Question

Kyle bought 2 snack boxes. Box A has a measure of $2in\times 3in\times 5in.$ Box B has a measure of $11in\times 4in\times 3in.$ Find the surface area of the boxes and their difference.

• A. $116{in}^2$
• B. $201{in}^2$
• C. $279{in}^2$
• D. $198{in}^2$

Answer 1

Answer: (A)

$116{in}^2$

Surface area of rectangle is 2 x (Length x width + Length x height + Width x height)

Here Box A has length $=2$ in, height $=3$ in and width $=5$ in

Therefore, the surface area of Box A is:

$A=2\left(\right(2\times 5)+(2\times 3)+(3\times 5\left)\right)=2(10+6+15)=2\times 31=62$ in${}^2$

Now, Box B has length $=11$ in, height $=4$ in and width $=3$ in

Therefore, the surface area of Box B is:

$A=2\left(\right(11\times 4)+(11\times 3)+(4\times 3\left)\right)=2(44+33+12)=2\times 89=178$ in${}^2$

Hence, the difference between the surface area of box A and box B is $178-62=116$ in${}^2$.