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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 25

Hello,

This is similar to a question I have asked before. My teacher says the formula I used was incorrect, so I was wondering if you could help me out. This is the question and what I have so far.

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

360 / 7 = 51.428

180 - 51.2 / 2 = 64.25

Tan (64.25) = opp. / adj.

Tan (64.25) = h (s/2)

My teacher wants me to solve this for h but leave s as a variable in your equation.

A = 1/2 bh (What is the area of one triangle? What is the expression for h in terms of s? What is the expression for b in terms of s?)

I am unsure of how you complete this, so if you could help that would be awesome.

Thank you again! You and your team have been very helpful!

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,274

hi Kayla,

Let's look at the maths involved here.

The angle at the centre is 360. If you divide by 7 you'll get angle EOD.

As the polygon is regular it consists of 7 identical triangles. I've made one of these yellow so it's clear which triangle I'm talking about.

To work with trig. you need a right angled triangle. The yellow triangle isn't suitable.

If H is the midpoint of AB then triangle AOH (green) is suitable. So you'll need angle AOH or OAH. Divide angle EOD by 2 to get AOH. You can work with that or subtract from 90 to get OAH.

If AB = s, then AH = s/2 and OH = s/2 x TAN(0AH). [opposite = adjacent x tan(angle)]

So the yellow area is half base x height = 0.5 x s x s/2 x TAN(OAH)

You know the total area of the heptagon. Dividing by 7 will give you the yellow area. So you can put that equal to the above and solve for s.

Hope that helps,

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: 25

Hello,

360 / 7 = 51.428 = angle EOD

51.428 / 2 = 25.714

25.714 = AOH

Now I am unsure of how you get AB with only knowing the area. AB is our base in the triangle AOB and the base and height are unknown and we only know the area. How can we find the side AB if we only know the area? After this step, I believe I know how to solve the rest, but I am unsure of how to solve for AB.

Thank you again,

Kayla

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 25

Hello,

Do you know what the next step is?

Thank you for all the help,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,274

hi Kayla,

So the yellow area is half base x height = 0.5 x s x s/2 x TAN(OAH)

So start with 130 for the whole polygon, divide by 7 to get one yellow triangle and set it equal to the above. Then you've got to re-arrange that equation to get s squared and then s.

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: 25

Hello,

Okay, this is what I got from the information you gave me.

130 / 7 = 18.571. (This is the area of one triangle)

51.428 / 2 = 25.714 = angle OAH

18.571 = 0.5 x s x s/2 x TAN (OAH)

18.571 = 0.5 x s x s/2 x TAN (25.714)

18.571 = 0.5 x s x s/s (0.481)

s^2= 18.571 x 0.5 x 0.481

s^2= 4.466 (I the took the square root)

s= 2.113

I'm not sure if this is correct. Thank you again for the help,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,274

hi Kayla,

130 / 7 = 18.571. (This is the area of one triangle)

51.428 / 2 = 25.714 = angle OAH

18.571 = 0.5 x s x s/2 x TAN (OAH)

18.571 = 0.5 x s x s/2 x TAN (25.714)

18.571 = 0.5 x s x s/s (0.481)s^2= 18.571 x 0.5 x 0.481

s^2= 4.466 (I the took the square root)

s= 2.113

Correct up to this line "18.571 = 0.5 x s x s/**s** (0.481)"

except that bold 's' should be a 2.

But then you haven't re-arranged it properly.

times by 4 to get

4x18.571 = s^2 x 0.481

So s = square root of (4x18.571/0.481)

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: 25

Hello,

This is what I got when I multiplied by 4.

4(18.571) = s^2 (0.481)

4(18.571) / 0.481

= 74.284 / 0.481

= 154.436

The square root of 154.436 = 12.427

s= 12.427 in

Is this the final answer? Also, why do you multiply by 4?

Thank you,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,274

hi Kayla,

That just seems too big for a total area of 130. A square with those sides would be more than 144.

So I checked this by constructing an accurate diagram and it didn't come out correct. I'm so sorry ... it's my fault for not spotting you made an earlier error. It would be correct but that for that.

You started with 360 degrees and divided by 7 to get 51.428

That's right.

Then find half of it to get 25.714. That also is correct.

But it is not the angle OAH. This is AOH, the angle at the top of the green triangle. So subtract from 90 to get OAH, then as before.

Nearly there.

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: 25

Hello,

No worries about not catching the mistake! Here is what I have...

360 / 7 = 51.428

51.428 / 2 = 25.714

90- 25.714 = 64.286

130 / 7 = 18.571

18.571 = 0.5 x s x s/2 x TAN(OAH)

18.571 = 0.5 x s x s/2 x TAN(64.286)

18.571 = 0.5 x s x s/2 (2.076)

4(18.571) = s^2 (2.076)

4(18.571) / 2.076

= 74.284 / 2.076

= 35.782

Then I took the square root of 35.782 and I got 5.981

s = 5.981

Is this correct?

Thanks again for all the help!

Kayla

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**kayla1dance****Member**- Registered: 2018-01-10
- Posts: 25

Hello,

I turned this into my teacher. I will let you know what she says!

Thank you,

Kayla

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,274

hi Kayla,

That answer looks good to me.

Bob

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

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