Straight Lines
Parallelogram and diagonals
Grade 11
Question:
<p>Let \(ABCD\) be a parallelogram whose equations for the diagonals \(AC\) and \(BD\) are \(x + 2y = 3\) and \(2x + y = 3\), respectively. If length of diagonal \(AC = 4\) units and area of parallelogram \(ABCD = 8\) sq. units, then the length of other diagonal \(BD\) is</p>
Step-by-Step Solution
Key Concept: In a parallelogram, diagonals bisect each other at a point whose coordinates can be found by solving the system of diagonal equations. Use the distance formula with the given diagonal length and the property that area = (1/2)d₁d₂sin(θ), where θ is the angle between diagonals.
<p><strong>Step 1:</strong> Find the intersection point of diagonals AC and BD by solving:</p><p>x + 2y = 3 ... (1)</p><p>2x + y = 3 ... (2)</p><p>From (1): x = 3 - 2y</p><p>Substituting in (2): 2(3 - 2y) + y = 3 → 6 - 4y + y = 3 → y = 1, x = 1</p><p>Diagonals intersect at O(1, 1).</p><p><strong>Step 2:</strong> Find the angle between the diagonals using their slopes.</p><p>Slope of AC: m₁ = -1/2 (from x + 2y = 3)</p><p>Slope of BD: m₂ = -2 (from 2x + y = 3)</p><p>tan(θ) = |m₁ - m₂|/(1 + m₁m₂) = |(-1/2) - (-2)|/(1 + (-1/2)(-2)) = |3/2|/(1 + 1) = 3/4</p><p>Therefore: sin(θ) = 3/5 (since sin²θ + cos²θ = 1 with tan(θ) = 3/4)</p><p><strong>Step 3:</strong> Use the area formula for a parallelogram in terms of its diagonals.</p><p>Area = (1/2)|d₁||d₂|sin(θ)</p><p>8 = (1/2) × 4 × |BD| × (3/5)</p><p>8 = (6/5)|BD|</p><p>|BD| = 40/6 = 20/3 units</p><p>∴ Answer: <strong>20/3 units</strong> (or 6⅔ units)</p>
Correct Answer: 20