The Area of a Triangle

Information:

• The side $CB$ has a length of $14$ cm.
• The height has a length of $17$ cm.

If we apply the formula, we multiply the height ($17$ cm) by the length of the base ($14$ cm).

By multiplying $17$ by $14$, we obtain $238$, a result that we must divide by $2$.

$238$ divided by $2$ equals $119$.

The area of this triangle is therefore: $119$.

$Area=\frac{14\times17}{2}=119$

Whether you are preparing for an test or are about to take your university entrance exams, it is essential to know how to calculate the area of a triangle, whether it is right angled, isosceles, etc.

So, how do you calculate a triangular area? This guide will clear up all of your doubts concerning one of the most frequently asked questions in geometry examinations.

A triangle is a geometric figure composed of three sides that form three angles and three vertices.

The vertices of the triangle are marked with the letters $A,B$ and $C$; together creating the sides: $AB, BC, CA$.

There are different types of triangles and some of them even share characteristics. We will deal with those later in this article!

Before addressing the different types of triangles that exist and how to calculate their areas,
we must first know the terms that are usually used when we talk about a triangular area.

• Straight line: A line extending in the same direction and having an infinite number of points.
• Segment: A fragment of a line between two points.
• Height: The height of a triangle is the length of a perpendicular line segment starting on one side and intersecting the angle opposite. The height is denoted by the letter 'h'.
• Median: The median is the segment extending from a given vertex to the midpoint of the side opposite that vertex.
• Bisector: A ray extending from a given vertex, dividing it into two equal angles.
• Perpendicular bisector: A line that is perpendicular to a side of the triangle and that passes through the midpoint of that side.
• Median segment: A median segment of a triangle is a segment that connects the midpoints of two sides of a triangle and is arranged parallel to the third side, its length being half of the third side.
• Opposite side: An opposite side is one that faces a given vertex.

One of the most useful tips when calculating the area of a triangle (and thus solving the problem) is to understand that a triangle is a half-square.

There are triangles that are easily distinguishable as "half-squares" because of their shape, such as the isosceles right triangle.

However, it is important to note that triangles that do not appear to be "half-squares" are also "half-squares", since this is one of the features that characterize them.

What is the next step?

Calculate the area of the triangle.
The formula used to calculate the area of a triangle is as follows:

height times base divided by $2$.

$Area=\frac{Base\times Height}{2}$

Information:

• The side $CB$ has a length of $15$ cm.
• The height has a length of $13$ cm.

Solution:

If we apply the formula, we multiply the height ($13$ cm) by the length of the base ($15$ cm).
By multiplying $13$ by $15$, we get $195$, a result that we must divide by $2$.

$192$ divided by $2$ equals $97.5$.

The area of this triangle is therefore: $97.5$.

$Area=\frac{13\times15}{2}=97.5$

Information:

• The side $CB$ has a length of $14$ cm.
• The height has a length of $17$ cm.

Solution:

Again, if we apply the formula, we multiply the height ($17$ cm) by the length of the base ($14$ cm).
By multiplying $17$ by $14$, we get $238$, a result that we must divide by $2$.

$238$ divided by $2$ equals $119$.

Therefore, the area of this triangle is: $119$.

$Area=\frac{14\times17}{2}=119$

Information:

• The side $CB$ has a length of $9$ cm.
• The height has a length of $10$ cm.

Solution:

If we apply the formula, we multiply the height ($10$ cm) by the length of the base ($9$ cm).
By multiplying $10$ by $9$ we get $90$, a result that we must divide by $2$.

$90$ by $2$ equals $45$.

Therefore, the area of this triangle is $45$ cm².

$Area=\frac{9\times10}{2}=45$

By clicking on the link you can find more information about a scalene triangle

Information:

• The side $CB$ has a length of $12$ cm.
• The height has a length of $14$ cm.

Note: in a right triangle or right-angled triangle, the base and the height correspond to the legs of the triangle.

Solution:

If we apply the formula, we multiply the height ($14$ cm) by the length of the base ($12$ cm).
By multiplying $14$ by $12$ we get $168$, a result that we must divide by $2$.

$168$ by $2$ equals $84$.

Therefore, the area of this triangle is: $84$.

$Area=\frac{12\times14}{2}=84$

Information:

• The side $CB$ has a length of $13$ cm.
• The height has a length of $16$ cm.

Please note:

In an obtuse triangle, the height is outside the triangle.
This means that we must extend the line of the base from point $C$ to the point $D$ to find the height.

In this way we create a right triangle $\triangle ABD$,where the height we are looking for is the side $AD$.

However, remember that since we are trying to calculate the area of the obtuse triangle, we only have to consider the side as the base. $CB$ is the base.

Solution:
In this case, if we apply the formula, we multiply the height ($16$ cm) by the length of the base of the triangle whose area we want to find.
By multiplying $16$ by $13$ we obtain $208$, a result that we must divide by $2$.

