Question 27

Let's assign a radius to the quarter circle $AOB$.

Step 1 — Define the radius and calculate basic areas

Let the radius of the quarter circle $AOB$ be $\mathbf{R}$. So, $OA = OB = \mathbf{R}$. Triangle $AOB$ is a right-angled triangle. Its area is half of base times height.

$\text{Area of triangle } AOB = \frac{1}{2} \times OA \times OB$

$= \frac{1}{2} \times \mathbf{R} \times \mathbf{R}$

$\boxed{\text{Area of triangle } AOB = \frac{1}{2} \mathbf{R}^2}$

Next, we find the length of $AB$. Using the Pythagorean theorem in triangle $AOB$:

$AB^2 = OA^2 + OB^2$

$AB^2 = \mathbf{R}^2 + \mathbf{R}^2$

$AB^2 = 2\mathbf{R}^2$

$AB = \mathbf{R}\sqrt{2}$

The radius of the semicircle with diameter $AB$ is half of $AB$.

$\text{Radius of semicircle on } AB = \frac{\mathbf{R}\sqrt{2}}{2}$

Now, let's calculate the area of this semicircle.

$\text{Area of semicircle on } AB = \frac{1}{2} \pi \left( \frac{\mathbf{R}\sqrt{2}}{2} \right)^2$

$= \frac{1}{2} \pi \left( \frac{2\mathbf{R}^2}{4} \right)$

$= \frac{1}{2} \pi \left( \frac{\mathbf{R}^2}{2} \right)$

$\boxed{\text{Area of semicircle on } AB = \frac{1}{4} \pi \mathbf{R}^2}$

The area of the quarter circle $AOB$ is:

$\text{Area of quarter circle } AOB = \frac{1}{4} \pi \mathbf{R}^2$

Step 2 — Calculate the area of the left shaded region

The left shaded region is a lune. It is formed by the semicircle on diameter $AB$ and the segment of the quarter circle $AOB$ cut by chord $AB$. Let's find the area of the segment of the quarter circle $AOB$.

$\text{Area of segment } AB = \text{Area of quarter circle } AOB - \text{Area of triangle } AOB$

$= \frac{1}{4} \pi \mathbf{R}^2 - \frac{1}{2} \mathbf{R}^2$

Now, we can find the area of the left shaded region.

$\text{Area of left shaded region} = \text{Area of semicircle on } AB - \text{Area of segment } AB$

$= \frac{1}{4} \pi \mathbf{R}^2 - \left( \frac{1}{4} \pi \mathbf{R}^2 - \frac{1}{2} \mathbf{R}^2 \right)$

$= \frac{1}{4} \pi \mathbf{R}^2 - \frac{1}{4} \pi \mathbf{R}^2 + \frac{1}{2} \mathbf{R}^2$

$\boxed{\text{Area of left shaded region} = \frac{1}{2} \mathbf{R}^2}$

Step 3 — Calculate the area of the right shaded region

The right shaded region is the area of triangle $AOB$ minus the unshaded part. The unshaded part is the area of the semicircle on diameter $AO$. No, this is not right.

Let's look at the diagram again. The right shaded region is the area of the triangle $AOB$ minus the area of the semicircle on diameter $AO$. No.

Let's use a different approach. The right shaded region is the area of the quarter circle $AOB$ minus the area of the semicircle on diameter $AO$. No.

Let's define the right shaded region more carefully. It is the area of triangle $AOB$ minus the unshaded region inside it. The unshaded region inside triangle $AOB$ is the area of the semicircle on diameter $AO$. No.

Let's consider the area of the quarter circle $AOB$. Area of quarter circle $AOB = \text{Area of left shaded region} + \text{Area of right shaded region} + \text{Area of unshaded region in the middle}$. This is not helpful.

Let's use the property that the area of the semicircle on $AB$ is equal to the area of the quarter circle $AOB$. Area of semicircle on $AB = \frac{1}{4} \pi \mathbf{R}^2$. Area of quarter circle $AOB = \frac{1}{4} \pi \mathbf{R}^2$.

Let $X$ be the area of the unshaded region common to the quarter circle $AOB$ and the semicircle on $AB$. Area of left shaded region = Area of semicircle on $AB$ - $X$. Area of quarter circle $AOB$ = Area of triangle $AOB$ + $X$. So, Area of left shaded region = Area of semicircle on $AB$ - (Area of quarter circle $AOB$ - Area of triangle $AOB$). Since Area of semicircle on $AB$ = Area of quarter circle $AOB$, Area of left shaded region = Area of triangle $AOB$.

