Peter's rectangle

Peter had a rectangle 2 cm wide and of unknown length. The line had a 2 cm rectangle whose length was equal to the perimeter of Peter's rectangle. When they put the rectangles together with their widths, they got a new rectangle with a circumference of 63 cm. Find the area of Peter's rectangle.

Correct answer:

S₁ =  17 cm²

Step-by-step explanation:

$$\begin{gathered} a=2\text{cm} \\ x=2\text{cm} \\ o=63\text{cm} \\ {o}_1=2(a+b) \\ y=o_1 \\ {o}_2=2(x+y) \\ o=2(x+c)=2(x+(b+y)) \\ o/2-x=b+2\cdot (a+b) \\ 63/2-2=b+2\cdot (2+b) \\ 6b=51 \\ b=\frac{51}{6}=8.5 \\ b=\frac{17}{2}=8.5 \\ {S}_1=a\cdot b=2\cdot 8.5=17=17{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ y=2\cdot (a+b)=2\cdot (2+8.5)=21\text{cm} \\ {o}_1=y=21\text{cm} \\ {o}_2=2\cdot (x+y)=2\cdot (2+21)=46\text{cm} \\ c=b+y=8.5+21=\frac{59}{2}=29.5\text{cm} \\ {o}_3=2\cdot (x+c)=2\cdot (2+29.5)=63\text{cm} \end{gathered}$$

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