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The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.

The picture is here as I can't upload it for some reason: https://cutt.ly/auSLG9T

 May 22, 2021
 #1
avatar+119820 
+3

I  think  there  might  be  some "formula" for solving this in an easier fashion, but, I believe  the Law of Cosines  might also work

 

Note  that angles  DCB  and DAB  are supplemental   so     cos ( DCB)   =  -cos (DAB)

 

Using the Law of  Cosines we have this system    (and subbing  -cos (DAB) for cos (DCB)  )

 

DB^2   =  3^2  + 2^2   - 2 ( 3 * 2)  (-cos DAB)

DB^2   =  1^2   + 4^2  - 2 ( 4 * 1)   (cos DAB)

 

DB^2  =  13  +  12 cos DAB

DB^2  =  17   -   8 cos DAB

 

Multiply  the  first equation  by     8    and the  second  by    12

 

8 DB^2   =  104  + 96  cos DAB

12 DB^2  = 204  -  96  cos DAB               add  these

 

20 DB^2  =   308

 

DB^2  =  308  / 20     =    15.4  

 

 

cool cool cool

 May 22, 2021
 #2
avatar+1293 
+3

The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

a = 1     b = 2     c = 3     d = 4      s = 5

Using the above formula, we can calculate the circumradius R.

 

R = 2.002602473

 

Now, we can calculate the measure of angle C.

 

Angle C = 101.53695902º

 

Finally, using the law of cosines, we can find the length of BD.

 

BD ≈ 3.924         BD2 = 15.4

 May 22, 2021

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