If the diagonal and the area of a rectangle are 25 m and 168 m$^2$, what is the length of the rectangle?

### Concept & Solving for Dimensions Using Identities When given the diagonal and area of a rectangle, we essentially have the sum of squares ($l^2 + b^2$) and the product ($l \times b$) of its dimensions. We can find the individual dimensions using algebraic identities. Formulas: $$(l + b)^2 = l^2 + b^2 + 2lb$$ $$(l - b)^2 = l^2 + b^2 - 2lb$$ ### Step-by-Step Solution * **Given:** Diagonal $D = 25$ m, Area $lb = 168$ m$^2$. * From the diagonal, we know $l^2 + b^2 = D^2 = 25^2 = 625$. * First, find $(l + b)^2$: $(l + b)^2 = (l^2 + b^2) + 2lb = 625 + 2(168) = 625 + 336 = 961$. * Taking the square root gives $l + b = 31$ (Equation 1). * Next, find $(l - b)^2$: $(l - b)^2 = (l^2 + b^2) - 2lb = 625 - 336 = 289$. * Taking the square root gives $l - b = 17$ (Equation 2). * Add Equation 1 and Equation 2: $(l + b) + (l - b) = 31 + 17 \Rightarrow 2l = 48$. * Solve for length: $l = 24$ m. ### Exam Strategy & Shortcut If you know your Pythagorean triples, this problem can be solved instantly. The diagonal is 25, suggesting a triangle with a hypotenuse of 25. The most common primitive triple is 7, 24, 25. Let's check: Area $= 7 \times 24 = 168$. This matches perfectly. The length (longer side) must be 24 m. ### Common Pitfall A common mistake is forming the equation $l^2 + (168/l)^2 = 625$ and trying to solve a complicated quartic equation ($l^4 - 625l^2 + 28224 = 0$). Always rely on $(l+b)$ and $(l-b)$ identities. ### Final Answer Therefore, the correct answer is **24 m**.