3 cubes of metal with edges 3cm,4cm and 5cm are melted to form a single cube. what is the ratio of the TSA of the smaller cubes and the bigger cubes
Answers
Corrected Question :-
3 cubes of metal with edges 3cm,4cm and 5cm are melted to form a single cube. what is the ratio of the sum of the TSA of the smaller cubes and the TSA of the bigger cube.
Given :-
3 cubes of metal with edges 3cm,4cm and 5cm are melted to form a single cube.
To find :-
The ratio of the TSA of the smaller cubes and the bigger cube.
Solution :-
Given that
The edges of the three cubes are 3 cm , 4 cm and 5 cm
We know that
Volume of a cube whose edge is 'a' units is a³ cubic units.
Volume of the cube whose edge is 3 cm
= 3³ cm³
= 3×3×3
= 27 cm³
Volume of the cube whose edge is 4 cm
= 4³ cm³
= 4×4×4
= 64 cm³
Volume of the cube whose edge is 5 cm
cm = 5³ cm³
= 5×5×5
= 125 cm³
Let the edge of the bigger cube be X cm
Volume of the cube = X³ cm³
We know that
If Solids are melted and recast into another solid then the sum of the all volumes is equal to the volume of the resultant solid.
Sum of the volumes of the small cubes = Volume of the bigger cube
=> 27+64+125 = X³
=> 216 = X³
=> X³ = 216
=> X³ = 6³
=> X = 6 cm
The edge of the bigger cube = 6 cm
We know that
Total Surface Area of a cube is 6a² sq.units
TSA of the cube of edge 3 cm
= 6(3)²
= 6(9)
= 54 cm²
TSA of the cube of edge 4 cm
= 6(4)²
= 6(16)
= 96 cm²
TSA of the cube of edge 5 cm
cm = 6(5)²
= 6(25)
= 150 cm²
TSA of the cube of edge 6 cm
cm = 6(6)²
= 6(36)
= 216 cm²
The sum of the TSA of the smaller cubes
= 54+96+150
= 300
The ratio of the sum of the TSA of smaller cubes and the TSA of the bigger cube
= 300:216
= (300/216)
= 50/36
= 25/18
= 25:18
Answer :-
The ratio of the sum of the TSA of smaller cubes and the TSA of the bigger cube = 25:18
Used formulae:-
♦Volume of a cube whose edge is 'a' units is a³ cubic units.
♦Total Surface Area of a cube is 6a² sq.units