Rectangle Geometry Problems with Solutions

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

Solution:

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

Problem Statement: The perimeter of a rectangle is 50 feet and its area is 150 ft\({}^2\). Find length \(L\) and width \(W\) where \(L > W\).

Solution:

1. Perimeter: \(2L + 2W = 50 \implies L + W = 25\)
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 \(= 15 \times 10 = 150\) ft\({}^2\), Perimeter \(= 2(15+10) = 50\) ft

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

Solution:

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

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

Solution:

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