A stadium has a a central rectangular area 125m long by 80m wide.there are two semi circular ends.a running track goes all the w
1 answer:
a)The total inside length of the curved ends of the track will be 25.13 meters
b)The total distance around the inside of the track will be 275.13 meters
c)The area inside the track is 10050 square meters.
<h3>What is an area?</h3>The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the circle in a two-dimensional plane is called the area of the circle.
Given that:-
A stadium has a central rectangular area 125m long by 80m wide. there are two semi-circular ends. a running track goes all the way around.
a) what is the total inside length of the curved ends of the track?
The length of the curved track means it is a perimeter of the circle so the radius of the semicircle will be 4 meters.
So the length will be = π r + πr = 2 πr
= 2 x π x 4 = 25.13
b) what is the total distance around the inside of the track?
The total distance of the track will be the sum of the length of the two semicircles and the longer side of the rectangle.
Total length = 125 + 125 + 25.13
= 275.13 meters
c) what is the area inside of the track?
The area will be equal to the area of the two semicircular tracks and the rectangular track.
Total Area = Area of semicircle + Area of semicircle + Area of rectangle
= ( π/2 ) r² + (π/2) r² + ( L x W )
= π r² + ( L x W)
= π ( 4 )² + ( 125 x 8 )
= 10050.26 square meters.
Therefore the total inside length of the curved ends of the track will be 25.13 meters. The total distance around the inside of the track will be 275.13 meters. The area inside the track is 10050 square meters.
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Answer:
(a)2(89x+84)
(b)
Step-by-step explanation:
The dimensions of the larger rectangular field are:
- Length=5x + 12; Width = 9x + 14.
The dimensions of the smaller rectangular soccer field are:
- Length=5x; Width = 9x.
(a)Area of the part of the field that is outside the soccer field
=Area of the larger rectangular field - Area of the Soccer Field
=(5x+12)(9x+14)-5x(9x)
=(5x)(9x)+70x+108x+168-5x(9x)
=178x+168
=2(89x+84)
(b)Radius of the Semicircular Fountain =2x
From Part (a),
Area of the larger rectangular field - Area of the Soccer Field=178x+168
Area of the Semicircular Fountain =
Area of the Field that does not include the soccer field or the fountain.
=Area of the larger rectangular field - Area of the Soccer Field-Area of the Semicircular Fountain