**Transcription for Educational Website**---### Homework Solutions**Solve the Following Problems:****17.** The area of the rectangle shown is $18a^2b$. Write an expression for the unknown dimension.[Diagram of Rectangle]---**18.** The width of a rectangle is 3 cm less than the length. The perimeter is 42 cm. Find the area of the rectangle.```[Diagram of Rectangle, labeled with "6a" for the width and "8b" for the length]```- Diagram Explanation: - The diagram presents a generic rectangle. One side (height) is labeled $6ab$ and the other side (length) is not labeled. ---### Solution Steps:For **Problem 17**:1. Let one dimension of the rectangle be $x$.2. The area of a rectangle is given by $ \text{Area} = \text{Length} \times \text{Width} $.3. Set up the equation based on the given area $18a^2b$: \[ \text{Length} \times \text{Width} = 18a^2b \]4. Solve for the unknown dimension.For **Problem 18**:1. Let the length of the rectangle be $L$ cm.2. The width is then $L - 3$ cm as it is 3 cm less than the length.3. The perimeter (P) of a rectangle is: \[ P = 2 \times (\text{Length} + \text{Width}) \]4. Substitute the given perimeter value and the expressions for length and width: \[ 42 = 2 \times (L + (L - 3)) \]5. Solve for $L$: \[ 42 = 2 \times (2L - 3) \] \[ 42 = 4L - 6 \] \[ 48 = 4L \] \[ L = 12 \]6. The width $W$ is: \[ W = L - 3 = 12 - 3 = 9 \text{ cm} \]7. Find the area (A) of the rectangle:

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The area of the rectangle shown is 18a?6. Write an expression for the unknown dimension.

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

**Transcription for Educational Website**

---

### Homework Solutions

**Solve the Following Problems:**

**17.** The area of the rectangle shown is $18a^2b$. Write an expression for the unknown dimension.

[Diagram of Rectangle]

---

**18.** The width of a rectangle is 3 cm less than the length. The perimeter is 42 cm. Find the area of the rectangle.

```
[Diagram of Rectangle, labeled with "6a" for the width and "8b" for the length]
```

- Diagram Explanation:
- The diagram presents a generic rectangle. One side (height) is labeled $6ab$ and the other side (length) is not labeled.

---

### Solution Steps:

For **Problem 17**:
1. Let one dimension of the rectangle be $x$.
2. The area of a rectangle is given by $ \text{Area} = \text{Length} \times \text{Width} $.
3. Set up the equation based on the given area $18a^2b$:
\[
\text{Length} \times \text{Width} = 18a^2b
\]
4. Solve for the unknown dimension.

For **Problem 18**:
1. Let the length of the rectangle be $L$ cm.
2. The width is then $L - 3$ cm as it is 3 cm less than the length.
3. The perimeter (P) of a rectangle is:
\[
P = 2 \times (\text{Length} + \text{Width})
\]
4. Substitute the given perimeter value and the expressions for length and width:
\[
42 = 2 \times (L + (L - 3))
\]
5. Solve for $L$:
\[
42 = 2 \times (2L - 3)
\]
\[
42 = 4L - 6
\]
\[
48 = 4L
\]
\[
L = 12
\]
6. The width $W$ is:
\[
W = L - 3 = 12 - 3 = 9 \text{ cm}
\]
7. Find the area (A) of the rectangle:

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