We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
245
2
avatar

In the diagram, \(AB = 13\text{ cm},\) \( DC = 20\text{ cm},\) and \( AD = 5\text{ cm}.\) What is the length of  to the nearest tenth of a centimeter?


 Apr 21, 2018
 #1
avatar+982 
+1

You didn’t specify what we need to find, but assuming you want BC, here is my solution:

 

Using the Pythagorean Thereom, BD^2 = AB^2 - AD^2 = 13^2 - 5^2 = 12^2 

 

BD = 12

 

Using the same method, BC^2 = DC^2 - BD^2 = 20^2 - 12^2

 

You take the square root of that and you have your answer.

 Apr 21, 2018
 #2
avatar+101088 
+1

And if you want to find AC   ...

 

cos BDC  =  12/20  = 3/5

cos BDA  = 5/13

sin BDC = 4/5

sin BDA = 12/13

 

So  cos  (BDC +  BDA)  =  cosBDCcosBDA - sinBDC sinBDA =

                                            (3/5)(5/13) - (4/5)(12/13)  =

                                             15/65  -  48/65 =

                                              -33/65

 

So....using the  Law of Cosines

 

AC^2  = AD^2 + DC^2  - 2(AD * DC) cos ( BDC + BDA)

 

AC^2  = 5^2   +  20^2  - 2(5 * 20)(-33/65)

 

AC^2  =  425   + 200(33/65)

 

AC^2  = 6845 / 13

 

AC  ≈ 22.946

 

 

cool cool cool

 Apr 22, 2018

14 Online Users