Question

if each edge of a cuboid of surface area S is doubled , the surface area of the new cuboid is​

Answer 1

Answer:

• The surface area of the new cuboid is 4S.

Step-by-step explanation:

Given that:

• Each edge of a cuboid of surface area S is doubled.

To Find:

• The surface area of the new cuboid.

We know that:

• T.S.A. = 2(LB + BH + HL)

Where,

• T.S.A. = Total surface area of a cuboid
• L = Length
• B = Breadth
• H = Height

Surface area of cuboid is S.

• S = 2(LB + BH + HL) ______(i)

When each edge of a cuboid of surface area S is doubled.

New edge will be,

• Length = 2L
• Breadth = 2B
• Height = 2H

⇢ New surface area = 2(2L•2B + 2B•2H + 2H•2L)

⇢ New surface area = 2(4LB + 4BH + 4HL)

Taking 4 common.

⇢ New surface area = 4 × 2(LB + BH + HL)

From equation (i).

⇢ New surface area = 4 × S

⇢ New surface area = 4S

Hence,

• The surface area of the new cuboid is 4S.

Answer 2

Step-by-step explanation:

Answer:✍️

The surface area of the new cuboid is 4S.

Step-by-step explanation:

Given that:

Each edge of a cuboid of surface area S is doubled.

To Find:

The surface area of the new cuboid.

We know that:

T.S.A. = 2(LB + BH + HL)

Where,

T.S.A. = Total surface area of a cuboid

L = Length

B = Breadth

H = Height

Surface area of cuboid is S.

S = 2(LB + BH + HL) ______(i)

When each edge of a cuboid of surface area S is doubled.

New edge will be,

Length = 2L

Breadth = 2B

Height = 2H

⇢ New surface area = 2(2L•2B + 2B•2H + 2H•2L)

⇢ New surface area = 2(4LB + 4BH + 4HL)

Taking 4 common.

⇢ New surface area = 4 × 2(LB + BH + HL)

From equation (i).

⇢ New surface area = 4 × S

⇢ New surface area = 4S

Hence,

The surface area of the new cuboid is 4S.

hope this helps you!!