Question

a wire in the form of square with side 22cm. it is reshaped and bent into the form of circle. calculate the area of this circle

Answer 1

Answer:

616 cm²

Step-by-step explanation:

No matter how many times it gets shaped(re-shaped), total length of the wire would never change.

Here,

length of the wire = perimeter of the square

Similarly, length of wire = circumference of circle

 Using,

   Perimeter of square = 4side

   Circumference of circle = 2πr

⇒ 4(22m) = 2πr

⇒ 88 = 2(22/7) r

⇒ 14 = r

 Hence,

area of circle = πr² = (22/7) (14)²

                      = 616 cm²

Answer 2

${\large{\sf{\pmb{\underline{Understanding \; the \; question...}}}}}$

★ This question says that we have to find out the area of the circle who is been reshaped and it is a square first but while bending it it come in form of circle. That wire was in the form of square(as we know) with the side of 22 cm. Let's do this question!

${\large{\sf{\pmb{\underline{Given \; that...}}}}}$

★ A wire in the form of square with side 22cm.

★ Square shaped wire is reshaped and bent into the form of circle.

${\large{\sf{\pmb{\underline{To \; find...}}}}}$

★ Area of the circle.

${\large{\sf{\pmb{\underline{Solution...}}}}}$

★ Area of the circle = 616 cm²

${\large{\sf{\pmb{\underline{Using \; concepts...}}}}}$

★ Formula to find perimeter of square.

★ Formula to find perimeter/circumference of the circle.

★ Formula to find area of circle.

${\large{\sf{\pmb{\underline{Using \; formulas...}}}}}$

★ Perimeter of square = 4 × a

★ Perimeter/circumference of circle = 2πr

★ Area of circle = πr²

${\large{\sf{\pmb{\underline{Where,}}}}}$

★ a denotes side of square.

★ π is pronounced as pi

★ The value of π is 22/7 or 3.14

★ r denotes radius.

${\large{\sf{\pmb{\underline{Full \; Solution...}}}}}$

~ Firstly let us find the perimeter of the square by using formula that is given below :-

${\small{\boxed{\sf{\rightarrow Perimeter \: of \: square \: = 4 \times a}}}}$

$:\implies \sf Perimeter \: of \: square \: = 4 \times a \\ \\ :\implies \sf Perimeter \: of \: square \: = 4 \times 22 \\ \\ :\implies \sf Perimeter \: of \: square \: = 88 \: cm \\ \\ \bf \underline {Henceforth, \: 88 \: cm \: is \: perimeter \: of \: square}$

~ Now let's see what to do : Now as it's given that "Square shaped wire is reshaped and bent into the form of circle" means perimeter of square is equal to perimeter of circle here! Henceforth,

${\small{\boxed{\sf{\rightarrow Perimeter \: of \: circle \: = 2 \pi r}}}}$

$:\implies \sf Perimeter \: of \: circle \: = Perimeter \: of \: square \\ \\ :\implies \sf 2 \pi r \: = 88 \: cm \\ \\ :\implies \sf 2 \times \dfrac{22}{7} \times r \: = 88 \\ \\ :\implies \sf r \: = \dfrac{88 \times 7}{2 \times 22} \\ \\ :\implies \sf r \: = 14 \: cm \\ \\ :\implies \sf Radius \: = 14 \: cm \\ \\ \bf \underline {Henceforth, \: 14 \: cm \: is \: radius \: of \: circle}$

~ Now let's find area of circle by using the given formula!

${\small{\boxed{\sf{\rightarrow Area \: of \: circle \: = \pi r^{2}}}}}$

$:\implies \sf Area \: of \: circle \: = \pi r^{2} \\ \\ :\implies \sf Area \: of \: circle \: = \dfrac{22}{7} \times 14^{2} \\ \\ :\implies \sf Area \: of \: circle \: = \dfrac{22}{7} \times 14 \times 14 \\ \\ :\implies \sf Area \: of \: circle \: = 616 \: cm^{2} \\ \\ \bf \underline {Henceforth, \: 616 \: cm^{2} \: is \: the \: area \: of \: circle}$

${\large{\sf{\pmb{\underline{Additional \; knowledge...}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$