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- In second quadrant sin and cosec are positive
Refer to the figure shown below.
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²