Ex 12.3, 8
Given figure depicts a racing track whose left and right ends are semi-circular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :
the distance around the track along its inner edge
It is given that two inner line segments are parallel,
So, EF II HG
Here, EF = AB = 106 m
HG = DC = 106 m
EH = FG = 60 m
AE = BF = HD = GC = 10 m
Distance around the track along inner edge
= Distance EF + Distance HG
+ Circumference of semicircle EH with centre O
+ Circumference of semicircle FG with centre O
Semicircle EOH with centre O
Diameter = EH = 60 m
So, radius = /2 = 60/2 = 30
Circumference of semicircle EOH = 1/2 Circumference of circle
= 1/2 (2 )
= 1/2 2 22/7 30
= 660/7 m
So, Circumference of semi circle EOH = 660/7 m
Similarly ,
As radius is same for both semi-circles EOH & FO G are same
Circumference of semi circle FO G = Circumference of semi circle EOH = 660/7 m
Distance around the track along inner edge
= Distance EF + Distance HG
+ Circumference of semicircle EH with centre O
+ Circumference of semicircle FG with centre O
= 106 + 106 + 660/7+660/7
= 212 + 660/7+660/7
= (7 212 + 660 + 660)/7
= (1484 + 1320)/7
= 2804/7 cm2
Hence , distance around track = 2804/7 cm2
Ex 12.3, 8
Given figure depicts a racing track whose left and right ends are semi-circular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :
(ii) the area of the track
Area of track
= Area ABFE + Area HGCD
+ (Area of semicircle with diameter AD
Area of semicircle with diameter EH)
+ (Area of semicircle with diameter BC
Area of semicircle with diameter FG)
Area ABFE
ABFE is a rectangle
with length = 106 m
& breadth = 10 m
Area of ABFE = Length Breadth
= 106 10
= 1060 m2
Area of track
= Area ABFE + Area HGCD
+ (Area of semicircle with diameter AD
Area of semicircle with diameter EH)
+ (Area of semicircle with diameter BC
Area of semicircle with diameter FG)
= (1060) + (1060) + ((17600 9900)/7) + ((17600 9900)/7)
= 2 1060 + 2((17600 9900)/7)
= 2120 + 2 ( 7700/7)
= 2120 + 2 (1100)
= 2120 + 2200
= 4320 m2

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.