The diagonal of a rectangular field is 15 metres and the difference between its length and width is 3 metres. The area of the rectangular field is (M.A.T., 2007)

### Concept & Quadratic Geometry This question combines the Pythagorean theorem with a linear constraint between the two sides, leading to a quadratic equation. Let the sides be expressed in terms of a single variable to solve for their exact values. ### Step-by-Step Solution 1. Let the width of the rectangular field be $w$. 2. The difference between the length ($l$) and width is $3 \text{ m}$, which means $l = w + 3$. 3. We are given the diagonal ($d$) $= 15 \text{ m}$. Apply the Pythagorean theorem: $$ l^2 + w^2 = d^2 $$ $$ (w + 3)^2 + w^2 = 15^2 $$ 4. Expand the squared binomial: $$ (w^2 + 6w + 9) + w^2 = 225 $$ $$ 2w^2 + 6w + 9 = 225 $$ 5. Form a standard quadratic equation by setting it to zero: $$ 2w^2 + 6w + 9 - 225 = 0 $$ $$ 2w^2 + 6w - 216 = 0 $$ 6. Simplify by dividing the entire equation by 2: $$ w^2 + 3w - 108 = 0 $$ 7. Factor the quadratic equation. We need two numbers that multiply to $-108$ and add to $3$. Those numbers are $12$ and $-9$. $$ (w + 12)(w - 9) = 0 $$ Since width cannot be negative, $w = 9 \text{ m}$. 8. Find the length: $$ l = w + 3 = 9 + 3 = 12 \text{ m} $$ 9. Calculate the area of the rectangular field: $$ \text{Area} = l \times w = 12 \times 9 = 108 \text{ m}^2 $$ ### Exam Strategy & Shortcut Instead of solving quadratics, look at the diagonal $15$. Recognize that $15$ is a multiple of the famous $3-4-5$ Pythagorean triple (multiplied by $3$). The sides would be $3 \times 3 = 9$ and $4 \times 3 = 12$. Check the condition: is the difference between these sides equal to $3$? Yes ($12 - 9 = 3$). The sides are definitively $12$ and $9$. Area $= 12 \times 9 = 108$. Solved in 5 seconds. ### Common Pitfall A recurring algebraic error is expanding $(w + 3)^2$ as $w^2 + 9$ instead of $w^2 + 6w + 9$. Missing the middle term completely changes the quadratic and leads to incorrect side lengths. ### Final Answer Therefore, the correct answer is **$108\text{ m}^2$**.