An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.

Guest Jun 15, 2020

#1

#3**+3 **

An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.

BC = 2 BQ = 1 ∠B = 60º

Height AQ = tan(B) * BQ = 1.732050808

AO = 2/3 * AQ = 1.154700537 AO = XO

AP = AQ / 2 = 0.866025402

PO = AO - AP = 0.288675134

XP = sqrt( XO² - PO² ) = 1.118033987

XY = 2 * XP = 2.236067975

Dragan
Jun 17, 2020

#2**+3 **

**An equilateral triangle of side length 2 units is inscribed in a circle. **

**Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.**

\(\begin{array}{|rcll|} \hline \text{In $\triangle COD$ } \\ \mathbf{r=\ ?} \\ \hline 2^2 &=& r^2+r^2-2rr\cos(120^\circ) \\ 4 &=& 2r^2\Big(1-\cos(120^\circ)\Big) \\ 2 &=& r^2\Big(1-\cos(120^\circ)\Big) \quad | \quad \cos(120^\circ) = -\dfrac{1}{2}\\ 2 &=& r^2\left(1+\dfrac{1}{2}\right) \\ 2 &=& \dfrac{3}{2}r^2 \\\\ \mathbf{r^2} &=& \mathbf{\dfrac{4}{3}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In $\triangle OED$ } \\ \mathbf{h=\ ?} \\ \hline 1+h^2 &=& r^2 \\ h^2 &=& r^2-1 \quad | \quad \mathbf{r^2=\dfrac{4}{3}} \\ h^2 &=& \dfrac{4}{3}-1 \\ h^2 &=& \dfrac{1}{3} \\\\ \mathbf{h^2} &=& \mathbf{\dfrac{1}{3}} \\ \mathbf{h} &=& \mathbf{\dfrac{1}{\sqrt{3}} } \\ \hline \end{array}\)

**cos-rule:**

\(\begin{array}{|rcll|} \hline \text{In $\triangle OEB$ } \\ \mathbf{x=\ ?} \\ \hline r^2 &=& h^2+x^2-2hx\cos(90^\circ+60^\circ) \\\\ r^2 &=& h^2+x^2-2hx\cos(150^\circ) \quad | \quad \mathbf{r^2=\dfrac{4}{3}},\ \mathbf{h^2=\dfrac{1}{3} } \\\\ \dfrac{4}{3} &=& \dfrac{1}{3}+x^2-2hx\cos(150^\circ) \quad | \quad \mathbf{h=\dfrac{1}{\sqrt{3}} },\ \cos(150^\circ)=-\dfrac{\sqrt{3}}{2} \\\\ 1 &=& x^2+\dfrac{2}{\sqrt{3}}x\dfrac{\sqrt{3}}{2} \\\\ 1 &=& x^2+x \\ \mathbf{x^2+x-1} &=& \mathbf{0} \\\\ x &=& \dfrac{-1\pm \sqrt{1^2-4(-1)} }{2} \\ x &=& \dfrac{-1\pm \sqrt{5} }{2} \\ x &=& \dfrac{-1 \mathbf{+} \sqrt{5} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{\sqrt{5}-1}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{length of chord $\mathbf{= 1+2x}$} \\ \hline && 1+2x \\\\ &=& 1+ \mathbf{\dfrac{2(\sqrt{5}-1)}{2}} \\\\ &=& 1+ \sqrt{5}-1 \\ &=& \mathbf{\sqrt{5}} \\ \hline \end{array}\)

The length of the chord is \(\mathbf{\sqrt{5}} \qquad (=2.23606797750)\)

heureka Jun 16, 2020