Students learn how to find the perimeter in elementary school. Then this information is constantly used throughout the course of mathematics and geometry.
Theory common to all figures
The parties are usually denoted in Latin letters. Moreover, they can be designated as segments. Then you will need two letters for each side and written in large letters. Or enter the designation with one letter, which will necessarily be small.
Letters are always chosen alphabetically. For a triangle, they will be the first three. The hexagon will have 6 of them - from a to f. This is useful for entering formulas.
Now about how to find the perimeter. It is the sum of the lengths of all sides of the figure. The number of terms depends on its type. The perimeter is denoted by the Latin letter P. The units of measurement are the same as those given for the sides.
Perimeter formulas for different shapes
For a triangle: P \u003d a + b + c. If it is isosceles, then the formula is converted: P \u003d 2a + c. How to find the perimeter of a triangle if it is equilateral? This will help: P \u003d 3a.
For an arbitrary quadrilateral: P=a+b+c+d. Its special case is the square, the perimeter formula: P=4a. There is also a rectangle, then the following equality is required: P \u003d 2 (a + b).
What if you don't know the length of one or more sides of a triangle?
Use the cosine theorem if there are two sides among the data and the angle between them, which is denoted by the letter A. Then, before finding the perimeter, you will have to calculate the third side. For this, the following formula is useful: c² \u003d a² + b² - 2 av cos (A).
A special case of this theorem is the one formulated by Pythagoras for a right triangle. In it, the value of the cosine of the right angle becomes equal to zero, which means that the last term simply disappears.
There are situations when you can find out how to find the perimeter of a triangle on one side. But at the same time, the angles of the figure are also known. Here the sine theorem comes to the rescue, when the ratios of the lengths of the sides to the sines of the corresponding opposite angles are equal.
In a situation where the perimeter of a figure needs to be found by area, other formulas will come in handy. For example, if the radius of the inscribed circle is known, then in the question of how to find the perimeter of a triangle, the following formula is useful: S \u003d p * r, here p is the semi-perimeter. It must be derived from this formula and multiplied by two.
Task examples
First condition. Find the perimeter of a triangle whose sides are 3, 4 and 5 cm.
Solution. You need to use the equality that is indicated above, and simply substitute the data in the value task into it. The calculations are easy, they lead to the number 12 cm.
Answer. The perimeter of a triangle is 12 cm.
Second condition. One side of the triangle is 10 cm. It is known that the second is 2 cm larger than the first, and the third is 1.5 times larger than the first. It is required to calculate its perimeter.
Solution. In order to find out, you need to count two sides. The second is defined as the sum of 10 and 2, the third is equal to the product of 10 and 1.5. Then it remains only to count the sum of three values: 10, 12 and 15. The result will be 37 cm.
Answer. The perimeter is 37 cm.
Third condition. There is a rectangle and a square. One side of the rectangle is 4 cm, and the other is 3 cm longer. It is necessary to calculate the value of the side of the square if its perimeter is 6 cm less than that of the rectangle.
Solution. The second side of the rectangle is 7. Knowing this, it is easy to calculate its perimeter. The calculation gives 22 cm.
To find out the side of the square, you must first subtract 6 from the perimeter of the rectangle, and then divide the resulting number by 4. As a result, we have the number 4.
Answer. The side of the square is 4 cm.
The ability to find the perimeter of a rectangle is very important for solving many geometric problems. Below is a detailed instruction on finding the perimeter of different rectangles.
How to find the perimeter of a regular rectangle
A regular rectangle is a quadrilateral whose parallel sides are equal and all angles = 90º. There are 2 ways to find its perimeter:
Add up all sides.
Calculate the perimeter of the rectangle, if its width is 3 cm, and its length is 6.
Solution (sequence of actions and reasoning):
- Since we know the width and length of the rectangle, finding its perimeter is not difficult. The width is parallel to the width, and the length is the length. Thus, in a regular rectangle, there are 2 widths and 2 lengths.
- Add up all sides (3 + 3 + 6 + 6) = 18 cm.
Answer: P = 18 cm.
