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Chapter 4: Inequalities

4.5 Geometric Word Problems

It is common to run into geometry-based word problems that look at either the interior angles, perimeter, or area of shapes. When looking at interior angles, the sum of the angles of any polygon can be found by taking the number of sides, subtracting 2, and then multiplying the result by 180°. In other words:

[latex]\text{sum of interior angles} = 180^{\circ} \times (\text{number of sides} - 2)[/latex]

This means the interior angles of a triangle add up to 180° × (3 − 2), or 180°. Any four-sided polygon will have interior angles adding to 180° × (4 − 2), or 360°. A chart can be made of these:

[latex]\begin{array}{rrrrrr} \text{3 sides:}&180^{\circ}&\times&(3-2)&=&180^{\circ} \\ \text{4 sides:}&180^{\circ}&\times&(4-2)&=&360^{\circ} \\ \text{5 sides:}&180^{\circ}&\times&(5-2)&=&540^{\circ} \\ \text{6 sides:}&180^{\circ}&\times&(6-2)&=&720^{\circ} \\ \text{7 sides:}&180^{\circ}&\times&(7-2)&=&900^{\circ} \\ \text{8 sides:}&180^{\circ}&\times&(8-2)&=&1080^{\circ} \\ \end{array}[/latex]

Example 4.5.1

The second angle [latex](A_2)[/latex] of a triangle is double the first [latex](A_1).[/latex] The third angle [latex](A_3)[/latex] is 40° less than the first [latex](A_1).[/latex] Find the three angles.

The relationships described in equation form are as follows:

[latex]A_2 = 2A_1 \text{ and } A_3 = A_1 - 40^{\circ}[/latex]

Because the shape in question is a triangle, the interior angles add up to 180°. Therefore:

[latex]A_1 + A_2 + A_3 = 180^{\circ}[/latex]

Which can be simplified using substitutions:

[latex]A_1 + (2A_1) + (A_1 - 40^{\circ}) = 180^{\circ}[/latex]

Which leaves:

[latex]\begin{array}{rrrrrrrrrrr} 2A_1&+&A_1&+&A_1&-&40^{\circ}&=&180^{\circ}&&&\\ &&&&4A_1&-&40^{\circ}&=&180^{\circ}&&\\ \\ &&&&&&4A_1&=&180^{\circ}&+&40^{\circ}\\ \\ &&&&&&A_1&=&\dfrac{220^{\circ}}{4}&\text{or}&55^{\circ} \end{array}[/latex]

This means [latex]A_2 = 2 (55^{\circ})[/latex] or 110° and [latex]A_3 = 55^{\circ}-40^{\circ}[/latex] or 15°.

Another common geometry word problem involves perimeter, or the distance around an object. For example, consider a rectangle, for which [latex]\text{perimeter} = 2l + 2w.[/latex]

Example 4.5.2

If the length of a rectangle is 5 m less than twice the width, and the perimeter is 44 m long, find its length and width.

[latex]L = 2W - 5 \text{ and } P = 44[/latex]

For a rectangle, the perimeter is defined by:

[latex]P = 2 W + 2 L[/latex]

Substituting for [latex]L[/latex] and the value for the perimeter yields:

[latex]44 = 2W + 2 (2W - 5)[/latex]

Which simplifies to:

[latex]44 = 2W + 4W - 10[/latex]

Further simplify to find the length and width:

[latex]\begin{array}{rrrrlrrrr} 44&+&10&=&6W&&&& \\ \\ &&54&=&6W&&&& \\ \\ &&W&=&\dfrac{54}{6}&\text{or}&9&& \\ \\ &\text{So}&L&=&2(9)&-&5&\text{or}&13 \\ \end{array}[/latex]

The width is 9 m and the length is 13 m.

Other common geometric problems are:

Example 4.5.3

A 15 m cable is cut into two pieces such that the first piece is four times larger than the second. Find the length of each piece.

[latex]P_1 + P_2 = 15 \text{ and } P_1 = 4P_2[/latex]

Combining these yields:

[latex]\begin{array}{rrrrrrr} 4P_2&+&P_2&=&15&& \\ \\ &&5P_2&=&15&& \\ \\ &&P_2&=&\dfrac{15}{5}&\text{or}&3 \end{array}[/latex]

This means that [latex]P_2 =[/latex] 3 m and [latex]P_1 = 4 (3),[/latex] or 12 m.

For questions 1 to 8, write the formula defining each relation. Do not solve.

  • The length of a rectangle is 3 cm less than double the width, and the perimeter is 54 cm.
  • The length of a rectangle is 8 cm less than double its width, and the perimeter is 64 cm.
  • The length of a rectangle is 4 cm more than double its width, and the perimeter is 32 cm.
  • The first angle of a triangle is twice as large as the second and 10° larger than the third.
  • The first angle of a triangle is half as large as the second and 20° larger than the third.
  • The sum of the first and second angles of a triangle is half the amount of the third angle.
  • A 140 cm cable is cut into two pieces. The first piece is five times as long as the second.
  • A 48 m piece of hose is to be cut into two pieces such that the second piece is 5 m longer than the first.

For questions 9 to 18, write and solve the equation describing each relationship.

  • The second angle of a triangle is the same size as the first angle. The third angle is 12° larger than the first angle. How large are the angles?
  • Two angles of a triangle are the same size. The third angle is 12° smaller than the first angle. Find the measure of the angles.
  • Two angles of a triangle are the same size. The third angle is three times as large as the first. How large are the angles?
  • The second angle of a triangle is twice as large as the first. The measure of the third angle is 20° greater than the first. How large are the angles?
  • Find the dimensions of a rectangle if the perimeter is 150 cm and the length is 15 cm greater than the width.
  • If the perimeter of a rectangle is 304 cm and the length is 40 cm longer than the width, find the length and width.
  • Find the length and width of a rectangular garden if the perimeter is 152 m and the width is 22 m less than the length.
  • If the perimeter of a rectangle is 280 m and the width is 26 m less than the length, find its length and width.
  • A lab technician cuts a 12 cm piece of tubing into two pieces such that one piece is two times longer than the other. How long are the pieces?
  • An electrician cuts a 30 m piece of cable into two pieces. One piece is 2 m longer than the other. How long are the pieces?

Answer Key 4.5

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Harder Geometric Word Problems

Intro Basic Problems Triangle Formulas Harder Problems The Box & The Goat Problems Max / Min Problems

Some geometrical exercises are a bit more difficult, owing to the context in which they're put (that is, because the math has been heavily wrapped in the "real world"); others are harder due to the complexity of the exercise statement (such as "two times that is three less than four times the other" kinds of phrasing).

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These are the kinds of word problems for which being neat and organized, labelling things clearly, and taking your time can make the difference between fairly quick success and really lengthy frustration and possible failure.

Just give yourself permission to step back, decompress, and then dive in. You can totally do these.

  • 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. Because he wants a nicer grade of fence along the narrow street-facing side of his plot, these dimensions will determine some of the details of the order, so you do need the information. But you don't want the customer to think that you're an idiot, so you need 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:

2 L + 2 w = 148

The area A is given by:

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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 or collection) of equations:

I can solve either one of these equations for either one of the variables, and then plug the result into the other equation for the other variable. I think I'll solve the addition equation for L in terms of w ; then I'll plug the result into the multiplication equation:

L = 74 − w

(74 − w ) × w = 1320

74 w − w 2 = 1320

0 = w 2 − 74 w + 1320

0 = ( w − 30)( w − 44)

Once again, I've come up with two valid solutions. Neither is negative, so I can't discard either of them. I'll check each.

If w = 30 , then:

= 74 − 30 = 44

If w  = 44 , then:

= 74 − 44 = 30

So, in this particular (and unusual) case, each of the solutions is valid; and taking one solution for the width will give the other solution as the length, and vice versa.

The important point is that the shorter side (whether I refer to it as the "width" or the "length") is the "narrow" side that is across the front of the lot.

garden dimensions: 44  ft by 30  ft

frontage length: 30  ft

Note that we cannot say which of the dimensions is the length or the width, since no information was provided regarding which was longer. So "this by that" is as accurate an answer as we can give.

  • Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeter. Find the dimensions of the rectangle.

The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width w . I'll start by doing things orderly, with clear and complete labelling:

the width: w

three times the width: 3 w

twice its length: 2 L

exceeds by three inches: + 3

equation: 3 w = 2 L + 3

The second statement, "four times its length is twelve more than its perimeter", compares the length L and the perimeter P . I will again be complete with my labelling:

four times its length: 4 L

perimeter: P = 2 L + 2 w

twelve more than: + 12

equation: 4 L = P + 12

I'll replace the perimeter variable with the expression (above) for the perimeter:

4 L = (2 L + 2 w ) + 12

So now I have my two equations:

3 w = 2 L + 3

4 L = 2 L + 2 w + 12

There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I'll take the first equation and solve for w :

w = ( 2 / 3 ) L + 1

Now I'll simplify the second equation, and then plug in this above expression for w :

2 L = 2 w + 12

2 L = 2[ ( 2 / 3 ) L + 1 ] + 12

2 L = ( 4 / 3 ) L + 2 + 12

2 L = ( 4 / 3 ) L + 14

2 L − ( 4 / 3 ) L = 14

( 6 / 3 ) L − ( 4 / 3 ) L = 14

( 2 / 3 ) L = 14

L = (14)×( 3 / 2 ) = 21

= ( 2 / 3 )×(21) + 1

= 14 + 1 = 15

The question didn't ask me to "Find the values of the variables L and w ". It asked me to "Find the dimensions of the rectangle," so my hand-in answer is:

length: 21 inches

width: 15 inches

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Geometric sequence word problems

This lesson will show you how to solve a variety of geometric sequence word problems.

Example #1:

The stock's price of a company is not doing well lately. Suppose the stock's price is 92% of its previous price each day. What is the stock's price after 10 days if the stock was worth $2500 right before it started to go down?  

Stock's price is declining

To solve this problem, we need the geometric sequence formula shown below.

a n  = a 1  × r (n - 1)

a 1  = original value of the stock  = 2500

a 2  = value of the stock after 1 day

a 11  = value of the stock after 10 days

a 11 = 2500 × (0.92) (11 - 1)

a 11  = 2500 × (0.92) 10

a 11  = 2500 × 0.434

a 11  = $1085

The stock's price is about 1085 dollars.

Example #2:

The third term of a geometric sequence is 45 and the fifth term of the geometric sequence is 405. If all the terms of the sequence are positive numbers, find the 15th term of the geometric sequence.

Solution To solve this problem, we need the geometric sequence formula shown below.

a n  = a 1  × r (n - 1)

Find the third term

a 3  = a 1  × r (3 - 1)

a 3  = a 1  × r 2

Since the third term is 45,  45 = a 1  × r 2 ( equation 1 )

Find the fifth term

a 5 = a 1  × r (5 - 1)

a 5  = a 1  × r 4

Since the fifth term is 405,  405 = a 1  × r 4 ( equation 2 )

Divide equation 2 by equation 1 .

(a 1  × r 4 ) / (a 1  × r 2 ) = 405 / 45

Cancel a 1 since it is both on top and at the bottom of the fraction.

r 4 / r 2 = 9

r = ±√9

r = ±3

Use r  = 3, and equation 1 to find a 1

45 = a 1  × (3) 2

45 = a 1  × 9

a 1 = 45 / 9 = 5

Since all the terms of the sequence are positive numbers, we must use r = 3 if we want all the terms to be positive numbers.

Let us now find a 15

a 15 = 5 × (3) (15 - 1)

a 15 = 5 × (3) 14  

a 15  = 5 × 4782969 

a 15  =  23914845

Challenging geometric sequence word problems

Example #3:

Suppose that the magnification of a PDF file on a desktop computer is increased by 15% for each level of zoom. Suppose also that the original length of the word " January " is 1.2 cm. Find the length of the word " January " after 6 magnifications.

a 1  = original length of the word  = 1.2 cm

a 2 = length of the word after 1 magnification

a 7 = length of the word after 6 magnifications

r = 1 + 0.15 = 1.15

a 7  = 1.2 × (1.15) (7 - 1)

a 7 = 1.2 × (1.15) 6

a 7 = 1.2 × 2.313

a 7  = 2.7756

After 6 magnifications, the length of the word "January" is 2.7756 cm.

Notice that we added 1 to 0.15. Why did we do that? Let us not use the formula directly so you can see the reason behind it. Study the following carefully !

Day 1 : a 1 = 1.2

Day 2 : a 2 = 1.2 + 1.2 (0.15) = 1.2 (1 + 0.15)

Day 3 : a 3 =  1.2(1 + 0.15) + [ 1.2(1 + 0.15) ]0.15 =  1.2(1 + 0.15) (1 + 0.15) = 1.2(1 + 0.15) 2

Day 7 : a 7 = 1.2(1 + 0.15) 6

Suppose that you want a reduced copy of a photograph. The actual length of the photograph is 10 inches. If each reduction is 64% of the original, how many reductions, will shrink the photograph to 1.07 inches.

a 1  = original length of the photograph  = 10 inches

a 2  = length of the photograph after 1 reduction

n = number of reductions = ?

1.07 = 10 × (0.64) (n - 1)

Divide both sides by 10

1.07 / 10 = [10 × (0.64) (n - 1) ] / 10

0.107 = (0.64) (n - 1)

Notice that you have an exponential equation to solve. The biggest challenge then is knowing how to solve exponential equations !

Take the natural log of both sides of the equation.

ln(0.107) = ln[(0.64) (n - 1) ]

Use the power property of logarithms .

ln(0.107) = (n - 1)ln(0.64)

Divide both sides of the equation by ln(0.64)

ln(0.107) / ln(0.64) = (n - 1)ln(0.64) / ln(0.64)

n - 1 = ln(0.107) / ln(0.64)

Use a calculator to find ln(0.107) and ln(0.64)

n - 1 = -2.23492644452 \ -0.44628710262

n - 1 = 5.0078

n = 1 + 5.0078

Therefore, you will need 6 reductions.

Geometric sequence

Arithmetic sequence word problems

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How to Solve Word Problems Requiring Quadratic Equations

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wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 14,258 times.

Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

Quadratic Equations

Step 1 Know what kind of problem you're tackling.

  • For the real life scenarios, factoring method is better.
  • In geometric problems, it is good to use the quadratic formula.

Real Life Scenario

Step 1 Ask to yourself,

  • In this problem, it asks for Kenny's birthday.

Step 2 Decide your variables.

  • Since negative month does not exist, 3 is the only one that makes sense.
  • Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

Geometric Problems

Step 1 Identify if it's a geometric problem.

  • In the problem above, it asks you only for the height of the triangle.

Step 3 Decide your variables.

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Algebra: Geometry Word Problems - Area

Related Topics: More Lessons for Algebra Math Worksheets

How to solve geometry word problems that involve geometric figures and angles described in words? You would need to be familiar with the formulas in geometry. Making a sketch of the geometric figure is often helpful.

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  • How do you solve word problems?
  • To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
  • How do you identify word problems in math?
  • Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
  • Is there a calculator that can solve word problems?
  • Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
  • What is an age problem?
  • An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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  2. Geometric Sequence Word Problems Worksheet

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  3. Geometric Sequence Word Problems Worksheets

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  4. Solving a Geometry Word Problem by Using Quadratic Equations

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  5. How to Solve Geometric Sequence Word Problems

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  6. Question Video: Using Geometric Sequences to Solve Word Problems

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VIDEO

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  5. Identify and solve problems involving geometric sequences LESSON 5

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COMMENTS

  1. Geometry Word Problems (video lessons, examples and solutions)

    Solution: Step 1: Assign variables: Let x = length of the equal side. Sketch the figure. Step 2: Write out the formula for perimeter of triangle. P = sum of the three sides Step 3: Plug in the values from the question and from the sketch. 50 = x + x + x + 5 Combine like terms 50 = 3x + 5 Isolate variable x 3x = 50 - 5 3x = 45 x =15 Be careful!

  2. 4.5 Geometric Word Problems

    ( A 1). Find the three angles. The relationships described in equation form are as follows: A2 = 2A1 and A3 = A1 −40∘ A 2 = 2 A 1 and A 3 = A 1 − 40 ∘ Because the shape in question is a triangle, the interior angles add up to 180°. Therefore: A1 + A2 +A3 = 180∘ A 1 + A 2 + A 3 = 180 ∘ Which can be simplified using substitutions:

  3. Geometry Word Problems

    A variety of geometry word problems along with step by step solutions will help you practice lots of skills in geometry. Word problem #1: The measure of one supplementary angle is twice the measure of the second. What is the measure of each angle? Let x be the measure of the first angle. Then, the second angle is 2x.

  4. Word Problems involving GEOMETRIC SEQUENCE

    25 Share 1.9K views 2 years ago Grade 10 Math Tutorials Math Learners! In this video, you will learn how to solve word problems involving Geometric Sequences. This video has 4 Word Problems...

  5. What is different about geometry word problems?

    In order to solve geometric word problems, you will need to have memorized some geometric formulas for at least the basic shapes (circles, squares, right triangles, etc). You will usually need to figure out from the word problems which formulas to use — and many times you will need more than one formula for one exercise. MathHelp.com

  6. How to do complicated geometric word problems

    The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width w. I'll start by doing things orderly, with clear and complete labelling: the width: w. three times the width: 3w. twice its length: 2L. exceeds by three inches: + 3. equation: 3w = 2L + 3.

  7. Geometry Word Problems

    K12 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Students learn to solve "Geometry" word problems involving perimeters of rectangles and measures of complementary and supplementary angles.

  8. Geometric Sequence Word Problems

    The third term of a geometric sequence is 45 and the fifth term of the geometric sequence is 405. If all the terms of the sequence are positive numbers, find the 15th term of the geometric sequence. Solution To solve this problem, we need the geometric sequence formula shown below. a n = a 1 × r (n - 1) Find the third term. a 3 = a 1 × r (3 - 1)

  9. Solve a scale drawing word problem

    See how we solve a word problem by using a scale drawing and finding the scale factor.Practice this lesson yourself on KhanAcademy.org right now: https://www...

  10. How to Solve Geometric Sequence Word Problems

    Visit http://www.mathtestace.com for more videos and math help!http://www.mathtestace.com/geometric-sequence-word-problems/

  11. Geometry Word Problems Worksheets

    Worksheet Students will write equations to detail the dimensions of the perimeter of the geometric shapes and solve for the variables. Ten problems are provided. Practice Write an equation and solve problems like: The difference of a number x and 12 whole multiplied by 6, the result is 18. Drill

  12. How to Solve a Geometry Word Problem

    In order to solve a geometry word problem, you need to read the problem carefully, recognize shapes in the drawing, pay attention to labels, and use whatever formulas you have to help you answer the question. In the following problem, you get to work with a picture. Mr. Dennis is a farmer with two teenage sons.

  13. How to Solve Word Problems in Geometry (How to Solve Word Problems

    Geometry word problems are abstract and especially hard to solve--this guide offers detailed, easy-to-follow solution procedures. Emphasizes the mechanics of problem-solving. Includes worked-out problems and a 50-question self-test with answers.

  14. Geometric Sequence word problems ! ! ! ! !

    PRECALCULUS PLAYLIST: https://goo.gl/zCAA7g_____In this video tutorial you will learn how to solve a Word Problem involving Geometric Sequences as ...

  15. 3 Ways to Solve Word Problems Requiring Quadratic Equations

    5. Write down any geometric formula that you need to solve the problem. Since the problem gives us the base, the height, and the area of a triangle, we can use the formula. a = b h 2 {\displaystyle a= {\frac {bh} {2}}} 6. Plug the values in the formula. Be sure to use the relationship you got in step three.

  16. Algebraic word problems

    To solve an algebraic word problem: Define a variable. Write an equation using the variable. Solve the equation. If the variable is not the answer to the word problem, use the variable to calculate the answer. It's important for us to keep in mind how we define our variables.

  17. Geometry Area Word Problems (solutions, videos, examples)

    Solution: Step 1: Assign variables: Let x = original width of rectangle Sketch the figure Step 2: Write out the formula for area of rectangle. A = lw Step 3: Plug in the values from the question and from the sketch. 60 = (4 x + 4) ( x -1) Use distributive property to remove brackets 60 = 4 x2 - 4 x + 4 x - 4 Put in Quadratic Form 4 x2 - 4 - 60 = 0

  18. Word Problems Calculator

    Pre Calculus Calculus Functions Trigonometry Full pad Examples Frequently Asked Questions (FAQ) How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking.

  19. Solving Quadratic Word Problems

    In this video we learn how to solve geometric word problems using quadratic equations. We use both factoring and the quadratic formula to solve. Algebra 1: ...

  20. Solving Word Problems in Geometry: Using Algebraic Principles ...

    Solving Word Problems in Geometry.pdf - Free download as PDF File (.pdf), Text File (.txt) or read online for free. This document provides examples of how to solve word problems involving geometry concepts using algebra. It explains that geometry word problems often require algebra to solve. Three examples are given that demonstrate solving for unknown side lengths or angles of shapes like ...