Question

1. find the probability that a randomly chosen point in each figure lies in the shaded region. hint: p(shaded region) = \\(\frac{\text{area of shaded region}}{\text{area of entire figure}}\\) a) figure: rectangle (35 ft length, 22 ft width) with a circle (16 ft diameter) inside b) figure: square (7 m side) with a smaller square (3 m side) inside c) figure: square (8 in side) with four circles inside

Explanation:

Part (a)

Step 1: Find area of entire figure (rectangle)

The formula for the area of a rectangle is \( A = l \times w \), where \( l = 35 \) ft and \( w = 22 \) ft.
\( A_{\text{rectangle}} = 35 \times 22 = 770 \) square feet.

Step 2: Find area of unshaded region (circle)

The diameter of the circle is 16 ft, so the radius \( r = \frac{16}{2} = 8 \) ft. The formula for the area of a circle is \( A = \pi r^2 \).
\( A_{\text{circle}} = \pi \times 8^2 = 64\pi \) square feet.

Step 3: Find area of shaded region

Subtract the area of the circle from the area of the rectangle:
\( A_{\text{shaded}} = 770 - 64\pi \).

Step 4: Calculate probability

Using the formula \( P(\text{shaded}) = \frac{\text{area of shaded region}}{\text{area of entire figure}} \):
\( P = \frac{770 - 64\pi}{770} \approx \frac{770 - 201.06}{770} \approx \frac{568.94}{770} \approx 0.74 \) (or simplify the fraction/keep in terms of \( \pi \)).

Part (b)

Step 1: Find area of entire figure (square)

The side length of the outer square is 7 m. Area of a square is \( A = s^2 \).
\( A_{\text{outer square}} = 7^2 = 49 \) square meters.

Step 2: Find area of unshaded region (inner square)

The side length of the inner square is 3 m.
\( A_{\text{inner square}} = 3^2 = 9 \) square meters.

Step 3: Find area of shaded region

Subtract the area of the inner square from the outer square:
\( A_{\text{shaded}} = 49 - 9 = 40 \) square meters.

Step 4: Calculate probability

\( P = \frac{40}{49} \approx 0.816 \).

Part (c)

Step 1: Find area of entire figure (square)

The side length of the square is 8 in.
\( A_{\text{square}} = 8^2 = 64 \) square inches.

Step 2: Find area of shaded regions (4 circles)

The diameter of each circle: since there are 2 circles along the side of the square (8 in), each diameter is \( \frac{8}{2} = 4 \) in, so radius \( r = \frac{4}{2} = 2 \) in.
Area of one circle: \( A_{\text{circle}} = \pi \times 2^2 = 4\pi \).
Area of 4 circles: \( 4 \times 4\pi = 16\pi \) square inches.

Step 3: Calculate probability

\( P = \frac{16\pi}{64} = \frac{\pi}{4} \approx 0.785 \).

Final Answers:

a) \( \boldsymbol{\frac{770 - 64\pi}{770}} \) (or approximately \( \boldsymbol{0.74} \))
b) \( \boldsymbol{\frac{40}{49}} \) (or approximately \( \boldsymbol{0.82} \))
c) \( \boldsymbol{\frac{\pi}{4}} \) (or approximately \( \boldsymbol{0.79} \))

Answer:

Step 1: Find area of entire figure (square)

The side length of the square is 8 in.
\( A_{\text{square}} = 8^2 = 64 \) square inches.

Step 2: Find area of shaded regions (4 circles)

The diameter of each circle: since there are 2 circles along the side of the square (8 in), each diameter is \( \frac{8}{2} = 4 \) in, so radius \( r = \frac{4}{2} = 2 \) in.
Area of one circle: \( A_{\text{circle}} = \pi \times 2^2 = 4\pi \).
Area of 4 circles: \( 4 \times 4\pi = 16\pi \) square inches.

Step 3: Calculate probability

\( P = \frac{16\pi}{64} = \frac{\pi}{4} \approx 0.785 \).

Final Answers:

a) \( \boldsymbol{\frac{770 - 64\pi}{770}} \) (or approximately \( \boldsymbol{0.74} \))
b) \( \boldsymbol{\frac{40}{49}} \) (or approximately \( \boldsymbol{0.82} \))
c) \( \boldsymbol{\frac{\pi}{4}} \) (or approximately \( \boldsymbol{0.79} \))