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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 27

Hello,

I need some help on these math problems. I have been stuck on them for awhile.

Thank you,

Kayla

2. If a heptagon has an area of 130 in2, what is the measure of one side?

3. An octagon’s radius (measure from center to vertex) measures 6 cm, what is the octagon’s area?

4. A regular hexagon rests on one of the flat sides and has a total height of 14 ft. What is the measure of one of the hexagon’s sides?

5. Problem solver (worth 6 points): 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.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 24,046

Hi Kayla,

Polygons may be of help to some extent.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 27

Hello,

Thank you for responding. My teacher wanted me to use trigonometry and the area of polygons to get these answers. This is what I have so far.

#2:

520/7 = s^2 * cot (180/ 7)

s^2 = 35. 774

s = 5.9811

The measure of one side of the heptagon with an area of 130 in^2 is 5.9811 in.

#3:

360 / 8 = 45

45 degrees / 2 = 22.5 degrees

This creates two 90°-22.5°-67.5° triangles.

R*Cos(22.5) = 0.92338*R = 0.92388*6 cm = 5.543 cm.

The shorter leg can be found by R*Sin(22.5) = 0.38268*R = 0.38268*6 cm = 2.296 cm.

1/2*b*h = 1/2*(5.543 cm)*(2.296 cm) = 6.364 cm^2

8 * 12.728 = 101.82 cm^2

The whole area of the octagon is 101.82 cm^2.

#4:

Tan (60) = opp. / adj.

Tan (60) = h (s/2)

7 = tan 60 * (s/2)

s/2 = tan 60 * 7

s/2 = 12.124

s = 6.062

(½) * base * height

(½) * 6.062 * 7

= 21.217 (7)

= 148.159 ft^2

= 148 ft ^2

148 / 7 / 6.062 / ½

= 6.975

The length of one side is 6.975 ft.

#5: My teacher said, "The two shapes must be REGULAR polygons. Two 6-sided figures will not have equal sides and equal angles." This is what I got for her to respond with that. I am confused on how to fix it.

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.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,282

hi Kayla,

I think your answers to (2) and (3) are correct.

I got a different result for (4). If the vertices are ABCDEF then AE = 14 and you want AF. So use COS not TAN.

I need to see the diagram for (5) although this sounds familiar. Maybe someone has posted this before so you could try a search on a key word in the problem.

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: 27

Hello,

This is #4:

360 / 6 = 60

180 - 60 / 2 = 60

Cos (60) = opp. / adj.

Cos (60) = h (s/2)

7 = Cos 60 * (s/2)

s/2 = Cos 60 * 7

s/2 = .5 * 7

s/2 = 3.5

s = 1.75

(½) * base * height

(½) * 1.75 * 7

= 6.125 (7)

= 42.875 ft^2

42.875 / 7 / 1.75 / ½

= 7

The length of one side is 7 ft.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,282

hi Kayla,

I'm not really following your working here. Only the side is asked for, so you can get this directly without finding the area first. Somewhere in that calculation you seem to have gone wrong with the trig..

Here's a start with my method:

Label the vertices A B C D E and F in order. The measurement you are given is AE = 14. You are asked for AF.

Let G be the midpoint of AE. In triangle AFG the angles are 30-60-90.

AF = AG/cos(angle FAH)

Can you finish it from here?

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: 27

Hello,

Would it be AF = 14/cos (60) and then solve?

I'm sorry this lesson was very hard for me to understand and master the concept. That's why I have lots of questions and confusion about it.

Thank you so much for all the help,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,282

hi Kayla,

Sorry, bad typo by me. The angle should be FAG which is 30 degrees, not 60. Don't know why I typed an H.

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: 27

Hello,

AF= 14/cos (30)

I did several different ways because I was unsure which one was correct.

14/cos = 0.970 (30) = 420

30/cos = 0.866 (14) = 420

cos 30 = 0.866 (14) = 12.124

Thank you again for the help,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,282

This sort of trigonometry only works for right angled triangles. (there are more complicated ways of dealing with other triangles). So we need to work on triangle FAG. In this triangle I'll use the 30 degree angle. AF is the hypotenuse, and AG is adjacent to the angle. The formula is

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: 27

Hello,

When I solved 7 / cos(30) I got 8.083. Is this correct? Is this the length of one of the hexagon's side?

Thank you so much for the help. You have really helped me understand the material.

Kayla

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 27

Hello,

Do you know if this is correct?

Thank you,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,282

hi Kayla,

That's the answer I get too.

Bob

ps. another member has posted the pool question and I've given some help here:

http://www.mathisfunforum.com/viewtopic … 81#p402481

I forgot to say that you'll need to calculate the side of the square, AJ. It's twice AH which you can get from triangle AHF split in two to make a right angle.

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

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