A,B and C are three unequal faces of a rectangular tank. the tank contains a certain amount of water. when the tank is based on the face A,the height of the water is half the height of the tank.the dimensions of the side B are 3ft x 4ft and the dimensions of the side C are 4ft x 5ft.what is the measure of the height of the water in the tank in feet?
a)2 b)3 c)4 d)5
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Dimensions of face B = 3' * 4'
Dimensions of face C = 4' * 5'
The common side /edge between the two faces = 4'
Hence volume of the tank is = 3' * 4 ' * 5' = 60 cubic feet
Dimensions of face A of the tank = 3' * 5' = 15 sq ft
Height of full tank with face A as the base = 60/15 = 4 feet
Water level = 4'/2= 2 feet
Dimensions of face C = 4' * 5'
The common side /edge between the two faces = 4'
Hence volume of the tank is = 3' * 4 ' * 5' = 60 cubic feet
Dimensions of face A of the tank = 3' * 5' = 15 sq ft
Height of full tank with face A as the base = 60/15 = 4 feet
Water level = 4'/2= 2 feet
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Given :-
Dimensions of face B = 3ft x 4ft
Dimensions of face C = 4ft x 5ftft
Common side between the two faces = 4ft
So volume of the tank is = 3ft × 4ft × 5ft = 60 cubic feet
Dimensions of face A of the tank = 3ft × 5ft = 15 sq ft
Height of tank with face A( as the base ) = 60/15 = 4ft
Water level = 4ft/2= 2ft.
Dimensions of face B = 3ft x 4ft
Dimensions of face C = 4ft x 5ftft
Common side between the two faces = 4ft
So volume of the tank is = 3ft × 4ft × 5ft = 60 cubic feet
Dimensions of face A of the tank = 3ft × 5ft = 15 sq ft
Height of tank with face A( as the base ) = 60/15 = 4ft
Water level = 4ft/2= 2ft.
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