The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.
Answers
Answer:
Area Of The Running Track is 6216 m² and the length of the outer running track is 488 m .
Step-by-step explanation:
Given :
Inside perimeter of a running track = 400 m
Width of the running track, = 14 m
Length of the each straight portion = 90 m
Inside perimeter of the running track = 2 × length of straight portion + 2 × circumferences of a semicircles
400 = 2 × 90 + 2 × πr
400 = 180 + 2 × 22/7 × r
400 - 180 = 44r/7
220 = 44r/7
r = (220 × 7 ) / 44
r = 5 × 7 = 35 m
Radius of inner semicircle, r = 35 m
Radius of outer semicircle,R = radius of semicircle + width
R = 35 m + 14 m = 49 m
R = 49 m
Radius of outer semicircle,R = 49 m
Area of the running track ,A = 2 × Area of the rectangle + 2 × Area of outer semicircle - 2 × Area of inner semicircle
A = 2 × (l ×b) + 2 × ½ πR² - 2 × ½ πr²
A = 2 × (l ×b) + πR² - πr²
A = 2 × (l ×b) + π(R² - r²)
A = 2 × 90 × 14 + 22/7 (49² - 35²)
A = 2520 + 22/7 ×(2401 - 1225)
A = 2520 + 22/7 × 1176
A = 2520 + 25872/7
A = 2520 + 3696
A = 6216 m²
Area Of The Running Track = 6216 m²
Length of the outer running track, L = 2 × l + 2πR
L = 2 × 90 + 2 × 22/7 × 49
L = 180 + 308
L = 488 m
Length of the outer running track = 488 m
Hence, the length of the outer running track is 488 m
HOPE THIS ANSWER WILL HELP YOU….
Here are more questions of the same chapter :
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.
https://brainly.in/question/9469383
A square water tank has its side equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plot at Rs. 1.25 per square metre (Take π = 3.14).
https://brainly.in/question/9470130
Answer:
Step-by-step explanation:
Let radius of the inner semi-circular ends = r m.
Inner perimeter of the track = 400 m
90 + πr + 90 + πr = 400 (Circumference of the semi-circle = πr)
2πr = 220 m
r = 35 m
Area of the track = Area of the ring AEHD + Area of rectangle ABFE + Area of ring BFGC + Area of rectangle CDHG
= 3696 + 2520
= 6216 sq. m.
Length of the outer running track = EF + Length of arc FG + GH + Length is arc HE
= 90 + [π × (35 + 14)] + 90 + [ π × (35 + 14)]
= [2π × 49] + 180
= 308 + 180
= 488 m