1) Greeting
2) Lesson motivation The teacher checks the class’s readiness for the lesson; motivates students to formulate a topic.
Read the definition on the board (topic sheet) and insert the concept in question:
The size of that part of the plane occupied by the polygon is ... (area)
A quadrilateral whose opposite sides are parallel in pairs - .... (parallelogram)
A figure made up of three points that do not lie on the same line and three segments that connect them is called .... (triangle)
A figure in which two sides are parallel and the other two are not parallel is called ... (trapezoid)
From the resulting words, try to create the topic of our today's lesson.
So, the topic of the lesson….Areas of a parallelogram, triangle, trapezoid.
Areas, what figures can we find and how?
Calculate the areas of the figures in Fig.
Are there other solutions?
What happened?
What attempts have been made to find the area?
Who tried to find the area of a parallelogram? Tell me.
Derivation of the formula for the area of a parallelogram.
Task.
How to “redraw” a parallelogram to get a rectangle with the same area?
The parallelogram was redrawn into a rectangle. This means that its area is equal to the area of the rectangle.
What are the length and width of a rectangle for a parallelogram?
The area of a parallelogram is equal to the product of its base and its height.
In a parallelogram, the base can be any side. And in order to apply the formula for finding the area, the height must be drawn to the base.
Let's calculate the area of this parallelogram.
Derivation of the formula for the area of a triangle.
How can you redraw or complete a triangle?
The area of a triangle is equal to half the product of its base and height.
What if the triangle is right-angled?
Look at fig.
It can be “redrawn” into a rectangle.
And we find its area using the formula
S =a *b . The length of the rectangle is half of the leg, and the width is the other leg.
The area of a right triangle is equal to half the product of its legs.
Derivation of the formula for the area of a trapezoid.
Look how the treapecia has been “reshaped” - into a triangle. And we find the area of the triangle using the formula:
The base of the triangle is the sum of the lengths of the upper and lower base, and the height of the triangle is the height of the trapezoid.
The area of a trapezoid is equal to the product of half the sum of its bases and its height.
1) Find S steam. , If A=5, h =4.
2) Find S triangle. , If A=3,5; h =2.
3) Find S ladder. , If A=4,5; b = 2,5; h =3.
Complete test tasks (see appendix)
Peer review of independent work.
Solving problems on a new topic:
No. 675(a,d), 676(a,b), 677(a,b)
For weak and underachieving students, individual work on cards has been prepared, which includes tasks in which there is an example of recording the solution.
The teacher offers to answer questions on a new topic.
Guys, let's sum it up!
What did you learn in class today?
What have you learned to do?
What was difficult to decide?
The teacher comments on the homework.
paragraph 23 No. 675(b,c), 676(c,d), 677(c,d)
Well done everyone!
The lesson is over. Goodbye!
Area of a geometric figure- a numerical characteristic of a geometric figure showing the size of this figure (part of the surface limited by the closed contour of this figure). The size of the area is expressed by the number of square units contained in it.
Triangle area formulas
- Formula for the area of a triangle by side and height
Area of a triangle equal to half the product of the length of a side of a triangle and the length of the altitude drawn to this side - Formula for the area of a triangle based on three sides and the radius of the circumcircle
- Formula for the area of a triangle based on three sides and the radius of the inscribed circle
Area of a triangle is equal to the product of the semi-perimeter of the triangle and the radius of the inscribed circle. where S is the area of the triangle,
- lengths of the sides of the triangle,
- height of the triangle,
- the angle between the sides and,
- radius of the inscribed circle,
R - radius of the circumscribed circle,
Square area formulas
- Formula for the area of a square by side length
Square area equal to the square of the length of its side. - Formula for the area of a square along the diagonal length
Square area equal to half the square of the length of its diagonal.S= 1 2 2 where S is the area of the square,
- length of the side of the square,
- length of the diagonal of the square.
Rectangle area formula
- Area of a rectangle equal to the product of the lengths of its two adjacent sides
where S is the area of the rectangle,
- lengths of the sides of the rectangle.
