+0  
 
0
76
9
avatar

As shown in the figure, an irregular quadrilateral is inscribed in a semicircle with its base being the diameter of the semicircle. Find the radius r of the semicircle.

 

 Dec 19, 2019
 #1
avatar+23788 
+2

As shown in the figure, an irregular quadrilateral is inscribed in a semicircle with its base being the diameter of the semicircle.

Find the radius r of the semicircle.

 

\(\begin{array}{|lrcll|} \hline (1) & 180^\circ &=& 4A+2B \quad | \quad :2 \\ & 90^\circ &=& 2A+B \\ & \mathbf{B} &=& \mathbf{90^\circ -2A} \\\\ (2) & \sin(B) &=& \dfrac{7}{r} \quad | \quad \mathbf{B=90^\circ -2A} \\\\ & \sin(90^\circ -2A) &=& \dfrac{7}{r} \\\\ & \cos(2A) &=& \dfrac{7}{r} \quad | \quad \cos(2A) = 1-2\sin^2(A) \\\\ & 1-2\sin^2(A) &=& \dfrac{7}{r} \quad | \quad \mathbf{\sin(A)=\dfrac{3}{r} } \\\\ & 1-2\left(\dfrac{3}{r}\right)^2 &=& \dfrac{7}{r} \\\\ & 1-\dfrac{2*9}{r^2} &=& \dfrac{7}{r} \\\\ & 1-\dfrac{18}{r^2} &=& \dfrac{7}{r} \quad | \quad * r^2 \\\\ & r^2-18 &=& 7r \\ & r^2-7r-18 &=& 0 \\ &(r-9)(r+2) &=& 0 \\ &\mathbf{r} &=& \mathbf{9} \\ \hline \end{array}\)

 

laugh

 Dec 19, 2019
edited by heureka  Dec 19, 2019
 #3
avatar+106516 
+2

Very nice, heureka!!!!

 

This one completely stumped me  !!!!

 

 

cool cool cool

CPhill  Dec 19, 2019
 #7
avatar+23788 
+2

Thank you, CPhill !

 

laugh

heureka  Dec 19, 2019
 #2
avatar+163 
+2

Woah!!! So concise. 

 Dec 19, 2019
 #4
avatar+106516 
+2

Heureka  is very good  !!!!

 

cool cool cool

CPhill  Dec 19, 2019
 #5
avatar+163 
+3

I agree!!! cool

 Dec 19, 2019
 #8
avatar+23788 
+3

Thank you, EpicWater !

 

laugh

heureka  Dec 19, 2019
 #6
avatar+163 
0

I kinda need help on a function question of my own. Help would be appreaciated. I kinda need it quick, but it is fine if you can't answer it now. Thanks for helping everyone of their questions, though!!!

 Dec 19, 2019
 #9
avatar+163 
0

you deserve it!!!

 Dec 19, 2019

13 Online Users

avatar
avatar