A cube is cut with 3 straight cuts . Either horizontally or vertically . What is the percent increased in the total surface area of the cube
Answers
Answer:
The percent increased in the total surface area of the cube is 66.67 %.
Step-by-step explanation:
Given that:
A cube is cut with 3 straight cuts, either horizontally or vertically.
To Find:
What is the percent increased in the total surface area of the cube.
Let us assume:
The edge of the cube be 3a.
We know that:
⍟ Total surface area of a cube = 6 × (Edge)² ⍟
Total surface area = 6 × (3a)²
Total surface area = 6 × 9a²
Total surface area = 54a²
But when a cube is cut with 3 straight cuts, then three cuboid will be formed having:
Dimension of each cuboid.
Length = 3a
Breadth = 3a/3 = a
Height = 3a
We know that:
⍟ T.S.A. of a cuboid = 2[(LB) + (BH) + (HL)] ⍟
Where,
T.S.A. = Total surface area
L = Length
B = Breadth
H = Height
T.S.A. of three cuboid = 3 × 2[(3a × a) + (a × 3a) + (3a × 3a)]
T.S.A. of three cuboid = 6[3a² + 3a² + 9a²]
T.S.A. of three cuboid = 6[15a²]
T.S.A. of three cuboid = 90a²
Finding the percent increased in the total surface area of the cube:
We have:
Initial area = 54a²
Final area = 90a²
Increased in area = 90a² - 54a²
Increased in area = 36a²
⍟ Increased percent = (Increased in area × 100)/Initial area % ⍟
Increased percent = (36a² × 100)/54a² %
Increased percent = 3600a²/54a² %
Increased percent = 66.67 % (approx.)
Hence,
The percent increased in the total surface area of the cube is 66.67 %.