Question

A box has a volume of 192 cubic inches, a length that is twice as long as its width, and a height that is 2 inches greater than the width. What are the dimensions of this box?

Answer 1

Answer:

Therefore the dimension of the cuboid is 8 inches×4 inches ×6 inches.

Step-by-step explanation:

Cuboid : A cuboid is a three dimension shape. The length ,breadth and height of a cuboid are not same.

• A cuboid has 6 faces.
• A cuboid contains 8 vertices.
• A cuboid contains 12 edges .
• The total surface area of a cuboid is

           = 2(length×breadth+breadth×height+length×height) square units

• The dimension of a cuboid is written as length×breadth×height.

• The volume is( length×breadth×height) cubic units

Given that the volume of the box is 192 cubic inches.

Let x inches be the width of the cuboid.

Since the length is twice as long as its width.

Then length = 2x inches

Again height is 2 inches longer than width.

Then height = (x+2) inches.

Therefore the volume of the cuboid is

$=[x\times 2x\times (x+2)]$   cubic inches

$=[2x^2(x+2)]$     cubic inches

$=(2x^3+4x^2)$     cubic inches

According to the problem,

$2x^3+4x^2=192$

$\Rightarrow 2x^3+4x^2-192=0$

$\Rightarrow 2(x^3+2x^2-96)=0$

$\Rightarrow (x^3+2x^2-96)=0$

$\Rightarrow x^3-4x^2+6x^2-24x+24x-96=0$

$\Rightarrow x^2(x-4) +6x(x-4)+24(x-4)=0$

$\Rightarrow (x-4)(x^2+6x+24)=0$

Therefore x=4

Since the all zeros of x²+6x+24 =0 is negative.

Therefore breadth = 4 inches

 length=(2×4) inches=8 inches

 and height = (4+2)inches = 6 inches.

Therefore the dimension of the cuboid is 8 inches×4 inches ×6 inches.