## GEOMETRY FORMULAS

No matter how many types of each question you will see, the use of geometry formulas has a lot of influence and contribution to the final answer.

Solving geometry questions; A certain level of knowledge, whose logic is comprehended, requires a very attentive and experienced perspective. In this, geometry formulas are used.Math expressions that will require many operations make it possible to answer questions quickly and practically by pouring into formulas.

Since the issues are related to each other, the solution ways of the problem can be seen more easily as a result of using formulas with different and different methods.In addition, the probability of making a transaction error is reduced.

In the math test; it is important to reach shorter, practical and solution-oriented results.The main thing is how clear the student or the candidate did from the math test.

What we mean is not a “memorization of formulas” like a pattern in itself.

However, the building blocks of the Geometry course, which is common in the problems, are known as the pythagorean relation, and the thales, basic similarity, euclidean and cosine formulas; allows practice.

By taking time on it, you should gain experience by progressing step by step.The experience you gained will inevitably lead you to use math-geometry formulas. By doing so, you will ensure that the formulas are memorable, that is, they both understand and memorize them.

While doing this, you should either solve as many questions as possible or be able to prove theorems very well.

As you progress to the next level, you will realize that the same geometry basic concepts, axioms, theorems are always used, such as the point within the repeating point.

The following images include geometry formulas.

Geometry Boomerang-Rocket Rule

It is formulated in a concave-concave rectangle (the measure of the angle formed when the edge of its triangle is bent in) as shown in the figure with x = a+b+c .

Two inner bisector angle formula

The measure of the angle formed by two inner angles in a triangle is 90 ° more than half the measure of the inside angle in the third corner.

Two Outer Bisector Angle Formula

The measure of the angle formed by two outer bishops in a triangle is all half of the measure of the angle in the third corner. In other words, the angle formed by two outer bishops in one triangle is half the angle of the angle in the third corner from 90 ° .

Inside and outside bisector angle formula

The measure of the angle formed by the angle angles of a non-adjacent inner angle and an outer angle in a triangle equals half the measure of the angle in the third corner.

Triangular Sided Inequalities formula

An edge length in a triangle; the lengths of the other two sides are smaller than their sum, greater than the absolute value of the difference.

If the length of the ABC triangle in the figure is a, b, c; | b-c |< a < b+c, | a-c |< b < a+c, | a-b |< c < a+b .

Triangular Sided Inequalities circumference formula

The sum of the lengths of any point taken in an ABC triangle to the corners of the triangle; is greater than half the circumference of its triangle, and smaller than its circumference.

In the ABC triangle in the figure, | BC | = a, | AC | = b, | AB | = c and a+b+c = 2u; u is < | KA | + | KB | + | KC | < 2u.

Thales theorem formula

On any two lines that cross parallel lines, the line segments that the parallel lines separate are proportional.

In the figure, if d1 parallel d2 parallel d3, m and n intersect the lines; | AB | / | BC | = | DE | / | EF |, | AB | / | AC | = | DE | / | DF | and | BC | / | AC | = | EF | / | DF | .

Right triangle formula hypotenuse

The median hypotenuse of the hypotenuse in a right triangle equals half the length. In other words, the middle point of the hypotenuse in a right triangle is equidistant from the corners.

In the ABC triangle in the figure, m (A) = 90 ° and | BD | = | DC | if; | BD | = | DC | = | AD | .

30 60 and 15 75 triangle rules

If the edge length under a 30 ° angle in a right triangle is x units; the hypotenuse length is 2x units, and the edge length below the 60 ° angle is xroot3 units.

In a right triangle of 15 °-75 °-90 °, the height of the hypotenuse is a quarter of the hypotenuse.

Isosceles triangle rules

In the isosceles triangle, the sum of the struts lowered from one point taken on the base to equal sides is equal to the height of the equal sides.

Form; if | AB | =| AC |, [DF] perpendicular [AC], [DE] perpendicular [AB]; | BN | = | ED | + | DF | .

Isosceles triangle rules

In an isosceles triangle, the sum of the lengths of parallel line segments drawn from equal points from the point taken from the base is equal to the length of one of the identical edges.

Isosceles triangle rules

The absolute value of the length difference of the struts descending from the point taken on the extension of the base of the isosceles triangle to the equal sides of the triangle is equal to a height of equal sides.

Form; if | AB | = | AC |, [EP] perpendicular [AC], [EF] perpendicular [AF] and [BN] perpendicular [AC]; | EP | – | EF | = | BN | .

Isosceles triangle rules

In an isosceles triangle, the square of the length of the line segment drawn from the vertex to the base is equal to the difference of the length of the sides separated by the square of the length of one of the identical edges.

