Question

math 10 unit 4
applications of expanding and simplifying polynomials
given the figure below, how would you find the area of the shaded region?
now lets find the area of the shaded region in this figure.
example 1: find a polynomial expression for the area of the shaded region.
simplify the polynomial.

Explanation:

To find the area of the shaded region in such problems, we typically use the subtraction method: find the area of the larger shape and subtract the area of the unshaded (inner) shape. Let’s apply this to the first figure (with numbers) and then to the algebraic figures.

First Figure (12 m × 10 m rectangle with a 7 m × 2 m inner rectangle)

Step 1: Area of the larger rectangle

The larger rectangle has length \( 12 \, \text{m} \) and width \( 10 \, \text{m} \).
Area formula for a rectangle: \( \text{Area} = \text{length} \times \text{width} \).
So, \( A_{\text{large}} = 12 \times 10 = 120 \, \text{m}^2 \).

Step 2: Area of the inner (unshaded) rectangle

The inner rectangle has length \( 7 \, \text{m} \) and width \( 2 \, \text{m} \).
Using the same formula: \( A_{\text{inner}} = 7 \times 2 = 14 \, \text{m}^2 \).

Step 3: Area of the shaded region

Subtract the inner area from the larger area:
\( A_{\text{shaded}} = A_{\text{large}} - A_{\text{inner}} = 120 - 14 = 106 \, \text{m}^2 \).

Second Figure (Algebraic: \( 2x \times (x+2) \) outer rectangle with a \( 4 \times x \) inner rectangle)

Step 1: Area of the outer rectangle

Length \( = 2x \), Width \( = x + 2 \).
\( A_{\text{outer}} = 2x \times (x + 2) = 2x^2 + 4x \).

Step 2: Area of the inner (unshaded) rectangle

Length \( = 4 \), Width \( = x \).
\( A_{\text{inner}} = 4 \times x = 4x \).

Step 3: Area of the shaded region

Subtract: \( A_{\text{shaded}} = (2x^2 + 4x) - 4x = 2x^2 \).

Example 1 (Polynomial: \( 3x \times (x+5) \) outer rectangle with a \( (x+2) \times (x+3) \) inner rectangle)

Step 1: Area of the outer rectangle

Length \( = 3x \), Width \( = x + 5 \).
\( A_{\text{outer}} = 3x \times (x + 5) = 3x^2 + 15x \).

Step 2: Area of the inner (unshaded) rectangle

Length \( = x + 2 \), Width \( = x + 3 \).
Expand: \( (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \).

Step 3: Area of the shaded region

Subtract: \( A_{\text{shaded}} = (3x^2 + 15x) - (x^2 + 5x + 6) \).
Simplify: \( 3x^2 + 15x - x^2 - 5x - 6 = 2x^2 + 10x - 6 \).

Key Takeaway

For any “shaded region” problem with a larger shape and a smaller inner shape, the formula is:
\( \boldsymbol{A_{\text{shaded}} = A_{\text{larger}} - A_{\text{inner}}} \).

(Adjust the steps for algebraic expressions by expanding and simplifying polynomials.)

Answer:

To find the area of the shaded region in such problems, we typically use the subtraction method: find the area of the larger shape and subtract the area of the unshaded (inner) shape. Let’s apply this to the first figure (with numbers) and then to the algebraic figures.

First Figure (12 m × 10 m rectangle with a 7 m × 2 m inner rectangle)

Step 1: Area of the larger rectangle

The larger rectangle has length \( 12 \, \text{m} \) and width \( 10 \, \text{m} \).
Area formula for a rectangle: \( \text{Area} = \text{length} \times \text{width} \).
So, \( A_{\text{large}} = 12 \times 10 = 120 \, \text{m}^2 \).

Step 2: Area of the inner (unshaded) rectangle

The inner rectangle has length \( 7 \, \text{m} \) and width \( 2 \, \text{m} \).
Using the same formula: \( A_{\text{inner}} = 7 \times 2 = 14 \, \text{m}^2 \).

Step 3: Area of the shaded region

Subtract the inner area from the larger area:
\( A_{\text{shaded}} = A_{\text{large}} - A_{\text{inner}} = 120 - 14 = 106 \, \text{m}^2 \).

Second Figure (Algebraic: \( 2x \times (x+2) \) outer rectangle with a \( 4 \times x \) inner rectangle)

Step 1: Area of the outer rectangle

Length \( = 2x \), Width \( = x + 2 \).
\( A_{\text{outer}} = 2x \times (x + 2) = 2x^2 + 4x \).

Step 2: Area of the inner (unshaded) rectangle

Length \( = 4 \), Width \( = x \).
\( A_{\text{inner}} = 4 \times x = 4x \).

Step 3: Area of the shaded region

Subtract: \( A_{\text{shaded}} = (2x^2 + 4x) - 4x = 2x^2 \).

Example 1 (Polynomial: \( 3x \times (x+5) \) outer rectangle with a \( (x+2) \times (x+3) \) inner rectangle)

Step 1: Area of the outer rectangle

Length \( = 3x \), Width \( = x + 5 \).
\( A_{\text{outer}} = 3x \times (x + 5) = 3x^2 + 15x \).

Step 2: Area of the inner (unshaded) rectangle

Length \( = x + 2 \), Width \( = x + 3 \).
Expand: \( (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \).

Step 3: Area of the shaded region

Subtract: \( A_{\text{shaded}} = (3x^2 + 15x) - (x^2 + 5x + 6) \).
Simplify: \( 3x^2 + 15x - x^2 - 5x - 6 = 2x^2 + 10x - 6 \).

Key Takeaway

For any “shaded region” problem with a larger shape and a smaller inner shape, the formula is:
\( \boldsymbol{A_{\text{shaded}} = A_{\text{larger}} - A_{\text{inner}}} \).

(Adjust the steps for algebraic expressions by expanding and simplifying polynomials.)