Find the area

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  • #1
anemone
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\begin{tikzpicture}
\draw (0,-3) rectangle (6,3);
\begin{scope} \draw (6,-3) arc (-90:-270:3cm); \end{scope}
\begin{scope} \draw (6,-3) arc (0:90:6cm); \end{scope}
\coordinate[label=above: A] (A) at (2,0);
\coordinate[label=above: B] (B) at (3.6,2.4);
\coordinate[label=above: C] (C) at (5.4,0.8);
\coordinate[label=above: D] (D) at (3.9,-1.2);
\end{tikzpicture}

The figure shows a square with a quarter circle and a semicircle inscribed in it. Given that the side of the square is 10 cm, find the combined areas of the regions labelled A and C.

Any hints, perhaps, on how to solve this seemingly simple geometry problem? This is a primary math problem, so one should refrain from using geometry or trigonometric method to solve it...any help is much appreciated!
 
Last edited:

Answers and Replies

  • #2
I like Serena
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Seemingly simple...
Also without calculus?

Without any of those methods, the best I can think of, is to:
  1. Draw it on paper.
  2. Use a scissors to cut out the pieces.
  3. Weigh the pieces $A+B+C+D$ together on a scale.
  4. Weigh the pieces $A+C$ together on the scale.
  5. Calculate $\frac{A+C}{A+B+C+D}\cdot 100\,\text{cm}^2$.
Using calculus and an online calculator, I can find the result $A+C = 100-\frac{75}{4}\pi + 25\tan^{-1}\frac{44}{117}\approx 50.0878\,\text{cm}^2$.

So using the paper and scissors method, we should find that $A+C$ are about $50\,\text{cm}^2$, or half of the square.
 
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  • #3
anemone
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Also without calculus?

Yes, this is an exam problem for kids of 12 years old. So they know nothing about calculus. Thanks Klaas for your reply! (Nod)
 
  • #4
I like Serena
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Yes, this is an exam problem for kids of 12 years old. So they know nothing about calculus. Thanks Klaas for your reply! (Nod)
Then perhaps they are merely supposed to estimate the area? (Wondering)
 
  • #5
jonah1
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Beer soaked ramblings follows.
Yes, this is an exam problem for kids of 12 years old. So they know nothing about calculus. Thanks Klaas for your reply! (Nod)
Surely 12 year old kids are already familiar with a circle's area formula.
 
  • #6
anemone
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Then perhaps they are merely supposed to estimate the area? (Wondering)
Nope, Klaas, the question asks the students to find the area of the shaded region, which are the combined area of the regions A and C in the diagram above... (Nod)
 
  • #7
Opalg
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\begin{tikzpicture}
\draw (0,-3) rectangle (6,3);
\begin{scope} \draw (6,-3) arc (-90:-270:3cm); \end{scope}
\begin{scope} \draw (6,-3) arc (0:90:6cm); \end{scope}
\coordinate[label=above: A] (A) at (2,0);
\coordinate[label=above: B] (B) at (3.6,2.4);
\coordinate[label=above: C] (C) at (5.4,0.8);
\coordinate[label=above: D] (D) at (3.9,-1.2);
\end{tikzpicture}

The figure shows a square with a quarter circle and a semicircle inscribed in it. Given that the side of the square is 10 cm, find the combined areas of the regions labelled A and C.

Any hints, perhaps, on how to solve this seemingly simple geometry problem? This is a primary math problem, so one should refrain from using geometry or trigonometric method to solve it...any help is much appreciated!
The area of the regions A+D is $25\pi$. The area of the regions C+D is $12.5\pi$. So to find the area of A+C we need to know the area of D.

[TIKZ] \draw (0,-3) rectangle (6,3);
\begin{scope} \draw (6,-3) arc (-90:-270:3cm); \end{scope}
\begin{scope} \draw (6,-3) arc (0:90:6cm); \end{scope}
\coordinate[label=above: A] (A) at (2,0);
\coordinate[label=above: B] (B) at (3.6,2.4);
\coordinate[label=above: C] (C) at (5.4,0.8);
\coordinate[label=above: D] (D) at (4,-1);
\coordinate[label=left: P] (P) at (0,-3);
\coordinate[label=above: Q] (Q) at (3.6,1.8);
\coordinate[label=right: R] (R) at (6,0);
\coordinate[label=right: S] (S) at (6,-3);
\draw (P) -- (Q) -- (R) -- (P) ;
\draw (Q) -- (S) ;
\draw (0.4,-2.8) node {$\theta$} ;
\draw (3,-3.2) node {$10$} ;
\draw (6.2,-1.5) node {$5$} ;[/TIKZ]

Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are congruent. In particular $\angle PQR$ is a right angle, so that $PQ$ is tangent to the semicircle and $QR$ is tangent to the quarter-circle. Also, $PR$ bisects the angle $\theta =\angle QPS$, so that $\angle RPS = \frac\theta2$ and $\tan\frac\theta2 = \frac{RS}{PS} = \frac12.$ Therefore $$\tan\theta = \frac{2\tan\frac\theta2}{1 - \tan^2\frac\theta2} = \frac1{1-\frac14} = \frac43.$$ It follows that $\sin\theta = \frac45$ and $\cos\theta = \frac35$.

