April 9th, 2018, 01:07 AM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 342 Thanks: 1  How CC` formula is written?
The lengths of the two parallel sides of an isosceles trapezium are 20 cm and 28 cm and the lengths of each of the non parallel sides is 5 cm. What is the distance between the parallel sides of the trapezium? Let ABCD be the given trapezium Let CC` be perpendicular drawn from C to B given that AB = 28 cm, CD = 20 cm BC` = 1 \ 2 (AB  CD) = 1\ 2 ( 8 ) = 4 cm CC` = $\sqrt (BC^2  BC`^2)$ $\sqrt (5^2  4^2)$ = 3 cm. CC` = 3 cm Last edited by skipjack; April 9th, 2018 at 09:02 AM. Reason: correct 25 to be 28. 
April 9th, 2018, 01:33 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,163 Thanks: 1135 
Is there a question here?

April 9th, 2018, 01:50 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 342 Thanks: 1 
How CC` formula is written?

April 9th, 2018, 04:44 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Draw the corresponding DD' perpendicular to AB from D. That divides the trapezoid into two right triangles and one rectangle. By "symmetry" (equivalently using triangle congruence theorems) the two triangles are congruent and length CC' is the same as length DD'. Length AB is 28 and Length CD is 20. So what are the lengths of AD' and BC' (which are equal? Lengths AD and BC are given as 5. Use the Pythagorean theorem to find length CC'. 
April 9th, 2018, 07:58 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,865 Thanks: 1833  
April 9th, 2018, 08:52 AM  #6 
Senior Member Joined: Aug 2014 From: India Posts: 342 Thanks: 1  
April 9th, 2018, 09:08 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,865 Thanks: 1833 
Let's assume that AB > CD, as in the diagram. By the symmetry of the isosceles trapezium, BC` = AB/2  CD/2. By Pythagoras, CC`² + BC`² = BC². 

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