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In triangle ABC, side AB = 20, AC = 11, and BC = 13.  Find the diameter of the semicircle inscribed in triangle ABC, whose diameter lies on AB, and that is tangent to AC and BC. 

 Nov 19, 2019
 #1
avatar+23542 
+4

In triangle ABC, side AB = 20, AC = 11, and BC = 13.  

Find the diameter of the semicircle inscribed in triangle ABC, whose diameter lies on AB, and that is tangent to AC and BC. 

 

1. sin-rule:

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{\sin(A)}{13} &=& \dfrac{\sin(B)}{11} \\ & \mathbf{\sin(A)} &=& \mathbf{\dfrac{13}{11} \sin(B)} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (2) & 20 &=& x+y+2r \\ & \mathbf{x} &=& \mathbf{20-y-2r} \\ \hline \end{array}\)

 

2. 

\(\begin{array}{|lrcll|} \hline (3) & \mathbf{\sin(B)} &=& \mathbf{\dfrac{r}{y+r}} \\\\ (4) & \sin(A) &=& \dfrac{r}{x+r} \quad | \quad \mathbf{ \sin(A)= \dfrac{13}{11} \sin(B) } \\\\ & \dfrac{13}{11} \sin(B) &=& \dfrac{r}{x+r} \quad | \quad \mathbf{x=20-y-2r} \\\\ & \dfrac{13}{11} \sin(B) &=& \dfrac{r}{20-y-2r+r} \\\\ & \mathbf{\dfrac{13}{11} \sin(B)} &=& \mathbf{\dfrac{r}{20-y-r}} \\ \hline \end{array}\)

 

3. cos-rule:

\(\begin{array}{|lrcll|} \hline (5) & 11^2 &=& 13^2+20^2-2*13*30*\cos(B) \\\\ & \cos(B) &=& \dfrac{13^2+20^2-11^2}{2*13*30} \\ & \cos(B) &=& \dfrac{56}{65} \\\\ & \sin(B) &=& \sqrt{1-\cos^2(B)} \\ & \sin(B) &=& \sqrt{1- \left(\dfrac{56}{65}\right)^2 } \\ & \mathbf{\sin(B)} &=& \mathbf{ \dfrac{33}{65}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (3) & \mathbf{\sin(B)} &=& \mathbf{\dfrac{r}{y+r}} \quad | \quad \mathbf{\sin(B)=\dfrac{33}{65}} \\ & \dfrac{33}{65} &=& \dfrac{r}{y+r} \\ & 33(y+r) &= 65r \\ & 33y &=& 65r - 33r \\ & 33y &=& 32r \\ (6) & \mathbf{y} &=& \mathbf{ \dfrac{32}{33}r } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (4) & \mathbf{\dfrac{13}{11} \sin(B)} &=& \mathbf{\dfrac{r}{20-y-r}} \quad | \quad \mathbf{\sin(B)=\dfrac{33}{65}},\ \mathbf{y=\dfrac{32}{33}r } \\\\ & \dfrac{13}{11}*\dfrac{33}{65} &=& \dfrac{r}{20-\dfrac{32}{33}r-r} \\\\ & \dfrac{13*3}{65} &=& \dfrac{r}{20-\dfrac{65}{33}r } \\\\ & \dfrac{39}{65} &=& \dfrac{r}{20-\dfrac{65}{33}r } \\\\ & \left( 20-\dfrac{65}{33}r\right)*39 &=& 65r \\\\ & 20*39 - \dfrac{65*39}{33}r &=& 65r \\\\ & 65r + \dfrac{65*39}{33}r &=& 20*39 \\ \\ & 65r*\left(1 + \dfrac{ 39}{33} \right) &=& 20*39 \\ \\ & \dfrac{65*72}{33}r &=& 20*39 \\ \\ & r &=& \dfrac{33*20*39 } {65*72}\\ \\ & \mathbf{r} &=& \mathbf{5.5} \\ \hline \end{array}\)

 

The diameter of the semicircle \(= 2r = 2*5.5 = \mathbf{11}\)

 

laugh

 Nov 19, 2019
 #2
avatar+105370 
+1

Very nice, heureka  !!!!

 

 

cool cool cool

CPhill  Nov 19, 2019
 #3
avatar+23542 
+2

Thank you, CPhill !

 

laugh

heureka  Nov 20, 2019

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