ABCD is a parallelogram. AD is produced to E, AF = 6 units, AF⊥DC and the area of
triangle BCE is 33 units2
.
1. Determine the area of ABCD
2. Determine the length AB.
1. Since AF is perpendicular to DC, triangle ADF is a right triangle. Since AF = 6 units and the area of triangle BCE is 33 units^2, the length of DC is also 6 units. Therefore, the area of parallelogram ABCD is equal to the base DC (6 units) multiplied by the height AE.
To find the height AE, we can use the fact that triangle ADF and triangle CEB are similar triangles.
Using the similarity ratio, we can write the following equation:
DF/CD = AE/EB
Since DF = CD and EB = BC, we have:
1 = AE/BC
Therefore, AE = BC.
Since the area of triangle BCE is given as 33 units^2, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
Substituting the values, we have:
33 = (1/2) * BC * BC
66 = BC^2
Taking the square root of both sides, we get:
BC = √66
Therefore, the height AE and the base DC are both equal to √66 units.
The area of parallelogram ABCD is:
Area = base * height = DC * AE = √66 * √66 = 66 units^2.
2. Since ABCD is a parallelogram, the opposite sides are equal in length. Therefore, AB = CD = √66 units.