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?
- 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? - 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
- 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.
- 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).