Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**jadewest****Member**- Registered: 2021-02-20
- Posts: 10

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.

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,950

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

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

Offline

**jadewest****Member**- Registered: 2021-02-20
- Posts: 10

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?

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,950

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

Offline

**jadewest****Member**- Registered: 2021-02-20
- Posts: 10

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?

Offline

**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,950

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

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 829

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

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

Offline

**jadewest****Member**- Registered: 2021-02-20
- Posts: 10

Hello,

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

Offline