The inside perimeter of a running track as shown in the adjoining figure is `340m`. The length of each straight portion is `60m` and the curved portion are semi-circles. If the track is `7m` wide, find area of te track. (Use `pi=22/7`)

The inside perimeter of a running track as shown in the adjoining figure is 340m . The length of each straight portion is 60m and the curved portion are semi-circles. If the track is 7m wide, find the outer perimeter of the track. (Use pi=22/7 )

The inside perimeter of a running track shown in the figure is 400m . The length of each of the straight portion is 90m and the ends are semi-circles. If the track is 14m wide everywhere, find the length of outer boundary of the track. ( pi=22/7 )

In Fig12.26 depicts a racing track whose left and right ends are semicircular.The distance between the two inner parallel line segments is 60 and they are each 106 m long.If the track is 10 m wide,find:(i)the distance around the track along its inner edge

A 7 m wide path runs round a circular garden whose circumference is 352m.Find the area of the path.

Find the perimeter and area of shaded region where ADC, AEB and BFC are semi-circles on the diameter AC, AB and BC respectively. ( pi=22/7 )

Find the area of the segment of a circle of radius 14 cm if the length of the corresponding arc APB is 22 cm.[use pi = 22/7 ]

In the given figure ABC is a right angled triangle where /_A=90^@ . AB=21cm and AC=28cm . Semicircles are described on AB, BC and CA as diameter. Find the area of the shaded portion. ( pi=22/7 )

A horse is tied to one corner of a field which is if the shape of an equilateral triangle of side 12 m . If the length of the rope is 7 m, find the area of the field which the horse cannot graze. ( pi = 3.14 , sqrt 3 = 1.732 )

In Fig 13,ABCD is a square of side 5.6 cm.As shown in the figure,four circles are drawn with centres A,B,C and D respectively.Find the area of the shaded region in the diagram. (pi=22/7)

In the adjoining figure the boundary of the shaded region consist of four semi-circular are, two smallest being equal. If diameter of the largest is 14cm and that of the smallest is 3.5cm . Calculate the area of the shaded region. (Use pi=22/7 )