area of a triangle

[Calculator]

Hello, I just have one (simple) question. I know that the area of a right triangle is a*b/2 where a and b are the legs(and b is like the height), and I know how we came to that formula( right triangle is half of a square) But, in an isosceles triangle A = b*h/2 (where b is the base and h is the height). I want to know how did we get that formula? I thought that the height on the base of an isosceles triangle divides that triangle into 2 right triangles so the area should be 2*(b*h/2) ?? I'm sorry if this is a stupid question, I've been trying to find the answer to this question elsewhere but did not succeed.

 

[stapel]

Hello, I just have one (simple) question. I know that the area of a right triangle is a*b/2 where a and b are the legs(and b is like the height), and I know how we came to that formula( right triangle is half of a square) But, in an isosceles triangle A = b*h/2 (where b is the base and h is the height). I want to know how did we get that formula?

If you're looking for a proof of the area formula, you could try, for instance, this article.

I thought that the height on the base of an isosceles triangle divides that triangle into 2 right triangles so the area should be 2*(b*h/2)

Each of the sub-triangles would have a base of length b/2, because you split the original triangle in half. So you'd have 2*[(1/2)(b/2)(h)], which equals...?

 

[Calculator]

If you're looking for a proof of the area formula, you could try, for instance, this article.


Each of the sub-triangles would have a base of length b/2, because you split the original triangle in half. So you'd have 2*[(1/2)(b/2)(h)], which equals...?

ok, thanks