+0

# Length of an equilateral traingle within a circle

0
1
2017
33

What is the side length of an equilateral triangle that fits exactly inside a circle with a 9 inch diameter?

Jan 28, 2015

#3
+10

The Law of Cosines is this

c^2  = a^2 + b^2 - 2(a)(b)cos(theta)

Where c is the opposite side to the angle "theta" ....and "theta" is the included angle between sides a and b.

So...... in the circle, a and b are radii = 9  ....and since there are three equal sides of the triangle, each side spans an arc of 120 degrees.   Jan 28, 2015

#1
+5

We can use the Law of Cosines to solve this..letting x be the side length, we have...

x^2 =  (4.5)^2 + (4.5)^2 - 2(4.5)(4.5)cos120

x= √60.75  = 7.7942286340599478  = 4.5√3   Thanks, Mathematician...for seeing my previous error...!!!

Jan 28, 2015
#2
+5

Can you give me the formula, and then the explanation? And also how to get an appriximation of the answer? I'm confused by your explanation (Probably because I'm unfamiliar with Cosines.)

Jan 28, 2015
#3
+10

The Law of Cosines is this

c^2  = a^2 + b^2 - 2(a)(b)cos(theta)

Where c is the opposite side to the angle "theta" ....and "theta" is the included angle between sides a and b.

So...... in the circle, a and b are radii = 9  ....and since there are three equal sides of the triangle, each side spans an arc of 120 degrees.   CPhill Jan 28, 2015
#4
+5

Just one final question. Can you explain it without Trigonmetry (cosines)? I haven't learned that yet.

Jan 28, 2015
#5
+5

I do not know Mathematician.

Maybe another mathematician can come up with something. Jan 28, 2015
#6
0

Sorry, Mathematician....I'm kinda' 'stuck" with using Trig.....if I think of some geometric way to do it...I'll let you know...!!!   Jan 28, 2015
#7
+5

I just realized that the answer doesn't make sense; shouldn't the side be less than the diameter to fit within the circle?

Jan 28, 2015
#8
0

Eh...you're right..I read "diameter" as "radius'.....let me go back and correct that....sorry!!!   Jan 28, 2015
#9
+5

Note in the pic below that FH = (1/2)s    where s is the side length

And also that DA =  AF = r = 4.5

So we have, by the Pythagorean Theorem

AF^2 =  (1/2)s^2 + AH^2

4.5^2 = 20.25 = (1/2)s^2 + AH^2

AH^2 = 20.25 - (s/2)^2

AH = √(20.25 - (s/2)^2 )

Therefore, again using the Pythagorean Theorem, we have

s^2 = (s/2)^2 + [DA + AH]^2

s^2 = (s/2)^2 + ((4.5 + √(20.25 - (s/2)^2 ))^2

(3/4)s^2 = 20.25 + 9 √(20.25 - (s/2)^2 ) + 20.25 - (s/4)^2   (add (s/4)^2 to each side)

s^2 = 40.5 + 9 √(20.25 - (s/2)^2 )  subtract 40.5 from each side

s^2 - 40.5  = 9 √(20.25 - (s/2)^2 )   square both sides

(s^2 - 40.5)^2 = 81 (20.25 - (s^2)/4)

s^4 - 81s^2 + 1640.25 = 1640.25 - (81/4)s^2

s^4 - 81s^2 + (81/4)s^2 = 0

s^4 - (243/4)s^2 = 0

Factor

(s^2) (s^2 - (243/4)) =0   one solution is s = 0 .... reject this

So we have

(s^2 - 243/4) = 0

s = (1/2)√243 = (1/2)(9)√3 = 4.5√3 = 7.7942286340599478

Here's a pic....(the angles are not exact..it's just to give you a feel for the geometry of the problem)    7.7942286340599478#sthash.QLAWaiN0.dpuf
7.7942286340599478#sthash.QLAWaiN0.dpuf
Jan 28, 2015
#10
+5

I had a thought Chris,  we (well Alan or Heureka) could probably do this with calculus as well. Jan 28, 2015
#11
+5

I've done it with Calculus, before......but...since "Mathematician" hasn't had trig......I doubt that Calculus would do him much good, either......LOL!!!   Jan 28, 2015
#12
0

Yes I was't thinking that it would be much use for Mathematician.  LOL

Jan 28, 2015
#13
+5

My solution below is basically the same as Chris's, but laid out slightly differently:  .

Jan 28, 2015
#14
+5

What is the side length of an equilateral triangle that fits exactly inside a circle with a 9 inch diameter ?

