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:

a=2 cm x=2 cm o=63 cm  o1=2(a+b) y = o1 o2 = 2(x+y) o = 2(x+c) = 2(x+(b+y))  o/2x=b+2 (a+b) 63/22=b+2 (2+b)  6b=51  b=651=8.5  b=217=8.5  S1=a b=2 8.5=17=17 cm2   Verifying Solution:  y=2 (a+b)=2 (2+8.5)=21 cm o1=y=21 cm o2=2 (x+y)=2 (2+21)=46 cm c=b+y=8.5+21=259=29.5 cm o3=2 (x+c)=2 (2+29.5)=63 cm

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