# Solving a Primary Math Problem: Find the Areas of A & C

• MHB
• anemone
In summary, the figure shows a square with a quarter circle and a semicircle inscribed in it, with the side of the square being 10 cm. The problem asks to find the combined areas of the regions A and C. One possible method is to use basic geometry and trigonometry to find the area of the region D, then calculate the combined areas of A and C using the areas of the regions A+D and C+D. However, this may be challenging for a 12-year-old and a simpler estimation method may be more suitable.
anemone
Gold Member
MHB
POTW Director
\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:
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.

Last edited:
Klaas van Aarsen said:
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)

anemone said:
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)

Beer soaked ramblings follows.
anemone said:
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.

Klaas van Aarsen said:
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)

anemone said:
\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:
Opalg said:
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

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)

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
5. Solve the system of equations

Palaguta said:
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.

## 1. How do I find the area of a shape?

The area of a shape is found by multiplying its length by its width. For example, if a rectangle has a length of 5 units and a width of 3 units, its area would be 5 x 3 = 15 square units.

## 2. What is the difference between area and perimeter?

Area is the measure of the surface of a shape, while perimeter is the measure of the distance around the shape. In other words, area is the amount of space inside a shape, while perimeter is the length of its boundary.

## 3. How do I find the area of a triangle?

The area of a triangle is found by multiplying its base by its height and then dividing by 2. So, if a triangle has a base of 6 units and a height of 4 units, its area would be (6 x 4) / 2 = 12 square units.

## 4. Can I use the same formula to find the area of any shape?

No, the formula for finding the area varies depending on the shape. For example, the formula for finding the area of a circle is different from the formula for finding the area of a rectangle or triangle.

## 5. How can I check if my answer for the area is correct?

You can check your answer by using a different formula for finding the area of the same shape or by using a calculator. Additionally, you can also use real-life objects to compare your answer to, such as using a ruler to measure the length and width of a rectangle and multiplying them to find the area.

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