Question

A teacher has two boxes. The first box has a height of 15 in., a width of 6 in., and a length of 8 in. The second box’s dimensions are twice that of the first box. How many times larger is the surface area of the second box than the first box?

A. The surface area of the second box is 6 times larger.
B. The surface area of the second box is 4 times larger.
C. The surface area of the second box is 3 times larger.
D. The surface area of the second box is 2 times larger.

Answer 1

The correct option is B: the surface area of the second box is 4 times larger.

Explanation:

The surface area of the first box is given by:

`SA_(1) = 2l_(1)h_(1) + 2w_(1)h_(1) + 2l_(1)w_(1)`

Where:

l₁: is the length = 8 in

h₁: is the height = 15 in

w₁: is the width = 6 in

`SA_(1) = 2*8*15 + 2*6*15 + 2*8*6 = 516 in^(2)`

The second box's dimensions are:

l₂ = 2*8 in = 16 in

h₂ = 2*15 in = 30 in

w₂ = 2*6 in = 12 in

Hence, the surface area of the second box is:

`SA_(2) = 2*2l_(1)*2h_(1) + 2*2w_(1)*2h_(1) + 2*2l_(1)*2w_(1) = 4(2l_(1)h_(1) + 2w_(1)h_(1) + 2l_(1)w_(1)) = 4*SA_(1) = 4*516 in^(2) = 2064 in^(2)`

Therefore, the correct option is B: the surface area of the second box is 4 times larger.

I hope it helps you!