+0  
 
0
98
2
avatar+183 

In right triangle ABC, AB=10, AC=6 and BC=8 units. What is the distance from C to the midpoint of segment AB ?

Confusedperson  Jun 2, 2018
 #1
avatar+9520 
+1

In right triangle ABC, AB=10, AC=6 and BC=8 units. What is the distance from C to the midpoint of segment AB ?

laugh

Omi67  Jun 2, 2018
 #2
avatar+87639 
+1

Here's another way

 

The area of the triangle  is   (1/2) product of the legs  = (1/2) (6) (8)  = 24 units^2

 

So....calling AB the base....the altitude, h,  is : 

24  = (1/2) (10) h

24/5  =  h

Call the altitude, CD

 

So.....

 

AD  = sqrt [ AC^2  - CD^2] =

sqrt (6^2  - (24/5)^2)   =

sqrt  (36  - 576/25 ]  =

sqrt [ 324/25]  =

18/5

 

So....DM  = 5 - 18/5  =  7/5

 

So  CM 

sqrt  ( DM^2  + CD^2 ] = 

sqrt [ (7/5)^2  + (24/ 5)^2 ]  = 

sqrt [ 49  + 576] / 5 =

sqrt [625] /5  =

25 / 5   =

5

 

 

cool cool cool

CPhill  Jun 2, 2018

19 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.