SOLUTION: A swimming pool is 50 m long and 25 m wide.It is 1 m deep at the shallow end and 8 m wide at the at other end.Find the volume of water in the pool when it is full and the total are

Click here to see ALL problems on Surface-area

Question 519238: A swimming pool is 50 m long and 25 m wide.It is 1 m deep at the shallow end and 8 m wide at the at other end.Find the volume of water in the pool when it is full and the total area of the pool which is in contact with the water ?
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
1 m shallow at one end and 8 m deep at other end. They are parallel sides.
(1/2 * (8+1)* 50)*25
4.5 *1250
5625 m^3 is the volume
You calculate the area of the trapezium and multiply by the width of the pool to get the volume.
length side area = 2 sides=> they are trapeziums
parallel sides are 1m & 8m perpendicular distance = 50
1/2 * (8+1)* 50=225 m^2
225*2 = 450 m^2
II side 25*1= 25m^2
III side25*8= 200m^2
bottom surface
The distance along the length of pool from 1m depth to 8m depth, length can be calculated by pythagoras theorem. This length is sqrt(50^2+7^2)=50.5 m
width = 25 m
Area of base = 50.5 *25= 1262 m^2
Total area = 450+25+200+1262 m^2
=1937m^2
m.ananth@hotmail.ca