For the school banner we have:
\(A = w * l
\)
Where,
w: width of the banner
l: length of the banner
Substituting values we have:
\(1496 = w * 44
\)
Clearing the width we have:
\(w = \frac{1496}{44}
w = 34
\)
Then, for the cheerleaders we have:
The ratio of measures is:
\(\frac{l}{l'} = \frac{44}{11} = 4
\)
Therefore, the width is given by:
\(w '= \frac{w}{4}
\)
\(w '= \frac{34}{4}
w '= 8.5
\)
Then, the perimeter is given by:
\(P = 2l '+ 2w'
\)
Substituting values we have
\(P = 2 (11) + 2 (8.5)
P = 22 + 17
P = 39
\)
The area is given by:
\(A = w '* l'
\)
Substituting values we have
\(A = (8.5) * (11)
A = 93.5
\)
Answer:
Its perimeter and its area are:
P = 39 inches
A = 93.5 square inches
