How to find the area between two overlapping quarter circles?

Question

Angle BAC = Angle YXZ = 90º. AB = XY = 6. Angle BXY = 60º. X is the midpoint of AB. Find the shaded areas.

I coloured the area I could get red. I'm asking on any advice/solutions to the blue area.

Solution for red area:

Angle XAS = 90º (supplementary angles) XS = 6 XA = 3 therefore triangle XAS is a scaled {1, √3, 2} triangle. therefore AS = 3√3

Angle PXA = 30º (supp. angles) XA = 3 therefore AP = 3/√3 = √3

SP = AS - AP = 3√3 - √3 = 2√3

red area = Sector XZS - triangle XPS Sector XZS = 36π(30º/360º) = 3π Triangle XPS = 2√3 * 3 / 2 = 3√3 therefore red area = 3(π - √3)

I don't know how to start on the blue area, as mentioned before. I don't know trig that well yet as I'm still in middle school, but I know enough to understand solutions using it.

Some ways that I tried are mirroring it on AB and solving for sector (XR'R - AQ'Q) adding and subtracting the other neccesary areas, and using a circumscribed square minus the unshaded areas, but I couldn't get far with those.

Answer 1

The blue region is given by $$|RQY| = |RXY| + |\triangle ARX| - |RAQ| - |\triangle AXQ|. \tag{1}$$

For the area of sector $RXY$, note that the intersection point $R$ depends only on the locations of the centers of each circular sector (in this case, $B$ and $X$), and the fact that the radii are equal. Since $X$ is the midpoint of radius $\overline{AB}$, it follows that the foot of the perpendicular from $R$ to $\overline{AB}$, which we will call $M$, must be the midpoint of $\overline{AX}$. Hence we must have $$\cos \angle RXA = \frac{AX/2}{RX} = \frac{1}{4}.$$ Hence $$\angle RXY = \angle YXA - \angle RXA = \frac{2\pi}{3} - \cos^{-1} \frac{1}{4}. \tag{2}$$

Next, by the Law of Sines, $$\frac{\sin \angle AQX}{AX} = \frac{\sin \angle AXQ}{AQ},$$ and since $AQ = 2AX$ and $\angle AXQ = \frac{2\pi}{3}$, it follows that $$\sin \angle AQX = \frac{\sqrt{3}}{4}.$$ Hence $$\angle QAX = \angle QXB - \angle AQX = \frac{\pi}{3} - \sin^{-1} \frac{\sqrt{3}}{4}$$ and $$\angle RAQ = \angle RAX - \angle QAX = \cos^{-1} \frac{1}{4} - \frac{\pi}{3} + \sin^{-1} \frac{\sqrt{3}}{4}. \tag{3}$$

Therefore, if the common radius is $XY = 6$, $$|RXY| - |RAQ| = 18 \left( \angle RXY - \angle RAQ \right) = 18 \left(\pi - 2 \cos^{-1} \frac{1}{4} - \sin^{-1} \frac{\sqrt{3}}{4}\right). \tag{4}$$

We now turn our attention to the triangles. We have

$$|\triangle ARX| = \frac{(AX)(RM)}{2} = \frac{3\sqrt{6^2 - (3/2)^2}}{2} = \frac{9\sqrt{15}}{4}, \tag{5}$$ where $M$ is the aforementioned foot of the perpendicular from $R$ to $\overline{AX}$, and $$\begin{align} |\triangle AXQ| &= \frac{1}{2} (AX)(AQ) \sin \angle QAX \\ &= \frac{1}{2} (3)(6) \sin \left(\frac{\pi}{3} - \sin^{-1} \frac{\sqrt{3}}{4}\right) \\ &= 9 \left( \sin \frac{\pi}{3} \cos \sin^{-1} \frac{\sqrt{3}}{4} - \cos \frac{\pi}{3} \sin \sin^{-1} \frac{\sqrt{3}}{4} \right) \\ &= 9 \left( \frac{\sqrt{3}}{2} \sqrt{1 - \sin^2 \sin^{-1} \frac{\sqrt{3}}{4}} - \frac{1}{2} \frac{\sqrt{3}}{4} \right) \\ &= 9 \left( \frac{\sqrt{3}}{2} \sqrt{1 - \frac{3}{16}} - \frac{\sqrt{3}}{8} \right) \\ &= \frac{9}{8} \sqrt{3} (\sqrt{13} - 1). \tag{6} \end{align}$$

Therefore,

$$|RQY| = 18 \left(\pi - 2 \cos^{-1} \frac{1}{4} - \sin^{-1} \frac{\sqrt{3}}{4}\right) + \frac{9\sqrt{15}}{4} - \frac{9}{8} \sqrt{3} (\sqrt{13} - 1) \approx 4.67265. \tag{7}$$

This answer is corroborated by the expression

$$|RQY| = \int_{x=-3}^{3/2} \sqrt{6^2 - x^2} \, dx - \int_{x=-3}^{Q_x} -\sqrt{3} x \, dx - \int_{x=Q_x}^{3/2} \sqrt{6^2 - (x-3)^2} \, dx, \tag{8}$$ where $Q_x = \frac{3}{4}(1-\sqrt{13})$ is the $x$-coordinate of point $Q$ when point $X$ is at the origin.

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Note that additional simplification of $(7)$ is possible and yields $$|RQY| = 18 \cos^{-1} \frac{3 \sqrt{5} + 7 \sqrt{13}}{32} + \frac{9}{8}(\sqrt{3} + 2\sqrt{15} - \sqrt{39}) \tag{7*}$$ but is probably not within the scope of a middle school student curriculum. Equation $(8)$ involves calculus which is clearly out of scope, but it is included to demonstrate that $(7)$ is the correct value.

Answer 2

This is a solution using trigonometry (for the purpose of this answer, all inverse trigonometric functions will be in degrees instead of radians):

The blue area $YQR$ equals the sector $YXR$ minus the curved area $XQR$. We shall calculate each of these areas.

For the curved area $XQR$, we note that $XQR$ equals the quarter circle $ABC$ minus the area $BXQ$, minus $\triangle XAR$, minus the sector $ARC$. The area $BXQ$ equals the sector $BAQ$ minus $\triangle XAQ$. So in total, $XQR = ABC - BAQ + XAQ - XAR - ARC$.

We first calculate $XAR$. $XR = AR = 6$ and $XA = 3$. So the height of this triangle is $\sqrt{6^2 - (3/2)^2} = \frac{3}{2}\sqrt{15}$. Thus, $\triangle XAR = \frac{9}{4}\sqrt{15}$.

This also means the $\angle AXR = \angle XAR = \cos^{-1}(\frac{3/2}{6}) = \cos^{-1}(1/4)$. So $\angle RAC = 90^\circ - \cos^{-1}(1/4)$. So the sector $ARC = 36\pi \times (90^\circ - \cos^{-1}(1/4))/360^\circ = 9\pi - \frac{\cos^{-1}(1/4)}{10^\circ}\pi$.

Now, let $QT$ be a line segment perpendicular to $AB$ and meeting $AB$ at $T$. Let $x = XQ$. Since $\angle BXQ = 60^\circ$, we have $XT = x/2$ and $QT = \frac{\sqrt{3}}{2}x$. Then $AQ = 6$, $AT = 3 + x/2$, and $QT = \frac{\sqrt{3}}{2}x$, so by the Pythagorean theorem, $36 = (3 + x/2)^2 + (3x^2)/4$. Solving this equation yields $x = \frac{3}{2}(\sqrt{13} - 1)$. Thus,

$$\triangle XAQ = \frac{1}{2}AX \times QT = \frac{1}{2} \times 3 \times \frac{\sqrt{3}}{2} \times \frac{3}{2}(\sqrt{13} - 1) = \frac{9\sqrt{3}}{8}(\sqrt{13} - 1)$$

We also have,

$$\sin(\angle XAQ) = \frac{QT}{AQ} = \frac{\sqrt{3}}{2} \times \frac{3}{2}(\sqrt{13} - 1) \times \frac{1}{6} = \frac{\sqrt{3}}{8}(\sqrt{13} - 1)$$

So $\angle XAQ = \sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))$ and,

$$BAQ = 36\pi \times \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{360^\circ} = \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{10^\circ}\pi$$

Since $ABC = 9\pi$, this gives,

$$XQR = ABC - BAQ + XAQ - XAR - ARC = 9\pi - \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{10^\circ}\pi + \frac{9\sqrt{3}}{8}(\sqrt{13} - 1) - \frac{9}{4}\sqrt{15} - (9\pi - \frac{\cos^{-1}(1/4)}{10^\circ}\pi) = \frac{9\sqrt{3}}{8}(\sqrt{13} - 1) + \frac{\cos^{-1}(1/4)}{10^\circ}\pi - \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{10^\circ}\pi - \frac{9}{4}\sqrt{15}$$

Recall that $\angle AXR = \cos^{-1}(1/4)$, so $\angle YXR = 180^\circ - 60^\circ - \cos^{-1}(1/4) = 120^\circ - \cos^{-1}(1/4)$, so the sector $YXR = 36\pi \times \frac{120^\circ - \cos^{-1}(1/4)}{360^\circ} = 12\pi - \frac{\cos^{-1}(1/4)}{10^\circ}\pi$. So, finally, putting everything together,

$$YQR = YXR - XQR = 12\pi - \frac{\cos^{-1}(1/4)}{10^\circ}\pi - (\frac{9\sqrt{3}}{8}(\sqrt{13} - 1) + \frac{\cos^{-1}(1/4)}{10^\circ}\pi - \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{10^\circ}\pi - \frac{9}{4}\sqrt{15}) = 12\pi - \frac{\cos^{-1}(1/4)}{5^\circ}\pi + \frac{\sin^{-1}(\frac{\sqrt{3}}{8}(\sqrt{13} - 1))}{10^\circ}\pi - \frac{9\sqrt{3}}{8}(\sqrt{13} - 1) + \frac{9}{4}\sqrt{15}$$

Approximately, $YQR \approx 4.6727$. For comparison, the red region has area $3(\pi - \sqrt{3}) \approx 4.2286$.

Answer 3

Hereby an analytical solution for the blue region:

Create a new axis system, based on points $XZ$ ($X'$-axis) and $XY$ ($Y'$-axis).
Calculate the projection of $R$ on the $X'$-axis, call this $R'$.

The circle, created by center $X$ and points $Y$ and $Z$, is obvious ($y' = ±C_1(x')$). Less obvious (but feasible) is the equation of the circle with center $A$ and points $B$ and $C$ ($y' = ±C_2(x')$), with the points $A$, $B$ and $C$ needing to be calculated first in the new axis system.

The blue region can be calculated as:

$$\int_0^{R'}{C_1(x') - C_2(x')dx'}$$

Hereby a screenshot: