**** A photograph is 18 cm by 12 cm. It is to be surrounded by a frame of uniform width whose area is

equal to the area of the photograph. Find the width of the frame.

Let's denote the width of the frame by \( x \) cm.

### Step 1: Determine the Area of the Photograph
The area of the photograph is:

\[
\text{Area of the photograph} = 18 \times 12 = 216 \text{ square cm}
\]

### Step 2: Determine the Total Area of the Photograph with the Frame
The total dimensions of the photograph plus the frame are:
- Length: \( 18 + 2x \) cm
- Width: \( 12 + 2x \) cm

The total area (including the photograph and the frame) is:

\[
\text{Total area} = (18 + 2x)(12 + 2x)
\]

### Step 3: Set Up the Equation
The problem states that the area of the frame is equal to the area of the photograph. Therefore, the area of the frame can be found by subtracting the area of the photograph from the total area:

\[
\text{Area of the frame} = \text{Total area} - \text{Area of the photograph}
\]

Since the area of the frame equals the area of the photograph, we have:

\[
(18 + 2x)(12 + 2x) - 216 = 216
\]

### Step 4: Solve the Quadratic Equation
1. Expand the expression on the left side:

\[
(18 + 2x)(12 + 2x) = 216 + 36x + 24x + 4x^2 = 216 + 60x + 4x^2
\]

2. Substitute into the equation:

\[
216 + 60x + 4x^2 - 216 = 216
\]

This simplifies to:

\[
60x + 4x^2 = 216
\]

3. Rearrange the equation into standard quadratic form:

\[
4x^2 + 60x - 216 = 0
\]

4. Divide the entire equation by 4 to simplify:

\[
x^2 + 15x - 54 = 0
\]

5. Factor the quadratic equation:

\[
(x + 18)(x - 3) = 0
\]

This gives two possible solutions:

\[
x = -18 \quad \text{or} \quad x = 3
\]

### Step 5: Interpret the Result
Since the width of the frame cannot be negative, we discard \( x = -18 \). Thus, the width of the frame is \( x = 3 \) cm.

### Final Answer:
The width of the frame is **3 cm**.

Would you like more details or have any further questions? Here are some related problems you might find interesting:

1. What would be the width of the frame if the area of the frame was double that of the photograph?
2. How would the solution change if the photograph had dimensions 20 cm by 15 cm?
3. What is the perimeter of the photograph including the frame?
4. If the frame was not uniform, but had different widths on each side, how would you set up the equations?
5. What if the frame was square, with a constant width of 2 cm all around?

**Tip:** When solving quadratic equations, always check if the equation can be factored before jumping to the quadratic formula. Factoring can often simplify the process.

Learn how to determine the width of a uniform frame surrounding a photograph such that its area equals the photograph's area. Step-by-step solution provided with detailed algebraic calculations and geometric insights. Suitable for Grades 7-9.

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