I.

$$\cos{(60\ensurement{^{\circ}})}=\frac{r-h}{r}=\frac{1}{2} \quad | \quad \small{\text{h = the height of the chord}} \\ \Rightarrow \boxed{h=\dfrac{r}{2}}$$

II.

Intersecting chord theorem:

$$\small{\text{  h*(d-h)= (\frac{x}{2})*(\frac{x}{2}) = (\frac{x}{2})^2  }}\\ \Rightarrow \boxed{x = d* \frac{ \sqrt{3} } {2}= 4.5* \sqrt{3}}$$ Jan 28, 2015
#15
+5

okay I have looked at CPhill's answer and I decided to reproduce it - but it is still CPhill's answer.

I left labelling the same as CPhill's except that I called the side length 2x inches. First I want to find AH in terms of x

$$\\AH^2=4.5^2-x^2\\ AH^2=20.25-x^2\\ AH=\sqrt{20.25-x^2}\\$$

Now consider the big right angled triangle DEH

$$\begin{array}{rll} DH^2&=&DE^2-EH^2\\ (DA+AH)^2&=&(2x)^2-x^2\\ (4.5+\sqrt{20.25-x^2})^2&=&4x^2-x^2\\ (4.5+\sqrt{20.25-x^2})^2&=&3x^2\\ \sqrt{(4.5+\sqrt{20.25-x^2})^2}&=&\sqrt{3x^2}\\ 4.5+\sqrt{20.25-x^2}&=&\sqrt{3}\;x\\ \sqrt{20.25-x^2}&=&\sqrt{3}\;x-4.5\\ (\sqrt{20.25-x^2})^2&=&(\sqrt{3}\;x-4.5)^2\\ 20.25-x^2&=&3x^2-2*4.5\sqrt{3}\;x+4.5^2\\ 20.25-x^2&=&3x^2-9\sqrt{3}\;x+20.25\\ -4x^2&=&-9\sqrt{3}\;x\\ x^2&=&\frac{9\sqrt{3}\;x}{4}\\ x&=&\frac{9\sqrt{3}}{4}\\ 2x&=&\frac{9\sqrt{3}}{2}\\ \end{array}$$

So each side is exactly   $$\frac{9\sqrt{3}}{2}$$    inches long

Jan 28, 2015
#16
+5

One can turn the equilateral triangle, thus one side is vertical, then the height (h) of the curve (chord) is half the radius. Jan 28, 2015
#17
0

I do not understand Heureka. Jan 28, 2015
#18
+5

+

|   .      +

|      .            +

x/2    |          .                +

|            r                      +

|                .                         +

|           60      .                              +

arc     (<----r/2---->+<----r/2----> C <------------r--------->) arc

<-      h   -> | <-   r - h    ->                            +

|                                          +

|                                   +

x/2       |                            +

|                     +

|            +

|      +

+

Jan 28, 2015
#19
+8

I was just looking at what Heureka was saying.  It took me a while to work it out. B

This is the triangle that Heureka was describing

The radius that bisects the chord is perpendicular to the chord so <AHE=90 degrees

which means <AEH=30 degrees (Angle sum of a triangle)

Using trig I know that AH=0.5*AE=2.25  but I am not allowed to use trig. Now Heureka says that AH=HB

Edited: that is because AB is one radius and AH is 1/2 radius so  HB must also be a 1.2 radius

Join EB forming a new triangle HEB

Now triangle HEB is congruent to triangle HEA because all the angles are congruent and they have a common corresponding side.

therefore triangle AEB is equilateral  and HA=HB

therefore all the sides are  4.5 inches

therefore HA=HB=2.25 inches

$$\\x^2=4.5^2-2.25^2\\ x^2=20.25-5.0625\\ x^2=15.1875\\ x^2=151875/10000\\ x=\sqrt{151875/10000}\\ x=\sqrt{25*25*81*3/10000}\\ x=(25*9/100)\sqrt{3}\\ x=(2.25)\sqrt{3}\\ side = 4.5\sqrt{3}\;\;inches$$

Thanks Heureka

Jan 28, 2015
#20
+5

Now Heureka says that AH=HB

This certainly looks right but at present I can't see how to prove it.

Given that AH = r/2, AD = r and DB = 2r  where r is the radius, then HB = DB - AH - AD = 2r - r/2 - r  = r/2 = AH

.

Jan 28, 2015
#21
+5

Hi Melody,

i thank you very much.  Sorry i  can't  Insert pictures. I don't know why.

bye. Jan 28, 2015
#22
+5

Heureka, have you tried the process explained here:  http://web2.0calc.com/questions/how-to-upload-an-image-nbsp_1 ?

