## Area of Triangles Formulas

Friends, in this topic, we will talk about “Area of Triangles Formulas Pdf”.

The altitude of a triangle is the perpendicular drawn from the vertex of the triangle to the opposite side. Also, known as the height of the triangle, the altitude makes a right angle triangle with the base.

### General Area Formula for a triangle with known height and base;

The area of a triangle is equal to half of multiplied the base by the height. This is the most commonly used formula to calculate the area of ​​the triangle.

! Regardless of the used edge and height that belongs to this edge, the area is always the same.

A, B, C: corner

a (|BC|), b(|AC|), c(|AB|): edge

h: height

NOTE:

In an ABC triangle, the height may not always be inside of the triangle. In other words, if the triangle is a wide-angle triangle, the height belongs to the triangle is taken from outside of the triangle as below.

In some special triangles (isosceles triangles, etc.), height information may not be provided. In such cases, we must first find the height. “Pythagoras relation” is used when finding the height.

According to the Pythagoras relation;
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two side lengths.

c: hypotenuse

a² + b² = c²

### Area of an equilateral triangle;

In special triangles such as equilateral triangles, we use some special formulas instead of the general area formula in triangle given above. This is because the equilateral triangle has three sides of equal length, so the lengths of the heights of these three sides are also equal. There is also a constant ratio between edge length and height. Therefore, the area of an equilateral triangle with only one edge can be easily found.

If ABC is an equilateral triangle;

a = b = c

$$​h_a = h_b = h_c$$

m(A) = m(B) = m(C) = 60 º

Area = ​$$\displaystyle\frac{a^2.\sqrt{3}}{4}$$

NOTE:
The area of the triangle “30º – 30º – 120º”, which is one of the special triangles, is calculated as the area of the equilateral triangle. If we say “a” to the sides opposite the 30º angle; the height is “a / 2” and the base length is “a√3”. In this case, the area ​$$\displaystyle\frac{a^2.\sqrt{3}}{4}$$​.

### Area formula of a triangle with known three side lengths (Heron’s Formula);

Three edge lengths are usually given when calculating the area of ​​triangles which all sides have been different lengths. In such questions, no height information is given. The area only needs to be calculated based on the length of the edges. In order to calculate the area of ​​the triangle;

– The triangle’s perimeter is found (sum of all three sides) ⇒ a + b + c
– The found perimeter is divided by 2 (This value is named an “u” value) ⇒ ​$$u = ​ \displaystyle\frac{a+b+c}{2}$$
– And the area information of the triangle is calculated with the area formula below. 🙂

Area = ​$$\sqrt{u. (u-a).(u-b).(u-c)}$$

### The area formula for a triangle with known perimeter and radius of the inscribed circle in a triangle;

If “O” is a centre of an inscribed circle in the triangle, ‘r’ is a radius of this circle and perimeter of the triangle is equal “a+b+c”, the area of the triangle is found with the formula which is below.

$$u = ​ \displaystyle\frac{a+b+c}{2}$$

Area = u.r

NOTE:

In a triangle, there is a relationship like given below between the radius of the inscribed circle and the edge heights of the triangle.

$$\displaystyle\frac{1}{r} = \frac{1}{h_a} +\frac{1}{h_b} +\frac{1}{h_c}$$

### The area of ​​a right-angle triangle;

The area of ​​a right-angle triangle is equal half of the multiply of perpendicular sides.

On the hypotenuse of the right tangent circle of a right triangle, the part lengths are m and n is the area with m.n.

In the triangle 15 – 75 – 90, which is one of the special right triangles, the height lowered from the right angle is equal to the length of the hypotenuse ¼.

#### Area of ​​triangles with equal height;

The ratio of the areas of the triangles with equal heights is equal to the ratio of the lengths of the floors from which the height is lowered.

! Since the bases of the ABC and ACD triangles are on the same line and the peaks are at the same point, their heights are equal.

#### Area of ​​triangles whose base lengths are equal;

The ratio of the areas of the triangles whose bases are equal is equal to the ratio of their height of the floor lengths that are equal.

! The bases of the ABC and DBC triangles are equal and overlapping.

When asked about the area of ​​the region whose bases are equal and between heights of two different triangles and shown as shaded, the area is calculated as follows.

#### Area of ​​equilateral triangles (Similar triangles);

Lines that are parallel to each other and base in a triangle and divide the edges they cut into equal lengths among themselves, divide the triangle proportionally into areas. The areas of the regions between these parallel lines are in direct proportion to the odd numbers. The main reason for this occurrence is similarity.

