a solid pyramid of height 40 cm with a square base of sides 30 cm each is put into a cubicle tank of sides 40 cm each. the tank is then filled with water. If

Volume of a pyramid = 1/3 x Area of base x height
= 1/3 x 900 x 40
= 12000 cm^3 

Volume of tank = length x breadth x height
= 40 x 40 x 40
= 64000 cm^3 
64000 - 12000 = 52000 cm^3 water left 
52000 cm^3 / 40 x 40 = h
h = 32.5cm

So, the depth of water in the tank is 32.5 cm. Hopefully that helped!

Hello Guest!

a solid pyramid of height 40 cm with a square base of sides 30 cm each is put into a cubicle tank of sides 40 cm each. the tank is then filled with water. If the pyramid is removed, find the depth of water in the tank.

V_Tank = $40^3 cm^3=64000cm^3$

V_Py = $\frac{1}{3}\times 30^2cm^2\times 40cm=12000cm^3$

V_Rest = V_Tank - V_Py = $64000cm^3-12000cm^3=52000cm^3$

h = V_Rest / A_Tank = $\frac{52000cm^3}{40^2cm^2}{\color{blue}=32,5cm}$

${\color{blue}After \ removing \ the \ pyramid \ from \ the \ tank, }$

${\color{blue}the \ water \ is \ 32.5cm \ high \ in \ the \ tank.}$

Greeting asinus :- )   !