.

Jan 28, 2015
#23
+5

Hi Alan, Jan 28, 2015
#24
0

Thank you Alan.  Yes is usual for two half radii to equal 1 radii.  I guess I had a brain freeze Jan 29, 2015
#25
0

Why doesn't some smart person show their skill and do this with calculus. Jan 29, 2015
#26
+5

Hi Melody,

my calculation with your picture of the triangle without trigonometry:

I.  The triangle BED is an right angle triangle ( Thales Theorem ). See : http://www.mathopenref.com/thalestheorem.html

II.  The triangle AEB is equilateral so the side EB = r ! ( You have worked out this )

III. We have the Altitude on Hypotenuse Theorem in a right triangle : EB * EB = BH * BD . See: http://www.dummies.com/how-to/content/how-to-solve-problems-with-the-altitude0nhypotenus.html

with EB = r and BD = d = 2r  we have: r*r = BH * 2r   or   r = BH * 2   or  $$\small{\text{  \boxed{BH = \frac{r}{2}}  }}$$

IV. Now we have the Geometric mean Theorem in a right triangle:  BH * HD = x * x

with BH = r/2 and HD = d - r/2 = 2r - r/2. See: https://en.wikipedia.org/wiki/Geometric_mean_theorem

$$\small{\text{  \dfrac{r}{2}\left( 2r-\dfrac{r}{2} \right)=x^2,\ \quad r^2-\dfrac{r^2}{4} = x^2 ,\ \quad \dfrac{3}{4} r^2 = x^2 ,\ \quad \dfrac{\sqrt{3}}{2}r = x  }}\\\\ \small{\text{ The side is  2x = \sqrt{3}*r  }} \small{\text{ and  r = \frac{9}{2} = 4.5  we have the side = 4.5*\sqrt{3}  }}$$ Jan 29, 2015
#27
0

Thank you Heureka Jan 29, 2015
#28
+5

This boils down to finding the side length of an equiateral triangle inscribed in a circle. Of course, the area is maximized when the side lengths are maximized....!!!

Applying the Calculus to our given problem where r = 4.5 and x = s

We have

A = .5x(4.5 + √(20.25 - .25x^2))

A = 2.25x + .5x√(20.25 - .25x^2)

dA/dx = 2.25 +.5(20.25 - .25x^2)^(.5) - (.125x^2)(20.25 - .25x^2)^(-.5) = 0

18 + 4(20.25 - .25x^2)^(.5) - x^2(20.25 - .25x^2)^(-.5) = 0  rearrange

18 + 4(20.25 - .25x^2)^(.5) = x^2(20.25 - .25x^2)^(-.5)

18 (20.25 - .25x^2)^(.5) + 4 (20.25 - .25x^2) = x^2

18(20.25- .25x^2)^(.5) = x^2 - 4 (20.25 - .25x^2)     square both sides

324(20.25- .25x^2) = x^4 - 8 (20.25 - .25x^2)x^2 + 16(20.25-.25x^2)^2

6561 - 81x^2 = x^4 - 162x^2 + 2x^4 + 16(6561/16 - 61/8x^2 + 1/16x^4)

6561 - 81x^2 = 4x^4 - 324x^2  +6561

4x^4 - 243x^2 = 0   factor

x^2 (4x^2 - 243) = 0      reject x =0

4x^2 - 243 - 0

4x^2 = 243

x^2 = 243/4

x = √(81 * 3) / 2 =   (9/2)√3  = 4.5√3 = s

And this result can be generalized to r√3 for any given radius   Jan 29, 2015
#29
0

Thank you Chris  :D

More for me to think about :)))

Jan 29, 2015
#30
+5

Mathematician, I bet you didn't think that your question would attract THIS much attention  LOL Jan 29, 2015
#31
+5

I don't think that "Fermat's Last Theorem" got as much attention as this one did...!!!

{But..it took a lot longer to prove....over 350 years....}

Regardless...it was a good one to play around with, Mathematician.....many different approaches by our "professionals"

And even some bumbling and stumbling by me, too!!!   Jan 29, 2015
#32
+5

Yes, a very interesting question. Someone commented I was a part of the Illuminati when they saw it on my project... it seems like that craze has taken over my school.

Feb 3, 2015
#33
0

The "Illuminati".......LOL!!! BTW....do you recognize this symbol????.....if so....you may REALLY be part of  "THE ILLUMINATI"   Feb 3, 2015