Ok, in geometry the altitude of a triange is ALT^2=b1*b2. Right? Ok then, if you have a triangle with a b1=5 ft and a b2=5 ft, the ALT=5 ft. Right? Ok, let's say i have a triangle that is isiolies (2 congruent side) with a b1 of 5ft and a b2 of 5ft and the congurent side are 28 ft. Alt= 5, using the formula at the top. And let's say i have another triangle that's is isiolies (2 congruent side) with a b1 of 5ft and a b2 of 5ft and the congurent side are 400 ft. Alt= 5, using the formula at the top. Why dosen't this formula give me the right answer? thanks in advance!

Your formula is not correct. That's the problem. To find the length of the altitude to a particular base of a triangle, multiply the length of one of the sides by the cosine of the angle that the side forms with the base. If you know the area of the triangle, multiply that area by 2 and divide by the length of the base to get the altitude. -Tony! Minored in math.

It's been a while since I took geometry, but I think that formula of altitude^2 = b1*b2 only applies to right triangles with the hypotenuse as the base of the triangle. The formula won't work on all triangles in general. [It can't work in general. Test it on an equilateral triangle with side lengths of 2. The formula would say the altitude has length one, but the altitude would really have a length of square root of 3 (which you can determine with the Pythagorean theorem).

I've been messing with it all day and i agree it only works on right triangles. So what is the formula for the height of a triangle? (i know it's simple, but i can't remember it)

for the side with a congruent side of 28, y can be solved by taking 28^2 - 5^2 = x^2. Solve for x. x = 27.5 that's If you are allowed to use pythagorem(sp?), seems like the easiest way

Ok thanks for all the help so far. The therom to find the alt of a scalene triangle is called Heron's Terom, whisch is crazy long. But, i still need to know how to find the length of the altitude in an isoselies triangle, and thankz again!

Well you could modify pythag by saying, the ALT = sqrt(Leg^2 - (base/2)^2) I don't know if there is a therom for that, I'm just using Trig

Both of those triangles you drew can be broken up into right angle triangles. Then simply observe: You've provided enough information in that scan to solve for the height of both triangles.

Exactly. If you have an isoceles triangle where you know the length of the congruent sides and the length of the base, you can use Pythagorus to find the altitude. If S is the length of one of the congruent sides, and B is the length of the base, and A is the altitude, then A^2 + (B/2)^2 = S^2. Solving for A results in A = \|(S^2-[B/2]^2). (Using my crappy text, ' \| ' means 'the square root of' the quantity following in parantheses. -Tony!