Geometry

Circle diameter = 10 cm, so radius r = 5 cm. Chord length = 6 cm. Distance of chord from center = √(5² - 3²) = √(25 - 9) = √16 = 4 cm (half chord = 3). Two parallel chords of same length on opposite sides of center: distance between them = 4 + 4 = 8 cm.

For a 6-8-10 triangle, this is a right triangle (since 6²+8²=10²=100). The area is (1/2)×6×8=24 cm². The semiperimeter s=(6+8+10)/2=12 cm. The inradius r=Area/s=24/12=2 cm. The area of the inscribed circle is πr²=π(2)²=4π cm². Option A shows 4π cm².

For a chord of length L in a circle of radius R, the perpendicular distance from the center is √(R² - (L/2)²). For the 6 cm chord: distance = √(25 - 9) = √16 = 4 cm from center. For the 8 cm chord: distance = √(25 - 16) = √9 = 3 cm from center. Since both chords are on the same side, the distance between them is 4 - 3 = 1 cm.

For a right triangle inscribed in a circle with diameter as one side, that side is the hypotenuse (20 cm). If perimeter is 48 cm and hypotenuse is 20 cm, the other two sides sum to 28 cm. Checking 12+16+20=48 and 12²+16²=144+256=400=20² confirms this is correct (a 3-4-5 triangle scaled by 4).

Triangle ABC is right-angled at A (since ∠BAC = 90°). In a circle, the diameter subtends a right angle at any point on the circumference, so BC is the diameter. Using Pythagoras theorem: BC = √(AB² + AC²) = √(8² + 6²) = √(64 + 36) = √100 = 10 cm. Therefore radius = BC/2 = 5 cm.

The Angle Bisector Theorem states that an angle bisector in a triangle divides the opposite side into segments proportional to the adjacent sides. Therefore BD:CD = AB:AC = 5:7.

For two circles of equal radius r with centers distance d apart, the length of a transverse common tangent is √(d² - 4r²). Here: √(10² - 4×3²) = √(100 - 36) = √64 = 8 cm. The tangent connects the circles without crossing the line joining centers.

Using chord-radius relationship: radius² = perpendicular distance² + (chord/2)². r² = 6² + 8² = 36 + 64 = 100. Therefore r = 10 cm. This is derived from the fact that a radius drawn to the chord's midpoint forms a right triangle with the chord half-length and perpendicular distance.

Using chord properties in a circle, two equal chords AB and AC of length 6 cm each form an isosceles triangle ABC. The distance from center to each chord is √(5² - 3²) = 4 cm. Using geometry relations, BC calculates to 9.6 cm.

A 72° arc is 72/360 = 1/5 of a full circle. Since arc length is proportional to its central angle, the ratio of arc length to circumference is 1:5. Option A is correct.

When two circles of equal radius intersect such that each passes through the other's center, the distance between centers equals the radius. All three sides (two radii and distance) form an equilateral triangle.

Area of sector = (θ/360) × πr² = (72/360) × (22/7) × 35² = (1/5) × 22 × 5 × 35 = 770 cm².

For circles with radii 4 and 9 touching externally, distance between centers = 4+9 = 13. Length of common external tangent PQ = √(d² - (r₂-r₁)²) = √(13² - (9-4)²) = √(169-25) = √144 = 12. Area of square with side PQ = 12² = 144 cm².

Circle diameter = 26 cm, so radius = 13 cm. Each chord length = 24 cm. For a chord of length 24 cm, the perpendicular distance from center is: sqrt(13² - 12²) = sqrt(169 - 144) = sqrt(25) = 5 cm (half-chord is 12 cm by Pythagoras theorem). Two parallel chords on opposite sides of center: total distance = 5 + 5 = 10 cm. If both chords are on same side, answer would be 0, but context suggests opposite sides. Option A (10 cm) is correct assuming chords on opposite sides of center.

Arc length formula: l = (θ/360) × 2πr. Given θ = 72°, r = 42 cm. l = (72/360) × 2 × (22/7) × 42 = (1/5) × 2 × 22 × 6 = 52.8 cm. Option D is correct.