300 algebra De mystif ieD
In the above examples and practice problems, we substituted the number for
t in the first distance equation to get d. It does not matter which equation we
us to find d, we would get the same value.
For distance problems in which the bodies are moving away from each other or
toward each other at right angles (for example, one heading east, the other north),
we use the Pythagorean theorem. This topic will be covered in the last chapter.Geometric Figures
Algebra problems involving geometric figures are very common. In algebra, we
normally work with rectangles, triangles, and circles. On occasion, we are asked
to solve problems involving other shapes such as right circular cylinders and
right circular cones. By mastering the more common types of geometric prob-
lems, you will find that the more exotic shapes are just as easy.
In many of these problems, we will have several unknowns which we again
must reduce to a single unknown. In the problems above, we reduced a prob-
lem of three unknowns to two unknowns by relating one quantity to another
(the time on one direction related to the time on the return trip) and by setting
the equal distances equal to each other. We use similar techniques here.PROBLEM
A rectangle is 112 times as long as it is wide. The perimeter is 100 cm. Find
the dimensions of the rectangle.
The formula for the perimeter of a rectangle is given by P = 2l + 2w. We are
told the perimeter is 100, so the equation becomes 100 = 2l + 2w. We are
also told that the length is 121 times the width, so l = 1.5w. We can substitute
1.5w for l into the equation:
100 = 2l + 2w = 2(1.5w) + 2w. We have reduced an equation with three
unknowns to an equation with a single unknown.
100 2152
100 32
100 5
100
5
20=+
=+
=
=
=(.ww)
ww
w
w
w
The width is 20 cm and the length is 1.5w = 1.5(20) = 30 cm.PROBLEM
A rectangle is