A parallelogram ABCD has perimeter equal to 124 . Let E be the foot of the perpendicular from A to BC, and let F be the foot of the perpendicular from A to CD. If AE= 7 and AF=20, what is the area of the parallelogram?
+113 Hm. Tried this one. by tinkering in GeoGebra. Not too easy. Maybe I'm getting a bit tired... By tinkering I found that AB is between 16.6, and that gives an area around 332, if all conditions are met. I am too tired right now to think out a way to calculate this.
+289 No need for any programs like geogebra. Impressive how you could do that though :D
Because opposite angles of a parallelogram are equal, angles ADF and ABE are congruent. Suppose that \(a\) represents the length of AB and b represents the length of AD. Using AA similarity, we can find that triangles ABE and ADF are similar. By the fact that the sides of similar triangles are in proportion, we can form the equation, 7/20 = a/b. In addition, because we know that the perimeter of the parallelogram is equal to 124, a + b = 62.
We have a system of equations!
a + b = 62
7b = 20a
b = 20/7a
(20/7)a + a = 62
(27/7)a = 62
a = 434/27
Multiplying this by 20 gives: 8680/27 as the area of the parallelogram.
+113 Thanks for that on. I'm obviously out of shape when it comes to geometry. Should have seen that \(\Delta ABE \cong \Delta ADF\). That is what solves this. Very good explanation of this.
