Solve the problem.
-The area of a picture projected on a wall varies directly as the square of the distance from the projector to the wall.
a. If a $12 - \mathrm { ft }$ distance produces a $81 \mathrm { ft } ^ { 2 }$ picture, what is the area of the picture when the projection unit is moved to a distance of $16 \mathrm { ft }$ from the wall?
b. If the projected image is $324 \mathrm { ft } ^ { 2 }$, how far is the projector from the wall?
A) a. $108 \mathrm { ft } ^ { 2 }$;
b. $48 \mathrm { ft }$
B) a. $196 \mathrm { ft } ^ { 2 }$;
b. $54 \mathrm { ft }$
C) a. $169 \mathrm { ft } ^ { 2 }$;
b. $36 \mathrm { ft }$
D) a. $144 \mathrm { ft } ^ { 2 }$;
b. $24 \mathrm { ft }$

Question

Quizplus · Accepted Answer

The area of the picture varies directly as the square of the distance from the projector to the wall, so we use the formula $A = kD^2$, where $A$ is the area, $D$ is the distance, and $k$ is the constant of variation. For part a, we find $k$ using the given values and then calculate the area for a 16 ft distance. For part b, we solve for $D$ given $A = 324 \, \text{ft}^2$.

Solve the Problem 12−ft12 - \mathrm { ft }12−ft Distance Produces A

Solve the Problem $12 - \mathrm { ft }$ Distance Produces A