Question

A giant 10 ft. tall Kaleidoscope is to be given as a present to the local astronomical society. The Kaleidoscope is in the shape of a triangular prism with sides 3 ft, 4 ft and 5 ft,. It is to be gift-wrapped for a present. The cost of gift-wrapping is $\$2/ft$^2$. Assuming that there is no overlap of wrapping paper, how much will it cost to wrap the Kaleidoscope?

• A. $ 248
• B. $ 252
• C. $ 260
• D. $ 264

Answer 1

The correct option is D $ 264

Rectangular prism-shaped Kaleidoscope


Cost $=\$2/{\text{ft}}^2$




$\text{Total Surface area}=2\times \text{Area of the floor}+\text{area of the walls}$

$\text{Area of the floor}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{12}{2}=6$
Since there is a triangular top also



$\therefore$ Total Surface area $=2\times 6{\text{ft}}^2+\text{area of the walls}$
$=12+\text{area of the walls}$

$\text{Area of the Walls = Sum of the area of all}\text{three rectangular faces}$
$=3\times 10+4\times 10+5\times 10$
$=30+40+50$



$\therefore \text{Total surface area}=12+\text{Area of the walls}$
$=12+30+40+50$
$=132$ ft${}^2$

Cost of wrapping the Kaleidoscope = ( Total Surface Area) $\times$ Cost
$=132\times 2$
$=\$264$