The Purplemath ForumsHelping students gain understanding and self-confidence in algebra powered by FreeFind

Geometry Word Problems: Triangles (page 3 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

• 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.

They have given me a relationship between the height and the base, and have given me the value of the area. So I'll need to use the formula for the area of a triangle with a given base and height, and I'll need to create an expression or equation relating the height and base.

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 the equation for their relationship is:

h = b – 5

Since I am given that the area is 52 square inches, I can then plug the base variable, the height expression, and the area value into the formula for the area of a triangle, and see where this leads:

(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 solution. (A solution which is "extraneous", pronounced "ek-STRAY-nee-uss", is a number that is a valid solution to the equation, but is not a relevant value in the context of the word problem. In this case, lengths cannot be negative.) This means that b = 13, so h = b – 5 = 13 – 5 = 8.

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

In the last exercise above, I solved for one value (the length of the base) and then back-solved for the other value (the length of the height). This other value turned out to be the same as the extraneous value, except for the sign change. Warning: Do not assume that you can get both of your answers by arbitrarily changing the sign on the extraneous solution. This does not always work, it is mathematically wrong, it annoys your teacher, and it can get you in trouble further down the line.

Another triangle 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 2000-2011 All Rights Reserved

a + b = 49

I can solve this for either one of the variables. I think I'll solve for a in terms of b:

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.

• A wood frame for pouring concrete has an interior perimeter of 14 meters. Its length is one meter greater than its width. The frame is to be braced with twelve-gauge steel cross-wires. Assuming an extra half-meter of wire is used at either end of a cross-wire for anchoring, what length of wire should be cut for each brace?
• I don't care that the wire is steel; I don't care that they're pouring concrete into a wood frame. All I need is the geometrical information: this is a rectangle with a certain perimeter and a certain relationship between the length and the width. They're asking me, effectively, to find the length of the diagonal. And this diagonal, together with the length and the width, will form a right triangle. So the perimeter formula for a rectangle may be useful, as may the Pythagorean Theorem.

width: w
length: w + 1
perimeter formula:
14 = 2(w + 1) + 2(w)

14 = 2w + 2 + 2w
14 = 4w + 2

12 = 4w

3 = w

Then the length, being one unit larger, is 4, and the Pythagorean Theorem lets me find the length of the diagonal d:

32 + 42 = d2
9 + 16 = 25 = d2

5 = d

Adding a half-meter at either end of the wire, I find that:

each wire should be cut to a length of six meters

Another useful triangle fact is 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 °.

<< Previous  Top  |  1 | 2 | 3 | 4 | 5 | 6  |  Return to Index  Next >>

 Cite this article as: Stapel, Elizabeth. "Geometry Word Problems: Triangles." Purplemath. Available from     http://www.purplemath.com/modules/perimetr3.htm. Accessed [Date] [Month] 2016

Purplemath:
Printing pages
School licensing

Reviews of
Internet Sites:
Free Help
Practice
Et Cetera

The "Homework
Guidelines"

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor

This lesson may be printed out for your personal use.