Construct an isosceles triangle a along the side. Problems about isosceles triangles

Isosceles is such triangle, which has the same lengths of its two sides.

When solving problems on the topic "Isosceles triangle" it is necessary to use the following known properties:

1. Angles opposite equal sides are equal.
2. Bisectors, medians and heights drawn from equal angles are equal to each other.
3. The bisector, median and height drawn to the base of an isosceles triangle coincide with each other.
4. The center of the inscribed and the center of the circumscribed circles lie at the height, and hence on the median and the bisector drawn to the base.
5. Angles that are equal in an isosceles triangle are always acute.

A triangle is isosceles if it has the following signs:

1. The two angles of a triangle are equal.
2. The height is the same as the median.
3. The bisector is the same as the median.
4. The height coincides with the bisector.
5. The two heights of a triangle are equal.
6. The two bisectors of a triangle are equal.
7. The two medians of a triangle are equal.

Consider a few tasks on the topic "Isosceles triangle" and give a detailed solution.

Task 1.

In an isosceles triangle, the height drawn to the base is 8, and the base is related to the side as 6: 5. Find how far from the vertex of the triangle is the point of intersection of its bisectors.

Decision.

Let an isosceles triangle ABC (Fig. 1).

1) Since AC: BC = 6: 5, then AC = 6x and BC = 5x. BH is the height drawn to the base AC of the triangle ABC.

Since the point H is the midpoint of AC (by the property of an isosceles triangle), then HC \u003d 1/2 AC \u003d 1/2 6x \u003d 3x.

BC 2 \u003d VN 2 + HC 2;

(5x) 2 \u003d 8 2 + (3x) 2;

x = 2, then

AC \u003d 6x \u003d 6 2 \u003d 12 and

BC \u003d 5x \u003d 5 2 \u003d 10.

3) Since the point of intersection of the bisectors of the triangle is the center of the circle inscribed in it, then
OH = r. The radius of the circle inscribed in the triangle ABC is found by the formula

4) S ABC = 1/2 (AC BH); S ABC = 1/2 (12 8) = 48;

p = 1/2 (AB + BC + AC); p = 1/2 (10 + 10 + 12) = 16, then OH = r = 48/16 = 3.

Hence VO \u003d VN - OH; VO \u003d 8 - 3 \u003d 5.

Answer: 5.

Task 2.

Bisector AD is drawn in an isosceles triangle ABC. The areas of triangles ABD and ADC are equal to 10 and 12. Find the area of the square built at the height of this triangle drawn to the base of AC by three times.

Decision.

Consider the triangle ABC - isosceles, AD - the bisector of angle A (Fig. 2).

1) Let's write the areas of triangles BAD and DAC:

S BAD = 1/2 AB AD sin α; S DAC = 1/2 AC AD sin α.

2) Find the ratio of areas:

S BAD /S DAC = (1/2 AB AD sin α) / (1/2 AC AD sin α) = AB/AC.

Since S BAD = 10, S DAC = 12, then 10/12 = AB/AC;

AB/AC = 5/6, then let AB = 5x and AC = 6x.

AN \u003d 1/2 AC \u003d 1/2 6x \u003d 3x.

3) From the triangle ABN - rectangular according to the Pythagorean theorem AB 2 \u003d AN 2 + VN 2;

25x 2 \u003d VN 2 + 9x 2;

4) S A BC = 1/2 AC HV; S A B C \u003d 1/2 6x 4x \u003d 12x 2.

Since S A BC \u003d S BAD + S DAC \u003d 10 + 12 \u003d 22, then 22 \u003d 12x 2;

x 2 \u003d 11/6; VN 2 \u003d 16x 2 \u003d 16 11/6 \u003d 1/3 8 11 \u003d 88/3.

5) The area of the square is equal to VN 2 = 88/3; 3 88/3 = 88.

Answer: 88.

Task 3.

In an isosceles triangle, the base is 4 and the side is 8. Find the square of the height dropped to the side.

Decision.

In the triangle ABC - isosceles BC \u003d 8, AC \u003d 4 (Fig. 3).

1) ВН - the height drawn to the base AC of the triangle ABC.

Since the point H is the midpoint of AC (by the property of an isosceles triangle), then HC \u003d 1/2 AC \u003d 1/2 4 \u003d 2.

2) From the VNS triangle - rectangular according to the Pythagorean theorem VS 2 \u003d VN 2 + NS 2;

64 = HH 2 + 4;

3) S ABC \u003d 1/2 (AC BH), as well as S ABC \u003d 1/2 (AM BC), then we equate the right parts of the formulas, we get

1/2 AC BH = 1/2 AM BC;

AM = (AC BH)/BC;

AM = (√60 4)/8 = (2√15 4)/8 = √15.

