Geometry

Using the arc length formula s = rθ, where θ is in radians: 60° = π/3 radians. Given s = 37.4 cm, we have r = s/θ = 37.4/(π/3) = 37.4×3/π ≈ 35.7 cm. This uses the relationship between arc length, radius, and central angle.

Angle between two radii = 130°. Tangent is perpendicular to radius at point of contact, so angle between radius and tangent at each end = 90°. Quadrilateral formed by two radii and two tangents has sum of angles = 360°. Angle between tangents = 360° - 90° - 90° - 130° = 50°. Option C is correct.

For a circle with radius r and an external point at distance d from center, the tangent length t satisfies t² + r² = d² (tangent-radius theorem). Here t² + 8² = 10², so t² = 100-64=36, giving t=6 cm. Option A is correct.

The angle between two tangents drawn from an external point to a circle equals the angle subtended by the line segment joining the points of tangency at the center, minus the central angle. Since angle POQ is 80°, the angle PTQ = 180° - 80° = 100° (as quadrilateral OPTQ has sum of angles = 360°, and angles at P and Q are each 90° since tangents are perpendicular to radii). 80° would be the central angle itself, not the exterior angle. 60° and 75° are not related to the tangent properties.

Two circles intersect, AC is tangent to circle B, BC is tangent to circle A. So AC ⊥ AB and BC ⊥ AB. Triangle ABC has AB = 3+4 = 7 cm, AC = 3 cm, BC = 4 cm (right triangle by Pythagoras: 3² + 4² = 5²). CD is common chord perpendicular to AB. Using chord properties: CD = 2 × √(3² - (3×5/7)²) ≈ 4.8 cm.

By Angle Bisector Theorem: AD bisects angle A, so BD/DC = AB/AC. Given BD = 3cm, AB = 6cm, AC = 5cm. So 3/DC = 6/5, therefore DC = 3×5/6 = 2.5cm. The angle bisector divides the opposite side proportionally to the adjacent sides.

Arc length formula: l = (θ/360) × 2πr. Here θ = 72°, r = 28 cm. So l = (72/360) × 2π × 28 = (1/5) × 56π = 11.2π ≈ 35.2 cm (using π ≈ 3.14). This applies the standard arc length formula relating central angle to circumference fraction.

For tangents PA and PB from point P to a circle with center O, angle between tangents ∠APB = 180° - ∠AOB = 180° - 120° = 60°. Since triangle APO is right-angled at A and ∠AOP = 60°/2 = 30°, we get ∠APO = 60°. Therefore ∠APB : ∠APO = 60° : 30° = 2:1.

The centroid divides each median in a 2:1 ratio, with the longer segment being from the vertex to the centroid. If the distance from vertex A to the centroid is 7 cm, then this represents 2/3 of the median. Therefore, the complete median length = 7 × (3/2) = 10.5 cm. This is a fundamental property of centroids in triangles.

For two concentric circles with radii 6 cm (inner) and 10 cm (outer), a tangent to the inner circle becomes a chord of the outer circle. The tangent length from the point of tangency to the intersection with the outer circle forms a right triangle with the line connecting to the center. Using Pythagoras theorem: tangent length = √(10² - 6²) = √(100 - 36) = √64 = 8 cm. Since the chord includes two equal tangents, total length = 2 × 8 = 16 cm.

For triangle with sides a=15, b=14, c=13: semi-perimeter s = (15+14+13)/2 = 21. Area using Heron's formula: √(s(s-a)(s-b)(s-c)) = √(21×6×7×8) = √7056 = 84. Inradius r = Area/s = 84/21 = 4 units. Option A (6 units) and D (7 units) are incorrect. Option B (5 units) doesn't satisfy the formula.

For a chord at distance d from center, half-chord length = √(r² - d²). Here half-chord = 3, d = 4. So 3 = √(r² - 16). r² - 16 = 9, r² = 25, r = 5. Diameter = 2r = 10 cm.

For similar triangles, (side ratio)² = (area ratio). Area ratio = 64/121, so side ratio = √(64/121) = 8/11. This means corresponding sides are in proportion 8:11. EF is from the larger triangle (area 121), BC from the smaller (area 64). So BC/EF = 8/11, giving BC = 15.4 × 8/11 = 11.2 cm.

For intersecting chords AB and CD meeting at P, the angle between the chords is half the sum of the arcs they intercept. Using the given central angles, angle BPD = 1/2 × (angle AOC + angle BOD) = 1/2 × (50° + 40°) = 45°. Options A (60°), B (40°), and D (75°) don't match this calculation.

For intersecting chords, PA × PB = PC × PD. Given CD = 3 cm and PD = 5 cm, then PC = CD - PD = -2 cm, which is impossible as lengths are positive. This suggests P lies outside the circle (external intersection). For external intersection: PA × PB = PC × PD. Let PB = x. Since P is external, PA = PB + AB = x + 6. Also PC = PD + CD = 5 + 3 = 8. So (x+6)x = 8×5 = 48. x² + 6x - 48 = 0. Solving: x = 6 cm.