$208$ divided by $2$ equals $104$.

Therefore, the area of this triangle is: $104$.

$Area=\frac{13\times16}{2}=104$

What is Heron's formula and what is it for?

Heron's formula, the invention of which is attributed to the Greek mathematician Heron of Alexandria, allows us to obtain the area of a triangle knowing the lengths of its three sides $a, b$ and $c$.

$A\text{rea}=\sqrt{s(s-a)(s-b)(s-c)}$

Where 's' is the perimeter of the triangle divided by $2$:

$s=\frac{a+b+c}{2}$

Task:

In the grounds of a hotel, they want to build a special triangular-shaped pool.

The length of the pool is $10$ m.

The width of the pool is $8$ m.

The pool is to be covered with tiles $2$ m long and $2$ m wide.

Question:

How many tiles are needed to cover the pool area?

Solution:

To find out how many tiles are needed, we must calculate the triangular area of the pool and the area of each tile, then divide these numbers.

$\frac{\text{S.triangle}}{S.tile}$

The result is equal to the number of tiles needed.

In a triangle, its length is equal to its height and its width is equal to the base of the triangle.

$\text{S.triangle=}\frac{10\cdot8}{2}=40$

h = length = $10$ meters.

base = width = $8$ meters.

Since the length is $2$ meters, the width is also $2$ meters.

Tile area: $2\cdot2=4$

$\frac{40}{4}=10$

Answer:

$10$ tiles

Task:

The triangle $\triangle ABC$ is a right-angled triangle.

The area of the triangle is equal to $6cm^2$.

Calculate $X$ and the length of the side $BC$.

Solution:

We use the formula to calculate the area of the right triangle:

$\frac{AC\cdot BC}{2}=\frac{leg\times leg}{2}$

Then compare the expression with the area of the triangle ($6$).

$\frac{4\cdot(X-1)}{2}=6$

Multiplying the equation by the common denominator means that we multiply by $2$.

$4(X-1)=12$

We open the parentheses before the distributive property:

$4X-4=12$ / $+4$

$4X=16$ / $:4$

$X=4$

Replace $X=4$ into the expression $BC$ and we find:

$BC=X-1=4-1=3$

Answer:

$BC=3$

$X=4$

Task:

Given the triangle $\triangle PRS$, calculate the height $PQ$.

The length of the side $SR$ is equal to $4\operatorname{cm}$.

The area of the triangle $PSR$ is equal to $30$ cm².

Solution:

We use the formula to calculate the area of the triangle.

Please note: in the obtuse triangle, the height is outside of the triangle!

$\frac{Side\cdot\text{Height}}{2}=Triangular Area$

Double the equation by a common denominator.

$\frac{4\cdot PQ}{2}=30$ / $\cdot2$

Divide the equation by the coefficient of $PQ$.

$4PQ=60$ / $:4$

$PQ=15$

Answer:

The length of the height $PQ$ is equal to $15 cm$.

Task:

Given the right triangle $\triangle ADB$, calculate the area of the triangle. $\triangle ABC$.

The perimeter of the triangle is equal to $30\operatorname{ cm}$.

Given: $AB=15, AC=13, DC=5, CB=4$.

Solution:

Given that the perimeter of the triangle $ΔADC$ is equal to $30 cm$, we can calculate $AD$

$AD+DC+AD=PerimeterΔADC$

$AD+5+13=30$

$AD+18=30$ /$-18$

$AD=12$

Now we can calculate the area of the triangle $ΔABC$.

Please note: we are talking about an obtuse triangle, so its height is $AD$.

We use the formula to calculate the area of the triangle:

$\frac{sideheight\times side}{2}=$

$\frac{AD\cdot BC}{2}=\frac{12\cdot4}{2}=\frac{48}{2}=24$

Answer:

The area of the triangle $ΔABC$ is equal to $24~cm²$.

The triangle $ΔABC$ is isosceles. Therefore, $AB=AC$.

$AD$ is the height from the side $BC$.

Since $DC=10$, the length of the height $AD$ is $20$ percent longer than the length of the side $BC$.

Task:

Calculate the area of the triangle $ΔABC$.

Solution:

Since it is an isosceles triangle (and, therefore, median), $DC=10$ and $BC=20$.

The height $AD$ is $20$ percent longer than the length of $BC$.

That is:

$AD=1.2\cdot BC$

$\frac{100}{100}+\frac{20}{100}=\frac{120}{100}=1.2$

$AD=1.2\cdot20=24$

Hence, the area of the triangle $ΔABC$:

$SΔ\text{ABC}=\frac{AD\cdot BC}{2}=\frac{24\cdot20}{2}=\frac{480}{2}=240$

Answer:

The area of the triangle $ΔABC$ is equal to $240~cm²$.

The rest of the terms, such as median, bisector, etc., are used when we are missing some data. These terms help us to find new data when we have to solve a problem in which we are missing information.

