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Geometry Word Problems: Basic Shapes (page 2 of 4)

Sections: Introduction, Basic shapes, The Pythagorean Theorem, Max/min problems

• A piece of wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.

Since the wire is 42 centimeters long, then the perimeter of the rectangle is 42 centimeters. That is:

2L + 2W = 42

I also know that the width is twice the length, so:

W = 2L

Then:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

2L + 2(2L) = 42    (by substitution for W from the above equation)
2L + 4L = 42
6L = 42
L = 7

Since W = 2L, then W = 14, and the rectangle is 7 centimeters long and 14 centimeters wide.

• A circular swimming pool with a diameter of 28 feet has a deck of uniform width built around it.  If the area of the deck is 60(pi) square feet, find its width.

If the diameter is 28, then the radius is 14. The area of the pool is then:

(pi)r² = (pi)(14)² = 196(pi)

Why do I need the area of the pool? Because, to find the area of the deck, I'll need to find the area enclosed by the circumference of the deck, and then subtract out the area enclosed by the pool. As is often true, a picture is helpful:

	The pool, surrounded by the deck. The pool has radius 14, and the deck has width "d".
	This is the area that I'm trying to find: the entire circular area, less the area of the pool.

Let d be the width of the deck. Then the radius of the entire deck-and-pool circle is 14 + d. Then the area of just the deck is given by:

deck area = (total circular area) – (pool area)

A = (pi)(14 + d)² – 196(pi)
    = (pi)(196 + 28d + d²) – 196(pi)
    = 196(pi) + 28(pi)d + (pi)d² – 196(pi)
    = (pi)d² + 28(pi)d

I am given that the area of the deck is 60(pi) square feet. Then:

(pi)d² + 28(pi)d = 60(pi)
(pi)d² + 28(pi)d – 60(pi) = 0
d² + 28d – 60 = 0
(d + 30)(d – 2) = 0
d = –30  or  d = 2

Since d stands for a distance, you can ignore the "d = –30" solution as not being useful; it is extraneous. ("Extraneous", pronounced "ek-STRAY-nee-uss", means that, while it is a valid answer to the mathematical equation, it is not relevant in the context of the word problem.)

Then the answer is that the deck is two feet wide.

Don't forget the units on your word-problem answers. The above question did not ask "What is the value of the variable d", but asked "How many feet wide is the deck?" Always remember to state your answer in terms of the actual question.

(Note: There can be more than one way to solve a given exercise. In the case of the above, once you've found the area of the central pool, you can add the two areas: 196(pi) + 60(pi) = 256(pi). Then you can solve for the total radius: 16(pi). Since the pool's radius was fourteen feet, then the deck's width must be two feet.)

• If the height of a triangle is five inches less than the length of its base, and if the area of the triangle is 52 square inches, find the base and the height.

The area of a triangle is given by:

A = ( ¹/₂ )bh

...where "b" is the base and "h" is the height (or "altitude"). I am given that the height is five less than the base, so:

h = b – 5

Since I am given that the area is 52 square inches, I have:

(1/2)(b)(b – 5) = 52   (by substitution for h from the above equation)
b(b – 5) = 104
b² – 5b = 104
b² – 5b – 104 = 0
(b – 13)(b + 8) = 0
b = 13  or  b = –8

I can safely ignore the extraneous negative result; this means that b = 13, so h = b – 5 = 13 – 5 = 8. Then:

The base is 13 inches, and the height is 8 inches.

(In this case, we solved for one value and then back-solved for the other value. This other value turned out to be the same as the extraneous value, except for the sign change. However, don't assume that you can get both of your answers by arbitrarily changing the sign on the extraneous solution. This does not always work, and it is mathematically wrong, it annoys your teacher, and it can get you in trouble further down the lline. Take the extra two seconds to derive the answer correctly.)

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