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A few days ago, I encountered the following problem:

enter image description here

After a little bit of thinking, I managed to come up with the following solution:

  1. Rotate the square $90^\circ$ clockwise and let the new bottom left corner of the square be $(0,0)$.
  2. The circle inscribed in the square is hence centered at $(5,5)$ with a radius of $5$. The circle equation thus becomes $(x-5)^{2} + (y-5)^{2} = 25 \Rightarrow y = 5 + \sqrt{25 - (x-5)^{2}}$ in the first quadrant.
  3. Similarly for the quarter circle, the equation becomes $y = \sqrt{100-x^2}$.

The graph hence looks like this:

enter image description here

My intention is to find the shaded area in the above graph. To do so, first I find $X$ by equating $5 + \sqrt{25 - (x-5)^{2}} = \sqrt{100-x^2} \Rightarrow x=\frac{25 - 5\sqrt{7}}{4}$.

From this, I calculate the area of the shaded region as follows: $$\text{Area} = (10 \cdot \frac{25 - 5\sqrt{7}}{4} - \int_0^\frac{25 - 5\sqrt{7}}{4} \sqrt{100-x^2} \,\mathrm{d}x) + (10 \cdot (5 - \frac{25 - 5\sqrt{7}}{4}) - \int_\frac{25 - 5\sqrt{7}}{4}^5 5 + \sqrt{25 - (x-5)^{2}} \,\mathrm{d}x) \approx 0.7285$$

Now, the diagram looks like this:

enter image description here

From here, I figured out the shaded area as follows: $$\text{Area} \approx 10^{2} - \frac{\pi(10^{2})}{4} - (\frac{10^{2} - \pi(5^{2})}{4} + 2 \times 0.7285) \approx \boxed{14.6 \: \text{cm}^{2}}$$

While I did figure out the correct solution, I find my approach to be rather lengthy. I was wondering if there is a quicker, simpler and more concise method (that probably does not require Calculus) that one can use and I would highly appreciate any answers pertaining to the same.

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1 Answer 1

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Here is an alternate solution,

enter image description here

We assume right bottom vertex to be the origin then, equations of circles are

Circle S: $x^2+y^2 = 100$
Circle T: $(x+5)^2+(y-5)^2 = 25$

Solving both equations, intersection points are $A \left(\frac{5}{4}\left(\sqrt7-5\right), \frac{5}{4}\left(\sqrt7+5\right)\right)$ and $B\left(-\frac{5}{4}\left(\sqrt7+5\right), -\frac{5}{4}\left(\sqrt7-5\right)\right)$

So length of chord $AB$ at intersection is $\frac{5 \sqrt7}{\sqrt2}$.

We note that this chord $AB$ is common chord of both circles.

Angle subtended by chord at the center is given by,

At $P$, $\angle APB = \alpha = 2 \arcsin \left(\frac{\sqrt7}{2\sqrt2}\right)$

At $O$, $\angle AOB = \beta = 2 \arcsin \left(\frac{\sqrt7}{4\sqrt2}\right)$

Shaded area is difference of area of two circular segments of this chord, which are $ATB$ and $ASB$.

$ = \left(circular \ sector \ PATB - \triangle PAB\right) - \left(circular \ sector \ OASB - \triangle OAB\right)$

$ = \left(25 \times \frac{\alpha}{2} - \frac{25 \sqrt7}{8}\right) - \left(100 \times \frac{\beta}{2} - \frac{125 \sqrt7}{8}\right) \approx 14.638 $

EDIT: to find $AB$ without coordinate geometry, we know that $OP = 5 \sqrt2$. If perp from $P$ to $AB$ is $x$ then,

$\left(5\sqrt2 + x\right)^2 + \left(\frac{AB}{2}\right)^2 = 10^2$ and $x^2 + \left(\frac{AB}{2}\right)^2 = 5^2$. Solving them, we get value of $AB$.

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  • $\begingroup$ Thanks for the answer! Still wondering, though, if there is a quicker way to find the length $AB$ i.e. without the need to get into coordinate geometry. $\endgroup$ Apr 2, 2021 at 13:26
  • $\begingroup$ Yes there is but it will still require to solve equations. I will edit. $\endgroup$
    – Math Lover
    Apr 2, 2021 at 13:29
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    $\begingroup$ Okay perfect, thanks 😊 $\endgroup$ Apr 2, 2021 at 13:39

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