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=24, what is the area of the parallelogram?
+773 Given parallelogram ABCD, whose perimeter is 124 units, E is the foot of the perpendicular from A to BC and F is the foot of the perpendicular from A to CD. Also, AE= 7 units and AF= 24 units. The area of the parallelogram ABCD is the sum of area of triangle ABC and triangle ACD
We have,
124=2(a+b)⇒a+b=62units Area of parallelogram=base∗height=7a=24b⇒7(62−b)=24b⇒b=43431=14units⇒a=62−14=48unitsTherefore area of ABCD=7a=7(48)=336sq.units
Hope this helps, whymenotsmart^m^.
