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Let $C$ and $D$ be points on the semicircle with diameter $\overline{AB}$. If $AD = BC = 3$ and $CD = 7$, then find the radius of the semicircle.

Guest May 8, 2018
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Let $C$ and $D$ be points on the semicircle with diameter $\overline{AB}$. If $AD = BC = 3$ and $CD = 7$,

then find the radius of the semicircle.

 

 

1. sin rule:

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(180^{\circ}-2\varphi) } {7} &=& \dfrac{ \sin(\varphi) }{r} \\ \dfrac{2\sin(\varphi)\cos(\varphi)}{7} &=& \dfrac{\sin(\varphi)}{r} \\ \dfrac{2 \cos(\varphi)}{7} &=& \dfrac{1}{r} \\ \cos(\varphi) &=& \dfrac{7}{2r} \\ \hline \end{array} \)

 

2. cos rule:

\(\begin{array}{|rcll|} \hline 3^2 &=& r^2+r^2-2rr\cos(\varphi) \\ 9 &=& 2r^2(1-\cos(\varphi)) \\ \dfrac{9}{2r^2} &=& 1-\cos(\varphi) \quad & | \quad \cos(\varphi) = \dfrac{7}{2r} \\ \dfrac{9}{2r^2} &=& 1-\dfrac{7}{2r}\quad & | \quad \cdot 2r^2 \\ 9 &=& 2r^2 -7r \\ 2r^2 -7r -9 &=& 0 \\ r &=& \dfrac{7 \pm \sqrt{49-4\cdot 2\cdot (-9) } } {2\cdot 2} \\ &=& \dfrac{7 \pm \sqrt{49+72 } } {4} \\ &=& \dfrac{7 \pm \sqrt{121 } } {4} \\ r &=& \dfrac{7 \pm 11 } {4} \quad & | \quad r > 0\ !\\ &=& \dfrac{7 +11 } {4} \\ &=& \dfrac{18 } {4} \\ \mathbf{r} &\mathbf{=}& \mathbf{4.5} \\ \hline \end{array}\)

 

The radius of the semicircle is 4.5

 

laugh

heureka  May 8, 2018
edited by heureka  May 8, 2018
edited by heureka  May 8, 2018

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