A string initially in the shape of a circle with a radius of 21 cm is cut into two distinct pieces. The first piece is reshaped into a rectangle where the length is twice its breadth, while the second piece is formed into a square. If the perimeter of the rectangle is 24 cm greater than the perimeter of the square, find the breadth of the rectangle. [Use \(\pi = \tfrac{22}{7}\)]
13 cm
Circumference of the string: \(2\times\tfrac{22}{7}\times21 = 132\) cm.
Let the rectangle's breadth be b, so its length is 2b and its perimeter is \(2(2b+b)=6b\). Let the square's side be s, with perimeter \(4s\).
Given \(6b = 4s+24 \Rightarrow s = \dfrac{6b-24}{4} = \dfrac{3b-12}{2}\).
Total string length: \(6b+4s = 132\). Substituting: \(6b+4\times\dfrac{3b-12}{2} = 132 \Rightarrow 6b+2(3b-12) = 132\).
\(12b-24 = 132 \Rightarrow 12b = 156 \Rightarrow b = 13\).
Hence, the breadth of the rectangle is 13 cm.
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