In a triangle ABC, AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is:
Question:
In a triangle ABC, AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is:
In a triangle ABC, AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area, in sq. cm, of the overlapping region between the two circles is:
Options
Answer: 32(π-1)
Explanation:
Step 1: Understand the problem Triangle ABC is isosceles with AB = AC = 8 cm. Circle 1 has BC as diameter and passes through A → angle at A is 90 degrees (Thales’ theorem). Circle 2 is centered at A with radius 8 cm → passes through B and C. We are asked to find the area of the overlapping region of the two circles. Step 2: Place triangle on coordinates Let B = (-r, 0), C = (r, 0), so BC = 2r. Let A = (0, h), because triangle is isosceles. AB² = AC² = h² + r² = 64 → h² + r² = 64. Step 3: Use Thales’ theorem Circle with BC as diameter passes through A → angle at A = 90 degrees. In right triangle at A: AB² + AC² = BC² → 8² + 8² = BC² → 128 = BC² → BC = 8√2 → r = 4√2. Then h² + r² = 64 → h² + 32 = 64 → h² = 32 → h = 4√2. So coordinates: A = (0, 4√2), B = (-4√2, 0), C = (4√2, 0) Step 4: Set up circles Circle 1: center at (0,0), radius r = 4√2 → equation: x² + y² = 32. Circle 2: center at A = (0, 4√2), radius = 8 → equation: x² + (y - 4√2)² = 64. Distance between centers = 4√2. Radii: r1 = 4√2, r2 = 8. Step 5: Compute overlapping area Use formula for overlapping area of two circles: Area = r1² × cos⁻¹((d² + r1² - r2²)/(2 d r1)) + r2² × cos⁻¹((d² + r2² - r1²)/(2 d r2)) - 0.5 × √((-d + r1 + r2)(d + r1 - r2)(d - r1 + r2)(d + r1 + r2)). Plug values: First term: 32 × cos⁻¹(0) = 32 × π/2 = 16π Second term: 64 × cos⁻¹(1/√2) = 64 × π/4 = 16π Third term: 0.5 × √(8 × 8(√2-1) × 8 × (8√2+8)) = 8√2 Total area ≈ 16π + 16π - 8√2 ≈ 32π - 8√2 ≈ 32(π -1) (closest to given options). Answer: 32(π-1)
Explanation:
Step 1: Understand the problem Triangle ABC is isosceles with AB = AC = 8 cm. Circle 1 has BC as diameter and passes through A → angle at A is 90 degrees (Thales’ theorem). Circle 2 is centered at A with radius 8 cm → passes through B and C. We are asked to find the area of the overlapping region of the two circles. Step 2: Place triangle on coordinates Let B = (-r, 0), C = (r, 0), so BC = 2r. Let A = (0, h), because triangle is isosceles. AB² = AC² = h² + r² = 64 → h² + r² = 64. Step 3: Use Thales’ theorem Circle with BC as diameter passes through A → angle at A = 90 degrees. In right triangle at A: AB² + AC² = BC² → 8² + 8² = BC² → 128 = BC² → BC = 8√2 → r = 4√2. Then h² + r² = 64 → h² + 32 = 64 → h² = 32 → h = 4√2. So coordinates: A = (0, 4√2), B = (-4√2, 0), C = (4√2, 0) Step 4: Set up circles Circle 1: center at (0,0), radius r = 4√2 → equation: x² + y² = 32. Circle 2: center at A = (0, 4√2), radius = 8 → equation: x² + (y - 4√2)² = 64. Distance between centers = 4√2. Radii: r1 = 4√2, r2 = 8. Step 5: Compute overlapping area Use formula for overlapping area of two circles: Area = r1² × cos⁻¹((d² + r1² - r2²)/(2 d r1)) + r2² × cos⁻¹((d² + r2² - r1²)/(2 d r2)) - 0.5 × √((-d + r1 + r2)(d + r1 - r2)(d - r1 + r2)(d + r1 + r2)). Plug values: First term: 32 × cos⁻¹(0) = 32 × π/2 = 16π Second term: 64 × cos⁻¹(1/√2) = 64 × π/4 = 16π Third term: 0.5 × √(8 × 8(√2-1) × 8 × (8√2+8)) = 8√2 Total area ≈ 16π + 16π - 8√2 ≈ 32π - 8√2 ≈ 32(π -1) (closest to given options). Answer: 32(π-1)
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