Question

A wire is bent in form of a rectangle having length twice the breadth . The same wire is bent in form of a circle . It is found that of the area of the circle is greater than that of the rectangle by 104.5 cm square. Find the length of the wire.​

Answer 1

Answer:

66 cm

Step-by-step explanation:

let the breadth of rectangle be x

length of rectangle is 2x

perimeter of rectangle = 2(l + b)

= 2(2x + x)

= 2(3x)

= 6x

perimeter of rectangle is equal to the perimeter of circle

2πr = 6x

$2\pi \: r = 6x \\ 2 \times \frac{22}{7} \times r = 6x \\ r = 6x \times \frac{7}{22 \times 2} \\ r = \frac{21x}{22}$

area of rectangle = l × b

= x × 2x

= 2x^2

area of circle = πr^2

area of circle is 104.5 sq cm more than area of square

$\pi {r}^{2} = 2 {x}^{2} + 104.5 \\ \frac{22}{7} \times {( \frac{21x}{22}) }^{2} = 2 {x}^{2} + 104.5 \\ \frac{22}{7} \times \frac{21 \times 21 \times {x}^{2} }{22 \times 22} = 2 {x}^{2} + 104.5 \\ \frac{63 {x}^{2} }{22} = 2 {x}^{2} + 104.5 \\ 63 {x}^{2} = 22(2 {x}^{2} + 104.5) \\ 63 {x}^{2} = 44 {x}^{2} + 2299 \\ 63 {x}^{2} - 44 {x}^{2} = 2299 \\ 19 {x}^{2} = 2299 \\ {x}^{2} = \frac{2299}{19} \\ {x}^{2} = 121 \\ x = \sqrt{121} \\ x = 11$

length of rectangle = 2x = 2 × 11 = 22 cm

breadth of rectangle = x = 11 cm

length of the wire = perimeter of rectangle

= 2(l + b)

= 2(22 + 11)

= 2 × 33

= 66 cm

therefore the length of wire = 66 cm

hope you get your answer