If AB and PQ are two chords of a circle with centre O. Both chords are the opposite side of the centre. If the length of AB is 16 cm and the length of PQ is 25% less than the length of AB and the distance between the chords are 12.5% less than the length of AB. Area of the circle is∶

AB = 16 cm, PQ = 25% less = 12 cm. Distance between chords = 12.5% less than AB = 14 cm. Since chords are on opposite sides of center, their distances from center sum to 14 cm. For chord length c at distance d from center: c = 2√(r² - d²). For AB: 16 = 2√(r² - d1²), giving d1 = √(r² - 64). For PQ: 12 = 2√(r² - d2²), giving d2 = √(r² - 36). Since d1 + d2 = 14 and both chords are on opposite sides, one is closer. Solving: r² - 64 + r² - 36 + 2√((r²-64)(r²-36)) = 196. This gives r = 10, so area = 100π.