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### Pentagon Area Using No Trig

```
Date: 05/14/2001 at 12:49:05
From: Sandra
Subject: Pentagon Area - Using NO Trig

Dr. Math,

I've been trying to find the area of a pentagon without the use of
trigonometry. Where I am stumped is in finding the area of one of the
five triangles. I can use trig to find the height of the triangle very
easily, but the challenge is finding the height and thus the area
without the use of trig. I also know the area of a pentagon is
1.720x^2, but I don't know the derivation of this equation either.

Thank you,
Sandra
```

```
Date: 05/14/2001 at 17:04:04
From: Doctor Rick
Subject: Re: Pentagon Area - Using NO Trig

Hi, Sandra, and thanks for writing to Ask Dr. Math.

Perhaps you've seen the item in our Dr. Math Archives about Finding
the Area of a Regular Pentagon using trigonometry:

http://mathforum.org/library/drmath/view/54071.html

Trigonometry isn't necessary, however. We can find the area without
it.

First draw the diagonals of the pentagon. Each of the five diagonals
is divided into three segments by the other diagonals. The outer two
segments are equal in length; call their length a. Call the length of
the center segment b. The length of a side of the pentagon is s.

You can prove that certain triangles formed by the pentagon and its
diagonals are similar. For one, the triangle formed by two adjacent
sides and one diagonal of the pentagon is similar to the smaller
triangle formed by one of these sides and two segments of length a.

Also, you can prove that the triangle formed by one side, one segment
of length a, and one segment of length a+b is isosceles. Therefore,
a + b = s.

Using these facts you can prove that the length of the diagonal of the
pentagon is (sqrt(5)+1)/2. I will call this length d.

Now we're ready to find the area of the pentagon without using
trigonometry. Draw the two diagonals from the top vertex of the
pentagon. These divide the pentagon into three isosceles triangles.

You can find the altitude of each of these triangles using the
Pythagorean theorem and the fact that the altitude of an isosceles
triangle bisects the base. Now you know the altitude and base of each
triangle, so you can find the area as half the product of the base and
the altitude.

Alternatively, you can use Heron's formula for the area of a triangle
in terms of the side lengths alone. I used this approach.

The sum of these areas is the area of the pentagon. I don't get a
really neat formula, however. I get this formula using the value of d,
which I'm leaving for you to calculate:

K = (sqrt(4d^2 - 1)/4 + d*sqrt(4 - d^2)/2)s^2
= 1.720477 s^2

Perhaps you could calculate d, plug the value of d into the formula
for K, and find a way to simplify the formula. I'll keep playing with
it myself.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/15/2001 at 18:36:33
From: Sandra Ward
Subject: Re: Pentagon Area - Using NO Trig

Dr. Math,

I wanted to thank you for your timely response and insightful
assistance to the pentagon problem. It certainly helped me out a
bunch!

Sandra
```

```Date: 08/01/2016 at 15:45:39
From: Thayer
Subject: Area of Regular Pentagon

Your FAQ gives the area of a regular pentagon as

(5s^2/4)sqrt(1 + 2/sqrt5)

Other sources on the Internet give it as

(s^2/4)sqrt(5(5 + 2sqrt5))

I'm trying to find the area by summing the areas of three isosceles
triangles: two with legs s and base (phi)s, and one with legs (phi)s
and base s.

I get the area of the former as

A = (s^2)/4 * sqrt(4 - phi^2)

For the latter, I get

A = (s^2)/4 * sqrt(4phi^2 - 1).

I can't figure out how to sum the areas of these three triangles to get
either of the published areas of a regular pentagon -- or even to show
that they are equivalent. I know that phi = [1 + sqrt(5)/2]. But I'm rusty

Can you help me sort this out?

```

```
Date: 08/02/2016 at 11:30:20
From: Doctor Peterson
Subject: Area of Regular Pentagon

Hi, Thayer.

Your work looks essentially like what Doctor Rick got, above. But he
appears to be multiplying one term by phi.

A = 2(s^2)/4 sqrt(4 - phi^2) + (s^2)/4 sqrt(4phi^2 - 1)

= (s^2)/4 [2sqrt(4 - phi^2) + sqrt(4phi^2 - 1)]

= (s^2)/4 [sqrt(16 - 4phi^2) + sqrt(4phi^2 - 1)]

You didn't show your work for me to check; so before we proceed, let's
compare your result to the two formulas you found, to see if they all
agree.

When s = 1, for example,

1/4 [sqrt(16 - 4phi^2) + sqrt(4phi^2 - 1)] = 1.3572

(5/4)sqrt(1 + 2/sqrt(5)) = 1.7204774

(1/4)sqrt(5(5 + 2sqrt(5))) = 1.7204774

So yours isn't right.

