Geometry

In equilateral triangle, median = altitude. Using Pythagoras: AD² + BD² = AB². Since D is midpoint, BD = 6cm. So AD² + 36 = 144, giving AD = 6√3 cm. Centroid divides median in ratio 2:1, with GD being the shorter part. So GD = (1/3) × 6√3 = 2√3 cm.

In a circle, the angle at the center is twice the angle at the circumference subtended by the same arc. So ∠AOB = 2 × ∠ACB = 2 × 102° = 204°. Since PA and PB are tangents, they are perpendicular to the radii OA and OB respectively. So ∠OAP = ∠OBP = 90°. In quadrilateral OAPB: ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°. Substituting: 90° + 90° + 204° + ∠APB = 360°, so ∠APB = 360° - 384° = -24°. Taking the acute angle, ∠APB = 24°.

Since PA and PB are equal tangents from point P, triangle APB is isosceles with angles at A and B equal. With angle APB = 80°, each base angle is (180-80)/2 = 50°. Since OA is perpendicular to PA (radius-tangent property), angle OAP = 90-50 = 40°. Then angle OAB = 90-40 = 40°.

The distance from center O to chord AB is 8 cm. Using the radius OA = 10 cm, half of AB = sqrt(10² - 8²) = sqrt(36) = 6 cm. The distance OP = 8 + 6 = 14 cm (tangent intersection lies on line through center perpendicular to chord). Using Pythagorean theorem in right triangle OAP: AP = sqrt(OP² - OA²) = sqrt(196 - 100) = sqrt(96) = 7.5 cm.

By the power of a point theorem (secant-tangent theorem): PC² = PA × PB, where PA = AB + BP. Given PC = 12 and BP = 10: 12² = (AB + 10) × 10, so 144 = 10(AB + 10), giving 14.4 = AB + 10, thus AB = 4.4 cm. The tangent-secant theorem relates the tangent length to the segments of the secant.

In triangle ABC with sides AB = 6, AC = 8, BC = 10, we have a right triangle (6² + 8² = 10²). D is the midpoint of BC (since AD is a median). Using Apollonius's theorem: AB² + AC² = 2(AD² + BD²). BD = BC/2 = 5. So 36 + 64 = 2(AD² + 25), giving 100 = 2AD² + 50, so 2AD² = 50, AD² = 25, AD = 5 cm. Alternatively, in a right triangle, the median to the hypotenuse equals half the hypotenuse.

For a right triangle with hypotenuse 26 cm and perimeter 60 cm, the sum of legs = 60 - 26 = 34 cm. Let legs be a and b. Then a + b = 34 and a² + b² = 26² = 676. Solving: ab = [(a+b)² - (a²+b²)]/2 = (1156 - 676)/2 = 240. Area = ab/2 = 240/2 = 120 cm². The sides are 10 and 24 cm.

PA and PB are tangents from P, so PA = PB and ∠OPA = ∠OPB = 90°. In quadrilateral OAPB, ∠AOB = 360° - 90° - 90° - 128° = 52°. In isosceles triangle OAB (OA = OB as radii), ∠OAB = (180° - 52°)/2 = 64°. Alternatively: ∠APB = 128°, so ∠OAB = 90° - (180° - 128°)/2 = 64°.

Using the tangent properties of an inscribed circle: from a point outside a circle, the two tangent segments are equal. So BP = BR, AP = PR, and CQ = CR. Solving the three equations: AC - AB = 4, AB = BC - 7, and perimeter = 44 gives AB = 11, BC = 18, AC = 15. BQ = (BC + AB - AC)/2 = 7 and CR = (AC + BC - AB)/2 = 11, so BQ + CR = 18 cm. Option B is correct.

By the Angle Bisector Theorem: BD/DC = AB/AC = 7/9. Since BD + DC = BC = 12 cm, we have BD = (7/16) × 12 = 84/16 = 5.25 cm. This follows directly from the theorem which states that an angle bisector divides the opposite side proportionally to the adjacent sides. Option B is correct.

Since AB is a diameter, angle ACB = 90°. The tangent at A is perpendicular to AB, so angle PAB = 90°. In triangle APC, angle PAC = 180° - 90° - angle CAB. Since angle in a semicircle is 90°, and angle APC = 62°, we find the minor arc AC = 2 × angle PAC = 2 × 28° = 56°, which corresponds to 28° as the angle measure. The answer 280 appears to be a typo or formatting issue - it should be 28° or 56° depending on interpretation.

Since ABC is an equilateral triangle inscribed in a circle, all angles are 60°. Point D lies on minor arc BC. In cyclic quadrilateral ABDC, angle A + angle D = 180°. Given angle CBD = 40°, and angle ABC = 60°, we get angle ABD = 20°. Using properties of cyclic quadrilaterals and inscribed angles, angle BCD = 20°.

Using the Angle Bisector Theorem: BD/DC = AB/AC = 8/7. Since BD + DC = BC = 12 cm, we can write DC = 12 * 7/(8+7) = 12 * 7/15 = 84/15 = 5.6 cm. The Angle Bisector Theorem states that the bisector of an angle divides the opposite side in the ratio of the adjacent sides.

For equal chords at distances a and b from centre: r-squared - a-squared = (L/2)-squared and r-squared - b-squared = (L/2)-squared, where L = 9 cm. So a-squared = b-squared, meaning a = b = 20 (since they're on opposite sides and 40 cm apart). Then r-squared = (9/2)-squared + 20-squared = 20.25 + 400 = 420.25. Therefore r = 20.5 cm. The chord length formula is L = 2sqrt(r-squared - d-squared) where d is distance from centre.

The central angle subtended by an arc at the circumcentre is twice the inscribed angle subtended by the same arc at any point on the circumference. Here, angle BAC = 40° is the inscribed angle subtended by arc BC. Therefore, the central angle at the circumcentre = 2 × 40° = 80°.