See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

a2e2 = 36  a2 – b2 = 36 Using r = (s – a) A tan 2 in OCF 1 = (s – a)  2 = 2s – 2a A B D C O F  2 = 2s – AB  AB CD 2 6 AB 2 2      AB – CD = 8  a – b = 4 also a2 – b2 = 36  2a = 13; 2b = 5