**Introduction to Geometric Figures in Word Problems**

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.

**Solving for Three Unknowns**

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 times as long as it is wide. The perimeter is 100 cm^{2} . Find the dimensions of the rectangle.

The formula for the perimeter of a rectangle is given by *P* = 2 *l* + 2 *w* . We are told the perimeter is 100, so the equation now becomes 100 = 2 *l* + 2 *w* . We are also told that the length is times the width, so *l* = 1.5 *w* . We can substitute this *l* into the equation: 100 = 2 *l* +2 *w* = 2(1.5 *w* ) + 2 *w* . 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.

Find practice problems and solutions at Geometric Figures Practice Problems - Set 1.

**Increasing and Decreasing Dimensions**

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.

**Rectangle 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* = 2 *w* and *A* = *lw* becomes *A* = 2 *ww* = 2 *w*^{2} . The new length is *l* – 4 = 2 *w* – 4 and the new width is *w* – 3, so the new area is (2 *w* – 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 = (2 *w* – 4)( *w* – 3). But the *A* can be replaced with 2 *w*^{2} . We now have the equation 2 *w*^{2} – 88 =(2 *w* – 4)( *w* – 3), an equation with one unknown.

The width of the original rectangle is 10 inches and its length is 2 *w* = 2(10) = 20 inches.

**Square Example**

A square’s length is increased by 3 cm, which causes its area to increase by 33 cm^{2} . 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* = *l* ^{2} . Let *l* represent the original length. The new length is *l* + 3. The original area is *A* = *l*^{2} 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 = *l*^{2} + 33. We now have the equation, with one unknown: ( *l* + 3)^{2} = *l*^{2} + 33.

The original length is 4 cm.

Find practice problems and solutions at Geometric Figures Practice Problems - Set 2.

**Circle Example **

The radius of a circle is increased by 3 cm. As a result, the area is increased by 45π cm^{2} . What was the original radius?

Remember that the area of a circle is *A* = π*r*^{2} , 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 45πcm^{2} . This gives us *A* + 45π = π( *r* + 3)^{2} . Because *A* = πr^{2} , *A* + 45π becomes πr^{2} + 45π. Our equation, then, is πr^{2} + 45π = π( *r* + 3) ^{2} .

The original radius was 6 cm.

Find practice problems and solutions at Geometric Figures Practice Problems - Set 3.

More practice problems for this concept can be found at: Algebra Word Problems Practice Test.

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