Question

shazli took a wire of length 44 cm and bent it into the shape of a circle.find it's radius.also find the area. if the same wire is bent into the shape of a square what will be the length of each of its side?which figure encloses more area ,the circle or square ​

Answer 1

Total length of the wire = 44cm

The circumference of the circle = 2πr = 44cm

$2 \times \frac{22}{7} \times r = 44$

$r \times \frac{44 \times 7}{2 \times 22} \: cm$

Now, area of the circle,

$\pi {r}^{2} = \frac{22}{7} \times 7 \times 7 = 154 {cm}^{2}$

Answer 2

★ Correct Question

Shazli took a wire of length 44 cm and bent it into the shape of a circle.Find the radius of that Circle.Also find its Area.If the same wire is bent into the shape of a square,what will be the length of each of its sides ?Which figure of encloses more Area,the circle of the square ? (Take π = 22/7)

★ Concept

In this question,We have given that shazli took a wire of length 44 cm and bent it into the shape of a circle.So,We have to Find the Radius of that Circle.And Also we have to Find Its Area.Then We have given that If the same wire is bent into the shape of a square,What will be the length of each of its sides.Which Figure of encloses more Area,the circle of the square.We have to Take π = 22/7 to solve this question.So,Firstly,We have to take Circumference of Circle.Then We have to Multiply 44 × 2 × 22/7 × r.Then We will cut 44 × 7/2 × 22 = r.Then We get 7 cm.Then we have to Find out the Area of circle.After finding the Area of circle we get 154 cm².Then We will find the same wire bent into the shape of square.Then We have to Find the Perimeter of square.After Finding the Perimeter of Square We get 11 cm.Then We have to Find the Area of square.After Finding the Area of square.We get 121 cm².

Now,

Let's Solve!

★ Solution

The length of wire is 44 cm αnd it is bent in the shαpe of circle.

We know thαt,

$\sf{Circumference \ of \ Circle = 2πr}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 44 = 2 × \dfrac{22}{7} × r}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { { \cancel\dfrac{44×7}{2×22} = r}}}$

$\sf{r(Radius) = 7 cm}$

$\sf{Area \ of \ Circle \ = πr^2}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 22/{ { \cancel{7} ×{ { \cancel 7 × 7}}}}}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 22×7 = 154 cm^2}$

$\sf{Now, \ the \ same \ wire \ bent \ into \ the \ shape \ to \ square.}$

$\sf{So,Perimeter \ of \ square \ = 44 cm}$

$\sf{Perimeter \ of \ square \ = 4 × side}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{44}{4} = 4 × side}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { { \cancel{ \dfrac{44}{4}}=side}}}$

$\sf{Side = 11 cm}$

$\sf{Area \ of \ square \ = side × side}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 11×11}$

$\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =121cm^2}$

${\underline{\bf{Hence, \ Circle \ encloses \ more \ Area.}}}$