Answer: 15.

Task 4.

In an isosceles triangle, the base and the height lowered onto it are equal to 16. Find the radius of the circle circumscribed about this triangle.

Decision.

In the triangle ABC - isosceles base AC \u003d 16, BH \u003d 16 - the height drawn to the base AC (Fig. 4).

1) AN \u003d HC \u003d 8 (by the property of an isosceles triangle).

2) From the VNS triangle - rectangular according to the Pythagorean theorem

BC 2 \u003d VN 2 + HC 2;

BC 2 \u003d 8 2 + 16 2 \u003d (8 2) 2 + 8 2 \u003d 8 2 4 + 8 2 \u003d 8 2 5;

3) Consider the triangle ABC: by the sine theorem 2R = AB/sin C, where R is the radius of the circumscribed circle around the triangle ABC.

sin C \u003d BH / BC (from the VNS triangle by definition of the sine).

sin C = 16/(8√5) = 2/√5, then 2R = 8√5/(2/√5);

2R = (8√5 √5)/2; R = 10.

Answer: 10.

Task 5.

The length of the height drawn to the base of an isosceles triangle is 36, and the radius of the inscribed circle is 10. Find the area of the triangle.

Decision.

Let it be given isosceles triangle ABC.

1) Since the center of the circle inscribed in the triangle is the point of intersection of its bisectors, then O ϵ VN and AO is the bisector of angle A, and the current is OH \u003d r \u003d 10 (Fig. 5).

2) VO \u003d VN - OH; VO \u003d 36 - 10 \u003d 26.

3) Consider the triangle ABH. By the triangle angle bisector theorem

AB/AN = BO/OH;

AB/AN = 26/10 = 13/5, then let AB = 13x and AH = 5x.

According to the Pythagorean theorem, AB 2 \u003d AN 2 + VN 2;

(13x) 2 \u003d 36 2 + (5x) 2;

169x 2 \u003d 25x 2 + 36 2;

144x 2 \u003d (12 3) 2;

144x2 = 144 9;

x \u003d 3, then AC \u003d 2 AN \u003d 10x \u003d 10 3 \u003d 30.

4) S ABC = 1/2 (AC BH); S ABC = 1/2 (36 30) = 540;

Answer: 540.

Task 6.

In an isosceles triangle, two sides are equal to 5 and 20. Find the bisector of the angle at the base of the triangle.

Decision.

1) Suppose the sides of the triangle are 5 and the base is 20.

Then 5 + 5< 20, т.е. такого треугольника не существует. Значит, АВ = ВС = 20, АС = 5 (Fig. 6).

2) Let LC = x, then BL = 20 – x. By the triangle angle bisector theorem

AB/AC = BL/LC;

20/5 \u003d (20 - x) / x,

then 4x \u003d 20 - x;

Thus, LC = 4; BL = 20 - 4 = 16.

3) Let's use the triangle angle bisector formula:

AL 2 \u003d AB AC - BL LC,

then AL 2 = 20 5 - 4 16 = 36;

Answer: 6.

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How to build an isosceles triangle? This is easy to do with a ruler, pencil and notebook cells.

We start building an isosceles triangle from the base. To make the drawing even, the number of cells at the base must be an even number.

We divide the segment - the base of the triangle - in half.

The vertex of the triangle can be chosen at any height from the base, but always exactly above the middle.

How to construct an acute isosceles triangle?

The angles at the base of an isosceles triangle can only be acute. In order for an isosceles triangle to turn out to be acute, the angle at the vertex must also be acute.

To do this, select the top of the triangle higher, away from the base.

The higher the top, the smaller the angle at the top. At the same time, the angles at the base increase accordingly.

How to construct an obtuse isosceles triangle?

As the apex of an isosceles triangle approaches the base, the degree measure of the angle at the apex increases.

So, to build an isosceles obtuse-angled triangle, we choose a vertex lower.

How to construct an isosceles right triangle?

To build an isosceles right triangle, you need to select the vertex at a distance equal to half the base (this is due to the properties of an isosceles right triangle).

For example, if the length of the base is 6 cells, then we place the top of the triangle at a height of 3 cells above the middle of the base. Please note: in this case, each cell at the corners at the base is divided diagonally.

The construction of an isosceles right triangle can be started from the top.

We select the top, from it at a right angle we set aside equal segments up and to the right. These are the sides of the triangle.

