To solve this problem, we need to find the area of the shaded region $WXYZ$. The shaded area is the difference between the areas of the two sectors, one with a radius of 34 cm (sector $OWX$) and the other with a radius of 23 cm (sector $OYZ$).

Step 1: Formula for the Area of a Sector

The formula for the area $A$ of a sector with radius $r$ and angle $\theta$ (in degrees) is given by: $A = \frac{\theta}{360} \times \pi r^2$ where:

• $r$ is the radius of the sector,
• $\theta$ is the central angle in degrees,
• $\pi$ is approximately 3.14159.

Step 2: Area of Sector $OWX$ (radius = 34 cm, angle = 75°)

$A_{\text{OWX}} = \frac{75}{360} \times \pi \times (34)^2$ $A_{\text{OWX}} = \frac{75}{360} \times \pi \times 1156$ $A_{\text{OWX}} \approx 0.2083 \times 3.14159 \times 1156$ $A_{\text{OWX}} \approx 757.92 \, \text{cm}^2$

Step 3: Area of Sector $OYZ$ (radius = 23 cm, angle = 75°)

$A_{\text{OYZ}} = \frac{75}{360} \times \pi \times (23)^2$ $A_{\text{OYZ}} = \frac{75}{360} \times \pi \times 529$ $A_{\text{OYZ}} \approx 0.2083 \times 3.14159 \times 529$ $A_{\text{OYZ}} \approx 346.36 \, \text{cm}^2$

Step 4: Shaded Area $WXYZ$

The shaded area is the difference between the two sector areas: $A_{\text{shaded}} = A_{\text{OWX}} - A_{\text{OYZ}}$ $A_{\text{shaded}} = 757.92 - 346.36$ $A_{\text{shaded}} \approx 411.56 \, \text{cm}^2$

Thus, the area of the shaded shape $WXYZ$ is approximately 411.6 cm² (rounded to 1 decimal place).

Would you like further details or have any questions?

Relative Questions:

1. How do we calculate the circumference of a sector?
2. What is the relationship between a sector's area and its central angle?
3. How would the solution change if the central angle was 90° instead of 75°?
4. Can the formula for a sector's area be extended to radians? How?
5. What if one sector was only partially shaded—how would you approach that calculation?

Tip:

For any geometric problem involving sectors, always ensure that the angle is in degrees or radians and is used correctly in formulas for area or arc length.