Question

how to prove that area of equilateral triangle is 1/2*b*h

Answer 1

First,make a parallelogram whose fornula is base×height.Divide it into two congruent parts.
They will be triangles.So,the formula of triangle will be 1/2×b×h.

Answer 2

Start with a rectangle ABCD and let h be the height and b be the base as shown below:

Rectangle ABCD


The area of this rectangle is b × h

However, if we draw a diagonal from one vertex, it will break the rectangle into two congruent or equal triangles

Rectangle ABCD turned into 2 triangles


The area of each triangle is half the area of the rectangle. For example, the area of triangle ABC is 1/2(b × h) Does that make sense?

Although it does make sense, the proof is incomplete because triangle ABC is a right triangle or what we can also call a special triangle

How do we know the formula is going to work for any triangle, such as isosceles, equilateral, or scalene triangles?

If we can proove that the formula is the same for any arbitrary triangle such as a scalene triangle which has nothing particular about it, then it will work for special triangles such as isosceles, equilateral, or right triangle.

So, let us start this time with a scalene triangle ABC which has nothing special about it

Triangle ABC


Then, draw the height from vertex B and label it as you see below:

Triangle ABC with height h


area of triangle ABC = area of triangle ABE + area of triangle CBE

area of triangle ABC = (y × h)/2 + (x × h)/2

area of triangle ABC = (y × h + x × h)/2

area of triangle ABC = h ×(y + x)/2

Notice that y + x is the length of the base of triangle ABC.

Thus, it is ok to say that y + x = b

Therefore, area of triangle ABC = (h × b)/2