Problem 49
Question
The perimeter of a rectangle is 70 and its diagonal is \(25 .\) Find its length and width.
Step-by-Step Solution
Verified Answer
The dimensions of the rectangle are 20 and 15.
1Step 1: Understand the problem
We need to find the length and the width of a rectangle. We know that its perimeter is 70, and the diagonal is 25. The perimeter formula for a rectangle is given by \(P = 2(l + w)\). The diagonal formula, by the Pythagorean theorem, is \(d = \sqrt{l^2 + w^2}\).
2Step 2: Set up the perimeter equation
Using the perimeter formula, we have: \(2(l + w) = 70\). Simplifying this gives us \(l + w = 35\).
3Step 3: Set up the diagonal equation
Using the diagonal formula, we write: \(\sqrt{l^2 + w^2} = 25\). Squaring both sides gives \(l^2 + w^2 = 625\).
4Step 4: Express one variable in terms of the other
Using the equation from Step 2, express \(w\) in terms of \(l\): \(w = 35 - l\).
5Step 5: Substitute into the diagonal equation
Substitute \(w = 35 - l\) into the diagonal equation \(l^2 + w^2 = 625\):\[ l^2 + (35 - l)^2 = 625 \]
6Step 6: Expand and simplify the equation
Expand \((35 - l)^2\) to get:\[ l^2 + (35^2 - 2\cdot 35 \cdot l + l^2) = 625 \]This simplifies to:\[ 2l^2 - 70l + 1225 = 625 \]
7Step 7: Rearrange into a quadratic equation
Bring all terms to one side:\[ 2l^2 - 70l + 600 = 0 \]Simplify by dividing the entire equation by 2:\[ l^2 - 35l + 300 = 0 \]
8Step 8: Use the quadratic formula to solve for l
Use the quadratic formula \(l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -35\), \(c = 300\):\[ l = \frac{35 \pm \sqrt{35^2 - 4 \cdot 1 \cdot 300}}{2} \]Calculate inside the square root:\(35^2 - 1200 = 25\).Hence \(l = \frac{35 \pm 5}{2}\) gives solutions \(20\) or \(15\).
9Step 9: Determine both dimensions
If \(l = 20\), then \(w = 35 - 20 = 15\). If \(l = 15\), then \(w = 35 - 15 = 20\). So the dimensions are either length 20 and width 15 or length 15 and width 20.
Key Concepts
Rectangle DimensionsQuadratic FormulaPythagorean TheoremDiagonal of a Rectangle
Rectangle Dimensions
When dealing with rectangles, dimensions refer to the length and width, which are the two sides of the rectangle. These dimensions are important because they determine other properties of the rectangle, such as the perimeter and area. The perimeter is the total distance around the rectangle, calculated by adding up the lengths of all four sides. The formula for the perimeter (P) of a rectangle is:
- \( P = 2(l + w) \)
Quadratic Formula
The quadratic formula is a powerful tool used to find the solutions of a quadratic equation. Quadratic equations are in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.Here's the quadratic formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Positive discriminant: Two real and distinct solutions
- Zero discriminant: One real solution (repeated)
- Negative discriminant: No real solutions (complex numbers)
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle used to relate the sides of right triangles. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.The theorem is represented by the equation:
- \( a^2 + b^2 = c^2 \)
Diagonal of a Rectangle
The diagonal of a rectangle connects opposite corners and divides the rectangle into two congruent right triangles. It's an important feature because Diagonals in geometry often help to reveal properties of shapes that aren't immediately obvious.Calculating the length of the diagonal involves using the Pythagorean theorem, as each diagonal forms the hypotenuse of a right triangle:
- \( d = \sqrt{l^2 + w^2} \)
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