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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 sum will be the same value as the value of the square of the longest side."

As a formula, the Pythagorean Theorem is often stated in the form "a² + b² = c²"., where a and b are the lengths of the two legs (the two shorter sides) and c is the length of the hypotenuse (being the longest side, opposite the right angle).

• 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 lengths of the two shorter sides". Letting "a" and "b" be the lengths of these 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 or variables 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. Or if b = 40, then a = 49 – b = 49 – 40 = 9. Since the problem didn't specify which of the two legs 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 the measures of any triangle's three angles add up to 180 degrees.

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

The smallest angle is defined in terms of the middle angle, but the middle angle is defined in terms of the largest angle. So it makes most sense to pick a variable for the measure of the largest angle, and then create expressions for the middle and then the smallest angles, using that variable.

I'll let "ß" stand for "beta", the largest angle, or, rather, for the measure of the largest angle. Then the middle angle has a measure of ( ³/₇ )ß. The smallest angle is two-thirds of the middle angle, so it has a measure of ( ²/₃ )( ³/₇ )ß = ( ²/₇ )ß. Then my angle-sum 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 °, 45 °, and 105 °.

• 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, and the formula for 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 must be 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, since (8 + 2)(6 + 2) = (10)(8) = 80, as required.

Each dimension was increased by two feet.

• You work for a fencing company. A customer called this morning, 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 want to figure out the length and width from the information the customer has already given you. What are the dimensions?

The perimeter P of this rectangular area with (as-yet unknown) length L and width W is given by:

2L + 2W = 148

The area A is given by:

L×W = 1320

I will divide my "perimeter" equation above by 2, so I am dealing with smaller numbers. This gives me 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. I think I'll solve the addition equation, and plug the result into the multiplication equation:

L = 74 – W   (solving the first equation for L)
(74 – W) × W = 1320    (substituting into the second 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. Warning: Don't make the mistake of thinking that, whenever you end up with two positive solutions, that one of the dimensions must be "width" and the other dimension must be "length". If this exercise 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. But this particular exercise did not give you enough information to make any "length" versus "width" determination. Keep these considerations in mind; you never know when technical considerations like this can make a big difference on a test.

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