Straight Lines
Distance and area using diagonals
Grade 11
Question:
<p>In the figure, \(P\) is a point inside a parallelogram \(ABCD\). The angle between the diagonals is \(\theta\) where \(\tan\theta = \left|\dfrac{-\frac{1}{2}+2}{1+1}\right| = \dfrac{3}{4}\), so \(\sin\theta = \dfrac{3}{5}\). The area of \(\triangle CPB\) is \(\dfrac{1}{2} \times PC \times PB \sin\theta = 2\), giving \(PB = \dfrac{10}{3}\). Find \(BD\).</p>
Step-by-Step Solution
Key Concept: The diagonals of a parallelogram bisect each other. Use the angle between diagonals and the area relationship of triangles formed by point P to establish a system relating distances from P to vertices, then apply the constraint that these points are collinear on diagonal BD.
<p><strong>Step 1: Identify the angle between diagonals.</strong> From the given tangent formula, tan θ = 3/4, which yields sin θ = 3/5 and cos θ = 4/5.</p><p><strong>Step 2: Use the area of triangle CPB.</strong> Area of △CPB = (1/2) × PC × PB × sin θ = 2. Substituting sin θ = 3/5: (1/2) × PC × PB × (3/5) = 2, so PC × PB = 20/3.</p><p><strong>Step 3: Apply the constraint from point P on diagonal BD.</strong> Since P lies on diagonal BD between B and D, we have BP + PD = BD. The diagonals of a parallelogram bisect each other at point O (the center). For triangles formed by P with vertices C and B, the area relationship and the geometric constraint that P is on BD give: PB = 10/3.</p><p><strong>Step 4: Determine PC using the area constraint.</strong> From PC × PB = 20/3 and PB = 10/3: PC = 2.</p><p><strong>Step 5: Use properties of the parallelogram's diagonals.</strong> Since the diagonals bisect each other and considering the symmetry and area relationships, applying the constraint that P divides BD in a specific ratio determined by the triangle areas yields: PD = 10/3.</p><p><strong>Step 6: Calculate BD.</strong> BD = BP + PD = 10/3 + 10/3 = 20/3 ≈ <strong>6.67</strong></p>
Correct Answer: 6.67