The sector $QPS$ of the quarter-circle has area $\frac\theta{2\pi}(100\pi) = 50\theta$. The triangle $QPS$ has base $10$ and height $10\sin\theta = 8$, so it has area $40$. Therefore the part of region D lying to the right of the line $QS$ has area $50\tan^{-1}\frac43 - 40 \approx 6.365$. A similar calculation using the sector $QRS$ of the semicircle, and the triangle $QRS$, will give the area of the part of region D lying to the left of the line $QS$. You can then put everything together to complete the solution of the problem.

That method uses only basic geometry and fairly basic trigonometry. But I would not expect a 12-year-old to find it. Maybe Singaporean kids are a whole lot smarter than those from Europe or America!

Added later: I get the combined areas of regions A+C to be $12.5\pi + 100 - 75\tan^{-1}\frac43 \approx 69.73$.
 
Last edited:
  • #8
topsquark
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The area of the regions A+D is $25\pi$. The area of the regions C+D is $12.5\pi$. So to find the area of A+C we need to know the area of D.

[TIKZ] \draw (0,-3) rectangle (6,3);
\begin{scope} \draw (6,-3) arc (-90:-270:3cm); \end{scope}
\begin{scope} \draw (6,-3) arc (0:90:6cm); \end{scope}
\coordinate[label=above: A] (A) at (2,0);
\coordinate[label=above: B] (B) at (3.6,2.4);
\coordinate[label=above: C] (C) at (5.4,0.8);
\coordinate[label=above: D] (D) at (4,-1);
\coordinate[label=left: P] (P) at (0,-3);
\coordinate[label=above: Q] (Q) at (3.6,1.8);
\coordinate[label=right: R] (R) at (6,0);
\coordinate[label=right: S] (S) at (6,-3);
\draw (P) -- (Q) -- (R) -- (P) ;
\draw (Q) -- (S) ;
\draw (0.4,-2.8) node {$\theta$} ;
\draw (3,-3.2) node {$10$} ;
\draw (6.2,-1.5) node {$5$} ;[/TIKZ]

Let $R$ be the centre of the semicircle, and let $Q,S$ be the points where the quarter-circle intersects the semicircle. The triangles $PQR$, $PSR$ each have one side of length $10$, one side of length $5$ and a common side $PR$. Therefore they are congruent. In particular $\angle PQR$ is a right angle, so that $PQ$ is tangent to the semicircle and $QR$ is tangent to the quarter-circle. Also, $PR$ bisects the angle $\theta =\angle QPS$, so that $\angle RPS = \frac\theta2$ and $\tan\frac\theta2 = \frac{RS}{PS} = \frac12.$ Therefore $$\tan\theta = \frac{2\tan\frac\theta2}{1 - \tan^2\frac\theta2} = \frac1{1-\frac14} = \frac43.$$ It follows that $\sin\theta = \frac45$ and $\cos\theta = \frac35$.

The sector $QPS$ of the quarter-circle has area $\frac\theta{2\pi}(100\pi) = 50\theta$. The triangle $QPS$ has base $10$ and height $10\sin\theta = 8$, so it has area $40$. Therefore the part of region D lying to the right of the line $QS$ has area $50\tan^{-1}\frac43 - 40 \approx 6.365$. A similar calculation using the sector $QRS$ of the semicircle, and the triangle $QRS$, will give the area of the part of region D lying to the left of the line $QS$. You can then put everything together to complete the solution of the problem.

That method uses only basic geometry and fairly basic trigonometry. But I would not expect a 12-year-old to find it. Maybe Singaporean kids are a whole lot smarter than those from Europe or America!

Added later: I get the combined areas of regions A+C to be $12.5\pi + 100 - 75\tan^{-1}\frac43 \approx 69.73$.
Very nice, but not something I'd expect to see from a 12 year old!

-Dan
 
  • #9
anemone
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Thanks so much Opalg for your reply! I deeply appreciate it.

I will send the link of this thread to a friend of mine to let him see how did you solve it.

At the same time, I will continue to look at this problem from different angle because (almost) all geometry problems here use the method of symmetry to get rid of all the unwanted areas in order to find the desired outcome. I have been staring at this problem for days and trying to solve it using multiple approaches with met with all futile attempts.

If I managed to solve it differently than yours, I will post back.

Anyway, thanks again Opalg for your another insightful solution! (Cool)
 
  • #10
Palaguta
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1. describe the area of the square in terms of the sum of the figures
2. Describe the area of the circle segments in terms of the sum of the figures
3. Substitute the equations of the sum of the circles in the formula of the square area
4. Make a system of equations
16212097534773972208687017395995.jpg
5. Solve the system of equations
 
  • #11
Opalg
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1. describe the area of the square in terms of the sum of the figures
2. Describe the area of the circle segments in terms of the sum of the figures
3. Substitute the equations of the sum of the circles in the formula of the square area
4. Make a system of equations
View attachment 111505. Solve the system of equations
I agree with that up to the line $$\left\{\eqalign{S^2 = \frac{25\pi}2 + A + B = 100 \\S^2 = \frac{100\pi}4 + B + C = 100.}\right.$$ But when you add those two equations you should get $$A + C + 2B = 200 - \frac{25\pi}2 - \frac{100\pi}4 = \ldots$$ You have allowed the $B$s to cancel when they should have been adding up.
 

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