A circle with center O has radius 10 cm. Chord AB is drawn such that its length is also 10 cm. C is a point on the major arc AB. Find: a) The special name of triangle AOB. b) The value of AOB. c) The

Welcome back RAMSY — missed you this week. Here are the solutions to your questions: a) State the special name of the triangle AOB. Step 1: Identify the sides of triangle AOB. OA and OB are radii of the circle. The problem states that the chord AB is equal in length to the radius. Therefore, OA = OB = AB. Step 2: Name the triangle based on its sides. A triangle with all three sides equal in length is an equilateral triangle. The special name of triangle AOB is an equilateral triangle. b) Find the value of AOB. Step 1: Use the property of an equilateral triangle. Since triangle AOB is an equilateral triangle (from part a), all its interior angles are equal to 60^. Therefore, AOB = 60^. AOB = 60^ c) Find the value of ACB. Step 1: Relate the angle at the center to the angle at the circumference. The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference. Here, AOB is the angle at the center subtended by arc AB, and ACB is the angle at the circumference subtended by the same arc AB. So, AOB = 2 × ACB. Step 2: Calculate ACB. We found AOB = 60^ in part b. 60^ = 2 × ACB ACB = (60^)/(2) ACB = 30^ ACB = 30^ d) If the radius of the circle is 7cm, calculate the length of the minor arc AB. (Take = (22)/(7)). Step 1: Identify the given values and the formula for arc length. Radius, r = 7 cm. Angle subtended by arc AB at the center, = AOB = 60^ (from part b). = (22)/(7). The formula for the length of an arc is L = ()/(360^) × 2 r. Step 2: Substitute the values into the formula. L = (60^)/(360^) × 2 × (22)/(7) × 7 cm Step 3: Simplify the expression. L = (1)/(6) × 2 × 22 cm L = (1)/(3) × 22 cm L = (22)/(3) cm L = 7(1)/(3) cm L = 7.33 cm (to 2 decimal places) Send me the next one 📸