Acircular wire of diameter 77 cm is bent in the form of a rectangle whose length is 142% of its breadth. What is the area of the rectangle? (Take $$\pi = \frac{22}{7}$$)

let breadth of rectangle be b cm.

Then, length=142%of b=142b/100

diameter of circle= 77cm,

radius of circle,r =77/2 cm.

perimeter of rectangle = circumference of circle

$$\rightarrow2( length + breadth)=2\pi r$$

$$\rightarrow 2(\dfrac{142b}{100}+b) = 2\times\dfrac{22}{7}\times \dfrac{77}{2})$$

$$\rightarrow 2\times \dfrac{100b+142b}{100}=11\times22$$

$$\rightarrow \dfrac{242b}{100}=\dfrac{242}{2}$$

$$\rightarrow 242b=121\times100$$

$$\rightarrow b=\dfrac{12100}{242}$$

$$\rightarrow b=50cm$$

$$\rightarrow area of rectangle = length \times breadth$$

$$\rightarrow area of rectangle =\dfrac{142b}{100}\times b$$

$$\rightarrow area of rectangle =\dfrac{142\times50}{100}\times50$$

$$\rightarrow area of rectangle =3550cm^2$$