2001 CEMC Gauss (Grade 8) Problems/Problem 14

The following problem is from both the 2001 CEMC Gauss (Grade 8) #14 and 2001 CEMC Gauss (Grade 7) #17, so both problems redirect to this page.

Problem

A rectangular sign that has dimensions $9 \text{ m}$ by $16 \text{ m}$ has a square advertisement painted on it. The border around the square is required to be at least $1.5 \text { m}$ wide. The area of the largest square advertisement that can be painted on the sign is

$\text{ (A) }\ 78 \text{ m}^{2} \qquad\text{ (B) }\ 144 \text{ m}^{2} \qquad\text{ (C) }\ 36 \text{ m}^{2} \qquad\text{ (D) }\ 9 \text{ m}^{2} \qquad\text{ (E) }\ 56.25 \text{ m}^{2}$

Solution 1

We want to use the minimum border to get the largest square advertisement. Since the border is required to be at least $1.5 \text{ m}$ wide, the minimum must be $1.5 \text { m}$.

Removing the border, we can see that the dimensions are $9 - 1.5 \times 2 = 9 - 3 = 6$ m by $16 - 1.5 \times 2 = 16 - 3 = 13$ m.

Because we cannot fit a $13$ m by $13$ m square on this, the maximum width of the square is $6$ m. Thus, the largest area is:

$6 \text{ m} \times 6 \text{ m} = \boxed {\textbf {(C) } 36 \text{ m}^{2}}$

~anabel.disher

2001 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
CEMC Gauss (Grade 8)
2001 CEMC Gauss (Grade 7) (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
CEMC Gauss (Grade 7)