle 67.5, AK is the hypotenuse and 2.5 is the adjacent. So you need to do AK = 2.5/COS(67.5) to get the exact value of AK.

Then work out the rectangle as you did and add on the octagon.

The solid for the volume is a prism. That means it has a constant cross section given by the surface shape. Volume of a prism is area of cross section times depth. You're told the depth is 6 ft.

Hope that helps you to finish this problem.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

Okay, I am unsure if this is correct, but I gave it my best shot. Please respond with any corrections needed, thank you.

AK = 2.5/cos (67.5)

AK = 0.999 (67.5)

AK = 168.75 (10)

= 1,687.5

120.7

= 1,808.2

Is this the area of the whole polygon including the square? If I made a mistake then it takes place during when I solved 2.5/cos (67.5) and I am confused on where I messed up.

Also, what is the cross section and how do you find it to solve for volume?

Thank you for all the help,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

You wrote:

AK = 2.5/cos (67.5)

AK = 0.999 (67.5)

AK = 168.75 (10)

= 1,687.5

120.7

= 1,808.2Is this the area of the whole polygon including the square?

Maths does try to model what's going on in the real world, so when you get an answer, you ought to ask yourself: "Does this look sensible for this problem?"

You'd already suggested that AK = 6.5 and I agreed that was nearly correct. So where did 168.75 come from ? I don't know what you did but it cannot be correct and you should have spotted this.

If you're using a calculator check first that it is in degrees mode. Calculators often have a radians mode and maybe even a gradians mode. You won't get the trig answers working correctly unless you're in degrees mode. There should be a little symbol showing deg. If it shows rad or grad you need to change to degrees. You can check it's doing the right thing by entering 60 and pressing cos. You should get 0.5 Some calculators expect you to press cos first and then the angle, so also try cos 60 That way you'll get to know how to do trig correctly on your device.

Allowing it may be cos 67.5 or 67.5 cos, the correct sequence will be 2.5 ÷ cos 67.5 or 2.5 ÷ 67.5 cos depending on the way your calculator does its calculations.

Are you looking at an answer of about 6.5 ? Keep experimenting until you get this right. Then, keep that value in the display, double it and times by 10. That is the answer for the rectangle (not square!). It is about the same size answer as the octagon. Think about the pool. The ends and the middle are roughly the same size.

Here's a link to cross sections:

http://www.mathsisfun.com/geometry/cross-sections.html

The volume of the pool is <area of cross section> times depth (6ft). The area of cross section is the answer you have just obtained for the area of the octagon plus rectangle.

Hope that helps,

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

I am very sorry. I was completing the trig problems on an iPhone calculator, which I now realize was not the best decision. I then plugged in the formula on a regular calculator and I got the answer that you were looking for.

2.5 / cos (67.5)

= 6.532 (2)

= 13.065 (10)

= 13.656 (This is the area of the rectangle.)

120.7 (area of octagon) + 130.656 (area of triangle)

= 251.356 (This is the area of the whole pool. It is also the area of the cross-section.)

251.356 (area of cross-section) (6)

= 1,503.136 (I believe this is the volume since I plugged in the numbers for the formula you gave me.)

Sorry for the inconvenience, but thank you for all the time and effort you have put into working out this problem with me.

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Kayla,

Answers correct. There are some minor corrections needed to your wording:

AK = 2.5 / cos (67.5)

= 6.532

so AJ = 13.065

= 13.656x you mean area of rectangle = 13.065 x 10 = 130.65 (This is the area of the rectangle.)

120.7 (area of octagon) + 130.656 (area of trianglex rectangle ... answer correct!

= 251.356 (This is the area of the whole pool. It is also the area of the cross-section.)

volume of pool = 251.356 (area of cross-section) (6)

= 1,503.136

I'm glad I could help you. I look forward to knowing how you got on with this work.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

My teacher has some revisions for me to do and I am a little confused about them. She wanted me to present my work in sections like I did below. Below that work, I will put what my teacher wanted me to elaborate that I am confused about.

Area of octagon: 120.7 ft.^2

Area of rectangle: 130.656 ft.^2

Other area needed for surface area:

360 / 8 = 45

180 - 45 / 2 = 67.5

h = tan (67.5) * s/2

h = 2.414 * s/2

h = 2.414 * 2.5

= 6.035 ft.

Surface area calculation:

(½)(5)(6.035)

= 15.087 (8)

= 120.7 ft.

2.5 / cos (67.5)

= 6.532 (2)

= 13.065 (10)

= 130.656 ft.

120.7 ft + 130.656 ft. = 251.356 ft.

Volume Calculation:

V = bh

251.356 (6)

= 1,503.136 ft.^2

My teacher said for the other area needed for surface area section she asked, "What other area were you calculating? You have the bottom of the pool. What area do you still need?"

Also, for the surface area section, she asked, "You will need all 3 of the answers above, so you need the other surface first."

I am confused because I thought that 120.7 ft^2 was the surface area. Was I mistaken?

My apologies if this is confusing. Let me know if you have any questions.

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

I did wonder earlier whether the question just wanted the area of the surface or are you expected to include the bottom and sides too ?

Sounds like it's the latter. Badly worded question. So add on another area for the bottom (same as the top) and then for the sides: every side is a rectangle 5 ft by 6 ft. So just add on 12 of these.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 38

Hello,

Okay, so the surface area would be 120.7 (2) = 241.4 ft ^2

For the sides, 5 (6) = 30. Would you do 30 (12) = 360 and add that to 241.4 ft ^2 to get 601.4 ft^2.

Thanks,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

Yes, I think so. I just wish the question was clearer.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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