Question

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​

Answer 1

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 %.

Answer 2

Question:-

• 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

To Find:-

• Find the percent increased in the total surface area of the cube.

Solution:-

First ,

We have to find the total surface area of a cube:-

$\sf\dashrightarrow \: Area = 6 \times { Edge }^{ 2 }$

$\sf\dashrightarrow \: Area = 6 \times { 9a }^{ 2 }$

$\sf\dashrightarrow \: Area = 54 { a }^{ 2 }$

Now ,

We have to find T.S.A of three cuboid:-

$\sf\dashrightarrow \: T.S.A = 3 \times 2( 3a \times a + a \times 3a + 3a \times 3a )$

$\sf\dashrightarrow \: T.S.A = 6( 15 { a }^{ 2 } )$

$\sf\dashrightarrow \: T.S.A = 90 { a }^{ 2 }$

Now ,

We have to find the percent increased in the total surface area of the cube:-

$\sf\dashrightarrow \: Percent = \dfrac { Increased \: in \: area \times 100 } { Initial \: percent }$

$\sf\dashrightarrow \: Percent = \dfrac { 3600 { a }^{ 2 } } { 54 { a }^{ 2 } }$

$\sf\dashrightarrow \: Percent = 66.6$

Hence ,

• Percentage increased is 66.6 %