A rectangular carpet has an area of 120 m$^2$ and a perimeter of 46 metre. The length of its diagonal is [SSC—CHSL (10+2) Exam, 2015]

### Concept & Algebraic Identity To find the diagonal of a rectangle, we need the sum of the squares of its length ($l$) and breadth ($b$), which is $\sqrt{l^2 + b^2}$. We can derive $l^2 + b^2$ using the algebraic identity $(l + b)^2 = l^2 + b^2 + 2lb$, substituting the given perimeter and area values. $$Area = l \times b$$ $$Perimeter = 2(l + b)$$ $$Diagonal = \sqrt{l^2 + b^2}$$ ### Step-by-Step Solution * Given: Area ($l \times b$) = 120 m$^2$. * Given: Perimeter $2(l + b)$ = 46 m. Therefore, $l + b = \frac{46}{2} = 23$. * Use the algebraic identity: $(l + b)^2 = l^2 + b^2 + 2lb$. * Substitute the known values into the identity: $(23)^2 = l^2 + b^2 + 2(120)$. * Calculate the squares and products: $529 = l^2 + b^2 + 240$. * Solve for $l^2 + b^2$: $l^2 + b^2 = 529 - 240 = 289$. * The diagonal is $\sqrt{l^2 + b^2}$. Therefore, Diagonal = $\sqrt{289} = 17$ metres. ### Exam Strategy & Shortcut If $l \times b = 120$ and $l + b = 23$, think of factor pairs of 120: (10, 12) - sum is 22; (8, 15) - sum is 23. Thus, $l = 15$ and $b = 8$. The diagonal forms a right-angled triangle with sides 8 and 15. The Pythagorean triplet for 8, 15 is 17. The answer is instantly 17. ### Common Pitfall Attempting to solve for $l$ and $b$ using complex quadratic equations instead of using the algebraic identity or Pythagorean triplets, which wastes valuable time. ### Final Answer Therefore, the correct answer is **17 metre**.