Let ABCD be a trapezium in which BC is parallel to AD , BC=a < AD=b, AD=d and CD=c. If I define x, y and h to be variables such that $ \sqrt{x^2+h^2} =d , \quad \sqrt{y^2+h^2}=c $ and $x+y+a=b $ , I obtain the simultaneous equations $$ x^2-y^2=d^2-c^2, \quad x+y =b-a $$.
My question: Is it possible for one to get the formula of height of this trapezium via this method in terms of a, b, c and d only? I know the Malaysian technique but I just wanted to try this one if it can work.
Thanks in anticipation.
Sure it is. Consider that $d^2 - c^2 = x^2 - y^2 = (x-y)(x+y) = (b-a)(x-y)$, so you have $$x - y = \frac{d^2-c^2}{b-a}, \quad x+y = b-a.$$ It's now easy to obtain $x$ by adding these equations, and then you can go back and get $h$ from $x^2 + h^2 = d^2$.