### Area of a Triangle

I was thinking about the area of a triangle and how one might show that area of a triangle is indeed half the product of breadth and height. Its simple to demonstrate when the triangle is right angled as the triangle is evidently half the rectangle and hence area is half the area of the rectangle.

When the triangle is acute or obtuse, the following demonstrates how the triangle area continues to be the same and is determined by the base and height (not the acuteness or obtuseness).

Take the case of an acute triangles:

Clearly the triangle ABC is a sum of triangles ABD and ACD and the areas of triangles ABD and ACD are easy to calculate (they being right angled triangles).

Therefore, area of ABC = 1/2*(x*h) + 1/2*(y*h) = 1/2*(x+y)*h = 1/2*b*h

Similarly , consider the obtuse angled triangle:

Now the triangle ABD is the sum of triangles ABC and ACD. Therefore, triangle ABC is the difference between triangles ABD and ACD.

Therefore area of ABC = 1/2*x*h - 1/2*y*h = 1/2*(x-y)*h = 1/2*b*h.

When the triangle is acute or obtuse, the following demonstrates how the triangle area continues to be the same and is determined by the base and height (not the acuteness or obtuseness).

Take the case of an acute triangles:

Clearly the triangle ABC is a sum of triangles ABD and ACD and the areas of triangles ABD and ACD are easy to calculate (they being right angled triangles).

Therefore, area of ABC = 1/2*(x*h) + 1/2*(y*h) = 1/2*(x+y)*h = 1/2*b*h

Similarly , consider the obtuse angled triangle:

Now the triangle ABD is the sum of triangles ABC and ACD. Therefore, triangle ABC is the difference between triangles ABD and ACD.

Therefore area of ABC = 1/2*x*h - 1/2*y*h = 1/2*(x-y)*h = 1/2*b*h.

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