Some geometric problems involve changing one or more dimensions. In the following problems, one or more dimensions are changed and you are given information about how this change has affected the figure’s area. Next you will decide how the two areas are related Then you will be able to reduce your problem from several unknowns to just one.
Example
A rectangle
is twice as long as it is wide. If the length is decreased by 4 inches and its
width is decreased by 3 inches, the area is decreased by 88 square inches. Find
the original dimensions.
The area formula for a rectangle is A = lw.
Let A represent the original area; l, the original length; and w, the original
width. We know that the original length is twice the original width, so l = 2w
and A = lw becomes A = 2ww = 2w2. The new length is l – 4 = 2w – 4
and the new width is w – 3, so the new area is (2w – 4)(w – 3). But the new
area is also 88 square inches less than the old area, so A – 88 represents the
new area, also. We then have for the new area, A – 88 = (2w – 4)(w – 3). But the
A can be replaced with 2w2. We now have the equation 2w2 –
88 = (2w – 4)(w – 3), an equation with one unknown.
The width of
the original rectangle is 10 inches and its length is 2w = 2(10) = 20 inches.
A square’s
length is increased by 3 cm, which causes its area to increase by 33 cm2.
What is the length of the original square?
A square’s length and width are the same, so
the area formula for the square is A = ll
= l2. Let l represent the original length. The new
length is l + 3. The original area is A = l2
and its new area is (l + 3)2.
The new area is also the original area plus 33, so (l + 3)2 = new area = A + 33 = l2 + 33. We now have the
equation, with one unknown: (l + 3)2 = l2 + 33.
The original
length is 4 cm.
Practice
1. A
rectangular piece of cardboard starts out with its width being three-fourths
its length. Four inches are cut off its length and two inches from its 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 one-and-a-half 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?
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 w = (¾)l. We will now
replace w with 
.
The
original length was 16 inches and the original width was 
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 þ 97 = (l + 4)(w + 3). But A = lw, so A + 97 becomes lw +
97.
We
now have lw + 97 = (l + 4)(w + 3).
Since
the original length is 1 ½ = 3/2 of the original width, l = 3/2 w.
Replace
each l by 3/2w.
The
original width is 10 inches and the original length is 
inches.
Example
The radius
of a circle is increased by 3 cm. As a result, the area is increased by 45 – cm2.
What was the original radius?
Remember that the area of a circle is A = πr2,
where r represents the radius. So, let r represent the original radius. The new
radius is then represented by r + 3. The new area is represented by π(r + 3)2.
But the new area is also the original area plus 45π cm2. This gives
us A + 45π = π(r + 3)2. Because A = πr2, A + 45π becomes πr2
+ 45π. Our equation, then, is πr2 + 45π = π(r + 3).
The original
radius was 6 cm.
Practice
A circle’s
radius is increased by 5 inches and as a result, its area is increased by 155π
square inches. What is the original radius?
Solution
Let r
represent the original radius. Then r + 5 represents the new radius, and A = πr2
represents the original area. The new area is 155π square inches more than the
original area, so 155π + A = π(r + 5)2 = 155 π + πr2.
The original
radius is 13 inches.
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