Question

A wire forming a rectangle of dimensions 9 cm and 6.4 cm, is stretched and bent again to form a circle. What would the diameter of the circle be?

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Answer 1

Answer:

Perimeter of rectangle = 2(l+b)

= 2( 9+6.4)

= 2 x 15.4

= 30.8 cm

then perimeter or circumference of circle is also 30.8cm

Circumference = 2πr

30.8 = 2 x 22/7 x r

30.8 = 44/7 x r

30.8 x 7/44 = r

4.9cm = r

Diameter = 2r

= 2 x 4.9

= 9.8 cm

Answer 2

Given:

• Length (l) = 9 cm
• Breadth (b) = 6.4 cm

To Find:

• What would the diameter of the circle be?

Solution:

$\\ \bigstar{\underline{\boxed{\tt\large{ \green{ Perimeter_{(Rectangle)} } = 2(l + b) }}}}$

Where

• l = Length
• b = Breadth

After substituting values,

$\implies$ p = 2(l + b)

$\implies$ p = 2(9 + 6.4)

$\implies$ p = 2( 15.4 )

$\implies$ p = 2 × 15.4

$\implies$ p = 30.8 cm

Hence,

• The Perimeter of the Rectangle is 30.8 cm

_________________________

The Rectangular Wire is re-drawn into Circular Shape.

So,

• Perimeter of Rectangle = Perimeter of Circle
• 30.8 cm = 30.8 cm

$\\ \bigstar{\underline{\boxed{\tt\large{ \purple{ Perimeter \ Or \ Circumference_{(Circle)} } = 2πr }}}}$

$\implies$ 2πr

$\implies$ 30.8 = 2 × 22/7 × r

$\implies$ 30.8 × 7/2 × 22 = r

$\implies$ r = 4.9 cm

D = 2r

D = 2 × 4.9

D = 9.8 cm

Hence,

• The Diameter of the Circle is 9.8 cm.