Math Is Fun Forum

jadewest

Member

Registered: 2021-02-20

Posts: 44

Geometry

Hello,

I'm having trouble solving this problem:

A pentagon’s radius (measure from center to vertex) measures 4 cm, what is the pentagon’s area?

I'm guessing that I'll have to solve this problem by using triangles. What is getting me confused is that I only have one measurement and can't think of how I can solve it? Would appreciate if someone could mentor me in solving it.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Geometry

hi jadewest

Welcome to the forum.

Questions like this have been asked many times on MIF, so you might be able to search for a solution.  But it won't take a moment to go through the steps anyway, so I will.

For this to work the pentagon has to be regular; so all it's sides are equal and all its internal angles.  As it has a 'radius' it sounds like this is the case here.

So let's label the vertices A,B,C,D and E and the centre, O.  We know that OA = 4cm.

Draw all the lines OA, OB etc so that the pentagon is divided into 5 identical triangles.  If we can calculate the area of one of these, it'll just be necessary to multiply that result by 5 to get the area of the whole pentagon.

Let M be the midpoint of AB.  Triangle OAB is isosceles, and OM is the height as angle OMA is 90.

The 5 lines meet at O so angle AOB = 360/5 = 72.  Hence angle AOM = 72/2 = 36 .

Using trigonometry, OM = 4 cos 36, and AM = 4 sin 36.

Area of AOB = half base x height = AM x OM.

Then times by 5 for the pentagon.

Hope that helps,

Bob

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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

jadewest

Member

Registered: 2021-02-20

Posts: 44

Re: Geometry

Hello,
Thank you for the explanation! I have written the solution on my notebook, is it okay if I post it here so that I can check it's correct?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Geometry

Yes, good idea!

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

jadewest

Member

Registered: 2021-02-20

Posts: 44

Re: Geometry

Hello Bob,
So, here is what I have figured out:

central angle = 360°/5=72°
2 * side angle + central angle = 180°
2SA + 72° = 180°
SA = 54°

sin = opposite/adjacent
sin 54° = h/4
h is approximately 3.2cm

cos = adjacent/hypotenuse
cos 54° = x/4
x is approximately 2.4cm

A = 1/2bh
A = 1/2 * 4.8 * 3.2
A = 7.68cm^2

Is that correct?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Geometry

hi jadewest

You have worked out angle OAM = 54 which is correct.  I suggested angle AOM = 36.  But this doesn't matter as you can do similar trig with either angle.  So I'll stick with 54.

In triangle OAM, OA is the hypotenuse, AM is adjacent to the angle and OM is opposite.

So h = OM = opposite = hypotenuse x sine angle.

And half base = AM = adjacent = hypotenuse x cosine angle.

I'm getting 4.7 for the area.

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

irspow

Member

Registered: 2005-11-24

Posts: 1,055

Re: Geometry

Last edited by irspow (2021-02-23 06:20:34)

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I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

jadewest

Member

Registered: 2021-02-20

Posts: 44

Re: Geometry

Hello,
Thank you for your help! I figured it out and waiting to see what my teacher will say.