In rectangle ABCD, the bisector of angle A intersects the side BC at point E, and the continuation of the side

In rectangle ABCD, the bisector of angle A intersects the side BC at point E, and the continuation of the side DC at point F. AE = 2 cm; EF = 10 cm. Find the area of the rectangle.

Since AE is a bisector, then ∠DAE = ∠BAE = 45 degrees.
1) Since point F belongs to the side CD, ∠DАЕ = 45 degrees, ∠Д = 90 degrees, then △ ADF is isosceles and rectangular. It follows that FD = AD are the legs of this triangle, and AF = 12 cm is the hypotenuse.
Let FD = AD = a. By the Pythagorean theorem (the sum of the squares of the legs is equal to the square of the hypotenuse):
a ^ 2 + a ^ 2 = 12 ^ 2,
2a ^ 2 = 144,
a ^ 2 = 72,
a = √72 (cm).
Hence, FD = AD = √72 cm.
2) Since ∠В = 90 gr., ∠ВАЕ = 45 gr., Then ∠BEA = 45 gr. (since the sum of the angles of a triangle is 180 degrees), which means △ ABE is isosceles and rectangular. It follows that AB = BE – his legs, and EA = 2 cm – hypotenuse.
Let AB = BE = m. Then, by the Pythagorean theorem:
m ^ 2 + m ^ 2 = 2 ^ 2,
2m ^ 2 = 4,
m ^ 2 = 2,
m = √2 (cm).
Hence, AB = BE = √2 cm.
3) We found the length and width of the given rectangle ABCD:
AD = √72 cm, AB = √2 cm.
Let’s calculate its area:
Sabcd = AD * AB = √72 * √2 = √144 = 12 (kV.cm).
Answer: 12 kV cm.