Question

A wire is bent in the form of a square of side 27.5 cm . It is straightened and then bent into a circle. What is the radius of the circle so formed?
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Answer 1

Answer:

radius=17.5cm

Step-by-step explanation:

The length of wire is the perimeter of square.

length of wire= 4x side of square

length= 4x27.5

= 110

then it is rebent into circle.

length of wire is the circumference of circle

110=2πr

110=2x22/7xr

110/2=22/7xr

55x7/22=r

r=17.5cm

radius= 17.5cm

Answer 2

EXPLANATION -:

Given :

First Condition -

• A wire is bent a square
• Side of a square = 27.5 cm

Second Condition -

• The wire is straightened and again bent into a circle

To Find :

• Radius of the circle

Solution :

First we will find perimeter of a square

$\star \: \small \boxed{\rm{ Perimeter \: of \: a \: square = 4 × Side }}$

$\small\sf{ Perimeter = (4 × 27.5) cm = 110 cm }$

Perimeter = 110 cm

Radius of the circle

Let us assume the radius as 'r'

Circumference of the circle = 110 cm

$\star \: \small \boxed{\rm {Circumference \: of \: a \: circle = 2\pi r}}$

$\small\sf{110 = 2 \times \frac{22}{7} \times r }$

$\leadsto \small\rm{110 \times 7 = 2 \times 22 \times r}$

$\leadsto \small\rm{770 = 2 \times 22 \times r}$

$\leadsto\small\rm{ \dfrac{ \cancel{770}}{\cancel{2}} = 22 \times r}$

$\leadsto\small \rm{385 \: = 22 \times r}$

$\leadsto\small \rm{ \dfrac{ \cancel{385}}{ \cancel{22}} = r}$

$\leadsto\small \rm{r = 17.5 \: cm}$

Radius = 17.5 cm

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