Geometry

For similar triangles, the ratio of areas is the square of the ratio of their perimeters. If the perimeter ratio is 2, the area ratio is 2^2 = 4.

The distance from the center to a chord is the leg of a right triangle where the hypotenuse is the radius (12.5 cm) and the other leg is half the chord length (10.5 cm). Using the Pythagorean theorem, d^2 = 12.5^2 - 10.5^2 = 156.25 - 110.25 = 46. Thus, the distance is sqrt(46) cm.

Diagonal of square = side * sqrt(2) = 10*sqrt(2). So side = 10. The diameter of the inscribed circle is equal to the side of the square, so diameter = 10. Circumference = pi * diameter = 3.14159 * 10 = 31.4159, which is approximately 31.43.

The number of chords formed by n points is given by the combination formula nC2. For 10 points, 10C2 = (10 * 9) / 2 = 45.

In the circle, let O be the center. Triangle OED is a right triangle with OE=6 and OD=6 (radius). However, the geometry description is complex; using the property of tangents and the right angle at E, AD can be calculated via the Pythagorean theorem in the resulting triangles as 6*sqrt(7).

For an equilateral triangle with side x, the inradius r = x / (2 * sqrt(3)). The area of the circle is pi * r^2 = pi * (x^2 / 12). Using pi approx 3.14, area = 3.14 * x^2 / 12 approx 0.2616x^2.

Using the law of cosines: c^2 = a^2 + b^2 - 2ab*cos(theta). 9.75^2 = 6.9^2 + 6.9^2 - 2 * 6.9 * 6.9 * cos(theta). 95.0625 = 47.61 + 47.61 - 95.22 * cos(theta). 95.0625 = 95.22 - 95.22 * cos(theta). 95.22 * cos(theta) = 0.1575. cos(theta) is near 0, so theta is approximately 90 degrees.

For a hyperbola x^2/a^2 - y^2/b^2 = 1, the locus of the midpoint (h, k) of a chord passing through (x1, y1) is given by T = S1, where T is the equation of the chord and S1 is the value of the hyperbola at the midpoint. This results in a hyperbola.

Area of a sector = (1/2) * arc_length * radius. Area = (1/2) * 55 * 21 = 577.5 cm^2.

Tangent-secant theorem: PT^2 = PA * PB. 12^2 = 9 * PB. 144 = 9 * PB. PB = 16. AB = PB - PA = 16 - 9 = 7.

Perimeter = a+b+c = 30. Hypotenuse c = 13. a+b = 17. a^2+b^2 = 13^2 = 169. (a+b)^2 = a^2+b^2+2ab. 17^2 = 169 + 2ab. 289 = 169 + 2ab. 2ab = 120. Area = 1/2 * ab = 1/2 * 60 = 30.

In a right-angled triangle, the circumradius R is half the hypotenuse (h). Thus, h = 2 * 5 = 10 cm. The area is (1/2) * base * height = (1/2) * hypotenuse * altitude_to_hypotenuse = (1/2) * 10 * 4 = 20 cm^2.

In a right triangle, the circumradius R = hypotenuse/2. So, hypotenuse = 20 cm. Area = 1/2 * base * height = 1/2 * hypotenuse * altitude = 1/2 * 20 * 8 = 80.