In the ABC triangle in the figure | AB | = | AC | = b, | AD | = x, | BD | = m and | DC | = n; x² = b² – m . n .

Equilateral triangle rules

If the figure is ABC equilateral triangle [DE] perpendicular [AE], [DT] perpendicular [BC], [DF] perpendicular [AF]; | AB | root3/2 = | ED | + | DF | – | DT | .

Angle bisector theorem formula

In a triangle, the bisector length divides the opposite edge in proportion to its neighboring edges.

In the ABC triangle in the figure, [AN] bisector is; | AB | / | BN | = | AC | / | CN | .

Exterior angle bisector theorem formula

In the ABC triangle in the figure [AD]; The bisector of m(CAE) is | CD | = x, | BC | = a, | AC | = b and | AB | = c; x / (x+a) = b/c.

Exterior angle bisector rules

In the ABC triangle in the figure [AN] angle bisector, [AD] exterior angle bisector, | CD | = x, | NC | = n and | BN | = m; x / (x+n+m) = n / m .

Exterior angle bisector rules

In the ABC triangle in the figure [AD] exterior angle bisector, | CD | = x, | BC | = a, | AC | = b, | AB | = c and | AD | = y; y² = x . (x+a) – bc .

Result for the area of the triangle

The height of the hypotenuse in a right triangle is the product of the lengths of the right sides divided by the length of the hypotenuse.

In the ABC triangle in the figure [BA] perpendicular [AC], [AD] perpendicular [BC], | AB | = c br, | AC | = b br, | BC | = a br and | AD | = h br; h = (c.b) / a .

Area rules in triangle

In any ABC triangle, the length drawn to the middle of an edge divides the area of the triangle into two equal areas.

In the triangle ABC of the figure | BF | = | FC | if; Area (ABF) = Area (AFC).

Angle bisector rules

If a triangle has two bisector angles, the third is the angle bisector.The incircle touches each line at one point, and the center point of the circle is the intersection of all the bisectors.

In the ABC triangle in the figure, if [BP], [CP] are bisector; [AP] becomes bisector.

Median of a triangle formula

In the ABC triangle in the figure, if | BC | = a br, | AC | = b br, | AB | = c br, | AP | = Va br, | BR |= Vb br, | CS | = Vc br; 2Va² = b²+c²-a² / 2, 2Vb² = a²+c²-b² / 2 and 2Vc² = a²+b²-c² / 2 .

Median rules and finding area with the inradius

The edge lengths of an ABC triangle are a, b, c and median length Va, Vb, Vc; 4 (Va²+Vb²+Vc²) = 3 (a²+b²+ c²) .

If the radius of the inner tangent circle of a triangle is r, the edge lengths are a, b, c and u = (a+b+c) / 2 ; Area (ABC) = u . r .

Area of a concave quadrilateral

In the figure, if m (ADB) = α in triangle ABC; The area of the concave rectangle is Area (ABEC) = (1/2) . | AE | . | BC | . sinα .

Area of a quadrilateral

In any quadrilateral; diagonals | CA | = e, | DB | = f and the angle between α; Area (ABCD) = (1/2) . e . f . sinα .

General quadrilateral formulas

If the diagonal quadrileteral cuts each other in a right angles, the sum of the squares of the opposite sides is equal.(a²+c² = b²+d²)

Height of right-angled trapezium

If diagonals are perpendicular to each other at right trapezoid, the height is the geometric middle of the lower and upper base.(the square of the height is equal to multiply of the bases.)

In the ABCD quadrileteral in the figure, [CD] perpendicular [DA], [DA] perpendicular [CD], [DB] perpendicular [AC], | AB |= a br, | CD | = c br and | DA | = h br; h = √ac .

Rule of a trapezoid area

The midpoint of one of the side edges in a trapezoid; the area of the triangle obtained, joining with the two opposite corners, is equal to half the area of the trapezoid.

If ABCD trapezoid and | CA | = | DB |; Area (DAE)=(1/2) . Area (ABCD) .

Rule of a trapezoid area

If the midpoints of the sides of a rectangle are combined, a parallelogram is created. The area of this parallelogram is equal to half the area of the given quadrileteral.

Also, the perimeter of the parallelogram is equal to the sum of the diagonal lengths of the quadrileteral.

In the ABCD quadrileteral in the figure, the middle points of N, P, K, L edges are; Area (NPKL) = (1/2) . Area (ABCD) .

An area feature in the parallelogram

Any point taken outside the parallelogram is combined with the corners of the parallelogram, the sum of the areas of the two triangles is equal to half the area of the parallelogram.

In the figure, ABCD is a parallelogram and a P point taken outside; Area (PDA) + Area (PBC) = (1/2) . Area (ABCD) .

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