# Thread: Isosceles triangle inscribed in a circle

1. ## Isosceles triangle inscribed in a circle

Determine the dimensions of the isosceles triangle inscribed in a circle of radius "r" that will give the triangle a maximum area. express the dimensions in terms of the circle's radius r.
figure:

determine the isosceles triangle's dimensions a, b,c
(sorry guys, hope someone can help me out here, i'm gonna fail if I can't solve this one and I really don't know how to..unlucky for me, I don't really have any friend at this clas..please help)

2. we need to find the area with respect of s and r then derive it to find the maximum values

ok

Attachment 14705

I divide the triangle to three equal triangles the hight of the small triangle is
using Pythagorean theorem

$h = \sqrt{r^2 - \frac{s^2}{4}}$

the area of every small triangle is

$\frac{h \cdot s }{2}$

so the area of the large triangle is

$A = \frac{3s\cdot \sqrt{r^2 - \frac{s^2}{4}}}{2}$

$A' = \left(\frac{3}{2}\right) \left( \sqrt{r^2 - \frac{s^2}{4}} - s\cdot \frac{s}{4} \cdot \frac{1}{\sqrt{r^2 - \frac{s^2}{4}}}\right)$

$A' = \left(\frac{3}{2}\right) \left( \sqrt{r^2 - \frac{s^2}{4}} - \frac{s^2}{4\sqrt{r^2 - \frac{s^2}{4}}}\right)$

$A' = \frac{4r^2 - 2s^2 }{4\sqrt{r^2 - \frac{s^2}{4}}}$

critical points

$4r^2 - s^2 = 0 \Rightarrow s = \mp r \sqrt{2}$ but we have length so ignore negatives

$r^2 - \frac{s^2}{4} \Rightarrow s = 2r$

we have two positive critical points study the sign of the Area around these two
the function is not define for s>=2r and s<=-2r

for s values -r sqrt(2) < s< r sqrt(2) the first derivative is positive so the function is increasing

for s> r sqrt(2) take for example 1.5r

$4r^2 - 2(1.5r)^2 = 4r^2 - 2(2.25 r^2) = 4r^2 - 4.5 r^2 <0$

for in this interval the first derivative is negative the function is decreasing

so you have s= r sqrt(2) as a maximum point

is there something you can't understand

3. Hello meggy garcia

Welcome to Math Help Forum!
Originally Posted by meggy garcia
Determine the dimensions of the isosceles triangle inscribed in a circle of radius "r" that will give the triangle a maximum area. express the dimensions in terms of the circle's radius r.
figure:

determine the isosceles triangle's dimensions a, b,c
(sorry guys, hope someone can help me out here, i'm gonna fail if I can't solve this one and I really don't know how to..unlucky for me, I don't really have any friend at this clas..please help)
In the attached diagram:
$AB = AC$

$\angle ABC = \angle ACB = \theta$

$AO = OB = r$, the radius of the circumcircle

$M$ is the mid-point of $BC$

$\angle AOB = 2\angle ACB = 2\theta$ (angle at centre)
So if $P$ is the mid-point of $AB$:
$AP = PB = r\sin\theta$, from $\triangle APO$

$\Rightarrow AM = 2r\sin^2\theta$ and $BM = 2r\sin\theta\cos\theta$, from $\triangle ABM$
Therefore the area of $\triangle ABC, A,$ is given by:
$A = \tfrac12$ base $\times$ height
$=BM\times AM$

$= 4r^2\sin^3\theta\cos\theta$
So we need to find the value of $\theta$ that makes $A$ a maximum. The simplest method uses calculus:
$\frac{dA}{d\theta}= 4r^2(-\sin^4\theta +3\sin^2\theta\cos^2\theta)$
$= 0$ when
$\sin^2\theta(-\sin^2\theta+3\cos^2\theta) = 0$

$\Rightarrow \sin^2\theta = 0$ or $-\sin^2\theta +3\cos^2\theta = 0$
$\sin^2\theta = 0$ gives a minimum value of $A$. The maximum occurs when
$\sin^2\theta = 3\cos^2\theta$

$\Rightarrow \tan^2\theta = 3$

$\Rightarrow \theta = \frac{\pi}{3}$
So (perhaps unsurprisingly) the area is maximum when the triangle is equilateral, with sides $2r\sin\left(\frac{\pi}{3}\right)=r\sqrt3$

4. ## thank you...

hi amer and grand dad..thanks a lot for the help..wow..you guys are really great,. this site is really helpful for someone like me,.

you guys just saved me from not graduating next month..to be honest i don't really understand the equation because this is the only thing i have to do for my trigo class, coz of exemptions and the fact that my school isnt that good..well, thanks again..you guys completed my day..

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# calculus - find the dimensions of isosceles triangle inscribe in a circle or radius r.

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