A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. The length of its diagonal is

### Concept & Algebraic Identities To find the diagonal without individually solving for length and width, we can manipulate standard algebraic identities. The diagonal is $\sqrt{l^2 + w^2}$. We can extract $(l^2 + w^2)$ from the expansion of $(l + w)^2$. $$ (l + w)^2 = l^2 + w^2 + 2lw $$ ### Step-by-Step Solution 1. We are given the area of the carpet: $$ \text{Area} = l \times w = 120 \text{ sq. m} $$ 2. We are given the perimeter of the carpet: $$ \text{Perimeter} = 2(l + w) = 46 \text{ m} $$ Divide by 2 to find the sum of length and width: $$ l + w = 23 \text{ m} $$ 3. We need to find the diagonal ($d$), where $d^2 = l^2 + w^2$. 4. Use the algebraic identity $(l + w)^2 = (l^2 + w^2) + 2lw$: Substitute the known values of $(l + w)$ and $(lw)$: $$ (23)^2 = (l^2 + w^2) + 2(120) $$ 5. Square 23 and multiply out the area term: $$ 529 = (l^2 + w^2) + 240 $$ 6. Isolate $(l^2 + w^2)$, which is $d^2$: $$ d^2 = 529 - 240 = 289 $$ 7. Take the square root to find the length of the diagonal: $$ d = \sqrt{289} = 17 \text{ m} $$ ### Exam Strategy & Shortcut You can also solve this by guessing factors of $120$ that add up to $23$. Pairs for $120$: $(10, 12) \rightarrow$ sum is $22$. $(8, 15) \rightarrow$ sum is $23$. Match! So sides are $8$ and $15$. The diagonal is the hypotenuse: $\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$. $8-15-17$ is a well-known Pythagorean triple. ### Common Pitfall A common pitfall is attempting to form a quadratic equation to find $l$ and $w$ individually. While this works, it takes significantly more time during a timed exam than utilizing the algebraic identity shortcut. ### Final Answer Therefore, the correct answer is **$17$ m**.