Geometry Math Problems - Perimeters

In these lessons, we will learn to solve geometry math problems that involve perimeter.

Related Pages Geometry math problems involving area Area Formula Geometry math problems involving angles More Algebra Word Problems

Geometry word problems involves 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.

Geometry Word Problems Involving Perimeter

Example: A triangle has a perimeter of 50. If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?

Solution: Step 1: Assign variables:

Let x = length of the equal sides 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! The question requires the length of the third side.

The length of third side = 15 + 5 = 20

Answer: The length of third side is 20.

Geometry Math Problem involving the perimeter of a rectangle

The following two videos give the perimeter of a rectangle, a relationship between the length and width of the rectangle, and use that information to find the exact value of the length and width.

Example: A rectangular garden is 2.5 times as long as it is wide. It has a perimeter of 168 ft. How long and wide is the garden?

Example: A rectangular landing strip for an airplane has perimeter 8000 ft. If the length is 10 ft longer than 35 times the width, what is the length and width?

Examples of perimeter geometry word problems This video shows how to write an equation and find the dimensions of a rectangle knowing the perimeter and some information about the about the length and width.

Example: The width of a rectangle is 3 ft less than its length. The perimeter of the rectangle is 110 ft. Find the dimensions.

Perimeter Word Problems

Example: The length of a rectangle is 7 cm more than 4 times its width. Its perimeter is 124 cm. Find its dimensions.

Geometry Math Problem involving the perimeter of a triangle

The following two videos give the perimeter of a triangle, a relationship between the sides of the triangle, and use that information to find the exact value or values of the required side or sides.

Example: Patrick’s bike ride follows a triangular path; two legs are equal, the third is 8 miles longer than the other legs. If Patrick rides 30 miles total, what is the length of the longest leg?

Example: The perimeter of a triangle is 56 cm. The first side is 6 cm shorter than the second side. The third side is 2 cm shorter than twice the length of the first side. What is the length of each side?

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Geometry Practice Questions

1. In a 30-60-90 triangle, the length of the hypotenuse is 6. What is the length of the shortest side?

problem solving geometry questions

2. What is the area of a circle with a diameter of 16?


3. The figure below contains only horizontal and vertical lines. Calculate its perimeter.

problem solving geometry questions

4. Find the measure of the missing angle in the triangle below.

problem solving geometry questions

6. What is the sum of the measures of the interior angles of a hexagon?

7. Find the area of the triangle below.

problem solving geometry questions

8. The figure below is a parallelogram with two angles given in terms of x . Determine the value of x .

problem solving geometry questions

9. Which of the following could be the side lengths of a right triangle?

10. The figure below is an equilateral triangle with sides of length 6. What is the area of the triangle?

problem solving geometry questions

12 + 3 + 9 + 9 + 6 + 3 = 42

4. D. The sum of the angles of a triangle is 180°. Therefore, if we subtract the two given angles from 180°, the result will be the missing angle.

180 – 95 – 35 = 50

Therefore, the missing angle is 50°.

problem solving geometry questions

Last Updated: June 4, 2019

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o Identify some critical steps of the process for solving practical geometry problems

o Apply geometry problem-solving techniques to practical situations

Geometry has a variety of real-life applications in everyday situations. In this article, we will learn to apply geometric principles and techniques to solve problems. The key to solving practical geometry problems is translation of the real-life situation into figures, measurements, and other information necessary to represent the situation conceptually. For instance, you already know how to calculate the area of a composite figure; if you were asked to determine how much floor space is available in a certain building with a composite shape, you would simply need to apply the same principles as you would use for calculating the area of a composite figure. Some measurements of the building might, of course, be required, but the same problem-solving techniques apply.

It behooves us to present a basic approach to solving practical geometry problems. This approach is similar to that for solving almost a word problem, but is geared slightly more toward the characteristics of geometry problems in particular.

1. Determine what you need to calculate to solve the problem. In some cases, you may need a length; in others, an area or angle measure. If you are conscious throughout the process of what you need to determine, you can save yourself a significant amount of time.

2. Draw a diagram. Sometimes a straightedge, compass, protractor, or some combination of these tools can be helpful. Even if you only use a rough sketch, however, making a visual representation of the problem can help you organize your thoughts and keep track of important information such as the relationship of line segments and angles as well as the measures thereof.

3. Record all appropriate measurements. If you are calculating an area, for instance, you may need to take measurements of certain lengths (alternatively, these may be provided to you). In either case, record them and mark them in some manner on your diagram.

4. Pay attention to units. Using units of square meters for a length or angle measure can be an embarrassing mistake! Keep careful track of the units you are using throughout the problem. If no units are given, simply use the generic term "units" in place of inches or meters, for example.

5. Divide the figure, if necessary, into manageable portions. If your diagram is a composite figure, it may help to divide the figure into bite-sized portions that you can handle.

6. Identify any appropriate geometric relationships. This step can greatly simplify the problem. Perhaps you can show two triangles to be congruent or similar, or perhaps you can identify congruent segments or angles. Use this step to fill in as much missing information in your diagram as you can.

7. Do the math. At this point, you need to apply what you've learned to analyze the figure and other data to solve the problem. You may, for example, need to apply the Pythagorean theorem, or you may need to calculate the perimeter of a figure. Whatever the details of the problem, you will need to apply your skills in geometry in an appropriate manner.

8. Check your results. Take a look at your answer in the context of your diagram-does your answer make sense? A result of millions of square meters for the area of a figure with dimensions in the range of a few meters should tell you that you've made an error at some point in your analysis.

Not every step of the approach outlined above will be needed in every problem. You must use your best judgment in determining what is necessary to solve the problem in a satisfactory and time-efficient manner. Also, you may not always think to use the exact progression of steps above; the outline is simply a way to describe a systematic approach to problem solving. The remainder of this article provides you the opportunity to test your geometry skills by way of several practice problems. Obviously, these problems do not require you to go out and make any measurements of lengths or angles, but keep in mind that problems you encounter in everyday life may require you to do so!

Practice Problem : The floor plan of a house is shown below. Determine the area covered by the house.

Solution : Let's first divide the diagram of the house into two rectangles and a trapezoid, since we can calculate the area of each of these figures. Using the properties of each figure, we can also fill in some of the unknown information.

Now, the area of the larger rectangle is the product of 40 feet and 20 feet, or 800 square feet. The area of the smaller rectangle is 25 feet times 6 feet, or 150 square feet. The area of the trapezoid is the following:

The height ( h ) is 6 feet, and the two bases ( b 1 and b 2 ) are 8 and 11 feet.

Adding all three areas gives us a total area of the house of 1,007 square feet.

Practice Problem : A hiker is walking up a steep hill. The slope of the hill between two trees is constant, and the base of one tree is 100 meters higher than the other. If the horizontal distance between the trees is 400 meters, how far must the hiker walk to get from one tree to the next?

Solution : Because this problem may be difficult to envision, a diagram is extremely helpful. Notice that the base of the trees differ in height by 100 meters--this is our vertical distance for the walk. The horizontal distance is 400 meters.

Note that we have shown the right angle because horizontal and vertical segments are perpendicular. We can now use the Pythagorean theorem to calculate the distance d the hiker must walk.

Thus, the hiker must walk about 412 meters. Note that although the hiker makes a significant (100 meter) change in elevation over this walk, the difference between the actual distance he walks and the horizontal distance is small--only about 12 meters.

Practice Problem : A homeowner has a rectangular fenced-in yard, and he wants to put mulch on his triangular gardens, as shown below. The inside border of each garden always meets the fence at the same angle. If a bag of mulch covers about 50 square feet, how many bags of mulch should the homeowner buy to cover his gardens?

Solution : We are told in the problem that the inside border of each garden meets the fence at the same angle in every case; thus, we can conclude (as shown below) that the triangles are all isosceles (and that the triangles with the same side lengths are congruent by the ASA condition). We can thus mark each side with an unknown variable x or y .

Recall that the fenced-in area is rectangular; thus the angle in each corner is 90°. We can then solve for x and y using the Pythagorean theorem. Notice first, however, that x and y are the height and base of their respective triangles.

Because the gardens include two of each triangle shape, the total garden area is simply the sum of x 2 and y 2 . (If you do not follow this point, simply use the triangle area formula in each case--you will get the same result.)

Thus, the homeowner needs six bags of mulch (for a total of 300 square feet) to cover his gardens. (Of course, we are assuming here that he must buy a whole number of bags.)

Using Classical Geometric Construction Techniques

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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.

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

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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problem solving geometry questions

Complete Test Preparation Inc.

Basic Geometry Practice Questions with Full Answer Key – Area, Perimeter, Volume & Angles

problem solving geometry questions

High School geometry questions similar to what you will find on a standardized test.

Most popular geometry questions.

Common geometry questions on on standardized tests :

Most Common Geometry Mistakes on a Test

Geometry Practice Test


1. What is measurement of the indicated angle assuming the figure is a square?

a. 45 o b. 90 o c. 60 o d. 30 o


2. What is the sum of all the angles in the rectangle above?

a. 180 o b. 360 o c. 90 o d. 120 o


3. What is the measurement of the indicated angle?

a. 45 o b. 90 o c. 60 o d. 50 o


4. If the line m is parallel to the side AB of ? ABC, what is angle a ?

a. 130 o b. 25 o c. 65 o d. 50 o


5. What is perimeter of the above shape?

a. 12 cm b. 16 cm c. 6 cm d. 20 cm


6. What is (area of large circle) – (area of small circle) in the figure above?

a. 8 п cm 2 b. 10 п cm 2 c. 12 п cm 2 d. 16 п cm 2


7. What is perimeter of ? ABC in the above shape?


8. What is the volume of the figure above?

a. 125 cm 3 b. 875 cm 3 c. 1000 cm 3 d. 500 cm 3


9. What is the volume of the above solid made by a hollow cylinder with half in size of the larger cylinder?

a. 1440 п in 3 b. 1260 п in 3 c. 1040 п in 3 d. 960 п in 3

1. B The diagonals of a square intersect perpendicularly with each other so each angle measures 90 o x =90 o

2. B a+b+c+d = ? The sum of angles around a point is 360 o    a+b+c+d = 360 o

Video Solution

3. C The sum of angles around a point is 360 o d+300 = 360 o d = 60 o

4. D Two parallel lines( m & side AB) intersected by side AC a= 50 o (interior angles)

5. B The square with 2 cm side common to the rectangle apart from 4cm side Perimeter = 2+2+2+4+2+4 = 16 cm

6. C In the figure, we are given a large circle and a small circle inside it; with the diameter equal to the radius of the large one. The diameter of the small circle is 4 cm. This means that its radius is 2 cm. Since the diameter of the small circle is the radius of the large circle, the radius of the large circle is 4 cm. The area of a circle is calculated by: пr 2 where r is the radius.  Area of the small circle: п(2)2 = 4п Area of the large circle: п(4) 2 2 = 16п

The difference area is found by: Area of the large circle – Area of the small circle = 16п – 4п = 12п

7. D Perimeter of a shape with two squares and triangle ABC. Perimeter = 8.5+8.5+6+6 Perimeter = 29 cm.

8. C Large cube is made up of 8 smaller cubes of 5 cm sides. Volume = Volume of small cube x 8 Volume = (5 x 5 x 5) x 8, 125 x 8 Volume = 1000cm 3

9. B Volume= Volume of large cylinder – Volume of small cylinder (Volume of cylinder = area of  base x height) Volume = (п 12 2 x 10) – (п 6 2 x 5), 1440? – 180п Volume= 1260п in 3

See also our tutorial on Complex Shapes

Tag: Basic Math , Geometry , Practice Questions

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answer to question five should be 18, you forgot to include the 2cm outer side of the rectangle

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Divide the figure into a square and a rectangle – perimeter of the square – 3 X 2 = 6 note 3 sides are included in the perimeter. Upper rectangle 2 X 4 = 8 + 2 = 10 total perimeter = 10 + 6 = 16

Try again – the question asks for the PERIMITER – 3 sides of the square – 2 + 2 + 2 = 6; 2 sides of the rectangle – 4 + 4 = 8; 1 side of the rectangle (top) = 2 6 + 8 + 2 = 16

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The line between the square and the rectangle isn’t counted for perimeter, perimeter is the amount of units around the OUTSIDE. Since the square and rectangle are connected, try thinking of them as one big rectangle and forget about the line in between.

Correct – the bottom square has 2 cm each side on 3 sides (2 + 2 + 2 = 6). The top rectangle has 2, 4 cm sides (4 + 4 = 8) and 1, 2 cm side. So the total is 6 + 8 + 2 = 16.

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Please in no 6 ,42 -22 is 20,so how comes about 16 and how did u get 122 in no 10

The answers are correct – I have updated and expanded the solutions – also the п symbol wasn’t displaying properly – thanks!

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102°-2x +78°=180°

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Love this. Super helpful!

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In #9, since it is a cylinder; shouldn’t you use Pi in calculating the area of the base??

hi – yes the answer does use pi – the web rendering is a little funny tho – looks like an ‘n’ but is is pi – e. g. Volume= (п 122x 10) – (п 62x 5), 1440? – 180п

in question #9 what is the formula of a cylinder if the diameter is given and not the radius?

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I understand how you got the answer in question 5, however the way it is displayed is unclear. It is not clear that the number 4 is attached to just the rectangle and not to the line as a whole. This means there are two interpretations of the figure.

1) 4*2+2*2 = 12 2) 2(4+2) + 2*2 = 16

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Thanks! this was super helpful!

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