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In an isosceles triangle, the sum of its base and its height is 10, and the radius of its circumscribed circle is 25/8.  Find the sides of the triangle.

 May 26, 2020
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In an isosceles triangle, the sum of its base and its height is 10,
and the radius of its circumscribed circle is 25/8.
Find the sides of the triangle.

 

\(\begin{array}{|lrcll|} \hline & c+h &=& 10 \\ \text{or} & \mathbf{h} &=& \mathbf{10-c} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(\dfrac{c}{2}\right)^2+h^2 &=& a^2 \\ \left(\dfrac{c}{2}\right)^2+(10-c)^2 &=& a^2 \\ \dfrac{c^2}{4}+(10-c)^2 &=& a^2 \\ \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline A &=& \dfrac{ch}{2} \\\\ \mathbf{A} &=& \mathbf{\dfrac{c(10-c)}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{abc}{4A}} \quad | \quad b=a \\\\ r &=& \dfrac{ca^2}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4*\dfrac{c(10-c)}{2}} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{2c(10-c)} \\\\ r &=& \dfrac{ \dfrac{c^2}{4}+(10-c)^2 }{2(10-c)} \\\\ 2(10-c)r &=& \dfrac{c^2}{4}+(10-c)^2 \\\\ (10-c)^2 - 2r(10-c) + \dfrac{c^2}{4} &=& 0 \\\\ 100-20c+c^2 -20r + 2rc - \dfrac{c^2}{4} &=& 0 \\ \ldots \\ \mathbf{\dfrac{5}{4}c^2-c(20-2r) + 100-20r} &=& \mathbf{0} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2-4*\dfrac{5}{4}*(100-20r)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2- 5(100-20r)} }{ 2*\dfrac{5}{4}} \\ && \ldots \\\\ c &=& \dfrac{20-2r \pm \sqrt{4(r^2+5r-25)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm 2\sqrt{ r^2+5r-25 } }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{10-r \pm \sqrt{ r^2+5r-25 } }{ \dfrac{5}{4}}\quad | \quad r=\dfrac{25}{8} \\\\ c &=& \dfrac{6.875 \pm \sqrt{ 0.390625 } }{ 1.25} \\\\ c &=& \dfrac{6.875 \pm 0.625 }{ 1.25} \\\\ c_1 = \dfrac{7.5}{1.25} && c_2 = \dfrac{6.25}{1.25} \\\\ \mathbf{c_1=6} && \mathbf{c_2=5} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \mathbf{a_1^2} &=& \mathbf{\dfrac{6^2}{4}+(10-6)^2} \quad | \quad c_1=6 \\ a_1^2 &=& \dfrac{36}{4}+4^2 \\ a_1^2 &=& 9+16 \\ a_1^2 &=& 25 \\ \mathbf{a_1} &=& \mathbf{5} \\ \hline \mathbf{a_2^2} &=& \mathbf{\dfrac{5^2}{4}+(10-5)^2} \quad | \quad c_2=5 \\ a_2^2 &=& \dfrac{25}{4}+5^2 \\ a_2^2 &=& \dfrac{25*5}{4} \\ \mathbf{a_2} &=& \mathbf{\dfrac{5}{2}\sqrt{5}} \\ \hline \end{array} \)

 

laugh

 May 26, 2020

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