The lengths of the two altitudes of a parallelogram are 4cm,6cm and the perimeter of the parallelogram is 40 cm,find the lengths of the sides of the parallelogram ?

1 Answer
May 1, 2018

The sides are 2 parallel sides of #8 " cm"# length and 2 parallel sides of #12" cm"# length.

Explanation:

Assume that the parallelogram is formed by two vectors, #veca#, and #vecb#.

Given:

#p = 2|veca|+2|vecb| = 40" cm [1]"#

The height where #vecb# is chosen to be the base

#h_a = |veca|sin(theta) = 4" cm [2]"#

where #theta# is the acute angle between #veca# and #vecb#

The property, adjacent angles of a parallelogram are supplementary, causes the equation for the height when #veca# is chosen to be the base to become:

#h_b = |vecb|sin(pi-theta)=6" cm [3]"#

You can use the identity #sin(A-B) = sin(A)cos(B)-cos(A)sin(B)# to prove that #sin(pi-theta) = sin(theta)#:

#h_b = |vecb|sin(theta)=6" cm [3.1]"#

Divide equation [2] by equation [3.1]:

#(|veca|sin(theta))/(|vecb|sin(theta)) = (4" cm")/(6" cm")#

The sine functions cancel and we have an equation for #|veca|# in terms of #|vecb|#:

#|veca| = 4/6|vecb|" [4]"#

Substitute equation [4] into equation [1]:

#2(4/6|vecb|)+2|vecb| = 40" cm"#

#20/6|vecb| = 40" cm"#

#|vecb| = 12" cm"#

#|veca| = 8" cm"#