ABCD is a 20 cm square and AMTD is a rectangle with a short side of 8 cm.

ABCD is a 20 cm square and AMTD is a rectangle with a short side of 8 cm. How much is the area of the MTD rectangle less than the area of the square?

Based on the condition of the problem, it is clear that:

the sides of the square ABCD will be called: AB, BC, CD and AD.

the sides of the AMTD rectangle will be named: AM, MT, TD and AD.

Since it is known that all sides of the square are equal and the side of the square is equal (from the conditions of the problem) 20 cm, we have AB = BC = CD = AD = 20 cm.

The formula for determining the area of ​​a square: S = axb, where a and b are the sides of the square, which means a = b, which means the area of ​​the square ABCD is:

S (ABCD) = AB x BC = 20 cm x 20 cm = 400 cm2 (square centimeters).

Based on the conditions of the problem and the names of the sides of the figures given to us: the square ABCD and the rectangle AMTD have one common side, namely AD, which is 20 cm.

Since the opposite sides of the rectangle are equal, we have:

AMTD: AM = TD, MT = AD = 20cm.

since the smaller side of the rectangle is 8 cm, then AM = TD = 8 cm.

The area of ​​the rectangle is equal:

S (AMTD) = AM x MT = 20 cm x 8 cm = 160 cm2 (square centimeters).

How much is the area of ​​the rectangle MTD less than the area of ​​the square (x-?):

x = S (ABCD) – S (AMTD) = 400 – 160 = 240 sq. see (square centimeters).

Answer: The area of ​​the rectangle AMTD is less than the area of ​​the square ABCD by 240 square centimeters.