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Find the ratio of the area of triangle BCX to the area of triangle ACX in the diagram if CX bisects the angle ACB. Express your answer as a common fraction. This is the image https://latex.artofproblemsolving.com/2/4/1/2415560a3ecd06119f3e9d785d5d655c158f158e.png

Guest Jul 6, 2017
 #1
avatar+87334 
+1

By Euclid, whenever an apex angle is bisected, we have the following relationship :

 

AX/AC  = BX/BC

 

AX/30  = BX/27   which implies that

 

27/30  = BX/AX 

 

9 /10  = BX/AX     →  BX = (9/10) AX

 

And triangle ACX  has the same height  as triangle BCX 

 

And again, by Euclid, triangles under the same height are to each other as their bases

 

So....the area of triangle BCX = (9/10)ACX

 

And the ratio of their areas is  :   

 

BCX :  ACX  =   9 : 10

 

 

cool cool cool

CPhill  Jul 7, 2017
 #2
avatar+19653 
+1

Find the ratio of the area of triangle BCX to the area of triangle ACX in the diagram if CX bisects the angle ACB. 

Express your answer as a common fraction.

 

 

Let \(A_1=\) area of BCX

Let \(A_2=\) area of ACX

Let \(\varphi = \angle ACB\)

Let ratio = \(\frac{A_1}{A_2}\)

 

\(\begin{array}{|lrcll|} \hline (1) & 2\cdot A_1 &=& \overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) \\ (2) & 2\cdot A_2 &=& \overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) \\ \hline \frac{(1)}{(2)}: & \dfrac{2\cdot A_1}{2\cdot A_2} &=& \dfrac{\overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) }{\overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) } \\ & \dfrac{ A_1}{ A_2} &=& \dfrac{ 27 }{ 30 } \\ & &=& \dfrac{ 3\cdot 9 }{ 3\cdot 10 } \\ & &=& \dfrac{ 9 }{ 10 } \\\\ &\mathbf{ ratio } & \mathbf{=} & \mathbf{ \dfrac{9}{10} } \\ \hline \end{array}\)

 

laugh

heureka  Jul 7, 2017

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