Doctor Rick's version is

1/4 [phi*sqrt(16 - 4phi^2) + sqrt(4phi^2 - 1)] = 1.7204774

I'll leave you to figure out where your work went wrong (I can easily
imagine some possibilities), and proceed with his version:

A = (s^2)/4 [phi * sqrt(16 - 4phi^2) + sqrt(4phi^2 - 1)]

We want to combine this sum into a single radical. We'll first have to

sqrt(16 - 4phi^2) = sqrt(16 - 4[(1 + sqrt(5))/2]^2)
= sqrt(16 - 4[1 + 2sqrt(5) + 5]/4)
= sqrt(16 - 1 - 2sqrt(5) - 5)
= sqrt(10 - 2sqrt(5))

phi * sqrt(16 - 4phi^2) = (1 + sqrt(5))/2 * sqrt(10 - 2sqrt(5))
= sqrt[(1 + sqrt(5))^2(10 - 2sqrt(5))]/2
= sqrt[(6 + 2sqrt(5))(10 - 2sqrt(5))]/2
= sqrt[60 - 12sqrt(5) + 20sqrt(5) - 20]/2
= sqrt[40 + 8sqrt(5)]/2
= sqrt[10 + 2sqrt(5)]

sqrt(4phi^2 - 1)  = sqrt(4[(1 + sqrt(5))/2]^2 - 1)
= sqrt(4[1 + 2sqrt(5) + 5]/4 - 1)
= sqrt(1 + 2sqrt(5) + 5 - 1)
= sqrt(5 + 2sqrt(5))

A = (s^2)/4 [sqrt(10 + 2sqrt(5)) + sqrt(5 + 2sqrt(5))]

Comparing this with one of the found formulas, it looks like the radicals
should be closely related even though they don't look it. Let's try seeing
if one is a multiple of the other:

sqrt(10 + 2sqrt(5))    sqrt(5 - 2sqrt(5))
------------------- *  ------------------
sqrt(5 + 2sqrt(5))     sqrt(5 - 2sqrt(5))

sqrt[(10 + 2sqrt(5))(5 - 2sqrt(5))]
= -----------------------------------
sqrt[(5 + 2sqrt(5))(5 - 2sqrt(5))]

sqrt[50 - 20sqrt(5) + 10sqrt(5) - 20]
= -------------------------------------
sqrt[25 - 20]

sqrt[30 - 10sqrt(5)]
= --------------------
sqrt(5)

= sqrt[6 - 2sqrt(5)]

So now we've shown that

sqrt(10 + 2sqrt(5))/sqrt(5 + 2sqrt(5)) = sqrt[6 - 2sqrt(5)]

That's not as simple as I was expecting, but let's continue.

A trick for simplifying expressions like this is to suppose it can be
written as a + b sqrt(5), where a and b are integers, and then equate the
squares of the two forms:

(a + b sqrt(5))^2 = 6 - 2sqrt(5)

a^2 + 5b^2 + 2ab sqrt(5) = 6 - 2sqrt(5)

Equating rational and irrational parts,

a^2 + 5b^2 = 6
2ab = -2

The latter gives us

b = -1/a

Back to the rational part:

a^2 + 5(-1/a)^2 = 6

a^4 + 5 = 6a^2

a^4 - 6a^2 + 5 = 0

(a^2 - 5)(a^2 - 1) = 0

If we take a = -1 (a choice needed to get a positive result), then b = 1.
Now we've found that

sqrt[6 - 2sqrt(5)] = sqrt(5) - 1

You can quickly check this by squaring the right side.

So now I know that

sqrt(10 + 2sqrt(5))/sqrt(5 + 2sqrt(5)) = sqrt(5) - 1

Equivalently,

sqrt(10 + 2sqrt(5)) = (sqrt(5) - 1) sqrt(5 + 2sqrt(5))

So, finally, we have

A = (s^2)/4 [sqrt(10 + 2sqrt(5)) + sqrt(5 + 2sqrt(5))]

= (s^2)/4 [(sqrt(5) - 1) sqrt(5 + 2sqrt(5)) + sqrt(5 + 2sqrt(5))]

= (s^2)/4 [sqrt(5)sqrt(5 + 2sqrt(5))]

= (s^2)/4 [sqrt(5(5 + 2sqrt(5)))]

This was one of the forms you had found. I'll leave it to you to show that
the other form is equivalent.

That was a bit more challenging than I expected; it used almost all the
tools I have for dealing with radical expressions. What a workout!

- Doctor Peterson, The Math Forum at NCTM
http://mathforum.org/dr.math/

```

```
Date: 08/02/2016 at 14:16:32
From: Thayer
Subject: Thank you (Area of Regular Pentagon)

Thayer
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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