Video Transcript
In the following figure, 𝐴𝐵𝐶𝐷
and 𝐸𝐹𝐺𝐻 are two parallelograms, and line 𝐴𝐹 is parallel to line 𝐷𝐺. If 𝐴𝐵 equals 𝐸𝐹, 𝐶𝐷 equals
five centimeters, and the sum of the areas of parallelogram 𝐸𝐹𝐺𝐻 and
parallelogram 𝐴𝐵𝐶𝐷 is 40 square centimeters, find the shortest distance between
line 𝐴𝐹 and line 𝐷𝐺.
The first thing we might note here
is that the shortest distance between the two lines 𝐴𝐹 and 𝐷𝐺 is the
perpendicular distance between these lines. Let’s define this perpendicular
distance or perpendicular height with the letter ℎ. From the diagram and the
information given in the question, we have that 𝐴𝐵 is congruent with 𝐸𝐹. So in fact, these two
parallelograms have congruent bases. And importantly, since the distance
between the parallel lines remains constant, this means that the areas of the two
parallelograms 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻 must be equal. And as we are given that the sum of
their areas is 40 square centimeters, then we can halve this value to see that the
area of each parallelogram must be 20 square centimeters.
Of course, we still need to find
this shortest distance, or the value of ℎ. And to do this, we need to recall
the formula to find the area of a parallelogram. This is equal to the base
multiplied by the perpendicular height. Given the information that the
length of the line segment 𝐶𝐷 is five centimeters, then we can work out the value
of ℎ by using the formula to find the area of 𝐴𝐵𝐶𝐷. So we know that 20 is equal to five
multiplied by ℎ. Dividing through by five, we have
that ℎ is equal to four centimeters. We can therefore conclude that the
shortest distance between the line 𝐴𝐹 and the line 𝐷𝐺 is equal to the
perpendicular height, and that’s four centimeters.
