The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be (Hotel Management, 2002)

### Concept & Algebraic Identities Inverse This problem reverses the logic of standard area/perimeter questions. We are given the diagonal ($d^2 = l^2 + w^2$) and the area ($lw$). We need the perimeter, which requires finding $(l + w)$. We use the algebraic identity: $$ (l + w)^2 = (l^2 + w^2) + 2lw $$ ### Step-by-Step Solution 1. We are given the diagonal ($d$) $= \sqrt{41}$ cm. From the Pythagorean theorem, $d^2 = l^2 + w^2$: $$ (\sqrt{41})^2 = l^2 + w^2 $$ $$ l^2 + w^2 = 41 $$ 2. We are given the area: $$ l \times w = 20 \text{ sq. cm} $$ 3. We need to find the perimeter, which is $2(l + w)$. First, find $(l + w)$ using the algebraic identity: $$ (l + w)^2 = (l^2 + w^2) + 2lw $$ 4. Substitute the known values for $(l^2 + w^2)$ and $lw$: $$ (l + w)^2 = 41 + 2(20) $$ $$ (l + w)^2 = 41 + 40 $$ $$ (l + w)^2 = 81 $$ 5. Take the square root of both sides to find $(l + w)$: $$ l + w = \sqrt{81} = 9 \text{ cm} $$ 6. Calculate the perimeter: $$ \text{Perimeter} = 2(l + w) = 2 \times 9 = 18 \text{ cm} $$ ### Exam Strategy & Shortcut Look at the area $20$. Factor pairs are $(1, 20), (2, 10), (4, 5)$. Test the Pythagorean theorem for these pairs to see which gives a hypotenuse squared of $41$. For $(4, 5)$: $4^2 + 5^2 = 16 + 25 = 41$. It matches! The sides are $4$ and $5$. Perimeter $= 2(4 + 5) = 18$. Mental math saves paper and time. ### Common Pitfall A frequent error is miscalculating $(l + w)$ as $81$ and subsequently giving the perimeter as $162$. Do not forget to take the square root of $(l + w)^2$ before substituting it into the perimeter formula. ### Final Answer Therefore, the correct answer is **$18$ cm**.