Parallelogram Geometry Problems with Solutions

Question: Prove that quadrilateral \(ABCD\) with vertices \(A(-2, 0)\), \(B(2, 4)\), \(C(4, 1)\), and \(D(0, -3)\) is a parallelogram.

Solution:

Approach: Verify two pairs of parallel and congruent sides.

Slope calculations:

• Slope \(AB = \dfrac{4 - 0}{2 - (-2)} = 1\)
• Slope \(BC = \dfrac{1 - 4}{4 - 2} = -\dfrac{3}{2}\)
• Slope \(CD = \dfrac{-3 - 1}{0 - 4} = 1\)
• Slope \(DA = \dfrac{-3 - 0}{0 - (-2)} = -\dfrac{3}{2}\)

Distance calculations:

• \(AB = \sqrt{(4)^2 + (4)^2} = 4\sqrt{2}\)
• \(BC = \sqrt{(2)^2 + (-3)^2} = \sqrt{13}\)
• \(CD = \sqrt{(4)^2 + (4)^2} = 4\sqrt{2}\)
• \(DA = \sqrt{(-2)^2 + (3)^2} = \sqrt{13}\)

Conclusion: \(AB \parallel CD\) (equal slopes) and \(AB = CD\). \(BC \parallel DA\) (equal slopes) and \(BC = DA\). Thus, \(ABCD\) is a parallelogram.

Question: Given side \(AB = 15\) ft, angle \(D = 135^\circ\), and area \(= 1000\) ft², find height \(h\) and side \(BC\).

Solution:

Step 1: Find angle A

Consecutive angles are supplementary: \(\angle A = 180^\circ - 135^\circ = 45^\circ\)

Step 2: Calculate height \(h\)

Using right triangle \(ABB'\): \(\sin 45^\circ = \dfrac{h}{15}\)

\(h = 15 \times \dfrac{\sqrt{2}}{2} = \dfrac{15\sqrt{2}}{2}\) ft

Step 3: Find side \(BC\)

Area formula: \(A = BC \times h\)

\(1000 = BC \times \dfrac{15\sqrt{2}}{2}\)

\(BC = \dfrac{1000 \times 2}{15\sqrt{2}} = \dfrac{2000}{15\sqrt{2}} \approx 94.28\) ft (rounded)

Question: In parallelogram \(ABCD\), \(BB'\) bisects angle \(B\) and \(CC'\) bisects angle \(C\). Given \(BC = 10\) m, find lengths \(x\) and \(y\).

Solution:

Step 1: Analyze angles

In triangle \(BOC\): \(\angle BCO = \dfrac{1}{2}\angle C\), \(\angle CBO = \dfrac{1}{2}\angle B\)

Since \(\angle B + \angle C = 180^\circ\), then \(\angle BCO + \angle CBO = 90^\circ\)

Thus \(\angle BOC = 90^\circ\) (triangle angle sum).

Step 2: Find angle BCO

In parallelogram, \(\angle C = \angle A = 60^\circ\) (from diagram)

\(\angle BCO = \dfrac{60^\circ}{2} = 30^\circ\)

Step 3: Calculate \(x\) and \(y\)

In right triangle \(BOC\):

• \(\sin 30^\circ = \dfrac{x}{10} \implies x = 10 \times \dfrac{1}{2} = 5\) m
• \(\cos 30^\circ = \dfrac{y}{10} \implies y = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66\) m

Question: Find all internal angles of a parallelogram with sides 20 ft and 30 ft and area 300 ft².

Solution:

Step 1: Area relationship

Area of parallelogram \(= 2 \times\) area of triangle \(ABC\)

Area of triangle \(ABC = \dfrac{1}{2} \times 30 \times 20 \times \sin(\angle B)\)

Thus: \(300 = 2 \times \dfrac{1}{2} \times 30 \times 20 \times \sin(\angle B)\)

Step 2: Solve for angle B

\(300 = 600 \times \sin(\angle B)\)

\(\sin(\angle B) = 0.5\)

\(\angle B = 150^\circ\) (since \(\sin 30^\circ = \sin 150^\circ = 0.5\), and angle \(B > 90^\circ\) in parallelogram)

Step 3: Find all angles

• \(\angle B = 150^\circ\)
• \(\angle D = \angle B = 150^\circ\) (opposite angles equal)
• \(\angle A = 180^\circ - 150^\circ = 30^\circ\) (consecutive angles supplementary)
• \(\angle C = \angle A = 30^\circ\)