A rectangular swimming pool is 48 m long and 20 m wide. The shallow edge of the pool is 1 m deep.
For every 2.6 m that one walks up the inclined base of the swimming pool, one gains an elevation of 1 m. What is the volume of water (in cubic meters), in the swimming pool? Assume that the pool is filled up to the brim.v

For every 2.6 m that one walks along the slanting part of the pool, there is a height of 1 m that is gained.

=> $$\frac{AC}{BC} = \frac{2.6}{1}$$

=> $$AC = 2.6 \times BC$$

Also, dimensions of cuboidal part = $$48 \times 20 \times 1$$

 In $$\triangle$$ ABC

=> $$(AC)^2 = (AB)^2 + (BC)^2$$

=> $$(2.6 \times BC)^2 = (48)^2 + (BC)^2$$

=> $$6.76 (BC)^2 - (BC)^2 = 2304$$

=> $$(BC)^2 = \frac{2304}{5.76} = 400$$

=> $$BC = \sqrt{400} = 20$$ m

$$\therefore$$ Volume of water in the pool = Volume of cuboid + Volume of triangle

= $$(l \times b \times h) + (\frac{1}{2} \times AB \times BC) \times height$$

= $$(48 \times 20 \times 1) + (\frac{1}{2} \times 48 \times 20 \times 20)$$

= $$960 + 9600 = 10560 m^3$$