<s>[INST] Trapezoid area and wrong answer Knowing that the acute angles of the trapezoid are $60^\circ$ and $45^\circ$ and the difference of the squares of base lenghts is equal to 100, calculate the area of this trapezoid.
Here's my solution: Let a be the shorter and b the longer base. By drawing two lines from the ends of a perpendicular to b, we form two right triangles. One of them has angles $45^\circ-45^\circ-90^\circ$ and the second is $60^\circ-30^\circ-90^\circ$. Using the properties of these triangles, we can see that the side of the first triangle lying on b is equal to h=height (both lie between 45 and 90) and the side of the latter which is lying on b is equal to $\frac{h\sqrt{3}}3$. Then, $b-a=h+\frac{h\sqrt{3}}3$ so $h=\frac{3(b-a)}{3+\sqrt{3}}=(b-a)(1-\frac{\sqrt{3}}{3})$. We just plug it into the initial equation and we get that the total area is $50(1-\frac{\sqrt{3}}{3})$.
And that's the problem. From what I read on the page I got this problem from, the result should be $25(3-\sqrt{3})$. Why is that so? What's wrong in my solution?
 [/INST] I noted things in the following order:


*

*$b^2-a^2=(b-a)(b+a)=100$ so $A=\frac{h(b+a)}{2}=\frac{50h}{b-a}$.

*Then I reasoned that $b-a=h+\frac{h}{\sqrt{3}}$.

*Combining them yields $A=\frac{50h}{h+\frac{h}{\sqrt{3}}}=\frac{50\sqrt{3}}{\sqrt{3}+1}=\frac{50\sqrt{3}(\sqrt{3}-1)}{2}=25(3-\sqrt{3})$.

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