Figure Q1 shows an equilateral triangle OPQ, of side 12 cm, and the point S such that OS = PS = QS. The arc PXQ has center O and radius 12 cm. Find the perimeter of the shaded region, giving your answer in terms of π and √3.

Here's the solution for Question 1: Step 1: Identify the components of the shaded region's perimeter. The problem describes an equilateral triangle OPQ with side length 12 cm. The arc PXQ has center O and radius 12 cm. This means P and Q are points on the arc, and since OP = OQ = 12 cm (sides of the equilateral triangle), the angle POQ is 60^. Point S is defined such that OS = PS = QS, meaning S is the circumcenter of OPQ. Given the requirement for the answer to be in terms of and sqrt(3), the most consistent interpretation of the "shaded region" is the area bounded by the arc PXQ and the line segments PS and QS. Step 2: Calculate the length of arc PXQ. The arc PXQ has center O and radius r = 12 cm. Since OPQ is equilateral, the angle POQ subtended by the arc at the center O is 60^. Convert this angle to radians: = 60^ × ()/(180^) = ()/(3) radians The length of an arc is given by the formula L = r. L_PXQ = 12 × ()/(3) = 4 cm Step 3: Calculate the lengths of segments PS and QS. S is the circumcenter of the equilateral triangle OPQ. The distance from any vertex of an equilateral triangle to its circumcenter is the circumradius, R. For an equilateral triangle with side length a, the circumradius is given by R = (a)/(sqrt(3)). Here, the side length a = 12 cm. R = (12)/(sqrt(3)) To rationalize the denominator, multiply the numerator and denominator by sqrt(3): R = 12sqrt(3)sqrt(3) × sqrt(3) = 12sqrt(3)3 = 4sqrt(3) cm Therefore, the lengths of segments PS and QS are both 4sqrt(3) cm. Step 4: Calculate the total perimeter of the shaded region. The perimeter of the shaded region is the sum of the arc length and the lengths of the two segments PS and QS. P = L_PXQ + PS + QS P = 4 + 4sqrt(3) + 4sqrt(3) P = (4 + 8sqrt(3)) cm The perimeter of the shaded region is (4 + 8sqrt(3)) cm. Send me the next one 📸