Question

A right rectangular prism has a length of 5 cm, a width of 4 cm, and a height of 3 cm. The dimensions of the prism are doubled. What is the surface area of the enlarged prism? Enter your answer in the box. ___cm²

Answer 1

Answer:

The surface area of the enlarged prism is, 376 cm²

Step-by-step explanation:

Surface area of the prism(S) is given by:

$S =2(lw+wh+hl)$

where, l is the length , w is the width and h is the height of the prism receptively.

Given that:

A right rectangular prism has a length of 5 cm, a width of 4 cm, and a height of 3 cm.

⇒ l = 5 cm , w = 4 cm and h = 3 cm

Since, the dimensions of the prism are doubled

⇒$l=2 \cdot 5 = 10 cm$

$w =2 \cdot 4 = 8 cm$ and

$h =2 \cdot 3 = 6 cm$

Substitute these in [1] we have;

$S = 2(10 \cdot 8 + 8 \cdot 6 + 6 \cdot 10) = 2(80+48+60) = 2(188) = 376 cm^2$

Therefore, the surface area of the enlarged prism is, 376 cm²

Answer 2

A=2(lw+lh+wh)  and if everything is doubled you have dimensions of:

10,8, and 6 so

A=2(10*8+10*6+8*6)

A=2(80+60+48)

A=2(188)

A=376 cm^2