Math worksheet for calculating the area of shaded sectors and segments of circles.

Here are the step-by-step solutions for each problem. Remember to keep your calculator in Degree mode, not Radians.

1) Two Sectors

The shaded area consists of two identical sectors.
* Radius ($r$): $4 \text{ cm}$
* Angle of unshaded part: $158^\circ$
* Angle of one shaded sector: The angles on a straight line add up to $180^\circ$. So, the angle for one orange slice is $180^\circ - 158^\circ = 22^\circ$.
* Total Angle: Since there are two slices, the total angle is $22^\circ + 22^\circ = 44^\circ$.

Calculation:
$$ \text{Area} = \frac{44}{360} \times \pi \times 4^2 $$
$$ \text{Area} = \frac{44}{360} \times \pi \times 16 \approx 6.143... $$

Answer: $6.14 \text{ cm}^2$

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2) Quarter Annulus (Ring)

This shape is a large quarter-circle with a smaller quarter-circle removed from the corner.
* Outer Radius ($R$): $14 \text{ cm}$
* Inner Radius ($r$): The vertical side is $10 \text{ cm}$, so the inner radius is $10 \text{ cm}$.
* Fraction: It is a quarter circle ($90^\circ$), so we divide by 4.

Calculation:
$$ \text{Area} = \frac{1}{4} \pi (R^2 - r^2) $$
$$ \text{Area} = \frac{1}{4} \pi (14^2 - 10^2) $$
$$ \text{Area} = \frac{1}{4} \pi (196 - 100) $$
$$ \text{Area} = \frac{1}{4} \pi (96) = 24\pi \approx 75.398... $$

Answer: $75.4 \text{ cm}^2$

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3) Segment

The shaded area is a segment. We find this by taking the area of the quarter-circle sector and subtracting the area of the triangle.
* Radius ($r$): $9 \text{ cm}$
* Angle: $90^\circ$ (indicated by the square symbol).

Step A: Area of Sector
$$ \text{Area}_{\text{sector}} = \frac{90}{360} \times \pi \times 9^2 = \frac{1}{4} \times \pi \times 81 \approx 63.617 \text{ cm}^2 $$

Step B: Area of Triangle
$$ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 9 = 40.5 \text{ cm}^2 $$

Step C: Subtract
$$ 63.617 - 40.5 = 23.117... $$

Answer: $23.1 \text{ cm}^2$

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4) Small Segment

Similar to question 3, we subtract the triangle from the sector.
* Radius ($r$): $16 \text{ cm}$
* Angle ($\theta$): $55^\circ$

Step A: Area of Sector
$$ \text{Area}_{\text{sector}} = \frac{55}{360} \times \pi \times 16^2 \approx 122.86 \text{ cm}^2 $$

Step B: Area of Triangle
Formula: $\frac{1}{2} r^2 \sin(\theta)$
$$ \text{Area}_{\text{triangle}} = \frac{1}{2} \times 16^2 \times \sin(55^\circ) = 128 \times 0.81915... \approx 104.85 \text{ cm}^2 $$

Step C: Subtract
$$ 122.86 - 104.85 = 18.01... $$

Answer: $18.0 \text{ cm}^2$

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5) Large Sector

This is a straightforward sector calculation.
* Radius ($r$): $7 \text{ cm}$
* Angle ($\theta$): $169^\circ$

Calculation:
$$ \text{Area} = \frac{169}{360} \times \pi \times 7^2 $$
$$ \text{Area} = \frac{169}{360} \times \pi \times 49 \approx 72.539... $$

Answer: $72.5 \text{ cm}^2$

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6) Rectangle inside a Circle

The shaded region is a rectangle. To find its area, we need its width and height.
* Radius ($r$): $6 \text{ cm}$
* Height of rectangle: The diagram shows distances of $2 \text{ cm}$ from the center to the top and bottom edges. Total height $= 2 + 2 = 4 \text{ cm}$.
* Width of rectangle: Imagine a right-angled triangle formed by the radius, the distance from the center to the edge ($2 \text{ cm}$), and half the width of the rectangle ($x$).

Using Pythagoras theorem ($a^2 + b^2 = c^2$):
$$ x^2 + 2^2 = 6^2 $$
$$ x^2 + 4 = 36 $$
$$ x^2 = 32 $$
$$ x = \sqrt{32} \approx 5.657 \text{ cm} $$

The full width is $2x$, so Width $= 2 \times \sqrt{32} \approx 11.314 \text{ cm}$.

Calculate Area:
$$ \text{Area} = \text{Width} \times \text{Height} $$
$$ \text{Area} = 11.314 \times 4 \approx 45.256... $$

Answer: $45.3 \text{ cm}^2$

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Final Answer:
1) 6.14 cm²
2) 75.4 cm²
3) 23.1 cm²
4) 18.0 cm²
5) 72.5 cm²
6) 45.3 cm²