Suppose lm ∥ kn , kl ∥ nm , h1=5, h2=6, and the perimeter of klmn is 42. find the area of klmn.

57 3/11 square units

Let x and y represent the short side and the long side of the parallelogram, and let α represent the acute angle. Then we have ...

... 2(x+y) = 42 . . . . the perimeter is 41

... x·sin(α) = 5 . . . . the height to the long side is 5

... y·sin(α) = 6 . . . . the height to the short side is 6

adding the last two equations gives ...

... (x +y)·sin(α) = 11

Dividing the first equation by 2 gives

... (x +y) = 21

So, we can find the value of sin(α) by substituting the latter equation into the one before:

... 21·sin(α) = 11

... sin(α) = 11/21

The area of the parallelogram will be the product of x, y, and sin(α).

... A = x·y·sin(α)

Using the first "sin(α)" equation, we can find x.

... x = 5/sin(α) = 5·21/11 = 105/11

Then the area is ...

... A = x·(y·sin(α)) = 105/11 · 6 = 630/11

... A = 57 3/11 . . . . square units

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>