#### The area of ​​the triangle between two parallel lines;

The distance between two parallel lines is the height of all triangles drawn with the corners on these lines. In this case, for the triangle with a point A on the line d1 in the figure; Regardless of where point A (points D, E) is, the base [BC] does not change at all, and since the height remains the same, the areas of the resulting triangles are equal.

#### Açıortayın oluşturduğu üçgenin alanı;

The proportion of the areas where the bisector divides the triangle drawn from one corner of the triangle is equal to the ratio of the edges adjacent to the corner where this bisector is located.

#### Kenarortay kullanarak üçgenin alanı;

The formula used when calculating area in triangles given Kenarortay lengths is called “Heron Formula”. As per the Kenarortay Heron formula;

Areas created by Kenarortaylar;

Kenar uzunluklarına göre üçgenin alanı;

## Circumference Of A Circle Formula

Circumference Of A Circle Formula: A circle consists of all points in a plane that are a given distance, called the radius, from a given point called the center.

#### A segment or line can intersect a circle in several ways.

A segment with endpoints that are the center of the circle and a point of the circle is a radius.
A segment with endpoints that lie on the circle is a chord.
A chord that contains the circle’s center is a diameter.

The circumference of a circle is the distance around the circle.

For a circumference of C units and a diameter of d units or a radius of r units,

C = π.d  or  C = 2πr

Example:  Find the circumference of the circle to the nearest hundredth.

Answer:  C = 2πr  ⇒ Circumference formula

⇒ 2π(13)

≈ 81.68  (Use a calculator)

The circumference is about 81.68 centimeters.

## Probability Formulas

Welcome to our article, where simple probability formulas with sample questions.

1. Probability deals with the measure or estimation of events that are likely to happen. In mathematics, probability is the calculation of uncertainty.

$$probability=\displaystyle\frac{Number \ of \ favorable \ outcomes}{Total \ numbers \ of \ outcomes}$$

● The probability of an event can only be between 0 and 1 and can also be written as a percentage.
● The probability of event A is often written as P(A).
● If P(A) > P(B), then event A has a higher chance of occurring than event B.
● If P(A) = P(B), then events A and B are equally likely to occur.

Exercise: What is the probability of getting head when tossing a coin?

Answer: Sample Space = {H, T}
Number of possible outcomes = 2
Number of favorable outcomes = 1(because of only one head).

$$probability=\displaystyle\frac{Number \ of \ favorable \ outcomes}{Total \ numbers \ of \ outcomes}$$

Probability of getting head is ​$$\displaystyle\frac{1}{2}$$

2. Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.

P(A or B) = P(A) + P(B)

Exercise: A single 6-sided dice is rolled. What is the probability of rolling a 2 or a 5?

Answer: Probability of getting 2, P(2)= 1/6

Probability of getting 6, P(6)= 1/6

Probability of getting 2 or 6,
P(2 or 6)=P(2)+P(6)

$$=\displaystyle\frac{1}{6}+\frac{1}{6}=\frac{2}{6}⇒\frac{1}{3}$$

3. Addition Rule 2: When two events, A and B, are non-mutually exclusive, the probability that A or B will occur is:

P(A or B) = P(A) + P(B) – P(A and B)

Exercise: In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4 boys and 5 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an A student?

P(girl or A)
= P(girl) + P(A) – P(girl and A)

$$\displaystyle\frac{13}{30}+\frac{9}{30}-\frac{5}{30}=\frac{17}{30}$$

4. Multiplication Rule 1:
If A and B are two independent events in a probability experiment, then the probability that both events occur simultaneously is:

P(A and B)=P(A).P(B)

Exercise: You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt?

Answer: The two events are independent events; the choice of hat has no effect on the choice of shirt. There are three different hats, so the probability of choosing the songkok is 1/3

There are four different shirts, so the probability of choosing the black shirt is 1/4 . So, by the Multiplication Rule:

P(songok and black shirt)= ​$$\displaystyle\frac{1}{3}.\frac{1}{4}=\frac{1}{12}$$

5. Multiplication Rule 2:
If A and B are two dependent events in a probability experiment, then the probability that both events occur simultaneously is:

P(A and B)=P(A).P(B | A)

(The notation P(B | A) means “the probability of B, given that A has happened.”)

Exercise: A bag has 4 white cards and 5 blue cards. We draw two cards from the bag one by one without replacement. Find the probability of getting both cards white.

Answer: Let A = event that first card is white and B = event that second card is white.

