Triangle Geometric Word Problems
Purplemath
Are triangles important in real life?
Triangles form very solid structures. These shapes are used in making bicycles, roofs, bridges, etc, and even play rôles in sport (and, in a bad way, in romantic situations). They are ubiquitous in our lives, supporting us unnoticed.
So spending a little time working with triangles' formulas is no bad thing.
- 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 of a triangle, and they've 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 = ½bh
...where b is the base and h is the height (also called the 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:
½(b)(b − 5) = 52
b(b − 5) = 104
b2 − 5b = 104
b2 − 5b − 104 = 0
(b − 13)(b + 8) = 0
b = 13, −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, the height being five less than the base, I get:
h = b − 5 = 13 − 5 = 8
Looking back at the exercise statement, I see that the units are inches, so my answer is:
base: 13 inches
height: 8 inches
In the exercise above, I solved for one value (being the length of the base) and then back-solved for the other value (being the length of the height). This other value turned out to be the same as the extraneous value, except for the sign change.
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. The Theorem (in one of its very many formulations) states:
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 equal the value of the square of the longest side.
As a formula, the Pythagorean Theorem is often stated in the following form:
a2 + b2 = c2
...where a and b are the lengths of the two legs (that is, they're the lengths of the two shorter sides) and c is the length of the hypotenuse (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 lengths of the two sides.
By "the sides", they mean "the lengths of the two shorter sides"; that is, the lengths of the two legs. Letting a and b be the lengths of these sides, the sum is:
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:
one leg: b
the other leg: a = 49 − b
the hypotenuse: c = 41
I'll plug these into the Pythagorean Theorem:
a2 + b2 = c2
(49 − b)2 + b2 = 412
2401 − 98b + b2 + b2 = 1681
2b2 − 98b + 720 = 0
b2 − 49b + 360 = 0
(b − 9)(b − 40) = 0
b = 9, 40
In this case, either solution will do. If b = 9, then:
a = 49 − b
= 49 − 9 = 40
On the other hand, if I choose b = 40, then:
a = 49 − b
= 49 − 40 = 9
Since the exercise statement didn't specify which of the two legs is to be assigned the longer length, it doesn't matter which one I call a and which one I call b. My answer remains the same, either way:
one side: 40 in.
the other side: 9 in.
- A rectangular wood frame built for a concrete pour 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 be the Pythagorean Theorem.
They've defined the length in terms of the width, so I'll pick a variable for the width, create an expression for the length, and see where that leads:
width: w
length: w + 1
perimeter: 14 = 2(w + 1) + 2(w)
14 = 2w + 2 + 2w
14 = 4w + 2
12 = 4w
3 = w
This is the value for the width. 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
Then I add a half-meter at either end of the wire:
5 + ½ + ½ = 6
Looking back at the exercise statement to be sure I've got the correct units, my answer then is:
each wire's length: 6 m
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 angles' 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 β (beta, pronounced BAY-tuh) stand for the measure of the largest angle. Then, the middle angle being three-sevenths the size of the largest angle, I get:
middle angle:
The smallest angle is two-thirds of the middle angle, so the expression for its measure is:
Then my angle-sum formula is:
7β + 3β + 2β = 1260
12β = 1260
β = 105
So the largest angle has a measure of 105 degrees. The middle angle is then:
...and the smallest angle is:
Checking the exercise statement, I confirm that "degrees" is the correct unit, and my answer is:
angle measures: 30°, 45°, and 105°
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