Q: 133 (IAS/1998)
question_subject:
Maths
question_exam:
IAS
stats:
0,1,5,1,1,2,2
keywords:
{'diameter lp': [0, 1, 0, 0], 'lsrqp': [0, 1, 0, 0], 'lmnop': [0, 1, 0, 0], 'perimeters': [0, 1, 0, 0], 'lsr': [0, 1, 0, 0], 'rqp': [0, 1, 0, 0], 'lp': [0, 1, 0, 0], 'ratio': [1, 0, 1, 12], 'centres': [0, 0, 0, 1], 'centre': [10, 3, 7, 19]}
The semi-circle LMNOP has a diameter LP. While, LSR and RQP are also semi-circles with their centers at T and U respectively, and with diameters half of LP. In other words, both LSR and RQP are half the size of LMNOP.
To find the ratio of their perimeters, we need to understand that the perimeter of a semi-circle depends on the diameter, as it is given by πD + D, where D is the diameter. The diameters LSR and RQP are combined equal to LP. So, the perimeters of both semi-circles combined would equal the perimeter of LMNOP.
Therefore, for every value of the perimeter of M, the perimeter of N will be the same, which is why the ratio is 1:1, corresponding to option 2.