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How do I find the height of a triangle without the area??

 

Thanks in advance!!

 

M

 Jan 14, 2015

Best Answer 

 #3
avatar+26387 
+10

How do I find the height of a triangle without the area??

$$\boxed{h_c=\dfrac{a*b}{c}}$$

 Jan 15, 2015
 #1
avatar+208 
+5
 Jan 14, 2015
 #2
avatar
0

u divide ur answer they gave you by the other amount on the triangle. then divide by 1/2

 Jan 14, 2015
 #3
avatar+26387 
+10
Best Answer

How do I find the height of a triangle without the area??

$$\boxed{h_c=\dfrac{a*b}{c}}$$

heureka Jan 15, 2015
 #4
avatar+118667 
+5

Really Heureka??  

 If I had more time I would try and work out how that could be demonstrated.

I did try a little but I had no success.  

I don't remember seeing anything like it before.   

Can you show me a proof of this?

 Jan 15, 2015
 #5
avatar+33657 
+5

Assuming you know the length of all three sides you could make use of Heron's formula.  If the sides are of length a, b and c, and b is the base then 

 

$$\frac{1}{2}\times b\times h=\sqrt{s(s-a)(s-b)(s-c)}$$

 

where h is the height and s is the semi-perimeter (=(a+b+c)/2).  This can be rearranged to find h.

.

 Jan 15, 2015
 #6
avatar+26387 
+5

See the question was: Height of a right triangle without area :

$$\\(1) \quad Area = \dfrac{h_c*c}{2} \\\\
(2) \quad Area = \dfrac{a*b}{2}\\\\
\dfrac{h_c*c}{2} = \dfrac{a*b}{2} \\\\
h_c*c = a*b \\\\
h_c= \dfrac{a*b}{c} \\\\$$

 Jan 15, 2015
 #7
avatar+118667 
+8

Thanks Heureka,

Sorry I did not notice that we were talking only of right triangles.  

 

In my defence there is no 'right' in the question it is only in the title.   

 

ALSO note:

The height that is being found here is the perpendicular distance from the hypotenuse to the right angle vertex.

 Jan 15, 2015
 #8
avatar+129849 
+5

Here's another proof of this....assuming that "c" is the hypoteneuse and the height is drawn from the right angle to the hypoteneuse ..  so we have

(1/2) ab sin(90)  = (1/2)ch

 ab   = ch

 h  = ab / c

 

 Jan 17, 2015

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