The areas of a square and a rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 5 cm and the breadth is less by 3 cm. Find the perimeter of the rectangle.

### Concept & Algebraic Modeling When given comparative dimensions, base all variables on a single unknown. Here, base the rectangle's dimensions on the square's side, then equate their areas. $$ \text{Area}_{\text{sq}} = \text{Area}_{\text{rect}} \Rightarrow s^2 = l \times b $$ $$ \text{Perimeter}_{\text{rect}} = 2(l + b) $$ ### Step-by-Step Solution 1. Let the side of the square be $x$ cm. 2. The area of the square is $x^2$. 3. The length of the rectangle is $l = (x + 5)$ cm. 4. The breadth of the rectangle is $b = (x - 3)$ cm. 5. The area of the rectangle is $l \times b = (x + 5)(x - 3)$. 6. Equate the areas: $x^2 = (x + 5)(x - 3)$ $x^2 = x^2 - 3x + 5x - 15$ $x^2 = x^2 + 2x - 15$ 7. Simplify the equation: $0 = 2x - 15$ $2x = 15 \Rightarrow x = 7.5$ cm. 8. Now, find the dimensions of the rectangle: Length $l = 7.5 + 5 = 12.5$ cm. Breadth $b = 7.5 - 3 = 4.5$ cm. 9. Calculate the perimeter of the rectangle: $\text{Perimeter} = 2(l + b) = 2(12.5 + 4.5) = 2(17) = 34$ cm. ### Exam Strategy & Shortcut Notice the equation $s^2 = (s+5)(s-3)$ always simplifies to $0 = 2s - 15$. You can jump straight to $2s = \text{diff of product} = 15$ for these specific "rectangle vs square area equal" structures. $s = 7.5$. Perimeter = $2( (s+5) + (s-3) ) = 2(2s + 2) = 4s + 4 = 4(7.5) + 4 = 30 + 4 = 34$. ### Common Pitfall A common mistake is finding the value of the square's side ($x = 7.5$) and trying to double it or miscalculating the perimeter formula as just $(l+b)$ instead of $2(l+b)$. ### Final Answer Therefore, the correct answer is **34 cm**.