https://file.aiquickdraw.com/mathbot/file/1730273910501vnephqj2.jpg

Let's tackle questions 12 and 13.

---

### Question 12
We have a rectangular swimming pool with dimensions \( 12 \) meters by \( 18 \) meters. Surrounding it is a pavement of uniform width \( x \) meters. We are asked to find the area of the pavement, \( A \), as a function of \( x \).

#### Solution:
1. The **outer dimensions** of the pavement (including the pool) will be:
   - Width: \( 12 + 2x \)
   - Length: \( 18 + 2x \)

2. The **area of the outer rectangle** (pool + pavement) is:
   \[
   \text{Outer area} = (12 + 2x)(18 + 2x)
   \]

3. The **area of the pool** (inner rectangle) is:
   \[
   \text{Pool area} = 12 \times 18 = 216 \, \text{m}^2
   \]

4. The **area of the pavement** (only) is the difference between the outer area and the pool area:
   \[
   A(x) = (12 + 2x)(18 + 2x) - 216
   \]

5. Expanding the terms:
   \[
   A(x) = (12 + 2x)(18 + 2x) - 216
   \]
   \[
   = (12 \cdot 18) + (12 \cdot 2x) + (2x \cdot 18) + (2x \cdot 2x) - 216
   \]
   \[
   = 216 + 24x + 36x + 4x^2 - 216
   \]
   \[
   = 60x + 4x^2
   \]

6. Therefore, the area of the pavement \( A \) as a function of \( x \) is:
   \[
   A(x) = 4x^2 + 60x
   \]

---

### Question 13
In this problem, we have a **right isosceles triangle** where the two equal sides are each \( x \) units long, and the hypotenuse has length \( h \) units. We need to write \( h \) as a function of \( x \).

#### Solution:
1. In a right isosceles triangle, the **Pythagorean theorem** states:
   \[
   h^2 = x^2 + x^2
   \]

2. Simplify the equation:
   \[
   h^2 = 2x^2
   \]

3. Taking the square root of both sides:
   \[
   h = \sqrt{2x^2}
   \]

4. Simplifying further:
   \[
   h = x\sqrt{2}
   \]

5. Thus, the hypotenuse \( h \) as a function of \( x \) is:
   \[
   h(x) = x\sqrt{2}
   \]

---

Let me know if you would like further clarification on any steps or have any questions!

---

### Related Questions for Further Practice:
1. What would the area formula in Question 12 look like if the pool were circular with radius \( 6 \) meters and a pavement width of \( x \)?
2. How would the solution to Question 13 change if the triangle was a right triangle with one side \( x \) and the other side \( y \), instead of an isosceles triangle?
3. In Question 12, find the perimeter of the outer rectangle (including the pavement) as a function of \( x \).
4. For Question 13, if \( x \) were doubled, what would be the effect on \( h \)?
5. For Question 12, calculate the area of the pavement if \( x = 1 \) meter.

### Quick Tip:
When dealing with right triangles, always apply the Pythagorean theorem carefully to ensure correct expressions for hypotenuse or leg lengths.

Learn how to find the area of pavement surrounding a rectangular pool and the hypotenuse length of a right isosceles triangle. This solution involves key geometric concepts, including the area formula for a rectangle and the Pythagorean theorem. Suitable for high school students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation