Thread: Trapezoid (Problem Solving)

The diagonals of a trapezoid divide a trapezoid into 4 triangles.


Find the area of the trapezoid in terms of the areas A and B of the shaded triangles.


I know that the area of a trapezoid is 1/2 (b1 + b2) h. But I am not sure how to use this in the problem.

I presume the shaded triangles are those that include the parallel sides of the trapezoid. Let the areas of the other two triangles are called C and D.

First, prove that C = D by considering the triangles with the areas A + C and A + D.

Next, prove that C / B = A / D, which implies AB = CD = C^2.

I think I understand where you are going with this.....but I do not know how to prove these. It has been so long since I took Number Theory I forgot how to assume and prove. Could you help me a little further?

Number theory is completely different from geometry...

Let's call the trapezoid ABCD where BC and AD are parallel sides. Let O be the intersection of the diagonals AC and BD. Below I'll be talking about areas of triangles, not triangles themselves, so if I say ABO = CDO, it means the areas of the triangles ABO and CDO are equal, not the triangles themselves are equal.

The key is that a triangle's area is one half of a height times the side to which the height is drawn. Therefore, if two triangles have the same height and side, their areas are equal. So, ABD = ACD, and by subtracting AOD we get AOB = COD.

Also, AOB / COB = AO / OC because these triangles have the same height and it cancels out. Likewise, AOD / COD = AO / OC. So, AOB / COB = AOD / COD, from where AOB * COD = COB * AOD, and since AOB = COD, AOB^2 = COB * AOD. Now, COB and AOD are known, so you can find AOB = COD. Obviously, the area of the trapezoid is the sum of the four triangles.