Given:
lm ∥ kn , kl ∥ nm ,lp = h­1 = 5, mq = h2 = 6,
perimeter of klmn = 42
find: area of klmn

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Given:
Perimeter of KLMN = 42
h1 = 5
h2 = 6

Based on the given figure here are my assumptions.
1) KLMN is a parallelogram
2) h1 and h2 are long legs of right triangles.
3) the right triangles are similar but not congruent.

Perimeter of a parallelogram = 2(a+b)
a = KL ; b = LM
Area of a parallelogram = bh
b = LM ; h = h1

We need to find the values of a and b to solve for the Area of KLMN.
LM and KL are hypotenuse of its own right triangle. Similar triangle means that their proportion are equal.

P = 2(a+b)
42 = 2(a+b) → 42/2 = a+b → 21 = a + b → 21 - a = b

hypotenuse/long leg

a/5 = b/6 → a/5 = (21-a)/6 → a*6 = 5(21-a) → 6a = 105 - 5a
6a + 5a = 105 → 11a = 105 → a = 105/11 → a = 9.55

b = 21 - a → b = 21 - 9.55 → b = 11.45 

Area of parallelogram = bh
A = 11.45 * 5 
A = 57.25 square units.

The answer is 26 to 1