Rectangle Geometry Problems

Problems involving rectangle dimensions, area, perimeter, and diagonal with comprehensive solutions.

Rectangle Formulas

Perimeter: \( P = 2W + 2L \)

Area: \( A = L \times W \)

Diagonal: \( d = \sqrt{L^2 + W^2} \)

where \( L \) = length and \( W \) = width.

Problems

Problem 1

A rectangle has perimeter of 320 meters. Its length \( L \) is three times its width \( W \). Find dimensions \( W \) and \( L \), and the area.

Problem 2

The perimeter of a rectangle is 50 feet and its area is 150 ft². Find length \( L \) and width \( W \) where \( L > W \).

Problem 3

The diagonal \( d \) of a rectangle measures 100 feet. The length \( y \) is twice the width \( x \). Find the area.

rectangle diagram

Problem 4

Determine if points \( A(-1, 0) \), \( B(5, 2) \), \( C(4, 5) \), and \( D(-2, 3) \) are vertices of a rectangle.

coordinate plane points

Solutions

Solution to Problem 1

  1. Perimeter formula: \( 2L + 2W = 320 \)
  2. Given \( L = 3W \)
  3. Substitute: \( 2(3W) + 2W = 320 \)
  4. Simplify: \( 6W + 2W = 320 \)
  5. Solve: \( 8W = 320 \) → \( W = 40 \) meters
  6. Then \( L = 3W = 120 \) meters
  7. Area: \( A = L \times W = 120 \times 40 = 4800 \) m²

Solution to Problem 2

  1. Perimeter: \( 2L + 2W = 50 \) → \( L + W = 25 \)
  2. Area: \( L \times W = 150 \)
  3. From (1): \( W = 25 - L \)
  4. Substitute into (2): \( L(25 - L) = 150 \)
  5. Expand: \( -L^2 + 25L - 150 = 0 \)
  6. Multiply by -1: \( L^2 - 25L + 150 = 0 \)
  7. Factor: \( (L - 10)(L - 15) = 0 \)
  8. Solutions: \( L = 10 \) or \( L = 15 \)
  9. Corresponding widths: \( W = 15 \) or \( W = 10 \)
  10. Since \( L > W \): \( L = 15 \) ft, \( W = 10 \) ft
  11. Verification: Area = \( 15 \times 10 = 150 \) ft², Perimeter = \( 2(15+10) = 50 \) ft

Solution to Problem 3

  1. Pythagorean theorem: \( x^2 + y^2 = 100^2 \)
  2. Given \( y = 2x \)
  3. Substitute: \( x^2 + (2x)^2 = 10000 \)
  4. Simplify: \( x^2 + 4x^2 = 10000 \) → \( 5x^2 = 10000 \)
  5. Solve: \( x^2 = 2000 \) → \( x = \sqrt{2000} = 20\sqrt{5} \) ft
  6. Then \( y = 2x = 40\sqrt{5} \) ft
  7. Area: \( A = x \times y = (20\sqrt{5}) \times (40\sqrt{5}) = 800 \times 5 = 4000 \) ft²

Solution to Problem 4

  1. Calculate slopes:
    • \( m_{AB} = \frac{2-0}{5-(-1)} = \frac{2}{6} = \frac{1}{3} \)
    • \( m_{BC} = \frac{5-2}{4-5} = \frac{3}{-1} = -3 \)
    • \( m_{CD} = \frac{3-5}{-2-4} = \frac{-2}{-6} = \frac{1}{3} \)
    • \( m_{DA} = \frac{0-3}{-1-(-2)} = \frac{-3}{1} = -3 \)
  2. Opposite sides parallel: \( AB \parallel CD \) (equal slopes \( \frac{1}{3} \)) and \( BC \parallel DA \) (equal slopes \( -3 \))
  3. Adjacent sides perpendicular: \( m_{AB} \times m_{BC} = \frac{1}{3} \times (-3) = -1 \)
  4. Thus quadrilateral has right angles and parallel opposite sides → rectangle.

Additional Resources

Geometry Tutorials and Interactive Problems