Point E is the midpoint of BC of a rectangle ABCD and F is the mid point of CD, The area of Triangle AEF is 3sq.units then the area of ABCD is?

Make your sketch.

let EC = x , FC = y
then (1/2)(x)(y) = 3
xy = 6

area of ABCD
= 2x(2y)
= 4xy
= 24

To find the area of rectangle ABCD, we need to first find the length and width of the rectangle.

Given that point E is the midpoint of BC, we can infer that BE = CE. Similarly, since F is the midpoint of CD, we can infer that DF = CF.

Since E is the midpoint of BC, we can also say that AE is parallel to BD. Similarly, since F is the midpoint of CD, we can say that AF is parallel to AC.

From this information, we can infer that triangle AEF is similar to triangle BCD by the Side-Angle-Side (SAS) similarity criterion.

Since the area of triangle AEF is given as 3 square units, we can denote the area of triangle BCD as x square units (as we do not know the actual value yet).

Now, since the triangles AEF and BCD are similar, the areas of similar figures are proportional to the squares of their corresponding sides.

Let's denote the lengths of sides BD and AF as a and b, respectively.

Therefore, we have the proportion: (AE / BD)² = (AE / AF)² = (EF / DF)² = (3 / x)²

Since AE is half the length of BD and EF is half the length of DF, we can write this proportion as: (1 / 2)² = (1 / 2)² = (3 / x)²

Simplifying the equation, we can write it as: 1/4 = 1/4 * 9 / x²

Cross-multiplying, we get: x² = 9

Taking the square root of both sides, we find: x = 3

Therefore, the area of triangle BCD is 3 square units.

Now, since triangle BCD is half of rectangle ABCD (as it is a right triangle), the area of rectangle ABCD is 2 times the area of triangle BCD.

Hence, the area of rectangle ABCD is 2 * 3 = 6 square units.