Equilateral Triange inside a circle

**Skilo** · February 15th 2010, 11:47 PM

Hello all,

I'm hoping someone can help with this one, I'm not sure why I'm not seeing it

. I've attached a picture from my text book and an answer I've found but don't quite understand.

Question:
An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:

$<br /> r^2=\frac{x^2}{3}<br />$

An answer I've found already but don't understand:

. Thus:

. Solve for r² and you'll get:

However, I'm not sure how one is supposed to know that $r=\frac{2}{3}\cdot h$ everything else made sense though...

Thanks,
Skilo

**Prove It** · February 16th 2010, 12:00 AM

Originally Posted by Skilo

Hello all,

I'm hoping someone can help with this one, I'm not sure why I'm not seeing it

. I've attached a picture from my text book and an answer I've found but don't quite understand.

Question:
An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:

$<br /> r^2=\frac{x^2}{3}<br />$

An answer I've found already but don't understand:

. Thus:

. Solve for r² and you'll get:

However, I'm not sure how one is supposed to know that $r=\frac{2}{3}\cdot h$ everything else made sense though...

Thanks,
Skilo

The area of the triangle can found using Heron's Formula:

$s = \frac{x + x + x}{2} = \frac{3x}{2}$

$A = \sqrt{s(s - x)(s - x)(s - x)}$

$= \sqrt{\frac{3x}{2}\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)}$

$= \sqrt{\frac{3x}{2}\left(\frac{x}{2}\right)\left(\f rac{x}{2}\right)\left(\frac{x}{2}\right)}$

$= \sqrt{\frac{3x^4}{16}}$

$= \frac{\sqrt{3}x^2}{4}$ .

We can also work out the area using the fact that the equilateral triangle is made up of three equal triangles of side lengths $r, r, x$ , with an angle of $120^{\circ}$ between the sides of lengths $r, r$ .

So the area of the equilateral triangle will be

$A = 3\cdot \frac{1}{2}\cdot r \cdot r \cdot \sin{120^\circ}$

$= \frac{3r^2}{2}\cdot \frac{\sqrt{3}}{2}$

$= \frac{3\sqrt{3}r^2}{4}$ .

Since the areas are equal, we have

$\frac{\sqrt{3}x^2}{4} = \frac{3\sqrt{3}r^2}{4}$

Therefore $x^2 = 3r^2$

or $r^2 = \frac{x^2}{3}$ .

**pragraphic** · February 16th 2010, 02:07 AM

for that h = (3/2) r :

First of all, the line from the center to the points of the equil triangle has 2 properties:
a. the length is equal to radius of circle (everybody knows)
b. the line bisect the angle of the equil triangle

So, consider h is the height of the equil triangle
h = r + s , assuming s is just the little portion extending from the line r.

do you see sin 30 = s / r ?
sin 30 = s / r
(1/2) = s / r
s = (1/2) r
thus h = r + (1/2) r
h = (3/2) r
and r = (2/3) h

To prove the properties above, it involves showing 2 trangles are SSS. Just try to join lines from the center and you will see how.

**Soroban** · February 16th 2010, 05:00 AM

Hello, Skilo!

However, I'm not sure how one is supposed to know that $r\,=\,\tfrac{2}{3}h$

Code:

                A
   -          * o *
   :      *    /|\    *
   :    *     / | \     *
   :   *     /  |r \     *
   h        /   |   \
   :  *    /    |    \    *
   :  *   /     oG    \   *
   :  *  /   *  |  *   \  *
   :    / *     |     * \
   - B o- - - - o - - - -o C
        *       M       *
          *           *
              * * *

We have an equilateral triangle inscribed in a circle.

The center of the circle $G$ is the centroid,
. . center of mass of the triangle,
. . which divides the median $AM$ in the ratio $2:1$

**bjhopper** · February 16th 2010, 08:27 AM

Posted by Skilo

Using Soroban's nice diagram apply the 30-60-90 triangle rule.
If hypothenuse is 2 the side opposite the 30 angle =1. the side opposite the 60 angle = radical 3 What is angle ACG .Can you prove it.

bjh

**Diagonal** · February 16th 2010, 06:13 PM

Drop a perpendicular form the center to one of the left or right sides, and form a 30, 60, 90 triangle whose sides are in the ratio 1:2:sqrt(3) Then you have...

r/(x/2) = 2/sqrt(3), and the result follows directly.

**Skilo** · February 16th 2010, 10:17 PM

Thanks everyone for all the help

!

**pragraphic** · February 16th 2010, 10:20 PM

Using the Soroban's diagram (thanks Soroban)

The proves for:
1. Why angles ABG, BAG, BCG, CBG, ACG, CAG are all 30 degree:
- triangle ABC is a equil trangle, so AB=BC=AC
- Angles ABC=CBA=CAB=60 degrees
- AG=BG=CG=radius
- So, triangles ABG, BCG, CAG are SSS
- thus angles ABG=BAG=BCG=CBG=ACG=CAG=30 degree
- so, BG bisect angle ABC, CG bisect angle BCA, AG bisect angle CAB

2. After extending the line AG to M becoming AM, why AM perp to BC:
conside triangle BAM and CAM
- AB=AC
- AM=AM
- angle BAM = angle CAM
so, triangle BAM and triangle CAM are SAS
thus angle AMB = angle AMC = 90 degree
so, AM perp to BC.

So, AM is the height of the triangle ABC, passing the centre of the circle.
let h = AM = AG + GM
let r = AG = BG = radius
let s = GM
h = r + s
sin 30 = GM / BG
1 / 2 = s / r
s = (1/2) r
so h = r + s
= r + (1/2) r
= (3/2) r
and r = (2/3) h

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