ABCD is a rectangle with AB = 12 cm and BC = 8 cm. P is a point inside the rectangle such that the area of triangle APB is twice the area of triangle DPC. Find the area of triangle APB.
Show Worked Solution
Worked Solution
Step 1: Let h₁ be the perpendicular distance from P to AB, and h₂ be the perpendicular distance from P to DC. Since AB and DC are parallel sides of the rectangle, h₁ + h₂ = BC = 8 cm.
Step 2: Area APB = ½ × 12 × h₁ = 6h₁. Area DPC = ½ × 12 × h₂ = 6h₂.
Step 3: 'APB is twice DPC' gives 6h₁ = 2 × 6h₂, so h₁ = 2h₂.
Step 4: From h₁ + h₂ = 8 and h₁ = 2h₂: 3h₂ = 8, so h₂ = 8/3, h₁ = 16/3.
Step 5: Area APB = 6 × 16/3 = 32 cm².
Answer: 32 cm²
Correct answer: 32 cm²
Want more questions like this? Superholic Lab has 10,000+ MOE-aligned questions with full worked solutions.
Start Free Trial — 7 Days Free