ABCD is a parallelogram in which BC is produced to E suh that CE=BC.AE intersects CD at F.if ar(DFB)=3cm^2.
find the area of the parallelogram ABCD.
by similar triangles,
FC/AB = CE/BE = 1/2
so, F is the midpoint of CD
if the altitude of ABCD from AB to CD is h,
area(ABCD) = h*CD
area(DFB) = 1/2 * h * DF = 1/2 * h * CD/2 = 1/4 h*CD
so, area(ABCD) = 12
You would try to find X first
To find the area of the parallelogram ABCD, we need to use the fact that triangles on the same base and between the same parallel lines are equal in area.
Given that F is the intersection point of AE and CD, we can see that triangle DFB and triangle CEF are on the same base DF and between the same parallel lines DF and BC. Therefore, their areas are equal.
We are given that the area of triangle DFB is 3 cm^2. So, the area of triangle CEF is also 3 cm^2.
Now, we know that the area of a parallelogram is equal to the product of its base and height. In this case, the base of the parallelogram is BC and the height is the perpendicular distance from BC to AE. Since AE intersects BC at F, this perpendicular distance is FE.
Since CE = BC, we can say that EF = BC (using the fact that CE and AE are parallel). Therefore, the height of the parallelogram is BC.
The area of the parallelogram ABCD is then given by:
Area = Base * Height = BC * BC = BC^2
So, to find the area of the parallelogram, we need to find the value of BC. Unfortunately, the problem statement does not provide any information about BC.
Therefore, without additional information, we cannot determine the exact area of the parallelogram ABCD.