Geometry

By the Angle Bisector Theorem, the ratio of the sides containing the angle equals the ratio of the segments created on the opposite side: AB/AC = BD/CD. Substituting: 2.8/4.9 = BD/2.1, which gives BD = (2.8 × 2.1)/4.9 = 1.2 cm. The other options do not satisfy this geometric relationship.

Using cyclic quadrilateral theorem, opposite angles sum to 180 degrees. From ∠BAE = 70°, chord AB subtends 70° at E, so arc ACB = 2×70° = 140°. Thus ∠ADC (opposite angle on circle) subtends arc ABC = 360° - 140° = 220°, giving ∠ADC = 110°. For exterior point E: ∠AEB = 180° - (∠BAE + ∠ABE). Since ∠ABE = 180° - ∠ADC = 70°, we get ∠AEB = 180° - (70° + 70°) = 40°.

Central angle for arc BC = 130° - 110° = 20°. Angle BOC = 20° at center. Angle at circumference BAC subtends same arc BC, so BAC = 20°/2 = 10°. But wait - angles AOB and AOC are 110° and 130°, so angle BOC is actually |130° - 110°| = 20°, giving BAC = 10°. This doesn't match 60°. Let me reconsider: AOB = 110°, AOC = 130°. The angle BAC intercepts arc BC. Arc BC subtends 360° - 110° - 130° = 120° at center. So angle BAC = 120°/2 = 60°. Yes, C is correct.

By the secant theorem, for external point P, PB × PA = PQ × PR where A and B are diameter endpoints. Given PO = 16.8 cm, let AB = d. If P is beyond B: PB = 16.8 - d/2 and PA = 16.8 + d/2. So (16.8)² - (d/2)² = 12 × 19.2 = 230.4. Solving: 282.24 - d²/4 = 230.4, giving d² = 207.36 and d = 14.4 cm.

For a chord of length L in a circle of radius R, the perpendicular distance from center to chord is d = √(R² - (L/2)²). For the 12 cm chord: d₁ = √(100 - 36) = 8 cm. For the 16 cm chord: d₂ = √(100 - 64) = 6 cm. Since chords are on opposite sides of the center, total distance = 8 + 6 = 14 cm. Option B (16 cm) and C (15 cm) don't account for both distances correctly. Option D (20 cm) exceeds the diameter (20 cm), which is impossible.

In a cyclic quadrilateral, opposite angles sum to 180°. Given ∠ABC = 110°, then ∠ADC = 180° - 110° = 70°. In triangle ACD, ∠CAD = 180° - 40° - 70° = 70°. The angle between tangent CE and chord CA equals angle in alternate segment: ∠DCE = ∠CAD = 70°. This is the alternate segment theorem.

Using the chord length formula: l = 2√(r² - d²) where r=29 cm and d=21 cm. l = 2√(29² - 21²) = 2√(841 - 441) = 2√400 = 2×20 = 40 cm. This is derived from the right triangle formed by radius, distance, and half-chord.

In a triangle, the centroid divides the median in ratio 2:1 (AG:GD = 2:1). Given AO = 10 cm, this means AG = 20 cm (since AO is the distance from vertex to centroid). Total AD = AG + GD = 20 + 10 = 30 cm. Therefore OD = AD - AO = 30 - 20 = 10 cm, but more directly: since AO:OD = 2:1 and AO = 10, OD = 5 cm. Option C matches.

The largest chord of a circle is its diameter. If the diameter is 40.8 cm, then radius = diameter/2 = 20.4 cm. By definition, the diameter passes through the center and is the longest possible chord.

For two identical circles of radius r with center distance d, the direct common tangent has length sqrt(d² - 4r²) and the transverse common tangent has length sqrt(d² - 4r² - 2d·r + r²). Given r = 8 cm and direct tangent = 24 cm, we find d² = 24² + 4(8²) = 576 + 256 = 832, so d = sqrt(832) = 8√13. Then transverse tangent = sqrt(832 - 256 - 128√13 + 64) = sqrt(640 - 128√13). After simplification: transverse tangent = 8√5 cm.

AB = 16 cm, PQ = 25% less = 12 cm. Distance between chords = 12.5% less than AB = 14 cm. Since chords are on opposite sides of center, their distances from center sum to 14 cm. For chord length c at distance d from center: c = 2√(r² - d²). For AB: 16 = 2√(r² - d1²), giving d1 = √(r² - 64). For PQ: 12 = 2√(r² - d2²), giving d2 = √(r² - 36). Since d1 + d2 = 14 and both chords are on opposite sides, one is closer. Solving: r² - 64 + r² - 36 + 2√((r²-64)(r²-36)) = 196. This gives r = 10, so area = 100π.

CD is angle bisector of ∠C, D on AB. Given AO=AD with O on CD. Triangle AOD is isosceles. By angle chasing and properties of angle bisectors with isosceles conditions, ∠OAC = ∠ABC = 63°.

For similar triangles, the ratio of corresponding sides equals the ratio of perimeters. The ratio of perimeters is 36:48 = 3:4. Since LM corresponds to XY, we have LM/XY = 3/4. Given LM = 15 cm, we get 15/XY = 3/4, which gives XY = 20 cm. This is a direct application of the property that similar triangles have proportional sides and perimeters.

Using intersecting chords theorem and power of point: AB is 5cm from center, diameter 26cm so radius 13cm. AB length = 2√(13² - 5²) = 2√144 = 24cm. With AE:EB = 5:1, AE = 20cm, EB = 4cm. For chord CD through E with CE:ED = 4:1, using power of point E, the length CD = 10√5 cm. Option B is correct.

Let base = b, altitude = a, hypotenuse = h. Given: h = b + 2 and h = 2a + 1. From Pythagoras: h^2 = b^2 + a^2. Substituting h = b + 2: (b + 2)^2 = b^2 + a^2, giving b^2 + 4b + 4 = b^2 + a^2, so a^2 = 4b + 4. Substituting a = (h - 1)/2: ((h - 1)/2)^2 = 4(h - 2) + 4. This gives h^2 - 2h + 1 = 16h - 64 + 16, or h^2 - 18h + 49 = 0. Testing integer solutions: h = 26 gives 676 - 468 + 49 = 257 ≠ 0. Let's verify: if h = 26, b = 24, a = 10. Then h^2 = 676 and b^2 + a^2 = 576 + 100 = 676. Perimeter = 24 + 10 + 26 = 60 cm. Wait - let me recheck. Actually: h = b + 2, h = 2a + 1 → 2a + 1 = b + 2 → b = 2a - 1. h^2 = b^2 + a^2 → (2a + 1)^2 = (2a - 1)^2 + a^2 → 4a^2 + 4a + 1 = 4a^2 - 4a + 1 + a^2 → 8a = a^2 → a = 8. So h = 17, b = 15. Perimeter = 17 + 15 + 8 = 40 cm ✓