LENGTH OF ARC GRADE 11 WORD PROBLEMS

Problem 1 :

What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?

Solution :

Distance covered by athlete = 1 km 

5 ⋅ 2πr = 1 km 

5 ⋅ 2 ⋅ (3.14)  r = 1000 m

r = 31.84 m

Problem 2 :

In a circle of diameter 40 cm, a chord is of length 20 cm. Find the length of the minor arc of the chord.

Solution :

Let arc AB  =  s. 

It is given that OA  =  20 cm and chord AB  =  20 cm.

Hence triangle OAB is an equilateral triangle.

<AOB  =  60 degree (or) (π/3) radians

 θ = arc length/radius

(π/3)  =  s/20

s  =  20π/3

π  =  3.14

s  =  20(3.14)/3  =  20.93

So, the required arc length is 20.93 cm.

Problem 3 :

Find the degree measure of the angle subtended at the center of circle of radius 100 cm by an arc of length 22 cm.

Solution :

Radius of the circle  =  100 cm

Length of arc  =  22 cm

 θ = arc length/radius

  =  22/100

Radian measure  =  (π/180) Degreee measure

22/100  =  (22/7)/180 ⋅ Degree measure

Degree measure =  (22 ⋅ 7 ⋅ 180)/(22 ⋅ 100)

=  126/10

12 (6/10)

Multiplying 6/10 by 60, we get 36'

So, the required angle is 12 36'

Problem 4 :

What is the length of the arc intercepted by a central angle of measure 41 in a circle of radius 10 ft?

Solution :

arc length   =   (θ/360) ⋅ 2 ⋅ π  r

=  (41/360) ⋅ 2 ⋅ (3.14)  r

=  7.16 feet

So, the required arc length is 7.16 feet.

Problem 5 :

If in two circles, arcs of the same length subtend angles 60 and 75 at the center, find the ratio of their radii.

Solution :

Let r1 and r2 be the radii of the given circles and let their arcs of same length subtend angles of 60 degree and 75 degree at their centers.

60◦  =  60 ⋅  (π /180) 

   =  π/3 

 θ = arc length/radius

π/3 = s/r

s  =  πr/3  -----(1)

75◦  =  75 ⋅ (π /180) 

   =  5π/12 

 θ = arc length/radius

5π/12 = s/r2

s  =  5πr2/12  -----(2)

(1)  =  (2)  

πr/3  =  5πr/12

r1/r=   (5π/π)(3/12)

r1/r2  =   5/4

r1 : r =  5 : 4

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