Question

In fig, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and angle AOB = 120°. Find the length of OAPBO.

Answer 1

SECTOR OF A CIRCLE:

The region enclosed by two radii and the corresponding arc of a circle is called the sector of a circle. The sector containing minor Arc is called minor sector and the sector containing major arc is called major sector. Angle of minor sector is less than 180° and Angle of major sector is more than 180°.The sum of angles of major and minor sector is  360°.

SOLUTION:

GIVEN :
Minor Sector angle (θ) = 120°
Radius of Circle =OA = OB =r =  3.5 cm

Major sector angle = 360 -120 = 240°

From the given figure, OAPB is a the major sector .
We have to find the Length of OAPBO, so θ for major Arc =240°

If the radius of a circle is r and length of the arc is l, then

Length of the arc, OAPBO (l) = (θ /180) × πr
l  =(240/180) × (22/7) × 3.5
l  = 4/3 × 22 × .5
l = (88×.5 )/3 = 44/3 = 14.67

l = 14.67 cm

Length of  OAPBO = Length of the arc (OAPBO) + OA +OB
= 14.67 + 3.5 +3.5
= 14.67 + 7
= 21.67 cm

Hence, the Length of  OAPBO is 21.67 cm.

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, r = 3.5 cm and ∠AOB = 120°

Length of OAPBO = Circumference of the circle + 2(r) - Length of the arc

$\sf =2\pi r+2r-\dfrac{\pi r \theta }{180}$

$\sf = r(2\pi+2-\dfrac{\pi \theta}{180})$

$\sf = 3.5(2\times\dfrac{22}{7}+2-\dfrac{\dfrac{22}{7}\times120}{180})$

$\sf =3.5(\dfrac{44}{7}+\dfrac{2}{1}-\dfrac{44}{21})$

$\sf =3.5(\dfrac{132+42-44}{21})$

$\sf =\dfrac{3.5\times130}{21}$

$\sf =\dfrac{65}{3}$

$\boxed{\bf Length \: of \: OAPBO \: =21.67 \: cm}$

Hence, the length of OAPBO is 21.67 cm.