1. **Problem statement:** Find the area of the corridor outside the square classroom ABCD, express it in terms of $x$, then find $x$ and the area of the classroom given the corridor area is 51 m². 2. **Given:** - Width of corridor = 3 m - EF = $x$ m - Area of corridor = 51 m² 3. **Understanding the problem:** The corridor surrounds the square classroom ABCD. The corridor forms a larger square with side length $(x + 6)$ because the corridor adds 3 m on each side (3 m on left + 3 m on right = 6 m total). 4. **Formula for area of corridor:** $$\text{Area of corridor} = \text{Area of larger square} - \text{Area of classroom}$$ 5. **Express areas:** - Area of larger square = $$(x + 6)^2$$ - Area of classroom = $$x^2$$ (since ABCD is a square with side $x$) 6. **Area of corridor in terms of $x$:** $$\text{Area of corridor} = (x + 6)^2 - x^2$$ 7. **Expand and simplify:** $$ (x + 6)^2 - x^2 = (x^2 + 12x + 36) - x^2 = 12x + 36 $$ 8. **Set equal to given corridor area:** $$ 12x + 36 = 51 $$ 9. **Solve for $x$:** $$ 12x = 51 - 36 $$ $$ 12x = 15 $$ $$ x = \frac{15}{12} = \frac{5}{4} = 1.25 $$ 10. **Find area of classroom:** $$ \text{Area} = x^2 = (1.25)^2 = 1.5625 \text{ m}^2 $$ **Final answers:** - $x = 1.25$ m - Area of classroom = $1.5625$ m²