Connect them and get an isosceles right triangle.

The construction of an isosceles triangle using a compass and a ruler without divisions will be considered in another topic.

VIII . Groups of tasks to build.

Solving groups of problems using an auxiliary triangle.

The essence of the method is the construction of auxiliary triangles and the use of their properties and newly obtained elements for the final solution of the problem.

Construction analysis consists of the following steps:

Look for an auxiliary triangle in the analysis.

If new elements appear, with the help of which it is already possible to construct a triangle ABC, then the goal has been achieved.

If this does not happen, then perhaps one more auxiliary triangle can be constructed, which will give the missing elements.

Let's analyze the essence of the method with examples.

Task 1. Construct an isosceles triangle ABC ( b= c) on a, h b .

We are looking for an auxiliary triangle. Obviously, it is convenient to consider the triangle CDB as such a triangle.

This will give angle C, hence angle ABC. So, there is a, angle B, angle C, so you can build a triangle ABC. Schematically, we will write it like this:

(a, h b) → Δ CDB →< C.

(a,< B, < C) → Δ ABC.

Tasks for independent solution:

Using reasoning similar to those given, we recommend constructing an isosceles triangle (b=c) according to the following data:

a)< А, h b ;

b)< В, h с;

G)< В, h b ;

e)< С, h b .

Task 2. Construct a triangle along the radius r of the inscribed circle, angle A and angle B.

Let I be the center of a circle inscribed in triangle ABC.

(r; ½< А) → Δ AID → |AD|;

(r; ½< В) → Δ ВID → |ВD|;

(|AD| + |ВD| = |AB|) → (s,< А, < В) → Δ ABC.

Tasks for independent solution:

Construct a triangle using the following elements:

a) a, h c , h b ; b) a, h a, h b; c) a, m a, m b;

G)< A, l A , b; д) R, h а, m a ; е) a, R, h b ;

g) b, h b , m b (where m are medians, l are bisectors, h are heights).

On one's own:

construct a rhombus ABCD along a diagonal BD and a height BM. (∆BHD →< BDH → равнобедренный Δ BDA → ABCD);

build a trapezoid on four sides.

Solving groups of problems based on the main one.
1. The main task:

Construct a triangle given two sides and an angle between them.

Construct a right triangle on two legs.

Construct a rhombus along two diagonals.

Construct a rectangle with two unequal sides.

Construct a parallelogram given two diagonals and the angle between them.

Construct a rectangle along the diagonals and the angle between them.

The main task:

Construct a triangle given a side and two adjacent angles.

Tasks for independent solution:

Construct an isosceles triangle given a base and an included angle.

Construct a right triangle given a leg and an adjacent acute angle.

Construct a rhombus given an angle and a diagonal passing through the vertex of that angle.

Construct an isosceles triangle by height and angle at the vertex.

Construct a square along the given diagonal.

The main task:

Construct a right triangle given the hypotenuse and the acute angle.

Tasks for independent solution:

Construct an isosceles triangle along the side and the angle at the base.

Construct an isosceles triangle on the side and the angle at the vertex.

The main task:

Construct a triangle with three sides.

Tasks for independent solution:

Construct an isosceles triangle with base and side.

Construct a rhombus along a side and a diagonal.

Construct a parallelogram given two unequal sides and a diagonal.

Construct a parallelogram given a side and two diagonals.

The main task:

Construct a right triangle given the leg and hypotenuse.

Tasks for independent solution:

Construct an isosceles triangle in height and side.

Construct an isosceles triangle given the base and the perpendicular dropped from the end of the base to the side.

Construct a parallelogram by base, height and diagonal.

Construct a rhombus in height and diagonal.

Construct an isosceles triangle given the lateral side and the height lowered from it.

Construct a triangle given its base, height and side.

Literature:

B. I. Argunov, M. B. Balk “Geometric constructions on the plane”, M, “Enlightenment”, 1955

Glazer G.I. “History of mathematics at school” IV - VI class., M, “Enlightenment”, 1981

I. Goldenblant “Experience in solving geometric construction problems” “Mathematics at school” No. 3, 1946

I. A. Kushnir “On one method for solving construction problems” “Mathematics at School” No. 2, 1984

A. I. Mostovoy “Apply different methods of solving construction problems” “Mathematics at school” No. 5, 1983

A. A. Popova “Mathematics” Textbook. "Chelyabinsk State Pedagogical University”, 2005

E. M. Selezneva, M. N. Serebryakova “ Geometric constructions in I - V classes high school” Methodical developments. Sverdlovsk, 1974