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

Solution:

$$\begin{gathered} o=52\text{ cm } \\ o=2\cdot (a+b) \\ b\mathrm{/}2=3+a\mathrm{/}2 \\ 52=2\cdot (a+b) \\ b\mathrm{/}2=3+a\mathrm{/}2 \\ 2a+2b=52 \\ a-b=-6 \\ a=10 \\ b=16 \\ S=a\cdot b=10\cdot 16=160{\text{cm}}^2 \end{gathered}$$

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