Question

A piece of a wire in the form of rectangle of 8.9 cm long and 54 mm broad is bent in the form of a circle. Find the radius of the circle in mm.
Plz solve.​

Answer 1

GiveN:-

A piece of a wire in the form of rectangle of 8.9 cm long and 54 mm broad is bent in the form of a circle.

To Find:-

Find the radius of the circle in mm.

SolutioN:-

• First we have to find the perimeter of the rectangle.

We know,

$\large{\green{\underline{\boxed{\bf{Perimeter=2(Length+Breadth)}}}}}$

where,

• Length = 8.9 cm
• 1 cm = 10 mm
• 8.9 cm = 8.9 × 10
• 89.0 mm = 89 mm
• Length = 89 mm
• Breadth = 54 mm

Putting the values,

$\large\implies{\sf{Perimeter=2(89+54)}}$

$\large\implies{\sf{Perimeter=2\times143}}$

$\large\therefore\boxed{\bf{Perimeter=286\:mm.}}$

Now,

• It is said that a piece of a wire in the form of rectangle is bent in the form of a circle.
• So, perimeter of rectangle is equal to the circumference of the circle.
• Therefore, the circumference of the circle is 286 mm.

As we know the circumference of the circle we can find the radius of the circle.

We know that,

$\large{\green{\underline{\boxed{\bf{Circumference=2\pi\:r}}}}}$

where,

• Circumference = 286 mm
• π = 22/7
• r is radius = ?

Putting the values,

$\large\implies{\sf{286=2\times\dfrac{22}{7}\times\:r}}$

$\large\implies{\sf{286=\dfrac{44}{7}\times\:r}}$

$\large\implies{\sf{\dfrac{286\times7}{44}=r}}$

$\large\implies{\sf{\dfrac{2002}{44}=r}}$

$\large\implies{\sf{45.5=r}}$

$\large\therefore\boxed{\bf{Radius=45.5\:mm.}}$

VerificatioN:-

$\large\implies{\sf{Circumference=2\pi\:r}}$

$\large\implies{\sf{286=2\times\dfrac{22}{7}\times45.5}}$

$\large\implies{\sf{286=\dfrac{44}{7}\times45.5}}$

$\large\implies{\sf{286=\dfrac{2002}{7}}}$

$\large\implies{\sf{286=\dfrac{\cancel{2002}}{\cancel{7}}}}$

$\large\implies{\sf{286=286}}$

$\large\therefore\boxed{\bf{LHS=RHS.}}$

• Hence verified.

The radius of the circle in mm is 45.5 mm.