Given: lm ∥ kn, kl ∥ nm

lp = h­1 = 5, mq = h2 = 6

perimeter of klmn = 42

find: area of klmn

57 3/11 square units

Step-by-step explanation:

The area of a parallelogram is the product of its height and the length of the perpendicular base. The given conditions allow us to find the area two ways. Of course, the area is the same in each case, so ...

area(KLMN) = KN·LP = KL·MQ

KN·5 = KL·6 . . . . . substituting the given numbers

KL = (5/6)·KN . . . . solve for one of the lengths in terms of the other

Now, the perimeter is the sum of the side lengths, and opposite sides are the same length, so we have the relation ...

perimeter(KLMN) = KN + KL + KN + KL = 2(KN +KL)

42 = 2(KN +(5/6)KN) = (11/3)KN . . . . . substitute for KL from above

KN = 42·(3/11) . . . . . . multiply by 3/11

area(KLMN) = KN·5 = (42·3/11)·5 = 630/11 = 57 3/11

Check

KN = 126/11

KL = 5/6·KN = 105/11

KN·5 = 630/11 = KL·6 = 630/11 . . . . . areas match

KL+KN = 231/11 = 21 = half the perimeter . . . . . perimeter agrees

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