Geometry

Find AC using Pythagoras: AC = √(12² + 5²) = 13 cm. Since triangles are similar, 4AC = PR means scale factor = PR/AC = 4. Area scales by the square of the scale factor: Area(PQR) = 4² × Area(ABC) = 16 × (1/2 × 12 × 5) = 16 × 30 = 480 cm².

PQ is the longest chord, so it's the diameter. Angle PTQ = 72° is in the semicircle (exterior point T). Using the property: angle subtended by arc at any point on circumference = half the angle at center. Arc PQ subtends 72° at T, so angle PRQ = 72°. At center O, angle ROS subtended by same arc RS = 2 × angle subtended at circumference = 2 × 18° = 36°.

When AB and AC are produced, O is intersection of external angle bisectors of ∠CBD and ∠BCE. Using exterior angle bisector theorem: ∠BOC = 90° - (∠A/2). Given ∠BOC = 69°, we have 69° = 90° - (∠A/2), so ∠A/2 = 21° and ∠A = 42°. This property relates intersection of external bisectors to the original vertex angle.

For two chords PQ and RS of a circle extended to meet at M, the power of point theorem states: MQ·MP = MR·MS. Given MR = 5.6 cm, QM = 6.3 cm, and PQ = 7.7 cm. We have MP = MQ + QP = 6.3 + 7.7 = 14 cm. So MQ·MP = 6.3 × 14 = 88.2. This equals MR·MS = 5.6 × MS, so MS = 88.2/5.6 = 15.75 cm. Therefore RS = MS - MR = 15.75 - 5.6 = 10.15 cm.

For similar triangles, ratio of perimeters equals ratio of corresponding sides. Perimeter PQR/Perimeter ABC = 46.8/78 = 0.6. So PQ/AB = 0.6, giving AB = 11.7/0.6 = 19.5cm.

In a right triangle with ∠BAC = 90°, BC = 25 cm is the hypotenuse. The median to the hypotenuse equals half the hypotenuse, so AD = BC/2 = 25/2 = 12.5 cm. This is a standard property: the median from the right angle to the hypotenuse is half the hypotenuse. Options B (24 cm), C (14.5 cm), and D (10 cm) don't match this property.

For two concentric circles, if a chord of the larger circle (length 10 cm) is tangent to the smaller circle, then the perpendicular from the center bisects the chord. This creates a right triangle with half the chord (5 cm) and the radius of the smaller circle as legs. By Pythagoras theorem: R² = 5² + r². The annular area = π(R² - r²) = π(25) = 25π cm².

The perpendicular from the center to a chord bisects the chord. So half the chord is 6 cm. Using Pythagorean theorem in the right triangle formed by the radius, perpendicular (8 cm), and half-chord (6 cm): radius² = 8² + 6² = 100, so radius = 10 cm. Diameter = 2 × radius = 20 cm. Option D is correct.

Radius r = 5√13 ≈ 18.03 cm. For 12 cm chord: distance from center d₁ = √((5√13)² - 6²) = √(325 - 36) = √289 = 17 cm. For 20 cm chord: d₂ = √((5√13)² - 10²) = √(325 - 100) = √225 = 15 cm. Distance between chords = d₁ - d₂ = 17 - 15 = 2 cm (assuming they're on same side of center in one semicircle).

Use the Angle Bisector Theorem, which states that the angle bisector divides the opposite side proportionally to the adjacent sides: BD/DC = AB/AC = 15/13. Since BC = 14 cm, we have BD + DC = 14. From the ratio, BD = (15/28) × 14 = 7.5 cm, and DC = (13/28) × 14 = 6.5 cm.

In triangle ABC, O is the orthocenter since BE ⟂ AC and CD ⟂ AB. P is the incenter of triangle OBC (intersection of angle bisectors). For any triangle, the angle formed by the intersection of two angle bisectors equals 90° plus half the remaining angle: ∠BPC = 90° + ∠BOC/2. Since quadrilateral AECB is cyclic (both ∠AEB and ∠ACB are right angles), ∠BOC = 180° - ∠A. Substituting: 148° = 90° + (180° - ∠A)/2, which gives ∠A = 64°.

In cyclic quadrilateral ABCD with AC as diameter, ∠ABC = 90° (angle in semicircle). In ΔABE: ∠ABE = 180° - 34° - 30° = 116°. Then ∠CBD = ∠ABD - ∠ABC. Since ∠ABD = 180° - ∠ABE = 64° (linear pair), we get ∠CBD = 64° - 90° = -26°. Taking magnitude: ∠CBD = 26°.

Since PQ is the diameter, ∠PRQ = 90° (angle in semicircle). Using Pythagoras: PQ² = PR² + QR². PQ = diameter = 2 × 14 = 28 cm. PR = 10 cm. QR² = 28² - 10² = 784 - 100 = 684, so QR = √684 = 6√19 cm. Area of right triangle PQR = 1/2 × base × height = 1/2 × 10 × 6√19 = 30√19 cm².

In any triangle, the angle formed at the incenter by two angle bisectors is given by BIC = 90° + A/2, where A is the angle at vertex A. Setting 90° + A/2 = 108° gives A/2 = 18°, so A = 36°. The key property is that the angle at the incenter is always 90° plus half of the opposite vertex angle.

Using the power of a point theorem for secant PAB: PA × PB = PT², where PT is the tangent length. Given PA = 10 cm and PB = 16 cm, we get PT² = 10 × 16 = 160. Since PO = 14 cm (distance from center to P), and PT² + r² = PO² for a tangent, we have r² = 14² - 160 = 36, so r = 6 cm. Therefore, the diameter is 12 cm.