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:



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