Q19
Q19 - In a trapezium ABCD, AB||CD and AB = 2CD. M, N are the midpoints of the diagonals AC and BD respectively. Let the perimeter of ABCD be L1, the perimeter of the quadrilateral CDMN be L2 and L1 = n.L2, find the value of n. Answer: 2 Solution: Put the co-ordinates like this: A (0,0), B (2CD,0) C (x+CD,y) D (x,y) M = (x+CD)/2, y/2 N = (x + 2CD)/2, y/2 Clearly MN is || to AB and CD. And MN = CD/2 Perimeter of MNCD = L1 = MN + NC + CD + DM CD = CD MN = CD/2 NC = sqrt(x^4 + y^2/4) DM = sqrt((x-CD)/2)^2 + y^4) L2 = AB + BC + CD + DA AB = 2CD CD = CD DA = sqrt(x^2 + y^2) BC = sqrt((x-CD)^2 + y^2) Clearly each segment in L2 is twice of that in L1. Hence n = 2. Another solution: