Area of triangle AQB

Question

How do we find the ratio of area of triangle AQB to area of square in this question?

Answer 1

$AQB$ is a right triangle and its legs are $|AQ|=a\sin(\phi)$, $|QB|=a\cos(\phi)$, hence $$\mbox{Area}(AQB)=\frac{1}{2}|AQ|\cdot|QB|=\frac{a^2}{2}\sin(\phi)\cos(\phi)=\frac{a^2}{4}\sin(2\phi)\\=\frac{a^2}{4}\cdot \frac{2\tan(\phi)}{1+\tan^2(\phi)}=\frac{a^2}{2}\cdot \frac{1/3}{1+1/9}=\frac{3a^2}{20}.$$

Answer 2

Since $\frac{1}{3} = \tan\phi = \frac{AQ}{QB}$, it follows that $QB = 3 AQ$. By the Pythagorean theorem, $$ AQ^2 + QB^2 = a^2\implies AQ^2 + (3AQ)^2 = a^2 \implies 10AQ^2 = a^2\implies AQ = \frac{a}{\sqrt{10}}.$$ Hence, $AQ = \frac{a}{\sqrt{10}}$ and $QB = \frac{3a}{\sqrt{10}}$. Can you finish from here?

Answer 3

You can't find this area, because you haven't any measure of sides. You only have ratios. If you have any similar figure, the area changes