Circles — Practice Quiz

90°

Step 1: Consider a semicircle with diameter AB and any point P on the semicircle. Step 2: The angle APB is inscribed in the semicircle. Step 3: The arc AB is a semicircle, so the central angle is 180°. Step 4: Using the theorem that inscribed angle = (1/2) × central angle, we get angle APB = (1/2) × 180° = 90°. This is Thales' theorem.

110°

Step 1: In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°). Step 2: ∠A and ∠C are opposite angles. Step 3: Therefore, ∠A + ∠C = 180°. Step 4: Given ∠A = 70°, so ∠C = 180° - 70° = 110°. This is a key property of cyclic quadrilaterals.

12 cm

Step 1: We use the theorem that chords equidistant from the center are equal in length. Step 2: Given that PQ and RS are equidistant from center O. Step 3: Therefore, PQ = RS (by the theorem). Step 4: Since PQ = 12 cm, we have RS = 12 cm. This theorem is the converse of 'equal chords are equidistant from center'.

Rhombus

Step 1: Let the two circles have centers O₁ and O₂, intersecting at points A and B. Step 2: Since A is on both circles, O₁A = r₁ and O₂A = r₂ (radii of respective circles). Step 3: Similarly, O₁B = r₁ and O₂B = r₂. Step 4: In quadrilateral O₁AO₂B, we have O₁A = O₁B = r₁ and O₂A = O₂B = r₂, making it a rhombus with equal opposite sides.