The second way is as follows:
You need to add the width and length, and multiply by 2. The formula for this method is as follows: 2 × (a + b), where a is the width, b is the length.
As part of this task, we get the following solution:
2x(3 + 6) = 2x9 = 18.
Answer: P = 18.
How to find the perimeter of a rectangle - square
A square is a regular quadrilateral. Correct because all its sides and angles are equal. There are two ways to find its perimeter:
- Add up all of its sides.
- Multiply its side by 4.
Example: Find the perimeter of a square if its side = 5 cm.
Since we know the side of the square, we can find its perimeter.
Add up all sides: 5 + 5 + 5 + 5 = 20.
Answer: P = 20 cm.
Multiply the side of the square by 4 (because everyone is equal): 4x5 = 20.
Answer: P = 20 cm.
How to Find the Perimeter of a Rectangle - Online Resources
While the steps above are easy to understand and master, there are several online calculators that can help you calculate the perimeters (area, volume) of different shapes. Just type in the required values and the mini-program will calculate the perimeter of the shape you need. Below is a short list.
Building a lesson:
- Organization and motivation of students to activities in the classroom.
- Organization of the perception of new material based on visual material
- Organization of comprehension.
- Primary check of understanding of new material.
- Organization of primary consolidation and independent analysis of educational information.
- Application of acquired knowledge in the workshop.
Lesson Objectives:
- Educational. Ensure that students learn to find the area and perimeter of geometric shapes;
visual perception of the material in the lesson; understand what area and perimeter are.
2. Developing. Use developmental exercises in the lesson, activate
mental activity of students.
3. Educational. Ensure the development of value-semantic culture of students;
motivation for the ability to correctly achieve the goal -
coincidence of expectation and result.
Equipment:
- M.I.Moro and others. “Mathematics” - a textbook for the 3rd grade of elementary school, part 1.
- Mathematics workbook.
- Pen, ruler, simple pencil, triangle, scissors.
- Models of geometric figures for finding the area.
- Above the board are posters with formulas for finding area and perimeter.
Means of education:
- didactic material.
- Visual aids.
Teaching methods:
- Comparison of items.
- Comparison of methods for finding the area of the same figure.
During the classes.
1. Organizational moment and message of the topic of the lesson.
Teacher: Hello guys. Today we will continue our study of a large topic called “Area and Perimeter”. The topic of our lesson today: “The ability to apply knowledge in finding the perimeter and area of a complex figure.” A complex figure is a geometric figure consisting of several simple figures. First, let's review what we have learned in the previous lessons.
II. Verbal counting.
Development tasks.
Teacher: Find the area of this figure if the side of the square is 1 cm.
The figure is shown on the board.
Student: If 1 square has an area of 1 cm 2, and 5 squares are shown, then the area of \u200b\u200bthis figure is 5 cm 2.
Teacher: Right. Next task. Remove 3 sticks to leave 3 such squares.
The student goes to the blackboard and removes 3 sticks.
Teacher: Remove 4 sticks so that 3 of the same squares remain.
The student goes to the blackboard and removes 4 sticks. Solution.
III. Work on the topic of the lesson
Teacher: What geometric shapes do you already know?
Student: Rectangle.
Student: Square.
Teacher: Right. What do we know about the square?
Student: A square has 4 sides and 4 corners.
Teacher: Right. What are the properties of the sides of a square?
Student: They are equal.
Teacher: Right. What are the angles of a square?
Student: They are straight.
Teacher: How can we build a right angle?
Student: With the help of a triangle.
Teacher: Let's build a square with a side of 4 cm in your notebook. What tools will we use to draw a square?
Student: With a ruler, a pencil and a triangle.
Pupils in notebooks build a square and color it.
Teacher: This is a geometric figure. How to find the perimeter and area of this square?
Student: The perimeter is the sum of all its sides. There are 4 sides to the square. So, add 4 4 times.
Teacher: How do you write it down?
Students write in their notebooks: Find the area of figure F1”.
The student is called to the board, and he writes: P \u003d 4 + 4 + 4 + 4 \u003d 16 (cm)
Students write in notebooks.