Parallelogram area formulas
- Formula for the area of a parallelogram based on side length and height
Area of a parallelogram - Formula for the area of a parallelogram based on two sides and the angle between them
Area of a parallelogram is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.a b sin α
where S is the area of the parallelogram,
- lengths of the sides of the parallelogram,
- length of parallelogram height,
- the angle between the sides of the parallelogram.
Formulas for the area of a rhombus
- Formula for the area of a rhombus based on side length and height
Area of a rhombus equal to the product of the length of its side and the length of the height lowered to this side. - Formula for the area of a rhombus based on side length and angle
Area of a rhombus is equal to the product of the square of the length of its side and the sine of the angle between the sides of the rhombus. - Formula for the area of a rhombus based on the lengths of its diagonals
Area of a rhombus equal to half the product of the lengths of its diagonals. where S is the area of the rhombus,
- length of the side of the rhombus,
- length of the height of the rhombus,
- the angle between the sides of the rhombus,
1, 2 - lengths of diagonals.
Trapezoid area formulas
- Heron's formula for trapezoid
Where S is the area of the trapezoid,
- lengths of the bases of the trapezoid,
- lengths of the sides of the trapezoid,
Area of a parallelogram
Theorem 1
The area of a parallelogram is defined as the product of the length of its side and the height drawn to it.
where $a$ is a side of the parallelogram, $h$ is the height drawn to this side.
Proof.
Let us be given a parallelogram $ABCD$ with $AD=BC=a$. Let us draw the heights $DF$ and $AE$ (Fig. 1).
Picture 1.
Obviously, the $FDAE$ figure is a rectangle.
\[\angle BAE=(90)^0-\angle A,\ \] \[\angle CDF=\angle D-(90)^0=(180)^0-\angle A-(90)^0 =(90)^0-\angle A=\angle BAE\]
Consequently, since $CD=AB,\ DF=AE=h$, by the $I$ criterion for the equality of triangles $\triangle BAE=\triangle CDF$. Then
So, according to the theorem on the area of a rectangle:
The theorem has been proven.
Theorem 2
The area of a parallelogram is defined as the product of the length of its adjacent sides times the sine of the angle between these sides.
Mathematically this can be written as follows
where $a,\ b$ are the sides of the parallelogram, $\alpha $ is the angle between them.
Proof.
Let us be given a parallelogram $ABCD$ with $BC=a,\ CD=b,\ \angle C=\alpha $. Let us draw the height $DF=h$ (Fig. 2).
Figure 2.
By definition of sine, we get
Hence
So, by Theorem $1$:
The theorem has been proven.
Area of a triangle
Theorem 3
The area of a triangle is defined as half the product of the length of its side and the altitude drawn to it.
Mathematically this can be written as follows
where $a$ is a side of the triangle, $h$ is the height drawn to this side.
Proof.
Figure 3.
So, by Theorem $1$:
The theorem has been proven.
Theorem 4
The area of a triangle is defined as half the product of the length of its adjacent sides and the sine of the angle between these sides.
Mathematically this can be written as follows
where $a,\b$ are the sides of the triangle, $\alpha$ is the angle between them.
Proof.
Let us be given a triangle $ABC$ with $AB=a$. Let's find the height $CH=h$. Let's build it up to a parallelogram $ABCD$ (Fig. 3).
Obviously, by the $I$ criterion for the equality of triangles, $\triangle ACB=\triangle CDB$. Then
So, by Theorem $1$:
The theorem has been proven.
Area of trapezoid
Theorem 5
The area of a trapezoid is defined as half the product of the sum of the lengths of its bases and its height.
Mathematically this can be written as follows
Proof.
Let us be given a trapezoid $ABCK$, where $AK=a,\ BC=b$. Let us draw in it the heights $BM=h$ and $KP=h$, as well as the diagonal $BK$ (Fig. 4).
Figure 4.
By Theorem $3$, we get
The theorem has been proven.
Sample task
Example 1
Find the area of an equilateral triangle if its side length is $a.$
Solution.
Since the triangle is equilateral, all its angles are equal to $(60)^0$.
Then, by Theorem $4$, we have
Answer:$\frac(a^2\sqrt(3))(4)$.
