Parallelogram Question? Should be easy for experts.?

Question

This is going to be hard to describe without an image but I'll try my best. Imagine there is a parallelogram ABCD. AB is the slant length on one side and CD is the slant length on the other.
So BD is the top and AC is the base.

The base (AC) is 10cm. The slant length is 8cm. The perpendicular height from base to top is 7 cm.

What is the area of the parallelogram?
I think its 84cm squared but would like confirmation.

Now the harder part (at least for me). What is the perpendicular distance from AB to CD?
Does it just remain 12 (the base)?

Answer 1

Draw a vertical height from C, meeting BD at point E. Imagine cutting off triangle CED and sliding it over onto the other end, with CD lying right on AB (which it will do, since CD is parallel to AB, and of the same length). You haven't changed the area, but now you have a 10 cm by 7 cm rectangle.

For the second question, draw a line from C to AB, meeting AB at point F, with CF perpendicular to AB. (Because AB is parallel to CD, CF is also perpendicular to CD.) The length of CF is the distance from AB to CD. Now, observe that triangles CED and CFA are similar. So

CF/CE = AC/CD

Alternatively, note that CF is a height for the parallelogram with AB serving as base, and so CF*AB = area of the parallelogram.