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Geometry Word Problems:
     The Pythagorean Theorem, etc. (page 3 of 4)

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

Another formula you should remember is the Pythagorean Theorem: "Take a right-angled triangle, and square the lengths of all three sides. If you add up the squares of the two shorter sides, this will give the same value as the square of the longest side (the hypoteneuse, the side opposite the right angle)." As a formula, the Pythagorean Theorem is often stated in the form "a² + b² = c²".

• If the sum of the sides of a right triangle is 49 inches and the hypoteneuse is 41 inches, find the two sides.

By "the sides", they mean "the two shorter sides". Letting "a" and "b" be the shorter sides, the sum is:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

a + b = 49

I can solve this for either one of the variables:

a = 49 – b

This gives me expressions for all three sides of the right triangle: a = 49 – b, b, and c = 41. I'll plug these into the Pythagorean Theorem:

a² + b² = c²
(49 – b)² + b² = 41²     (by substitution)
2401 – 98b + b² + b² = 1681
2b² – 98b + 720 = 0
b² – 49b + 360 = 0
(b – 9)(b – 40) = 0
b = 9  or  b = 40

In this case, either solution will do. If b = 9, then a = 49 – b = 49 – 9 = 40. If b = 40, then a = 49 – b = 49 – 40 = 9. Since the problem didn't specify which of the two sides is longer, it doesn't matter which one I call "a" and which one I call "b". The answer is:

One side is forty inches long, and the other side is nine inches long.

On a related note, remember that a triangle's three angles add up to 180 degrees.

• The smallest angle of a triangle is two-thirds of the middle angle, and the middle angle is three-sevenths of the largest angle. Find all three angle measures.

Let "ß" stand for "beta", the largest angle, or, rather, for the measure of the largest angle. Then the middle angle has measure ( ³/₇ )ß. The smallest angle is two-thirds of the midle angle, so it has measure ( ²/₃ )( ³/₇ )ß = ( ²/₇ )ß. The formula is:

ß + ( ³/₇ )ß + ( ²/₇ )ß = 180
7ß + 3ß + 2ß = 1260
12ß = 1260
ß = 105

So the largest angle has a measure of 105 degrees. The middle angle is then:

( ³/₇ )(105) = 45

...or 45 degrees, and the smallest angle is:

( ²/₃ )(45) = 30

...or 30 degrees.

The angle measures are 30 degrees, 45 degrees, and 105 degrees.

• A rectangle is 8 feet long and 6 feet wide. If each dimension is increased by the same number of feet, the area of the new rectangle formed is 32 square feet more than the area of the original rectangle. By how many feet was each dimension increased?

The area of the original rectangle is 8×6 = 48 square feet. Suppose I add x feet to each dimension. Then the length will be 8 + x, and the width will be 6 + x. The new area will be:

A = (8 + x)(6 + x)

This new area is also 32 square feet more than the old area, so the new area is 48 + 32 = 80. Then I'll have:

A = (8 + x)(6 + x) = 80

I need to solve this quadratic for the value of x:

(8 + x)(6 + x) = 80
48 + 8x + 6x + x² = 80
x² + 14x – 32 = 0
(x – 2)(x + 16) = 0
x = 2  or  x = –16

Since I'm looking for a length, which is a positive value, I have no use for the "x = –16" solution, so I'll ignore it. On the other hand, "x = 2" works fine.

Each dimension was increased by two feet.

• You work for a fencing company. A customer called, wanting to fence in his 1,320 square-foot garden. He ordered 148 feet of fencing, but you forgot to ask him for the width and length of the garden (these dimensions will determine some of the details of the order, so you need the information). You don't want the customer to think that you're incompetent, so you figure out the length and width from the information the customer has already given you. What are the dimensions?

The perimeter is given by:

2L + 2W = 148

The area is given by:

L×W = 1320

Dividing the first equation by 2 (so I am dealing with smaller numbers), I get the following system (or "set of equations"):

L + W = 74
L×W = 1320

I can solve either one of these equations for either one of the variables, and then plug this into the other equation:

L = 74 – W   (solving for L)
(74 – W) × W = 1320    (substituting into the area equation)
74W – W ² = 1320
0 = W ² – 74W + 1320
0 = (W – 30)(W – 44)
W = 30  or  W = 44

Once again, I've come up with two valid solutions. If W = 30, then L = 74 – W = 74 – 30 = 44.  If W = 44, then L = 74 – W = 74 – 44 = 30. In either case, the dimensions work out the same, and the answer is:

The garden is 44 feet by 30 feet.

In the above problem, there were two equally-valid answers (the width could be either thirty feet or forty-four feet). Don't mess this up by deciding that one of the dimensions must be "width" and the other dimension must be "length", because you haven't been given enough information to make that determination. On the other hand, if this problem had said something like "the length is greater than the width", then you could have said that the length was forty-four feet and the width was thirty feet. Keep these considerations in mind; you never know when fiddly points like this can make a difference on a test.

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