Teacher: In what units is the perimeter still measured?
Student: In centimeters, in millimeters, in meters, in decimeters, in kilometers.
Teacher: Well done! How else can you write the perimeter?
Student: By multiplication.
The student writes on the board: P \u003d 4 4 \u003d 16 (cm)
Students write in notebooks.
Teacher: What is the area of the square?
Student: Multiply the length of the square by its width. Since the sides of a square are equal, then
S \u003d 4 4 \u003d 16 (cm 2)
Pupils make an entry in a notebook and write down - “ Answer: S = 16 cm 2”.
Teacher: What other units of area do you know?
Student: square centimeter, square decimeter, square meter, square millimeter.
Teacher: And now let's complicate the task. There is a card in front of you.
This card shows a square the same as in your notebook. In the middle of this square is another square with a side of 2 cm. Now you will take scissors and carefully cut out this small square.
Pupils do this work and write in a notebook: “ Find the area of figure F2”.
Teacher: We got a figure “with a window” - F2. How can you find the area of this interesting figure? The area of the square is already known and is equal to 16 cm 2.
Student: You need to find the area of a small square with a side of 2 cm.
The student goes to the blackboard and writes down - S2 = 2 2 = 4 (cm 2)
Students write in notebook
Student: Subtract the area of the small square from the area of the large square.
Teacher: Right.
The student writes on the board - S = S1 - S2 = 16 - 4 = 12 (cm 2)
Students make notes in their notebooks.
Teacher: Look carefully at this figure and tell me, how else can you measure the area? Is it possible to somehow cut this figure to get the shapes you already know?
Students think and say different options.
One of the options turned out to be very interesting.
Student: You can cut it so that you get rectangles and shows on the board how this can be done. 
The students cut the figure as shown on the board.
Teacher: What is the area of a rectangle?
Student: You need to multiply the length by the width.
Teacher: You got four figures. What can be said about them?
Student: Two figures, like twins, are the same, and the second two are also the same.
You can find the area of one figure and multiply by 2.
The student decides on the board: S1 = 1 4 = 4 (cm 2)
S2 = 1 2 = 2 (cm2)
S \u003d 2 S1 + 2 S2 \u003d 2 4 + 2 2 \u003d 8 + 4 \u003d 12 (cm 2)
Teacher: Well done! We got the same area value as before.
Pupils write in a notebook - " Answer: S = 12 cm2.”
Teacher: Are you tired?
It's time to rest.
suggest fatigue
Remove physical education.
IV. Fizkultminutka.
Every day in the morning
We do exercises (walking in place).
We like to do it in order:
It's fun to walk (walking),
Hands up (hands up)
Squat and stand up (squat 4-6 times),
Jump and jump (10 jumps).
Teacher: And now sit down at the desks and
look at the next model. Figure F3
How to find the area of this interesting figure?
Student: A triangle that protrudes
can be cut off and substituted in the part where
the triangle "goes" inward.
Teacher: Let's take scissors, cut off a triangle and substitute it in the upper part.
What kind of figure do we have?
Student: Rectangle!
Teacher: How to find the area of this rectangle,
If the parties are unknown to us.
Student: We can take a ruler and measure
the length and width of the rectangle.
Students write down - Find the area of figure F3”.
Students measure the length and width with a ruler. It turns out the length, a \u003d 6 cm, width b \u003d 2 cm.
Student: The area of this figure is S = 6 2 = 12 (cm 2).
Pupils make a note in a notebook and write down - “ Answer: S \u003d 12 cm 2.
Teacher: But that's not all. Here is the next figure. We need to find its area.
What is the figure in front of you?
Student: Triangle. But the area of the triangle
we can't find it!
Teacher: It's true. From this triangle
let's make a rectangle. I'll give you a hint. Figure F4
First, we will fold this triangle in half
Students: We got it! right
flip the side.
Get a rectangle.
Student: Measure with a ruler
length a and width b, and by S = a b,
find the area.
Teacher: If we are measuring, we
we get that the length
will be expressed in mm and the width in cm,
what should we do?
