304 algebra De mystif ieD

width. The area of the cardboard is 72 square inches smaller than before

it was trimmed. What was its original length and width?

2. A rectangle’s length is 1^12 times its width. The length is increased by
   4 inches and its width by 3 inches. The resulting area is 97 square inches
   more than the original rectangle. What were the original dimensions?


3. A circle’s radius is increased by 5 inches and as a result, its area is
   increased by 155o square inches. What is the original radius?

✔SOLUTIONS

1. Let l represent the original length and w, the original width. The original
   area is A = lw. The new length is l – 4 and the new width is w – 2. The new
   area is then (l – 4)(w – 2). But the new area is 72 square inches smaller
   than the original area, so (l – 4)(w – 2) = A – 72 = lw – 72. So far, we have
   (l – 4) (w – 2) = lw – 72.
   The original width is three-fourths its length, so wl = ^34. We will now
   replace w with ()^34 l = ^34 l.

()l l l l

l l l

−−







= 







−

−−





4 3

4

2 3

4

72

3

4

243

4

(^2) 

+= −






−
8 3
4
72 3
4
2
ll^2
l
oneachsidecancels.
−−+=−
−+=−
−−
−=−
=−
−

3872
5872
88
580
80
5
16
l
l
l
l
l
The original length was 16 inches and the original width was
3
4
3
l (=  416 ) =  12 inches.

2. Let l represent the original length and w, the original width. The original
   area is then given by A = lw. The new length is l + 4 and the new width is
   w + 3. The new area is now (l + 4)(w + 3). But the new area is also the old
   area plus 97 square inches, so A + 97 = (l + 4)(w + 3). But A = Iw, so A + 97
   becomes lw + 97. We now have
   lw + 97 = (l + 4)(w + 3).