Note that the result of this problem can be used to find the area of any equilateral triangle with a given side.
Let us agree to call one of the sides of the parallelogram basis, and the perpendicular drawn from any point on the opposite side to the line containing the base is parallelogram height.
Theorem
Proof
Let's consider a parallelogram ABCD with area S. Let's take side AD as the base and draw the heights ВН and СК (Fig. 182). Let us prove that S = AD VN.
Rice. 182
Let us first prove that the area of the rectangle ABCD is also equal to S. The trapezoid ABCD is composed of a parallelogram ABCD and a triangle DCK. On the other hand, it is composed of a rectangle НВСК and a triangle АВН. But right triangles DCK and ABH are equal in hypotenuse and acute angle (their hypotenuses AB and CD are equal as opposite sides of a parallelogram, and angles 1 and 2 are equal as the corresponding angles when parallel lines AB and CD intersect with secant AD), so their areas are equal.
Consequently, the areas of the parallelogram ABCD and the rectangle NVSK are also equal, i.e., the area of the rectangle NVSK is equal to S. By the theorem on the area of the rectangle, S = BC BN, and since BC = AD, then S = AD BN. The theorem has been proven.
Area of a triangle
One of the sides of a triangle is often called it basis. If the base is selected, then the word “height” means the height of the triangle drawn to the base. Theorem
Proof
Let S be the area of triangle ABC (Fig. 183). Let's take side AB as the base of the triangle and draw the height CH. Let's prove that 
.
Rice. 183
Let's complete the triangle ABC to the parallelogram ABDC as shown in Figure 183. Triangles ABC and DCB are equal on three sides (BC is their common side, AB = CD and AC = BD as opposite sides of the parallelogram ABDC), so their areas are equal. Therefore, the area S of triangle ABC is equal to half the area of parallelogram ABDC, i.e. . The theorem has been proven.
Corollary 1
Corollary 2
Let us use Corollary 2 to prove the theorem on the ratio of the areas of triangles having equal angles.
Theorem
Proof
Let S and S 1 be the areas of triangles ABC and A 1 B 1 C 1, for which ∠A = ∠A 1 (Fig. 184, a). Let's prove that 
.
Rice. 184
Let's superimpose triangle A 1 B 1 C 1 on triangle ABC so that vertex A 1 aligns with vertex A, and sides A 1 B 1 and A 1 C 1 overlap rays AB and AC, respectively (Fig. 184, b). Triangles ABC and AB 1 C have a common height - CH, therefore 
.
Triangles AB 1 C and AB 1 C 1 also have a common height - B 1 H 1, therefore 
. Multiplying the resulting equalities, we find:
The theorem has been proven.
Area of trapezoid
To calculate the area of an arbitrary polygon, you usually do this: divide the polygon into triangles and find the area of each triangle. The sum of the areas of these triangles is equal to the area of the given polygon (Fig. 185, a). Using this technique, we will derive a formula for calculating the area of a trapezoid. Let us agree to call the altitude of a trapezoid a perpendicular drawn from any point of one of the bases to a line containing the other base. In Figure 185, b, segment BH (as well as segment DH 1) is the height of the trapezoid ABCD.
Rice. 185
Theorem
Proof
Consider the trapezoid ABCD with bases AD and BC, height BH and area S (see Fig. 185, b).
Let's prove that
The diagonal BD divides the trapezoid into two triangles ABD and BCD, so S = S ABD + S BCD.
Let us take segments AD and ВН as the base and height of triangle ABD, and segments ВС and DH 1 as the base and height of triangle BCD. Then
.
The theorem has been proven.
Tasks
459. Let a be the base, h the height, and S the area of the parallelogram. Find: a) S, if a = 15 cm, h = 12 cm; b) a, if S = 34 cm 2, h = 8.5 cm; c) a, if S = 162 cm 2, h = 1/2a; d) h, if h = 3a, S = 27.
460. The diagonal of a parallelogram, equal to 13 cm, is perpendicular to the side of the parallelogram, equal to 12 cm. Find the area of the parallelogram.
