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**KatieWind5223****Member**- Registered: 2019-03-25
- Posts: 15

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

For this one, I started by dividing the area, 130, by the amount of triangles I split the heptagon into, being 7. From that, I got 18.5 which is the area of each of the individual triangles. I am having trouble finding the sides and the height though. Please help.

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

I got an area of 80in^2. Is this correct?

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?

One of the sides is 4.6 feet long? Please check and help.

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.

I haven't started this yet and I would rather go one by one, please.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 28,456

Hi,

Area of a Regular Heptagon = The area (A) of a regular heptagon of side length a is given by:

Given, the Area is 130 square inches, the side length can be found.

Similarly, see the links:

.

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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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,541

Q5 has recently and many times earlier been answered in other posts. Use the search to find these.

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**KatieWind5223****Member**- Registered: 2019-03-25
- Posts: 15

My teacher told me to set up a trig equation using the tangent ratio to be equivalent to the height. How would I do that?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,541

hi KatieWind5223

Let's say you have a regular polygon with n sides. Centre point A.

Draw lines out from A to each of the vertices of the polygon. This makes n identical isosceles triangles. Label one with points B and C. Mark the midpoint of BC as point D.

Angle BAC = 360/n, and angle BAD = (360/n)/2 = 180/n

Angle ADB = 90 so we can use trig. on the triangle ADB. AD = height of a triangle.

BD/AD = tan(BAD) , so AD = BD / tan (BAD) (note BD = half a side = s/2)

area of one triangle = 1/2 side x height = 1/2 . s . s/2 .1/tan(BAD)

area of polygon = n . 1/2 . s . s/2 / tan(180/n)

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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