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February 27th, 2017, 09:47 AM  #1 
Newbie Joined: Feb 2017 From: uk Posts: 6 Thanks: 0  Height and area of a triangle  surd
Hey, How do I work out the height of an isosceles triangle as a surd? I have the base length and the length of the side. Then how do I work out the area of a triangle as a surd? Thanks for any help. Last edited by skipjack; February 28th, 2017 at 07:04 AM. 
February 27th, 2017, 09:59 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
$h$ = triangle height $s$ = side length $b$ = base length $h^2 = s^2  \left(\dfrac{b}{2}\right)^2 \implies h = \sqrt{s^2  \left(\dfrac{b}{2}\right)^2}$ 
February 27th, 2017, 10:06 AM  #3 
Newbie Joined: Feb 2017 From: uk Posts: 6 Thanks: 0 
so if my base was 20cm and my side was 30cm. height (2) = 30(2)  (20/2) (2) height (2) = 900  100 sorry I'm really bad at this. 
February 27th, 2017, 10:16 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
$h = \sqrt{30^2  \left(\dfrac{20}{2}\right)^2}$ $h = \sqrt{900  100}$ $h = \sqrt{800} = \sqrt{400} \cdot \sqrt{2} = 20\sqrt{2}$ 
February 27th, 2017, 10:25 AM  #5 
Newbie Joined: Feb 2017 From: uk Posts: 6 Thanks: 0 
Thank you! So the area of the triangle will be 282.8? How do I convert that to a surd? 
February 27th, 2017, 01:29 PM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  Quote:
$A = \dfrac{1}{2} \cdot b \cdot h$ $A = \dfrac{1}{2} \cdot 20 \cdot 20\sqrt{2} = 10 \cdot 20\sqrt{2} = 200\sqrt{2} \, \,\text{cm}^2$ Last edited by skipjack; February 28th, 2017 at 07:06 AM.  
February 27th, 2017, 01:33 PM  #7 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  

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