461. The adjacent sides of a parallelogram are 12 cm and 14 cm, and its acute angle is 30°. Find the area of the parallelogram.
462. The side of a rhombus is 6 cm, and one of the angles is 150°. Find the area of the rhombus.
463. The side of a parallelogram is 8.1 cm, and the diagonal, equal to 14 cm, forms an angle of 30° with it. Find the area of the parallelogram.
464. Let a and b be the adjacent sides of the parallelogram, S the area, a h 1 and h 2 its heights. Find: a) h 2 if a = 18 cm, b = 30 cm, h 1 = 6 cm, h 2 > h 1 ; b) h 1, if a = 10 cm, 6 = 15 cm, h 2 = 6 cm, h 2 > h 1 c) h 1 and h 2, if S = 54 cm 2, a = 4.5 cm, b = 6 cm.
465. The acute angle of the parallelogram is 30°, and the heights drawn from the vertex of the obtuse angle are 2 cm and 3 cm. Find the area of the parallelogram.
466. The diagonal of a parallelogram is equal to its side. Find the area of a parallelogram if its longest side is 15.2 cm and one of its angles is 45°.
467. A square and a rhombus that is not a square have the same perimeters. Compare the areas of these figures.
468. Let a be the base, h the height, and S the area of the triangle. Find: a) S, if a = 7 cm, h = 11 cm; b) S, if a = 2√3 cm, h = 5 cm; c) h, if S = 37.8 cm 2, a - 14 cm; d) a, if S = 12 cm 2, h = 3√2 cm.
469. Sides AB and BC of triangle ABC are equal to 16 cm and 22 cm, respectively, and the height drawn to side AB is equal to 11 cm. Find the height drawn to side BC.
470. Two sides of a triangle are equal to 7.5 cm and 3.2 cm. The height drawn to the larger side is 2.4 cm. Find the height drawn to the smaller of these sides.
471. D Find the area of a right triangle if its legs are equal: a) 4 cm and 11 cm; b) 1.2 dm and 3 dm.
472. The area of a right triangle is 168 cm 2. Find its legs if the ratio of their lengths is 7/12.
473. Through vertex C of triangle ABC, a straight line m is drawn parallel to side AB. Prove that all triangles with vertices on line m and base AB have equal areas.
474. Compare the areas of two triangles into which a given triangle is divided by its median.
475. Draw triangle ABC. Draw two straight lines through vertex A so that they divide this triangle into three triangles having equal areas.
476. Prove that the area of a rhombus is equal to half the product of its diagonals. Calculate the area of a rhombus if its diagonals are equal to: a) 3.2 dm and 14 cm; b) 4.6 dm and 2 dm.
477. Find the diagonals of a rhombus if one of them is 1.5 times larger than the other, and the area of the rhombus is 27 cm 2.
478. In a convex quadrilateral, the diagonals are mutually perpendicular. Prove that the area of a quadrilateral is equal to half the product of its diagonals.
479. Points D and E lie on sides AB and AC of triangle ABC. Find: a) S ADE, if AB = 5 cm, AC = 6 cm, AD = 3 cm, AE = 2 cm, S ABC = 10 cm 2 ; b) AD, if AB = 8 cm, AC = 3 cm, AE = 2 cm, S ABC = 10 cm 2, S ADE = 2 cm 2.
480. Find the area of trapezoid ABCD with bases AB and CD if:
a) AB = 21 cm, CD = 17 cm, height BH is 7 cm;
b) ∠D = 30°, AB = 2 cm, CD = 10 cm, DA = 8 cm;
c) BC ⊥ AB, AB = 5 cm, BC = 8 cm, CD = 13 cm.
481. Find the area of a rectangular trapezoid whose two smaller sides are 6 cm and the larger angle is 135°.
482. The obtuse angle of an isosceles trapezoid is 135°, and the altitude drawn from the vertex of this angle divides the larger base into segments of 1.4 cm and 3.4 cm. Find the area of the trapezoid.
Answers to problems
459. a) 180 cm 2; b) 4 cm; c) 18 cm; d) 9.
460. 156 cm 2.
