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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?

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You made a computational error when you rationalized the denominator to get $h$:

$$\frac3{3+\sqrt3}=\frac3{3+\sqrt3}\cdot\frac{3-\sqrt3}{3-\sqrt3}=\frac{3(3-\sqrt3)}6=\frac12(3-\sqrt3)\;,$$ so $h=\frac12(b-a)(3-\sqrt3)$, and the area is $$\frac12(b+a)h=\frac14(b^2-a^2)(3-\sqrt3)=25(3-\sqrt3)\;.$$

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  • $\begingroup$ That's right, thank you! :) $\endgroup$ – Straightfw May 3 '12 at 17:34
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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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