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Height of a right triangle without area

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

M

Jan 14, 2015

#3
+20831
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How do I find the height of a triangle without the area??

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

.
Jan 15, 2015

#1
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Jan 14, 2015
#2
0

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

Jan 14, 2015
#3
+20831
+10

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

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

heureka Jan 15, 2015
#4
+95177
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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
+27336
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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
+20831
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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
+95177
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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
+94526
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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