Geometry worksheet on calculating the area of circles and sectors, including shaded regions, with answers to be left in terms of π.

Problem Analysis:

The worksheet asks us to calculate the areas of circles and sectors. Here's a breakdown of the tasks:

1. Find the area of each circle (Questions 1 and 2).
2. Find the area of each shaded sector (Questions 3, 4, 5, and 6).

We will solve each part step by step.

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Formula Recap:

1. Area of a Circle:
The formula for the area of a circle is:
\[
A = \pi r^2
\]
where \( r \) is the radius of the circle.

2. Area of a Sector:
The formula for the area of a sector is:
\[
A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2
\]
where \( \theta \) is the central angle in degrees and \( r \) is the radius of the circle.

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Solutions:

#### 1. Find the area of each circle. Leave answers in terms of \( \pi \).

##### Question 1:
- The radius \( r \) is given as 15.
- Using the formula for the area of a circle:
\[
A = \pi r^2 = \pi (15)^2 = \pi \cdot 225 = 225\pi
\]
- Answer: \( 225\pi \)

##### Question 2:
- The diameter is given as 10. The radius \( r \) is half of the diameter:
\[
r = \frac{10}{2} = 5
\]
- Using the formula for the area of a circle:
\[
A = \pi r^2 = \pi (5)^2 = \pi \cdot 25 = 25\pi
\]
- Answer: \( 25\pi \)

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#### 2. Find the area of each shaded sector. Leave answers in terms of \( \pi \).

##### Question 3:
- The radius \( r \) is 15, and the central angle \( \theta \) is \( 40^\circ \).
- Using the formula for the area of a sector:
\[
A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{40^\circ}{360^\circ} \times \pi (15)^2
\]
Simplify the fraction:
\[
\frac{40^\circ}{360^\circ} = \frac{1}{9}
\]
Substitute \( r = 15 \):
\[
A_{\text{sector}} = \frac{1}{9} \times \pi (15)^2 = \frac{1}{9} \times \pi \cdot 225 = \frac{225\pi}{9} = 25\pi
\]
- Answer: \( 25\pi \)

##### Question 4:
- The radius \( r \) is 12, and the central angle \( \theta \) is \( 40^\circ \).
- Using the formula for the area of a sector:
\[
A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{40^\circ}{360^\circ} \times \pi (12)^2
\]
Simplify the fraction:
\[
\frac{40^\circ}{360^\circ} = \frac{1}{9}
\]
Substitute \( r = 12 \):
\[
A_{\text{sector}} = \frac{1}{9} \times \pi (12)^2 = \frac{1}{9} \times \pi \cdot 144 = \frac{144\pi}{9} = 16\pi
\]
- Answer: \( 16\pi \)

##### Question 5:
- The radius \( r \) is 8, and the central angle \( \theta \) is \( 90^\circ \) (since it forms a right angle).
- Using the formula for the area of a sector:
\[
A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi (8)^2
\]
Simplify the fraction:
\[
\frac{90^\circ}{360^\circ} = \frac{1}{4}
\]
Substitute \( r = 8 \):
\[
A_{\text{sector}} = \frac{1}{4} \times \pi (8)^2 = \frac{1}{4} \times \pi \cdot 64 = \frac{64\pi}{4} = 16\pi
\]
- Answer: \( 16\pi \)

##### Question 6:
- The radius \( r \) is 6, and the central angle \( \theta \) is \( 36^\circ \).
- Using the formula for the area of a sector:
\[
A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{36^\circ}{360^\circ} \times \pi (6)^2
\]
Simplify the fraction:
\[
\frac{36^\circ}{360^\circ} = \frac{1}{10}
\]
Substitute \( r = 6 \):
\[
A_{\text{sector}} = \frac{1}{10} \times \pi (6)^2 = \frac{1}{10} \times \pi \cdot 36 = \frac{36\pi}{10} = 3.6\pi
\]
- Answer: \( 3.6\pi \)

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Final Answers:

1. \( 225\pi \)
2. \( 25\pi \)
3. \( 25\pi \)
4. \( 16\pi \)
5. \( 16\pi \)
6. \( 3.6\pi \)

\[
\boxed{225\pi, 25\pi, 25\pi, 16\pi, 16\pi, 3.6\pi}
\]