Q18

Q18

---

In the trapezium $ABCD$, $AD \parallel BC$, $\angle B = 30^\circ$ & $\angle C = 60^\circ$, $E, M, F, N$ are the midpoints of $AB, BC, DC, DA$ respectively. Given that $BC = 7$, $MN = 3$. Find $EF$.

---

Solution:
Set up the co-ordinates.
B = (0,0)
C = (7,0)
M = (7/2,0)
Let the height be h.
So y co-ordinates of A,D = h.
Let CD = c, since angle at C is 60 degree, we can use that to get x coord of D.
Also h = c.sqrt(3)/2
D = (7-c/2,h)
Similarly using the angle at B which is 30 degree,
h = AB.sin(30) => AB = 2h = c.sqrt(3)
x coord of A = AB.cos(30) = 3c/2
A = (3c/2,h)
N = midpoint of AD = ((7+c)/2,h)
MN^2 = 9 = h^2 + [(7+c)/2 - 7/2]^2 = h^2 + c^2/4 = c^2 (replace h)
So MN  = c = CD = 3

Now,
E = AB's midpoint = (3c/4,h/2)
F = DC's midpoint = (7-c/4,h/2)
EF = 7 - c/4 - 3c/4 = 7 - c = 4 = Answer.

Another Solution: