If the length of the diagonal of a cube is 4cm then let us calculate total surface area
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Solution:
Given: The length of the diagonal of a cube is 4 cm
To find: Total Surface Area of the cube.
Let us consider a dimension of the cube(a square).
A B
l------------l
l / l
l / l
l / l
l----------Γl
C D
In Δ BCD,
∠D=90°
Hence,Δ BCD is right angled triangle.
CD=BD=4 (length of a dimension of a cube(a square) are equal)
The diagonal of the square (by pythagoras theorem): √a²+a² =√2a²= a√2 cm.
The top dimension of the cube(where we can use diagonal as base) will have:
The diagonal of top dimension of the cube= dimension of cube
4 = √a²+(a√2)²
4 =√a²+2a²
4 =√3a²
4 =a√3
4/√3 =a,.
we know that,TSA of cube = 6a²
= 6(4/√3)
=3 x 2 x(4/√3)
=√3 x √3 x 2 x (4/√3)
=√3 x 2 x4
=8√3 cm²
Given: The length of the diagonal of a cube is 4 cm
To find: Total Surface Area of the cube.
Let us consider a dimension of the cube(a square).
A B
l------------l
l / l
l / l
l / l
l----------Γl
C D
In Δ BCD,
∠D=90°
Hence,Δ BCD is right angled triangle.
CD=BD=4 (length of a dimension of a cube(a square) are equal)
The diagonal of the square (by pythagoras theorem): √a²+a² =√2a²= a√2 cm.
The top dimension of the cube(where we can use diagonal as base) will have:
The diagonal of top dimension of the cube= dimension of cube
4 = √a²+(a√2)²
4 =√a²+2a²
4 =√3a²
4 =a√3
4/√3 =a,.
we know that,TSA of cube = 6a²
= 6(4/√3)
=3 x 2 x(4/√3)
=√3 x √3 x 2 x (4/√3)
=√3 x 2 x4
=8√3 cm²
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