Many students experience a sense of failure when they don't do as well in their exams as they would have liked. However, success in this regard is subjective. Instead of comparing yourself with your peers, you must take into consideration your own achievements and leave aside the results of your classmates. Often, the problem is not that you did not know something or that you did not understand how to calculate the area of a triangle, but rather that you did not prepare well for the exam.

To give an example: imagine an excellent pastry chef who knows many recipes, knows the products and succeeds in creating truly delicious pastries. If he had not practiced and had not prepared well (buying the products and appliances he needs, finding good recipes and having patience with the right timing, etc.), he would not have achieved good results. This is also true for students—good preparation is key!

• Studying too intensively. Some students study for a week before the exam, perhaps ten hours a day or so. It is obvious that they are motivated and have good intentions, but the problem here is that sometimes this causes them to run out of energy before they even get to the exam. As a result, they are tired and burned out without enough time to go over all the material they are going to be tested on.
• Too much self-confidence. Did you get a 10 on the mock exam on triangular areas? If so, this does not mean that you should study two days before the exam. An exam requires preparation, both mental and practical. To prepare for an exam, it is recommended to study for at least a week.
• Stress and nerves. If exams make you anxious, it is best to start working on it beforehand. Students who suffer from this anxiety are usually those who know the subject matter inside out, but their self-confidence affects them negatively. Prepare yourself mentally because, otherwise, you may draw a blank on the exam and this will be reflected in your results.

Your study skills are as important as learning the subject matter. For this reason, a stopwatch can be your great ally. We recommend that you adopt its use from now on. The one that often comes with a cell phone is more than enough!

Once you have studied a subject area, it is advisable to practice and solve problems with a stopwatch, even when you are not preparing for exams. Why?

• The stopwatch gives you an indication of how long it takes you to solve a problem.
• The stopwatch lets you know which subject areas are your weakest.
• Thanks to using a stopwatch, you can also measure how fast you're progressing—if you're taking less time, you've made progress!

Many students have a hard time digesting the fact that they got a low grade and, more importantly, the disappointment that often accompanies it. It is very important for you to know that the more you let this affect you, the worse the impact on your academic achievement will be.

The grade you get on an exam is a kind of feedback that tells you what you are doing well and what you could improve. How can this feedback change the way you study?

• Read the test questions several times before answering.
• Use private tutors to reinforce your knowledge.
• Include extra study days before each exam.
• Study on your own instead of with friends.

We do not recommend it and the reason is very simple: a private lesson is "private", that is to say, it is adapted to your needs. When two friends study in the same private class, one of them will have to adapt to the rhythm of the other.

Thus, the idea of tailoring a "private lesson" to the student's needs is diluted. That said, if both you and your friend find the same topic difficult (for example, how to calculate the area of a triangle), you can take a joint private lesson.

If the question asks about a right triangle, you can multiply the legs (the sides of the triangle that are not the base) and divide by 2. This method is often a great shortcut to reaching the solution. That is why it is important that you know the formula and this specific feature of this triangle.

Also, if you are asked about an isosceles triangle, you should know that both the bisector and the median are considered the height of the triangle. With this knowledge, you can quickly work out the area of the triangle.

If you are interested in learning how to calculate areas of other geometric shapes, you can enter one of the following articles:

• How to calculate the area of a trapezoid?
• The area of a parallelogram: what is it and how is it calculated?
• Circular area
• Surface area of triangular prisms
• How to calculate the area of a rhombus?
• How to calculate the area of a regular hexagon?
• Area of a rectangle
• How to calculate the area of an orthohedron
• Acute triangle
• Obtuse triangle
• Scalene triangle
• Equilateral triangle
• Isosceles triangle
• The edges of a triangle
• Area of a right triangle
• Height of a triangle
• How to calculate the perimeter of a triangle?

In Tutorela you will find a variety of articles about mathematics!

Many times, the study of mathematics arouses some anxiety among students in high school and further education. A private math class is ideal for those who want to get good grades on their exams, but don't know how.

A private lesson focuses on a certain aspect and includes more than just doing exercises, like:

• Learning to read the statement and understand what is being asked of you.
• An emphasis on understanding what is being asked of us and how we should answer.
• Finding information that can help us solve the problem.
• Studying formulas and tricks that can help us when it comes to finding the solution to a problem or exercise.

In the past, private lessons in mathematics or any other subject took place at the student's or teacher's home. Nowadays, it is also possible to have private lessons online. This offers a great way to learn advanced subject from the comfort of your own home, at the hours that best suit both the student and the teacher.

Triangles and other geometric shapes are aspects that students are exposed to as early as their first years in high school. Their grades in mathematics are what set the pace of their learning and can affect whether or not a student chooses to continue studying the subject at later on. Often, what holds people back when studying mathematics is not based on intelligence or aptitude, but rather arises from erroneous learning methods that do not help the student understand the subject matter effectively. A private mathematics teacher works side by side with the student, ensuring that in the end the student has understood their lessons.