Geometric Figures
Algebra problems involving geometric figures are very common. In algebra, you normally deal with rectangles, triangles, and circles. On occasion, you will be asked to solve problems involving other shapes like right circular cylinders and right circular cones. If you master solving the more common types of geometric problems, you will find that the more exotic shapes are just as easy.
In many of these problems, you will have
several unknowns which you must reduce to one unknown. In the problems above,
you reduced a problem of three unknowns to one unknown 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 will use similar
techniques here.
Example
A rectangle
is 1 ½ times as long as it is wide. The perimeter is 100 cm2.
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 now
becomes 100 = 2l + 2w. We are also told that the length is 1 ½ times the width,
so l = 1.5w. We can substitute this l into the equation: 100 = 2l + 2w = 2(1.5w)
+ 2w. We have reduced an equation with three unknowns to one with a single
unknown.
The width is
20 cm and the length is 1.5w = 1.5(20) = 30 cm.
Practice
1. A
box’s width is ⅔ its length. The perimeter of the box is 40 inches. What are
the box’s length and width?
2. A
rectangular yard is twice as long as it is wide. The perimeter is 120 feet.
What are the yard’s dimensions?
Solutions
1. The
perimeter of the box is 40 inches, so P = 2l
+ 2w become 40 = 2l + 2w. The width is ⅔ its length, and w = (2l/3), so 40 = 2l + 2w becomes 40 = 2l + 2(2l/3) = 2l + (4l/3).
The
length of the box is 12 inches and its width is ⅔ l = ⅔ (12) = 8 inches.
2. The
perimeter of the yard is 120 feet, so P = 2l
+ 2w becomes 120 = 2l + 2w. The
length is twice the width, so l = 2w,
and 120 = 2l + 2w becomes 120 = 2(2w) + 2w.
The
yard’s width is 20 feet and its length is 2l = 2(20) = 40 feet.
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