461.84 cm 2.
462. 18 cm 2.
463.56.7 cm2.
464. a) 10 cm; b) 4 cm; c) 12 cm and 9 cm.
465. 12 cm 2.
466. 115.52 cm 2.
467. The area of a square is greater.
468. a) 38.5 cm 2; b) 5√3 cm 2; c) d) 4√2 cm.
470.5.625 cm.
471. a) 22 cm 2; b) 1.8 dm 2.
472. 14 cm and 24 cm.
473. Instruction. Use Theorem 38.
474. The areas of the triangles are equal.
475. Instruction. First, divide side BC into three equal parts.
476. a) 224 cm 2; b) 4.6 dm 2. Note. Note that the diagonals of a rhombus are mutually perpendicular.
477. 6 cm and 9 cm.
479. a) 2 cm 2; b) 2.4 cm. Instruction. Use the second theorem of paragraph 53.
480. a) 133 cm 2; b) 24 cm 2; c) 72 cm 2.
481.54 cm 2.
Area of a parallelogram
Theorem 1
The area of a parallelogram is defined as the product of the length of its side and the height drawn to it.
where $a$ is a side of the parallelogram, $h$ is the height drawn to this side.
Proof.
Let us be given a parallelogram $ABCD$ with $AD=BC=a$. Let us draw the heights $DF$ and $AE$ (Fig. 1).
Picture 1.
Obviously, the $FDAE$ figure is a rectangle.
\[\angle BAE=(90)^0-\angle A,\ \] \[\angle CDF=\angle D-(90)^0=(180)^0-\angle A-(90)^0 =(90)^0-\angle A=\angle BAE\]
Consequently, since $CD=AB,\ DF=AE=h$, by the $I$ criterion for the equality of triangles $\triangle BAE=\triangle CDF$. Then
So, according to the theorem on the area of a rectangle:
The theorem has been proven.
Theorem 2
The area of a parallelogram is defined as the product of the length of its adjacent sides times the sine of the angle between these sides.
Mathematically this can be written as follows
where $a,\ b$ are the sides of the parallelogram, $\alpha $ is the angle between them.
Proof.
Let us be given a parallelogram $ABCD$ with $BC=a,\ CD=b,\ \angle C=\alpha $. Let us draw the height $DF=h$ (Fig. 2).
Figure 2.
By definition of sine, we get
Hence
So, by Theorem $1$:
The theorem has been proven.
Area of a triangle
Theorem 3
The area of a triangle is defined as half the product of the length of its side and the altitude drawn to it.
Mathematically this can be written as follows
where $a$ is a side of the triangle, $h$ is the height drawn to this side.
Proof.
Figure 3.
So, by Theorem $1$:
The theorem has been proven.
Theorem 4
The area of a triangle is defined as half the product of the length of its adjacent sides and the sine of the angle between these sides.
Mathematically this can be written as follows
where $a,\b$ are the sides of the triangle, $\alpha$ is the angle between them.
Proof.
Let us be given a triangle $ABC$ with $AB=a$. Let's find the height $CH=h$. Let's build it up to a parallelogram $ABCD$ (Fig. 3).
Obviously, by the $I$ criterion for the equality of triangles, $\triangle ACB=\triangle CDB$. Then
So, by Theorem $1$:
The theorem has been proven.
Area of trapezoid
Theorem 5
The area of a trapezoid is defined as half the product of the sum of the lengths of its bases and its height.
Mathematically this can be written as follows
Proof.
Let us be given a trapezoid $ABCK$, where $AK=a,\ BC=b$. Let us draw in it the heights $BM=h$ and $KP=h$, as well as the diagonal $BK$ (Fig. 4).
Figure 4.
By Theorem $3$, we get
The theorem has been proven.
Sample task
Example 1
Find the area of an equilateral triangle if its side length is $a.$
Solution.
Since the triangle is equilateral, all its angles are equal to $(60)^0$.
Then, by Theorem $4$, we have
Answer:$\frac(a^2\sqrt(3))(4)$.
Note that the result of this problem can be used to find the area of any equilateral triangle with a given side.