In the

In the rectangle ABCD, the distance of its center from the line AB is 3 cm greater than from the line BC. The circumference of the rectangle is 52 cm. Calculate the contents of the rectangle. Express the result in cm2.

Correct result:

S =  160 cm2


o=52 cm   o=2 (a+b) b/2=3+a/2 52=2 (a+b) b/2=3+a/2  2a+2b=52 ab=6  a=10 b=16   S=a b=10 16=160 cm2o=52 \ \text{cm} \ \\ \ \\ \ \\ o=2 \cdot \ (a+b) \ \\ b/2=3 + a/2 \ \\ 52=2 \cdot \ (a+b) \ \\ b/2=3 + a/2 \ \\ \ \\ 2a+2b=52 \ \\ a-b=-6 \ \\ \ \\ a=10 \ \\ b=16 \ \\ \ \\ \ \\ S=a \cdot \ b=10 \cdot \ 16=160 \ \text{cm}^2

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