Mathematics College

You want to install a 1 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the​ walk? Use 3.14 for pi π.

Answers

Answer 1

Complete Question:

You want to install a 1 yd wide walk around a circular swimming pool. The diameter of the pool is 23 yd. What is the area of the​ walk? Use 3.14 for pi π.

Answer:

75.36 square yard

Step-by-step explanation:

From the question,

The diameter of this circular pool inside is 23 yd.

This means that the radius = Diameter/2 = 23yd/2 = 11.5 yd.

The formula for the area of a circle =

A = πr²

A = π(11.5)²

A =3.14 × 11.5²

A = 415.265 yd²

This is the Area of the inner circle.

We were told in the question also that he wants to install a walk of 1 yard

Hence, the radius of outer circle =

radius of inner circle +length of the walk

11.5yard + 1 yard

= 12.5 yard

A = πr²

A = 3.14 × (12.5)²

A = 490.625yd²

Area of the walk = Area of the Outer circle - Area of the inner circle

= (490.625 - 415.265)yd = 75.36 yd²

Therefore, the area of the walk is 75.36 square yards.

Related Questions

Cheryl bought 3.4 pounds of coffee that cost $6.95 per pound . How many did she spend on coffee

Answers

Answer:

23.63

Step-by-step explanation:

multiply the cost by the pounds

Answer:

$23.63

Step-by-step explanation:

3.4 X 6.95 = 23.63

A scooter runs 40 km using 1 litre of petrol tje distance covered by it using 15/4 litres of petrol is

Answers

Answer:

150 km

Step-by-step explanation:

1 liter ............ 40 km

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x = 15/4×40/1 = 600/4 = 150 km

A 12 sided die is rolled the set of equally likely outcomes is 123 456-789-10 11 and 12 find the probability of rolling a number greater than three

Answers

Answer:

6

Step-by-step explanation:

nerd physics

CAN SOMEONE HELP ME ASAP







A. 5
B. 53‾√53
C. 10
D. 103√3

Answers

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 30 = n/ 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

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