Student: Be sure to convert the length and width into one unit of measurement.
Students write in their notebooks: Find the area of figure F4”.
V. Work in pairs.
Teacher: And now I propose to work in pairs. There are two of you at the desk. One student (Option I) finds the perimeter of this figure, and the second (Option II) finds the area.
To do this, draw this figure in a notebook. After you complete the task, exchange notebooks and check the results with each other. 
Students complete the task and the results
write down in a notebook.
Teacher: What did you get?
Student: A square with a side of 3 cm. P \u003d 3 4 \u003d 12 (cm)
S \u003d 3 3 \u003d 9 (cm 2) 3 cm
Students write down: Answer: P \u003d 12 cm, S \u003d 9 cm 2.
Teacher: Well done! And now I suggest you work on your own.
Find the area of the next figure. She lies in front of you.
VI. Independent work to consolidate the studied material.
The teacher distributes pre-prepared figures.
Students independently, without the help of a teacher, cut this figure, get three rectangles. 
Students write down: Find the area of figure F5”.
Students find S1 = 4 3 = 12 (cm 2), S2 = 2 1 = 2 (cm 2), then find the area of this figure: S = S1 + S2 + S2 = 12 + 2 + 2 = 16 (cm 2 ) and make an entry in a notebook, then
write: " Answer: S = 16 cm 2”.
Teacher: Did you like the lesson?
Students: Yes.
Teacher: What have you learned in this lesson?
Student: We have learned how to find the area and perimeter of complex shapes. It turned out to be very simple. You need to think a little and rebuild or remake this figure into the one, the perimeter and area, which we already know how to find.
Teacher: I'm very glad you liked it. At home, repeat the formulas for finding the perimeter and area of \u200b\u200ba square and rectangle; remember how to translate one unit
to another. The following students answered well today. . .
The teacher gives grades.
VII. Homework: textbook p. 77 No. 8.
It is enough to find the length of all its sides and find their sum. The perimeter is the total length of the boundaries of a flat figure. In other words, it is the sum of the lengths of its sides. The unit of measurement of the perimeter must match the unit of measurement of its sides. The formula for the perimeter of a polygon is P \u003d a + b + c ... + n, where P is the perimeter, but a, b, c and n are the length of each side. Otherwise, (or the perimeter of a circle) is calculated: the formula p \u003d 2 * π * r is used, where r is the radius and π is a constant number, approximately equal to 3.14. Let's look at a few simple examples that clearly demonstrate how to find the perimeter. As an example, we take such figures as a square, a parallelogram and a circle.
How to find the perimeter of a square
A square is a regular quadrilateral in which all sides and angles are equal. Since all sides of a square are equal, the sum of the lengths of its sides can be calculated using the formula P = 4 * a, where a is the length of one of the sides. Thus, with a side of 16.5 cm it is equal to P \u003d 4 * 16.5 \u003d 66 cm. You can also calculate the perimeter of an equilateral rhombus.
How to find the perimeter of a rectangle
A rectangle is a quadrilateral with all angles equal to 90 degrees. It is known that in such a figure as a rectangle, the lengths of the sides are equal in pairs. If the width and height of a rectangle are the same length, then it is called a square. Usually, the length of a rectangle is called the largest of the sides, and the width is the smallest. Thus, to get the perimeter of a rectangle, you need to double the sum of its width and height: P = 2 * (a + b), where a is the height and b is the width. Given a rectangle with one side 15 cm long and the other side set to 5 cm wide, we get a perimeter equal to P = 2 * (15 + 5) = 40 cm.
How to find the perimeter of a triangle
A triangle is formed by three line segments that join at points (triangle vertices) that do not lie on the same line. A triangle is called equilateral if all three of its sides are equal, and isosceles if there are two equal sides. To find out the perimeter, you need to multiply the length of its side by 3: P \u003d 3 * a, where a is one of its sides. If the sides of the triangle are not equal to each other, it is necessary to carry out the addition operation: P \u003d a + b + c. The perimeter of an isosceles triangle with sides 33, 33 and 44, respectively, will be equal to: P \u003d 33 + 33 + 44 \u003d 110 cm.
