Answer:
Area of the portion outside the circle but inside the square is reduced by 114 m² after the third day, reduced by 239.14284 in the fourth day.
Step-by-step explanation:
First day
Radius of the circle = 3 meters
Length of sides of the square = 21 meters
Area of the circle = π R2 = 22/7 x 32 = 28.285714 m²
Area of the square = L x B = 21 x 21 = 441 m²
Area of area outside the circle but inside the square = 412.71429 m²
Second day
Radius of the circle = 8 meters
Length of sides of the square = 25 meters
Area of the circle = π R2 = 22/7 x 82 = 201.14286 m²
Area of the square = L x B = 25 x 25 = 625 m²
Area of area outside the circle but inside the square = 423.85714 m²
Third day
Radius of the circle =13 meters
Length of sides of the square = 29 meters
Area of the circle = π R2 = 22/7 x 132 = 531.14286 m²
Area of the square = L x B = 29 x 29 = 841 m²
Area of area outside the circle but inside the square = 309.85714 m²
Fourth day
Radius of the circle =18 meters
Length of sides of the square = 33 meters
Area of the circle = π R² = 22/7 x 182 =1018.2857 m²
Area of the square = L x B = 33 x 33 = 1089 m²
Area of area outside the circle but inside the square = 70.7143 m²
From the second and third day, it can be noticed that the area of the portion outside the circle but inside the square is reducing changing by half.
Answer:
The area outside the circle but inside the square is changing at the speed of (168 - 30π)m^2/day or 73.8m^2 per day
Step-by-step explanation:
Let us denote the radius of the circle by "r" and the length of a side of the square by "L".
Then given that the rate of change of the radius of the circle is 5 metres per day;
then dr/dt = 5
Again, given that the rate of change of a side of the square is 4 metres per day;
then dL/dt = 4
Let the area outside the circle but inside the square be "A".
Therefore, A = (area of the square) - (area of circle)
= L^2 - πr^2
We are now required to calculate dA/dt If the initial radius of the circle was 3 metres and each side of the square was 21 metres initially.
i.e find dA/dt
When r = 3 and L = 21
dA/dt = 2L(dL/dt) - 2πr(dr/dt)
= [2×(21)×(4)] - [2π×(3)×(5)]
= (168 - 30π)m^2/day
OR
If π = 3.14
Then: 168 - (30 × 3.14)
=168 - 94.2
= 73.8m^2 per day
The area outside the circle but inside the square is changing at the speed of 73.8m^2 per day
