I was given the following assignment:
A trapezoid's one angle is right and other is 45°. The length of the shorter parallel side is a. The length of the longer diagonal is 4a. Define the area of the trapezoid.
(Please mind the possible mistakes in the text. I translated the assignment from Finnish and I'm not too familiar with English math vocabulary)
I have usually been able to solve problems like this, but now it feels like I've hit a wall. Could someone explain step by step how to solve this? There's obviously something basic I'm just missing here. According to my book, the answer is supposed to be $\frac{1}{4}a^2(14+\sqrt{31})$.
Best Answer
Hint:
Let $b$ the longer basis $AD$. Than, given that the angle $\angle BAH$ is $45°$, the height of the trapezoid is $h= BH=AH=AD-BC=b-a$, and from the diagonal $AC=4a$ ( as hipothenuse of the triangle $ACD$ ) we have:
$$ AC^2=AD^2+CD^2 \qquad \iff \qquad b^2+(b-a)^2=16a^2 $$
solve this equation for $b>0$ and you have done.
The equation becomes: $$ b^2-2ab-15a^2=0 $$ with solutions : $$ b=\frac{a\pm\sqrt{a^2+30a^2}}{2} $$ and, for $b>0$: $$ b=\frac{a}{2}(1+\sqrt{31}) $$ so the area is: $$ A=\frac{1}{2}(b+a)h=\frac{1}{2}(b+a)(b-a)=\frac{1}{2}\left(\frac{a}{2}(1+\sqrt{31})+a \right)\left(\frac{a}{2}(1+\sqrt{31})-a \right)= $$ $$ =\frac{1}{2}\left(\frac{a^2}{4}(1+\sqrt{31})^2-a^2 \right)=\frac{1}{4}a^2(14+\sqrt{31}) $$