+0  
 
0
35
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?


Guest Apr 21, 2018
Sort: 

2+0 Answers

 #1
avatar+646 
+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.

GYanggg  Apr 21, 2018
 #2
avatar+86613 
+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

CPhill  Apr 22, 2018

28 Online Users

avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy