The rectangle

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

Correct answer:

S =  160 cm²

Step-by-step explanation:

$$\begin{gathered} o=52\text{cm} \\ o=2\cdot (a+b) \\ b/2=3+a/2 \\ 52=2\cdot (a+b) \\ b/2=3+a/2 \\ 2a+2b=52 \\ a-b=-6 \\ Row2-\frac{1}{2}\cdot Row1\to Row2 \\ 2a+2b=52 \\ -2b=-32 \\ b=\frac{-32}{-2}=16 \\ a=\frac{52-2b}{2}=\frac{52-2\cdot 16}{2}=10 \\ a=10 \\ b=16 \\ S=a\cdot b=10\cdot 16=160{\text{cm}}^2 \end{gathered}$$

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