1. Problem 33: Describe the opposite angles of a quadrilateral inscribed in a circle. In a cyclic quadrilateral (one inscribed in a circle), the sum of each pair of opposite angles is 180°. This means the opposite angles are supplementary. Answer: C. supplementary 2. Problem 34: Given a circle centered at C with radius 6 cm and arc AB measuring 60°, find the area of the shaded sector. The formula for the area of a sector is $$\text{Area} = \frac{\theta}{360} \times \pi r^2$$ where $\theta$ is the central angle in degrees. Here, $\theta=60^\circ$, $r=6$ cm. Calculate: $$\frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times \pi \times 36 = 6 \pi$$ cm² Answer: A. 6 π cm² 3. Problem 35: Name the shaded region in problem 34. The shaded region defined by two radii and the included arc is called a sector of a circle. Answer: B. Sector of a circle 4. Problem 36: In triangle with center A and arc DSY a semicircle, given $m\angle SAD=70^\circ$, find the measure of $\angle SAY$. Since arc DSY is a semicircle, angle $\angle SAY$ subtended by this arc at the circumference is a right angle, i.e., $90^\circ$. Angles in triangle SAD sum to 180°: $$m\angle SAY = 180^\circ - m\angle SAD - m\angle ADS$$ But $m\angle ADS = 90^\circ$ (angle in semicircle). Thus: $$m\angle SAY = 180^\circ - 70^\circ - 90^\circ = 20^\circ$$ Answer: A. 20° 5. Problem 37: Name the shaded region which is bounded by a chord and the arc. This shape is known as a segment of a circle. Answer: D. Segment of a circle 6. Problem 38: Arc PQ measures 60°, radius 5 cm; find the area of the shaded segment. First find sector area: $$\text{Sector area} = \frac{60}{360} \times \pi \times 5^2 = \frac{1}{6} \times 25\pi = \frac{25\pi}{6}$$ Next find area of triangle formed by radii and chord: Using formula for triangle area with central angle $60^\circ$: $$\text{Triangle area} = \frac{1}{2} r^2 \sin 60^\circ = \frac{1}{2} \times 25 \times \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4}$$ Area of segment = sector area - triangle area Approximate numeric values (using $\pi \approx 3.1416$, $\sqrt{3} \approx 1.732$): $$\text{Sector area} \approx \frac{25\times 3.1416}{6} = 13.09$$ $$\text{Triangle area} \approx \frac{25 \times 1.732}{4} = 10.825$$ Segment area: $$13.09 - 10.825 = 2.265$$ cm² None of the options fit this number exactly; options are quite small, so let's check if radius might have been 5 cm as in the question. If radius is 5 cm, yes, above holds. Among options, closest is C. 0.583 cm², but calculation shows larger. Possibly, the answer choices imply radius different from original question or meant in different units. We trust the method and numerical estimate. 7. Problem 39: A dart board diameter 40 cm, divided into 20 sectors; find area of one sector. Radius: $$r= \frac{40}{2} = 20$$ cm Area of full circle: $$\pi r^2 = \pi \times 20^2 = 400\pi$$ Area of one sector: $$\frac{1}{20} \times 400 \pi = 20 \pi$$ cm² Answer: A. 20 π cm² 8. Problem 40: Pendant is a regular octagon inscribed in a circle; lines connect opposite vertices meeting at center. Find measure of each angle at center formed. Regular octagon has 8 vertices; central angles between vertices are: $$\frac{360^\circ}{8} = 45^\circ$$ Lines connecting opposite vertices pass through center, forming angles equal to $45^\circ$ at center. Answer: B. 45°