1. **State the problem:** A rectangular picture frame measures 20 cm by 30 cm. A mat of uniform width $x$ cm is placed inside the frame, creating a border. The area of the mat is equal to the area of the picture. We need to find the width $x$ of the mat. 2. **Identify known values and variables:** - Outer frame dimensions: 20 cm by 30 cm - Width of mat: $x$ cm (unknown) - Area of picture: $20 \times 30 = 600$ cm$^2$ - Area of mat: equal to area of picture = 600 cm$^2$ 3. **Express the area of the mat in terms of $x$:** - The mat forms a border inside the frame, so the inner rectangle (picture area) is reduced by $2x$ in each dimension (width and height). - Inner rectangle dimensions: $(20 - 2x)$ by $(30 - 2x)$ - Area of inner rectangle (picture): $(20 - 2x)(30 - 2x)$ 4. **Write the equation for the mat area:** - Area of mat = Area of frame - Area of picture - Given mat area = picture area = 600 - So, $600 = 600 - (20 - 2x)(30 - 2x)$ 5. **Set up the equation and solve:** $$ (20 - 2x)(30 - 2x) = 600 - 600 = 0 $$ This is incorrect because mat area equals picture area, so mat area = 600, and mat area = frame area - inner area. Correct equation: $$ \text{Mat area} = \text{Frame area} - \text{Inner area} = 600 $$ $$ 600 = 600 - (20 - 2x)(30 - 2x) $$ Rearranged: $$ (20 - 2x)(30 - 2x) = 600 - 600 = 0 $$ This implies inner area is zero, which is impossible. Re-examining problem: The mat area equals the picture area, so mat area = 600. Since mat area = frame area - inner area, $$ 600 = 600 - (20 - 2x)(30 - 2x) $$ Rearranged: $$ (20 - 2x)(30 - 2x) = 600 - 600 = 0 $$ Again zero inner area, impossible. The problem states mat area equals picture area, so mat area = 600. Frame area = 20 \times 30 = 600 Mat area = frame area - inner area So, $$ 600 = 600 - (20 - 2x)(30 - 2x) $$ $$ (20 - 2x)(30 - 2x) = 600 - 600 = 0 $$ This is contradictory. Therefore, the problem likely means the mat area equals the picture area inside the mat, i.e., the inner rectangle area. Assuming mat area = picture area inside mat = inner area Then, $$ \text{Mat area} = \text{Frame area} - \text{Inner area} $$ Given mat area = inner area, $$ \text{Mat area} = \text{Inner area} = A $$ So, $$ \text{Frame area} = \text{Mat area} + \text{Inner area} = A + A = 2A $$ $$ 600 = 2A \Rightarrow A = 300 $$ So inner area = 300 6. **Set up equation for inner area:** $$ (20 - 2x)(30 - 2x) = 300 $$ 7. **Expand and simplify:** $$ 600 - 40x - 60x + 4x^2 = 300 $$ $$ 600 - 100x + 4x^2 = 300 $$ 8. **Bring all terms to one side:** $$ 4x^2 - 100x + 600 - 300 = 0 $$ $$ 4x^2 - 100x + 300 = 0 $$ 9. **Simplify by dividing entire equation by 4:** $$ \cancel{4}x^2 - \cancel{100}x + \cancel{300} = 0 \Rightarrow x^2 - 25x + 75 = 0 $$ 10. **Solve quadratic equation using quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a=1, b=-25, c=75 $$ $$ x = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(1)(75)}}{2(1)} = \frac{25 \pm \sqrt{625 - 300}}{2} = \frac{25 \pm \sqrt{325}}{2} $$ 11. **Calculate square root and approximate:** $$ \sqrt{325} \approx 18.03 $$ $$ x = \frac{25 \pm 18.03}{2} $$ 12. **Find two possible solutions:** $$ x_1 = \frac{25 + 18.03}{2} = \frac{43.03}{2} = 21.52 $$ $$ x_2 = \frac{25 - 18.03}{2} = \frac{6.97}{2} = 3.49 $$ 13. **Check for valid solution:** Width $x$ must be less than half the smaller frame dimension (20 cm), so $x < 10$ cm. Therefore, $x = 3.5$ cm (rounded to nearest tenth) is the valid width of the mat. **Final answer:** The width of the mat is approximately **3.5 cm**.