In the trapezium ABCD where AB is parallel to CD, AB is 3 units longer than CD, BC = 5, and AD is perpendicular to AB. Given that the area of the trapezium is 34, find the perimeter of the trapezium.
The perimeter of the trapezium is 26 units.
Since AB is 3 units longer than CD.....then the height of the trapezium = 4 =AD
We have that
34 = (1/2)(4) ( CD + 3 + CD)
17 = 3 + 2CD
14 = 2CD
7 = CD
10 = AB
4 = AD
5 = BC
So.....the perimeter = CD + AB + AD + BC = 7 + 10 + 4 + 5 = 26
