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Question 62
The surface of the water in a swimming pool is a rectangle 26 m long and 10 m wide and the depth of water increases uniformly from 1.6 m at one end to 4.4 m at the other end. Volume (in m$$^3$$ ) of the water in the pool is
Solution
The given shape will essentially have one cuboid with dimensions 26*10*1.6 and a right-angle triangular shape with a height of 26 m, a base of 10 m, and a width of 2.8 m.
Now, the volume of the cuboid is 26*10*1.6 $$m^3$$, which is 416$$m^3$$.
The volume of the triangular shape is (Area)*(width), which is 1/2*26*10*2.8= 364$$m^3$$.
Adding the two, we get a total volume of 780$$m^3$$.
