A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.

$\frac{2 π r}{2} + d = 108$

Area enclosed $= \frac{π r^{2}}{2}$

Open in App

Solution

Lenth of wire =108 cm

Let r be the radius of the semicircle =r

$∴ π r + 2 r = 108$

$\Rightarrow r (π + 2) = 108 \Rightarrow r (\frac{22}{7} + 2) = 108$

$\Rightarrow \frac{36}{7} r = 108 \Rightarrow r = \frac{108 \times 7}{36} = 21 c m$

Area $= \frac{π r^{2}}{2} = \frac{22}{7 \times 2} = 21 \times 21 = \frac{1386}{2} c m^{2}$

$= 693 c m^{2}$

Suggest Corrections

Similar questions

Q. A piece of wire of length 108 cm is bent to form a semicircle arc bounded by its diameter. Find its radius and area enclosed.

Question 108

Priyanka took a wire and bent it to form a circle of radius 14 cm. She bent it into a rectangle with one side 24 cm long. What is the length of the wire? Which figure encloses more area, the circle or the rectangle?

Q. A copper wire when bent in the form of a square encloses an area of

484 c m^{2}

. If the same wire is bent into the form of a circle, find the area enclosed by it.

Q. A wire when bent in the form of an equilateral triangle encloses an area of

121 \sqrt{3} {cm}^{2}

. The same wire is bent to form a circle. Find the area enclosed by the circle.

Q. A wire when bent in the form of a square encloses an area

=

576 cm

^{2}

. Find the largest area enclosed by the same wire when bent to form an equilateral triangle.

Join BYJU'S Learning Program

A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed. 2πr2+d=108 Area enclosed =πr22

Lenth of wire =108 cm Let r be the radius of the semicircle =r ∴πr+2r=108 ⇒r(π+2)=108⇒r(227+2)=108 ⇒367r=108⇒r=108×736=21cm Area =πr22=227×2=21×21=13862cm2 =693cm2

A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.

$\frac{2 π r}{2} + d = 108$

Area enclosed $= \frac{π r^{2}}{2}$