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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 enter image description here

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 image2

$OC = r$ 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 -

enter image description here

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – TheSimpliFire
    Jan 16 at 11:06
  • $\begingroup$ Your second diagram is pure fantasy. Why should the top and left sides of the square be tangent to the circle? (Hint: they are not.) $\endgroup$
    – TonyK
    Jan 18 at 21:29
  • $\begingroup$ @TonyK The figure was wrong.. see that the center is in AB and should be in AE $\endgroup$ Jan 18 at 22:04
  • 1
    $\begingroup$ So what's your final diagram then? I think you have to delete your question and start again. $\endgroup$
    – TonyK
    Jan 18 at 22:10
  • 1
    $\begingroup$ Why don't you just edit your question to remove the incorrect figure? $\endgroup$
    – TonyK
    Jan 19 at 13:16

3 Answers 3

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enter image description here

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.

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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.

enter image description here

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)$

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  • $\begingroup$ I don't understand your first paragraph. The problem is independent of the specific value of $a$, isn't it? In other words, if you have a solution for $a=2$, then you can just scale the diagram to get a solution for any value of $a$. $\endgroup$
    – TonyK
    Jan 18 at 21:24
  • $\begingroup$ excellent...As I already said..the questions are the same..only the first one, the radius is literal but the solution is the same..I found it strange to have two questions with the same wording and there is no solution... $\endgroup$ Jan 18 at 21:28
  • $\begingroup$ @TonyK That's what I would think but if you take $a$ instead of $2$ and try to solve the quartic to find the relationship between $R$ and $a$, you do not get a solution other than trivial one's or complex roots. So unless I do more work on it, I cannot confirm either way. $\endgroup$
    – Math Lover
    Jan 18 at 21:31
  • $\begingroup$ That just means you must be doing it wrong. Clearly the problem is independent of $a$. $\endgroup$
    – TonyK
    Jan 18 at 21:33
  • $\begingroup$ @TonyK I usually don't make mistakes in my work but as I said I would check tomorrow and if you are interested, I would give you the quartic equation that I get. $\endgroup$
    – Math Lover
    Jan 18 at 21:36
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enter image description here

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.

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    $\begingroup$ Proving $\angle HOF=90^\circ$ is the task. Did you notice that $O$ lies on $AE$? So it defines a unique figure and not necessary to state $AQ=0.9a$ in the problem. $\endgroup$
    – ACB
    Jan 18 at 15:25
  • $\begingroup$ @ACB, I found it due to given answer. Suppose you do not know the answer, then how would you find this ratio? The information is not enough. $\endgroup$
    – sirous
    Jan 18 at 15:31
  • $\begingroup$ @ sirous, which ratio? By the way, I focus on proving $\angle HOF=90^\circ$. $\endgroup$
    – ACB
    Jan 18 at 15:34
  • $\begingroup$ @ACB Strange that there is another exercise like it, just providing the radius and the image does not show any detail.(see the figure) $\endgroup$ Jan 18 at 15:35
  • $\begingroup$ @ACB, It does not work if O lies on AE. The figure is made by computer giving various values to diameter of circle keeping BQ constant. $\endgroup$
    – sirous
    Jan 18 at 15:49

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