1 Find the probability that a non leap year selected at random has 53 Sundays 2 Two - Maths -

Question

1 Find the probability that a non leap year selected at random
has 53 Sundays
2 Two concentric circles are of radii 5cm and 3cm Find the
length of the chord of the larger circle which touches the smaller
circle
3 A circle touches the side BC of triangle ABC at P and touches
AB and AC produced at Q and R repectively If AQ=5cm, find the
perimeter of triangle ABC

Rayya Halim · Accepted Answer

Ans 1 In a non-leap year there are 365 days We have, 365 days = 364 days + 1 days 365 days = (52 Ã— 7) days + 1 day 365 days = 52 weeks + 1 day Thus a non-leap year will always have 52 Sundays The remaining one day can be (i) Sunday (ii) Monday (iii) Tuesday (iv) Wednesday (v) Thursday (vi) Friday (viii) Saturday âˆ´ Total number of possible outcomes = 7 ( Number of days in a week ) The number of favourable outcome = 1 ( since Sunday comes once in a week ) Therefore the probability of 53 sundays= $\frac{f a v o u r a b l e o u t c o m e}{t o t a l n u m b e r o f o u t c o m e s} = \frac{1}{7}$ Ans 2

In the diagram AB is the chord touching the smaller circle we have the right angled triangle OO'B using Pythagorean theorem we have $O' B = \sqrt{{5^{2} - 3}^{2}} = 4 c m$ Now since the chord of the larger circle which touches the smaller circle is bisected at the point of contact, we have AB=2Ã—4 cm = 8 cm Ans 3

consider the diagram Since tangents from a point to a circle are equal in length Therefore, tangents from A,B,C to the circle are equal in length AQ=AR , BP=BP, CP=CR Consider the segment AQ and AR AQ=AR AB+BQ=AC+CR AB+BP=AC+CP (1) (since BQ=BP and CR=CP) The perimeter of the triangle ABC is given as Perimeter=AB+BC+CA = AB+(BP+PC)+CA =(AB+BP)+(PC+CA) = 2(AB+BP) from (1) =2(AB+BQ) ( since BP=BQ) =2AQ =2Ã—5 cm perimeter= 10 cm