Find the area of ​the shaded region in the figure below

Question

For reference:

In the figure $O$ is the center of the circle and its radius measures $a$ and $AQ = QB$. Calculate the area of ​​the shaded region.(Answer: $\frac{a^2}{4}(\pi-2)$)

correct figure

My progress:

If $AQ = BQ \implies \angle AQB=90^\circ$

Complete the square $AQBD$.

incorrect figure. incorrect figure, please do not consider it for any effect

$OC = r$ and $QC =R = AC.$

$O$ is centre of square.

$QO$ is angle bisector, therefore $\angle AQO$ is $45^\circ.$

$QD = R\sqrt2$

Considering $\triangle OCQ$,

$\displaystyle r^2+\left(\frac{R}{2}\right)^2=OQ^2\implies r^2+\frac{R^2}{4}=(R\sqrt2)^2$

$\therefore R = \dfrac{2r\sqrt7}{7}$

I don't see a solution...is it missing some information?

The book has another similar question but in this question $a = 2$ and answers match if we replace $a$ with $2$. Diagram below -

Answer 1

Considering $\angle AQB = 90^\circ$ as shown in the last diagram of the question,

If $\angle BQE = 2\theta~, ~\angle AEQ = \angle EAQ = 45^\circ + \theta, ~$ given $AQ = QE$.

Also, $\angle OFE = \angle AEQ = 45^\circ + \theta$

That leads to $\angle FOE = 90^\circ - 2\theta$. As $\angle AQF = \angle FOE$, quadrilateral $AOFQ~$ is cyclic.

So we have, $\angle AFO = \angle AQO = \angle 45^\circ$ but as $OH = OF$, $\angle HOF = 90^\circ$.

From here, it is straightforward to find the shaded area.

Answer 2

The answer given is possible only if $\angle FOH=90^o$, because the area of shaded area will be:

$s=\frac {\pi a^2}4-\frac{a\times a}2=\frac {a^2}4(\pi-2)$

As can be seen in above optimized figure the measure of AQ=BQ can not arbitrary , it is:

$AQ=0.9 a$

For example if $a=10$ then $AQ=9$. This relation must be included in statement of question.

Answer 3

Here is a solution that I have so far. It does turn out after all the lengthy work that $\angle HOF \approx 90^\circ$. So there should be a way to show $\angle HOF = 90^\circ$ when $O$ lies on $AE$. But I have not been able to see a geometric solution yet.

If $R$ is the radius of the quarter circle and radius of the smaller circle is $a$,

$AS^2 = AG \cdot AE~$ i.e. $~(R-a)^2 = (AO + a) \cdot (AO-a)$ $\implies AO^2 = R^2 - 2aR + 2a^2$

Also, $~QF \cdot QE = QT^2~$ or $~(R-FE) \cdot R = a^2 $
$\implies FE = \dfrac{R^2-a^2}{R}, ~QF = \dfrac{a^2}{R}$

If $M$ is the midpoint of $FE$, $OM \perp FE$. Notice that $\triangle OEM \sim \triangle OAS~$. So,

$\displaystyle \frac{ME^2}{OE^2} = \frac{AS^2}{AO^2} \implies \frac{(R^2-a^2)^2}{4 a^2 R^2} = \frac{(R-a)^2}{R^2-2aR+2a^2}$

Simplifying, $2a^4 + 2a^3R + R^4 - 5a^2R^2 = 0$

$(R-a) (2a^3 + 4a^2R - aR^2 - R^3) = 0$

As $R = a$ is not a solution that we are interested in, we solve $2a^3 + 4a^2R - aR^2 - R^3 = 0$. WolframAlpha gives an approximate form solution of $R \approx 1.81361 a$. We then find $AO \approx 1.28917a, QF \approx 0.551386 a$

Next, $ \displaystyle AK = AS \cdot \frac{AE}{AO} = (1.81361 a - a) \cdot \frac {AO + a}{AO} \approx 1.44472a$

So, $QK = R - AK \approx 0.36889a$

$ \displaystyle \cos \angle AQE = \frac{QK}{QE} \approx 0.2034$

Applying law of cosine, $AF^2 = AQ^2 + QF^2 - 2 AQ \cdot QF \cdot \cos \angle AQE \approx 3.18641 a^2$

$AS^2 = AH \cdot AF = (AF - HF) \cdot AF$

We obtain $~HF \approx 1.41422 a, ~$ which is approximately $a \sqrt2$. With $HF \approx a \sqrt2$ and $OF = OH = a$, we have $\angle HOF \approx 90^\circ$.

That leads to the shaded area $ \displaystyle A \approx \frac{\pi}{4} \cdot a^2 - \frac 12 \cdot a^2 = \frac {a^2}{4} (\pi - 2)$