The Area of a Triangle

Whether you are preparing for a 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.

Key characteristics:

• The three interior angles always sum to $180°$
• The vertices are typically labeled with capital letters (such as $A,B$ and $C$)
• The sides are named by their endpoint vertices: $AB, BC, CA$
• Each side connects two vertices

There are several types of triangles, and some share common characteristics. Before exploring how to calculate the area of different triangle types, let's review the important terms used when working with triangular areas.

• Base: Any side of the triangle that you choose to use in the area formula.
• Height: A perpendicular line segment from a vertex to the opposite side (or the extension of that side). The height forms a 90° angle with the base.
• Opposite side: The side of the triangle that faces a given vertex.

Additional geometric terms:

• Straight line: A straight line is a one-dimensional figure that extends infinitely in both directions with no curves.
• Segment: A fragment of a line between two points.
• Median: A line segment from a vertex to the midpoint of the opposite side.
• Angle bisector: A ray from a vertex that divides the angle at that vertex into two equal angles.
• Perpendicular bisector: A line perpendicular to a side that passes through the midpoint of that side.
• Midsegment: A segment connecting the midpoints of two sides of a triangle. It is parallel to the third side and half its length.

The area formula for a triangle comes from the relationship between triangles and parallelograms. Here's why we divide by 2:

• If you take any triangle and create a copy, then flip and attach the copy, you form a parallelogram.
• A parallelogram's $area = base × height$.
• Since the triangle is half of this parallelogram, we divide by $2$.

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 $2$.

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

Given:

• 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 $2$.

$195$ divided by $2$ equals $97.5$.

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

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

Method 2: Special formula for equilateral triangles

When you know only the side length \(s\), you can use:

$Area = \frac{s^2\sqrt{3}}{4}$

For $s = 15 cm: Area = \frac{15^2\sqrt{3}}{4} = \frac{225\sqrt{3}}{4} \approx 97.43 \text{ cm}^2$

Note: In an equilateral triangle with side s, the height is always $h = \frac{s\sqrt{3}}{2}$

given:

• 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: $119$ cm².

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

Given:

• 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.

$Area=\frac{9\times10}{2}=45 \text{ cm}^2$

Note: For scalene triangles where the height is not given, you may need to:

• Use the Pythagorean theorem to find the height (if other information is available)
• Use Heron's formula if you know all three side lengths.

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

Given:

• The leg $CB$ has a length of $12$ cm.
• The other leg has a length of $14$ cm.

Important: in a right triangle or right-angled triangle, the two legs (the sides that form the 90° angle) are perpendicular to each other. This means:

• You can use one leg as the base
• You can use the other leg as the height
• No additional height calculation is needed!

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: $84$.

$Area=\frac{12\times14}{2}=84 \text{ cm}^2$

Given:

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

Important concept:

In an obtuse triangle, When drawing a height to the side opposite the obtuse angle, the height falls outside the triangle. To measure this height:

1. Extend the base line beyond the triangle
2. Draw a perpendicular from the opposite vertex to this extended line
3. This perpendicular is the height

Even though the height is outside the triangle, you use the same area formula - nothing changes in the calculation!

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: $104$.

$Area=\frac{13\times16}{2}=104 \text{ cm}^2$

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

Heron's formula, attributed to the Greek mathematician Heron of Alexandria, allows you to calculate the area of a triangle when you know the lengths of all three sides ($a, b$and $c$) but don't know the height.

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

Where 's' is the perimeter of the triangle divided by $2$ (the semi-perimeter):

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

Example:

Find the area of a triangle with sides $a = 5 cm$, $b = 6 cm$, and $c = 7 cm$.

Step 1: Calculate the semi-perimeter: $s=\frac{5+6+7}{2}=\frac{18}{2}=9 \text{ cm}$

Step 2: Apply Heron's formula: $Area=\sqrt{9(9-5)(9-6)(9-7)} \\ Area=\sqrt{9 \times 4 \times 3 \times 2} \\ Area=\sqrt{216} \approx 14.7 \text{ cm}^2$

When to use Heron's formula: This formula is especially useful when you know all three side lengths but the height is not given and would be difficult to calculate.

Task:

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

The base of the pool is $10$ m.

The height 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 apply 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.

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!

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