Question

a wire in the form of rectangle 18.7 cm long and 14.3 wide in reshaped and bent into the form of a circle find the radius of the circle so formed​

Answer 1

AnswEr :

• CONCEPT⠀BEHIND⠀THIS :

◗ Wire is in the form of Rectangle and, then reshaped into the form of a Circle.

◗ Length of wire won't change ever, that's why Perimeter of Rectangle and, Circumference of the Circle so formed will be Equal.

$\rule{100}{2}$

⋆ Refrence of Image is in the Diagram :

• Length of Rectangle = 18.7 cm
• Breadth of Rectangle = 14.3 cm

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(7,2){\mathsf{\large{14.3 cm}}}\put(7.7,1){\large{B}}\put(9.2,0.7){\matsf{\large{18.7 cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,3){\large{D}}\end{picture}$

• Perimeter of the Rectangle :

$\longrightarrow \sf Perimeter = 2(Length + Breadth) \\ \\\longrightarrow \sf Perimeter = 2(18.7\:cm + 14.3\:cm) \\ \\\longrightarrow \sf Perimeter = 2 \times 33\:cm \\ \\\longrightarrow \green{\sf Perimeter = 66 \:cm}$

$\rule{200}{1}$

• Circumference of the Circle so formed :

$\implies \sf Circumference = 2\pi r \\ \\\implies \sf Perimeter = 2\pi r \\ \\\implies \sf 66 \:cm = 2 \times \dfrac{22}{7} \times r \\ \\\implies \sf \cancel\dfrac{66 \:cm \times 7}{2 \times 22} = r \\ \\\implies \boxed{\red{\sf Radius = 10.5 \:cm}}$

⠀

∴ Radius of Circle so formed is 10.5 cm

$\rule{200}{2}$

• S H O R T C U T⠀T R I C K :

Once you got to know the Perimeter i.e. Circumference for the Circle. we can simply find Radius from this following table :

$\begin{tabular}{|c |c | c|}\cline{1-3}Radius & Circumference & Area \\\cline{1-3}3.5 & 22 & 38.5 \\7 & 44 &154\\10.5 & 66&346.5 \\14 &88&616\cline{1-2}\cline{1-3}\end{tabular}$

The Circle whose Circumference is 66 cm, Radius will be 10.5 cm.

⠀

∴ Radius of Circle so formed is 10.5 cm

Answer 2

❏ Question:-

A wire in the form of rectangle 18.7 cm long and 14.3 wide in reshaped and bent into the form of a circle find the radius of the circle so formed.

❏ Solution:-

•Given➔

length (l)=18.7 cm

breadth (b)=14.3 cm

•To find➔

Radius of the circular ring or wire=?

•Ans:-

$\therefore$ Perimeter of the rectangular wire is

=2×(length+breadth)

=2×(18.7+14.3) cm

=(2×33) cm.

=66 cm

Now, according the question the wire is reshaped in the form of a circle,

Let, the radius of the circular wire (reshaped) is

= r cm.

$\therefore$ Perimeter of the circular wire

=2πr cm

$\therefore$ According to the given condition,

$\sf\longrightarrow$ perimeter of the rectangular wire = perimeter of the circular wire.

$\sf\longrightarrow 2 \pi r=66$

$\sf\longrightarrow 2 \times\frac{22}{7}\times r=66$

$\sf\longrightarrow r=66\times\frac{7}{22\times2}$

$\sf\longrightarrow r=\cancel{66}\times\frac{7}{\cancel{22}\times2}$

$\sf\longrightarrow r=\frac{7\times\cancel3}{\cancel2}$

$\sf\longrightarrow r=(7\times1.5)\:\:cm$

$\sf\longrightarrow \boxed{\large{\red{r=10.5 \:\:cm}}}$

$\bf\therefore$ Radius of the circle= 10.5 cm.

━━━━━━━━━━━━━━━━━━━━━━━

❏ Useful Formulas:-

✦CIRCLE✦

❚ For a Circle of diameter r ,

(1)Area is given by,

$\sf\longrightarrow \boxed{Area=\pi r{}^{2}}$

(2) Circumference is given by,

$\sf\longrightarrow\boxed{ Circumference=2\pi r=\pi d}$

✦RECTANGLE✦

For a rectangle of length l and breadth b,

$\sf\longrightarrow \boxed{Area=length\times breadth}$

$\sf\longrightarrow \boxed{Perimeter=2\times(length+Breadth)}$

$\sf\longrightarrow \boxed{Diagonal=\sqrt{length{}^{2}+Breadth{}^{2}}}$

✦SQUARE✦

For a square of side a ,

$\sf\longrightarrow \boxed{Area=Side{}^{2}}$

$\sf\longrightarrow \boxed{Perimeter=4\times side}$

$\sf\longrightarrow \boxed{Diagonal=\sqrt{2}\times side}$

✦TRIANGLE✦

Triangle of sides a, b and c

$\sf\longrightarrow \boxed{Perimeter=a+b+c}$

$\sf\longrightarrow \boxed{S=\frac{a+b+c}{2}}$

[Where, S= Half of Perimeter]

$\sf\longrightarrow \boxed{Area=\sqrt{S(S-a)(S-b)(S-c)}}$

[ using Heron's Formula]

✰For a Right angled triangle of base b and height h,

$\sf\longrightarrow \boxed{Area=\frac{1}{2}\times base\times height}$

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$