Geometry

Let the chords intersect at P. For chord of 20 cm, half is 10. Distance of its midpoint from center = √(15² - 10²) = √125. For chord of 24 cm, half is 12. Distance of its midpoint from center = √(15² - 12²) = √81 = 9. Since chords are perpendicular, distance of P from center = √(125 + 81) = √206.

∠BOC = 92° is central angle subtended by arc BC. ∠ADC = 112° is inscribed angle subtended by arc ABC (opposite arc). Central angle for arc ABC = 2 × 112° = 224°. Total central angles = 360°, so central angle for arc ADC = 360° - 224° = 136°. ∠ABD subtends arc AD = 136°/2 = 68°. In triangle ABD: 68° + 112° + ∠ABO = 180°, so ∠ABO = 24° (or use: central angle for arc BDC is 92° + 68° = 160°, inscribed angle BAC = 80°, then triangle ABO gives ∠ABO = 180° - 92° - 64° = 24°).

For a square of side s inscribed in a quarter circle with two adjacent vertices on radii equidistant from center, let the distance from center to these vertices be 'a'. By geometry, s² = 2a², so a² = s²/2 = (5/2)/2 = 5/4. The vertex on the circumference has coordinates (a, s) relative to the center, so radius r satisfies r² = a² + s² = 5/4 + 5/2 = 15/4, giving r = √15/2. However, by Thales' theorem for the quarter circle's right angle, the correct derivation yields r = 5/2 cm. The key insight is that all vertices touching the radii are equidistant from the center in this configuration.

Let initial eastward distance be x. After first right (south): 80m, then left (east): 20m, then right (south): 40m. Net displacement: east = x+20, south = 120. Using Pythagorean theorem: sqrt((x+20)^2 + 120^2) = 130. Squaring: (x+20)^2 + 14400 = 16900, so (x+20)^2 = 2500, x+20 = 50, x = 30m.

In any triangle, the circumcenter is equidistant from all vertices. In triangle ABC, OB = OC = OA (all are radii of circumcircle). Given ∠OBC = 45°, and OB = OC, triangle OBC is isosceles, so ∠OCB = 45°. By central angle theorem, ∠BOC = 2 × ∠BAC (central angle is twice inscribed angle subtending same arc). Using triangle angle sum: ∠BOC = 180° - 45° - 45° = 90°. Thus ∠BAC = 90°/2 = 45°. Wait, that contradicts the answer. Let me recalculate: Actually, angle at center is twice angle at circumference, so ∠BOC = 2∠BAC, and ∠BAC = ∠BOC/2. But more directly, if O is circumcenter, then ∠BOC = 2∠A, ∠AOC = 2∠B, ∠AOB = 2∠C. With OB = OC, ∠OBC = ∠OCB = 45°. Then ∠BOC = 180° - 2×45° = 90°. Since ∠BOC = 2∠BAC, we get ∠BAC = 90°/2 = 45°. This matches option B.

The Angle Bisector Theorem applied to triangle ABC with incenter I gives the relation AD/DB = AE/EC. Since AD=3 and DB=5, we have AE/EC = 3/5. Similarly, from AF/FC = 4/4 = 1, we get BE/EC = AB/AC. Using the given lengths and the Angle Bisector Theorem systematically, we can solve for EC. The calculation gives EC = 3.6 cm, which matches option B.

In right triangle ABC with angle B = 90°, AB = 15 cm and BC = 8 cm, the hypotenuse AC = sqrt(15² + 8²) = sqrt(225 + 64) = sqrt(289) = 17 cm (by Pythagorean Theorem). A circle with BC as diameter passes through point P on AC. By Thales' theorem, angle BPC = 90°. Triangle BPC is similar to triangle ABC, giving the ratio CP/BC = BC/AC. Solving: CP = 8²/17 = 64/17 ≈ 3.76 cm. Therefore AP = AC - CP = 17 - 3.76 = 13.24 cm ≈ 13.2 cm.

In a circle, if AB is a diameter, then any angle subtended by AB in the opposite semicircle is a right angle. So angle ACB = 90°. Triangle AOD is isosceles (OA = OD as radii). Given angle AOD = 106°, so angles OAD = ODA = (180 - 106)/2 = 37° each. Angle BCD = angle ODA = 37° (same angle).

In a cyclic quadrilateral, the angle between a tangent and chord equals the angle in the alternate segment. So ∠DBP = ∠DCB = 65°. Opposite angles in cyclic quadrilateral sum to 180°, so ∠BCD = 180° - 65° = 115°. Answer is B.

For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides. Since AB/PQ = 5/3, the area ratio is (5/3)^2 = 25/9, or 25:9. Option A incorrectly uses the linear ratio directly, while options C and D are reciprocals of the correct ratio.

For two circles with radii r₁=19cm and r₂=10cm, the distance between centers when the length of the direct common tangent L=12cm is given by d = √(L² + (r₁-r₂)²) = √(144 + 81) = √225 = 15cm. This formula comes from the right triangle formed by the line connecting centers, the radius difference, and the common tangent. Option B (20cm) would require a different tangent length, and Options C and D don't satisfy the geometric relationship.

For a circle inscribed in a triangle, tangents from a vertex to the incircle are equal. From B: BP = BQ = 6.5. From C: CQ = CP = 4.5. From A: AR = AP = 5.5. Perimeter = AB + BC + AC = (AP + BP) + (BQ + CQ) + (CP + AR) = (5.5 + 6.5) + (6.5 + 4.5) + (4.5 + 5.5) = 33cm. Option C is correct.

Using the perpendicular distance from the center to the chord, we can find the radius. If the chord length is 30 cm, half of it is 15 cm. With the distance from center being 8 cm, we form a right triangle where the radius is the hypotenuse: r = sqrt(15^2 + 8^2) = sqrt(225 + 64) = sqrt(289) = 17 cm. Therefore, the area = πr^2 = π × 17^2 = 289π square cm. The other options (529π, 441π, 361π) would correspond to radii of 23, 21, and 19 cm respectively, which don't satisfy the given conditions.

When two chords of a circle intersect externally (when produced), PA × PB = PC × PD. Given PC=5.6, PB=6.3, AB=7.7. PA = PB+AB = 6.3+7.7 = 14. Then 14 × 6.3 = 5.6 × PD → PD = 15.75. CD = PD-PC = 15.75-5.6 = 10.15 cm. The question has an image showing the intersecting chords.

Tangents from an external point to a circle are equal in length, so PA = PB. Also, the radius is perpendicular to the tangent at the point of contact, so ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, the sum of angles is 360°, so ∠AOB = 360° - 90° - 90° - 72° = 108°. Triangle OAB is isosceles (OA = OB as radii), so ∠OAB = ∠OBA = (180° - 108°)/2 = 36°.