Geometry

The diameter is 20 cm, so the radius is 10 cm. A right angle at the centre gives a sector of 90 degrees, whose area is (90/360) × pi × 10^2 = 25pi = 550/7 cm^2 using pi = 22/7. The numerical answer matches option D, although its unit should be cm^2.

In a parallelogram, the centroid of a triangle divides the median in a 2:1 ratio. The distance between centroids of two triangles sharing a diagonal is 1/3 of the diagonal length. PQ = (1/3) * BD = 18 / 3 = 6 cm.

A square inscribed in a circle has a diagonal equal to the diameter of the circle. Since the diagonal is 8 inches, the diameter is 8 inches. The perimeter (circumference) of a circle is pi * diameter, which is 8 * pi.

The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. Since 16:25 = (4:5)^2, the sides must be in the ratio 4:5. Option D (16:25) is not in the ratio 4:5.

By the alternate segment theorem, the angle between the tangent CD and chord AB (angle CAD = 70) is equal to the angle subtended by the chord in the alternate segment. Thus, the angle at the circumference is 70 degrees. The angle subtended at the center (AOB) is twice the angle at the circumference, so 2 * 70 = 140 degrees.

A chord, the radius, and the distance from the center form a right-angled triangle. The distance from the center bisects the chord, so the base of the triangle is 8 cm. Using the Pythagorean theorem, radius^2 = 6^2 + 8^2 = 36 + 64 = 100, so the radius is 10 cm.

Area of triangle = sqrt(s(s-a)(s-b)(s-c)) where s = (13+14+15)/2 = 21. Area = sqrt(21 * 8 * 7 * 6) = sqrt(7056) = 84. Inradius r = Area / s = 84 / 21 = 4.

For intersecting chords AB and CD at E, the property is AE * EB = CE * ED. 4 * 9 = 6 * ED. 36 = 6 * ED, so ED = 6.

For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. Area ratio = 9:16, so side ratio = sqrt(9):sqrt(16) = 3:4.

In a circle, if two chords intersect at right angles, the angle between them relates to the arcs they subtend. Given the geometry of the circle and the intersection, angle NST = angle SMO = 34 degrees.

For a right triangle with legs a and b and hypotenuse c, the inradius r = (a + b - c) / 2. Here, a = 6, b = 8, so c = sqrt(6^2 + 8^2) = 10. r = (6 + 8 - 10) / 2 = 4 / 2 = 2 cm.

The distance from the external point to the center forms a right triangle with the tangent and the radius. Distance^2 = 10^2 + 6^2 = 100 + 36 = 136. Distance = sqrt(136) approx 11.6619.