A fountain is enclosed by a circular fence of circumference 11m and is surrounded by a circular path. The circumference of the outer boundary of the path is 16m. A gardener increase the area of the path to decrease the area enclosed by the fence such that the length of the fence decrease by 3m. The path is covered by bricks which cost Rs.125 per m². What will be the total cost to the nearest whole number required to cover the area by the bricks.
Answers
Answer:
we have,
Circumference of the outer boundary of the path = 16m
=>2πr1 = 16 => 8/πm
and, length of decreased fence = 11 - 3 = 8m
2πr2 = 8 => r2 = 4/πm
.°. Area of path = π{r1^2 - r2^2} = { 64 - 16 / π^2 } =48/πm^2
.°. Total cost to cover the path by bricks
= ₹ 48/π × 125
= ₹ 48/22 × 7 × 125
= ₹ 1909.09
= ₹ 1910
we have,
Circumference of the outer boundary of the path = 16m
2πr1 = 16 => 8/πm
and, length of decreased fence = 11 - 3 = 8m
2πr2 = 8 => r2 = 4/πm
Area of path = π{r1^2 - r2^2} = { 64 - 16 / π^2 } =48/πm^2
Total cost to cover the path by bricks
= ₹ 48/π × 125
= ₹ 48/22 × 7 × 125
= ₹ 1909.09
= ₹ 1910