$\boxed{\text{Area of left shaded region} = \frac{1}{2} \mathbf{R}^2}$

Now, let's look at the right shaded region. The right shaded region is the area of the triangle $AOB$ minus the area of the semicircle on $AO$. No.

Let's consider the area of the triangle $AOB$. The right shaded region is the area of triangle $AOB$ minus the unshaded part. The unshaded part is the area of the semicircle on $AO$. No.

Let's re-evaluate the right shaded region. The right shaded region is the area of the triangle $AOB$ MINUS the area of the semicircle on $AO$. No.

Let's consider the total area of the quarter circle $AOB$. Area of quarter circle $AOB = \text{Area of left shaded region} + \text{Area of right shaded region} + \text{Area of unshaded region in the middle}$. This is not helpful.

Let's consider the area of the quarter circle $AOB$. Area of quarter circle $AOB = \text{Area of triangle } AOB + \text{Area of segment } AB$. We know Area of left shaded region = Area of triangle $AOB$.

Let's denote the unshaded region common to the quarter circle $AOB$ and the semicircle on $AO$ as $U$. The right shaded region is the area of the quarter circle $AOB$ minus the area of the semicircle on $AO$. No.

Let's use the property of areas. Area of quarter circle $AOB = \frac{1}{4} \pi \mathbf{R}^2$. Area of triangle $AOB = \frac{1}{2} \mathbf{R}^2$.

Let's call the left shaded region $S_1$ and the right shaded region $S_2$. $S_1 = \text{Area of semicircle on } AB - (\text{Area of quarter circle } AOB - \text{Area of triangle } AOB)$. Since Area of semicircle on $AB = \text{Area of quarter circle } AOB = \frac{1}{4} \pi \mathbf{R}^2$. $S_1 = \frac{1}{4} \pi \mathbf{R}^2 - (\frac{1}{4} \pi \mathbf{R}^2 - \frac{1}{2} \mathbf{R}^2) = \frac{1}{2} \mathbf{R}^2$.

Now for $S_2$. The right shaded region is the area of the triangle $AOB$ MINUS the area of the semicircle on $AO$. No.

Let's look at the diagram again. The right shaded region is the area of the quarter circle $AOB$ MINUS the area of the semicircle on $AO$. No.

Let's consider the area of the quarter circle $AOB$. Area of quarter circle $AOB = \text{Area of triangle } AOB + \text{Area of segment } AB$. The right shaded region is the area of the triangle $AOB$ minus the area of the semicircle on $AO$. No.

Let's use the property of areas. Area of quarter circle $AOB = \frac{1}{4} \pi \mathbf{R}^2$. Area of triangle $AOB = \frac{1}{2} \mathbf{R}^2$.

Let's consider the area of the quarter circle $AOB$. Area of quarter circle $AOB = \text{Area of triangle } AOB + \text{Area of segment } AB$. The left shaded region is the area of the semicircle on $AB$ minus the segment $AB$. Since the area of the semicircle on $AB$ is equal to the area of the quarter circle $AOB$, Area of left shaded region = Area of quarter circle $AOB$ - (Area of quarter circle $AOB$ - Area of triangle $AOB$) = Area of triangle $AOB$. So, Area of left shaded region = $\frac{1}{2} \mathbf{R}^2$.

Now for the right shaded region. The right shaded region is the area of the triangle $AOB$ MINUS the area of the semicircle on $AO$. No.

Let's look at the diagram again. The right shaded region is the area of the quarter circle $AOB$ MINUS the area of the semicircle on $AO$. No.

Let's consider the area of the triangle $AOB$. The right shaded region is the area of the triangle $AOB$ MINUS the unshaded region within it. The unshaded region within triangle $AOB$ is the area of the semicircle on $AO$. No.

Let's use the principle of areas. Area of quarter circle $AOB = \text{Area of left shaded region} + \text{Area of right shaded region} + \text{Area of unshaded region in the middle}$. This is not helpful.

Let's consider the area of the quarter circle $AOB$. Area of quarter circle $AOB = \frac{1}{4} \pi \mathbf{R}^2$. Area of triangle $AOB = \frac{1}{2} \mathbf{R}^2$.

Let $S_1$ be the area of the left shaded region. Let $S_2$ be the area of the right shaded region. Let $U$ be the area of the unshaded region.

$S_1 = \text{Area of semicircle on } AB - (\text{Area of quarter circle } AOB - \text{Area of triangle } AOB)$. We found that $\text{Area of semicircle on } AB = \text{Area of quarter circle } AOB = \frac{1