Dado l = 3w y P = 64 cm, sustituyendo obtenemos 64 = 2(3w + w).
["Title: How to Solve for Width When Given Area and Perimeter: A Step-by-Step Example", "When working with geometric shapes like rectangles, understanding the relationship between dimensions, area, and perimeter is essential. In this article, we’ll explore a practical problem involving a rectangle with a given perimeter and another expression for area, then solve for the width and length using algebraic substitution.", "---", "### Problem Statement", "We are given that the perimeter ( P = 64 ) cm and that the length ( l ) is three times the width ( w ), expressed as:", "[\nl = 3w\n]", "Also, the area of the rectangle is given by:", "[\nP = 2(l + w)\n]", "But in this case, the problem presents a substitute form:", "[\n64 = 2(3w + w)\n]", "This equation results from substituting ( l = 3w ) into the perimeter formula.", "---", "### Step-by-Step Solution", "Start with the perimeter formula for a rectangle:", "[\nP = 2(l + w)\n]", "Substitute the known perimeter ( P = 64 ) cm and replace ( l ) with ( 3w ):", "[\n64 = 2(3w + w)\n]", "Simplify inside the parentheses:", "[\n64 = 2(4w)\n]", "Multiply:", "[\n64 = 8w\n]", "Solve for ( w ) by dividing both sides by 8:", "[\nw = \frac{64}{8} = 8 , \ ext{cm}\n]", "Now, find the length ( l ) using ( l = 3w ):", "[\nl = 3 \ imes 8 = 24 , \ ext{cm}\n]", "---", "### Verification", "Double-check the perimeter:", "[\nl + w = 24 + 8 = 32 , \ ext{cm} \quad \Rightarrow \quad 2 \ imes 32 = 64 , \ ext{cm}\n]", "Verify the area:", "[\n\ ext{Area} = l \ imes w = 24 \ imes 8 = 192 , \ ext{cm}^2\n]", "Using the alternative formula:", "[\n2(l + w) = 2(24 + 8) = 2(32) = 64 , \ ext{cm}\n]", "Both methods confirm consistency.", "---", "### Summary", "Given a rectangle with perimeter ( P = 64 ) cm and length equal to three times width (( l = 3w )), substituting into the perimeter equation yields:", "[\n64 = 2(3w + w) = 8w \Rightarrow w = 8 , \ ext{cm}\n]", "Then, ( l = 24 , \ ext{cm} ), confirming the rectangle dimensions.", "This example illustrates how algebraic substitution connects geometric formulas and helps solve real-world measurement problems efficiently.", "---", "### Key Takeaways", "- Use ( P = 2(l + w) ) to relate area and perimeter.\n- Always substitute known relationships—like ( l = 3w )—into the perimeter equation.\n- Simplify carefully and solve step-by-step to avoid errors.\n- Verify your solution by checking both perimeter and area.", "For more tips on solving rectangle geometry problems and optimizing measurements, stay tuned to our ongoing geometry guides!", "---", "Keywords: rectangle area formula, perimeter substitution, solve for width, geometry problems, algebra and geometry, Dado l = 3w, P = 64 cm, length and width, perimeter equation, 2(3w + w)", "Meta Description: Solve for width when perimeter is 64 cm and length is three times width. Step-by-step algebra using 64 = 2(3w + w).", "---", "Curated for educators, students, and DIY problem solvers—mastering geometry through substitution made easy."]
