91Ó°ÊÓ

Understanding the dimensions of a rectangle is key to solving any geometry problem involving this shape. A rectangle has two pairs of equal opposite sides. The longer side is called the length (denoted as \(l\)) and the shorter side is called the width (denoted as \(w\)). Correctly identifying and differentiating these sides helps when applying specific geometric formulas. Remember to use consistent units (e.g., meters, centimeters) throughout your calculations to avoid any confusion. Whether you're calculating the perimeter or area, knowing the length and width is a fundamental step.

The perimeter of a rectangle is the total distance around the outside of the shape. It is calculated by adding up all four sides. The simple formula for the perimeter \(P\) of a rectangle is: \[ P = 2l + 2w \] This means you add the lengths of all the sides: the two lengths (\(2l\)) and the two widths (\(2w\)). For our problem, since the perimeter is given as 26 meters, you can set up the equation: \[ 2l + 2w = 26 \] This equation is then simplified to easily find one variable in terms of the other (e.g., \(l + w = 13\)). This simplification makes it easier to solve for the dimensions.

The area of a rectangle refers to the amount of space enclosed within its boundaries. It's calculated by multiplying the length and width. The formula for the area \(A\) of a rectangle is: \[ A = l \times w \] For our problem, the area is given as 40 square meters. So, you can set up the equation: \[ l \times w = 40 \] This equation expresses the interdependence of length and width. When one dimension changes, the other must adjust to maintain the given area. Substituting the width from the simplified perimeter equation into the area equation allows you to solve for the length (or vice versa).

In geometry problems, when you substitute one variable into another equation, you often end up with a quadratic equation. A quadratic equation has the standard form: \[ ax^2 + bx + c = 0 \] In our problem, substituting the width \(w = 13 - l\) into the area equation \( l(13 - l) = 40 \) simplifies to: \[ l^2 - 13l + 40 = 0 \] This is a standard quadratic equation where \(a = 1\), \(b = -13\), and \(c = 40\). You solve quadratic equations using the quadratic formula: \[ l = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \] Applying this formula helps find the possible values for \(l\). In our problem, both solutions \(l = 8\) and \(l = 5\) are tested to find corresponding widths ensuring the final dimensions align with the problem's conditions.

Understanding these core concepts makes solving similar problems straightforward and builds a solid foundation in geometry.