From question, P(A) = 4/9

Now P (B) = P(B|A) because the events given are dependent on each other.
P (B) = 3/8

So, P(A and B) = P(A) × P(B|A)
= ​$$\displaystyle\frac{4}{9}x\frac{3}{8}=\frac{1}{6}$$

6. Bayes’ Theorem:
Bayes’ theorem is stated mathematically as the following equation:

$$P(A|B)=\displaystyle\frac{P(B|A)P(A)}{P(B)}$$

where A and B are events and P(B) ≠ 0.

● P(A | B) is a conditional probability: the likelihood of event A occuring that B is true.
●P(B | A) is also a conditional probability: the likelihood of event B occuring that A is true.
●P(A) and P(B) are probabilities of observing A and B independently of each other; this is known as marginal probability.

Exercise: If dangerous fires are rare (2%) but smoke is fairly common (20%) due to barbecues, and 80% of dangerous fires make smoke. Then what is the Probability of dangerous Fire when there is Smoke?

Answer: ​$$P(Fire|Smoke)=\displaystyle\frac{P(Fire|Smoke)P(Fire)}{P(Smoke)}$$

$$\displaystyle\frac{2\%x80\%}{20\%}=8\%$$

So the “Probability of dangerous Fire when there is Smoke” is 8%

## Trigonometric Formulas

There are some special formulas in Trigonometry such as “Transformation Formulas, Inverse Transformation Formulas, Trigonometric Sum and Difference Formulas, Half Angle Formulas”. In this article, we will tell about these special formulas, friends.

Sum and Difference Formulas in Trigonometry;

♦  sin (x + y) = sin x. cos y + cos x. sin y

♦  cos (x + y) = cos x. cos y – sin x. sin y

♦  tan (x + y) = ​$$\displaystyle\frac{tan~x + tan~y}{1- tan~x. tan~y}$$

♦  cot (x + y) = ​​$$\displaystyle\frac{1}{tan(x+y)}​$$

♦  sin (x – y) = sin x. cos y – cos x. sin y

♦  cos (x – y) = cos x. cos y + sin x. sin y

♦  tan (x – y) = ​​$$\displaystyle\frac{tan~x – tan~y}{1+ tan~x. tan~y}$$

♦  cot (x – y) = ​$$\displaystyle\frac{1}{tan(x-y)}$$

NOTE : If a, b ∈ ​$$\mathbb{R}$$​;

$$-\sqrt{a^2 + b^2}​≤a.sinx +b.cosx≤\sqrt{a^2 + b^2}​$$

Half Angle Formulas;

♦  sin 2x = 2. sin x. cos x

♦  cos 2x = cos²x – sin²x

♦  cos 2x = 2. cos²x – 1

♦  cos 2x = 1 – 2. sin²x

♦  tan 2x  = ​$$\displaystyle\frac{2. tan x}{1 – tan^2 x}$$

Pythagoras Formulas;

♦  sin²x + cos²x = 1

♦  tan²x + 1 =  sec²x

♦  cot²x + 1 =  cosec²x

Transformation Formulas in Trigonometry;

♦  ​​$$sin x + sin y = 2.sin\displaystyle\frac{x + y}{2}. cos\displaystyle\frac{x – y}{2}​$$

♦  ​$$sin x – sin y = 2.cos\displaystyle\frac{x + y}{2}. sin\displaystyle\frac{x – y}{2} ​$$

♦ ​$$cos x + cos y = 2.cos\displaystyle\frac{x + y}{2}. cos\displaystyle\frac{x – y}{2}$$

♦  ​​$$cos x – cos y = – 2.sin\displaystyle\frac{x + y}{2}. sin\displaystyle\frac{x – y}{2}$$

♦  ​​$$\displaystyle\frac{sin x + sin \displaystyle\frac{x + y}{2} + sin y}{cos x + cos \displaystyle\frac{x + y}{2} + cos y} = tan\displaystyle\frac{x + y}{2}$$

♦  ​​$$\displaystyle\frac{cos x + cos \displaystyle\frac{x + y}{2} + cos y}{sin x + sin \displaystyle\frac{x + y}{2} + sin y} = cot\displaystyle\frac{x + y}{2}$$​​

Inverse Transformation Formulas in Trigonometry;

♦  ​​$$cos x. cos y = \displaystyle\frac{1}{2}[cos (x + y) + cos (x – y)]$$

♦  ​​$$sin x. cos y = \displaystyle\frac{1}{2}[sin (x + y) + sin (x – y)]$$

♦  ​​$$sin x. sin y = – \displaystyle\frac{1}{2}[cos (x + y) – cos (x – y)]$$

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