Find the area of the given hexagon $A B C D E F$ in which each one of $B J, C L, E M$ and $F K$ is perpendicular to $A D$ and it is being given that $A J = 6 c m, A K = 10 c m, A L = 18 c m, A M = 21 c m, A D = 27 c m, B J = 5 c m, C L = 6 c m, E M = 4 c m$ and $F K = 6 c m$ .

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Solution

$A J = 6 c m; A K = 10 c m; A L = 18 c m; A M = 21 c m; A D = 27 c m; B J = 5 c m; C L = 6 c m; E M = 4 c m$
and $F K = 6 c m$ .
$L D = A D - A L = 27 - 18 = 9 c m$
$M D = A D - A M = 27 - 21 = 6 c m$
$J L = A D - A J - L D = 27 - 6 - 9 = 12 c m$
$K M = A D - A K - M D = 26 - 10 - 6 = 10 c m$
Area of triangles $A B C, C L D, D M E, A K F = (A J \times B J) / 2 + (C L \times L D) / 2 + (A K \times K F) / 2 + (M D \times M E) / 2$
Area of trapezium $B J L C$ and trapezium $K M E F = J L (B J + C L) / 2 + K M (M E + K F) / 2$
Total area $= (A J \times B J) / 2 + (C L \times L D) / 2 + (A K \times K F) / 2 + (M D \times M E) / 2 + J L (B J + C L) / 2 + K M (M E + K F) / 2$
$A = (6 \times 5) / 2 + (6 \times 9) / 2 + (10 \times 6) / 2 + (6 \times 4) / 2 + 12 (5 + 6) / 2 + 10 (4 + 6) / 2$
$A = 30 / 2 + 54 / 2 + 60 / 2 + 24 / 2 + 132 / 2 + 100 / 2$
$A = 15 + 27 + 30 + 12 + 66 + 50$
$A = 200 c m^{2}$

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