How to find the perimeter of a parallelogram
A parallelogram is a quadrilateral with opposite sides parallel in pairs. Square, rhombus and rectangle are special cases of the figure. The opposite sides of any parallelogram are equal, therefore, to calculate its perimeter, we use the formula P \u003d 2 (a + b). In a parallelogram with sides of 16 cm and 17 cm, the sum of the sides, or perimeter, is equal to P \u003d 2 * (16 + 17) \u003d 66 cm.
How to find the circumference of a circle
The circle is a closed straight line, all points of which are located at an equal distance from the center. The circumference of a circle and its diameter always have the same ratio. This ratio is expressed as a constant, written with the letter π, and equals approximately 3.14159. You can find the perimeter of a circle by multiplying the radius times 2 times π. It turns out that the circumference of a circle with a radius of 15 cm will be equal to P \u003d 2 * 3.14159 * 15 \u003d 94.2477
In the following test tasks, you need to find the perimeter of the figure shown in the figure.
There are many ways to find the perimeter of a shape. You can transform the original shape in such a way that the perimeter of the new shape can be easily calculated (for example, change to a rectangle).
Another solution is to look for the perimeter of the figure directly (as the sum of the lengths of all its sides). But in this case, one cannot rely only on the drawing, but find the lengths of the segments based on the data of the problem.
I want to warn you: in one of the tasks, among the proposed answers, I did not find the one that turned out for me.
c) .
Let's move the sides of the small rectangles from the inner area to the outer one. As a result, the large rectangle is closed. Formula for Finding the Perimeter of a Rectangle
In this case, a=9a, b=3a+a=4a. Thus P=2(9a+4a)=26a. To the perimeter of the large rectangle we add the sum of the lengths of four segments, each of which is equal to 3a. As a result, P=26a+4∙3a= 38a .
c) .
After transferring the inner sides of the small rectangles to the outer area, we get a large rectangle, the perimeter of which is P=2(10x+6x)=32x, and four segments, two of x length, two of 2x length.
Total, P=32x+2∙2x+2∙x= 38x .
?) .
Let's move 6 horizontal "steps" from the inside to the outside. The perimeter of the resulting large rectangle is P=2(6y+8y)=28y. It remains to find the sum of the lengths of the segments inside the rectangle 4y+6∙y=10y. Thus, the perimeter of the figure is P=28y+10y= 38y .
D) .
Let's move the vertical segments from the inner area of the figure to the left, to the outer area. To get a big rectangle, move one of the 4x lengths to the bottom left corner.
We find the perimeter of the original figure as the sum of the perimeter of this large rectangle and the lengths of the remaining three segments P=2(10x+8x)+6x+4x+2x= 48x .
e) .
Moving the inner sides of the small rectangles to the outer area, we get a large square. Its perimeter is P=4∙10x=40x. To get the perimeter of the original figure, you need to add the sum of the lengths of eight segments, each 3x long, to the perimeter of the square. Total, P=40x+8∙3x= 64x .
b) .
Let's move all horizontal "steps" and vertical upper segments to the outer area. The perimeter of the resulting rectangle is P=2(7y+4y)=22y. To find the perimeter of the original figure, you need to add to the perimeter of the rectangle the sum of the lengths of four segments, each with a length of y: P=22y+4∙y= 26y .
D) .
Move all horizontal lines from the inner area to the outer area and move the two vertical outer lines in the left and right corners, respectively, z to the left and right. As a result, we get a large rectangle, the perimeter of which is P=2(11z+3z)=28z.
The perimeter of the original figure is equal to the sum of the perimeter of the large rectangle and the lengths of six segments in z: P=28z+6∙z= 34z .
b) .
The solution is completely similar to the solution of the previous example. After transforming the figure, we find the perimeter of the large rectangle:
P=2(5z+3z)=16z. To the perimeter of the rectangle we add the sum of the lengths of the remaining six segments, each of which is equal to z: P=16z+6∙z= 22z .
