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Two triangles are congruent if all their corresponding angles have the same measure and all their corresponding sides have the same length.

A triangle is plane **shape** called a polygon that is formed by three edges and three vertices.

**TIP!** Firstly you need to read **Polygons**

Properties of Triangles

1. The sum of all internal angles of a triangle is 180°

2. The sum of any two side lengths in any triangle is always greater than the length of the third side.

3. The difference between the two sides of a triangle is less than the length of the third side.

4. The side opposite to the largest angle in a triangle it’s always the largest side.

5. Any exterior angle in a triangle is always equal to the sum of the opposite interior angles.

6. Two triangles are similar if the corresponding angles are congruent and the length of the sides are proportional.

7. The perimeter of a triangle is equal to the sum of all the side lengths of the triangle.

8. The area of a triangle is equal half the base times the height. A=1/2(bxh)

Triangles are grouped based on their angles or side lengths

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But, what if it’s impossible to find the height in a triangle but we are given all the side lengths.

Then an easy way is to apply the “Heron Formula”

The area of a triangle whose side lengths are a, b and c is given

$\displaystyle A=\sqrt{p(p-a)(p-b)(p-c)}$

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

Example 1: Find the area of the triangle where we are given the sides 13 cm, 14 cm and 15 cm.

Solution: Using the Heron Formula we are given above we write:

Firstly we find the value of the perimeter and then we divide it by 2.

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

$\displaystyle p=\frac{13+14+15}{2}$

$\displaystyle p=\frac{42}{2}=21$

Then we find the area:

$\displaystyle A=\sqrt{p(p-a)(p-b)(p-c)}$

$\displaystyle A=\sqrt{21(21-13)(21-14)(21-15)}$

$\displaystyle A=\sqrt{21\cdot 8\cdot 7\cdot 6}$

$\displaystyle A=\sqrt{7056}$

**$\displaystyle A=84c{{m}^{2}}$**

Since we know that the perimeter of a triangle is the sum of all the side lengths, then in the equilateral triangle:

P = a + a + a = 3a

First Method: Draw one height of the triangle.

Find the height by using “Pythagoras Theorem”

In an equilateral triangle the height is perpendicular to the base and always divides the base in two equal parts.

So, in our figure BH=HC=12a

We know that AC = CB = AB = a

To find the height (AH) we explore the right angle triangle AHC or ABH since they are congruent based on the SAS rule.

Then we apply the Pythagoras Theorem in the Right angle triangle AHC

$\displaystyle A{{H}^{2}}=A{{B}^{2}}-H{{B}^{2}}$

$\displaystyle {{h}^{2}}={{a}^{2}}-{{\left( \frac{a}{2} \right)}^{2}}$

$\displaystyle {{h}^{2}}=\frac{4{{a}^{2}}}{4}-\frac{{{a}^{2}}}{4}$

$\displaystyle {{h}^{2}}=\frac{3{{a}^{2}}}{4}$

$\displaystyle h=\sqrt{\frac{3{{a}^{2}}}{4}}$

**$\displaystyle h=\frac{a\sqrt{3}}{2}$**

So, we found the height of our equilateral triangle that is always like this only the value of a changes depending on the side length of the triangle.

You can memorize it as a formula or find it the long way as we did above.

Apply the formula $\displaystyle A=\frac{1}{2}\left( b\cdot h \right)$

Now we know the base and also the height of our equilateral triangle so it’s easy to apply the formula.

$\displaystyle A=\frac{1}{2}\left( b\cdot h \right)$

$\displaystyle A=\frac{1}{2}\left( a\cdot \frac{a\sqrt{3}}{2} \right)$

**$\displaystyle A=\frac{{{a}^{2}}\sqrt{3}}{4}$**

Second method: Using Heron Formula

Firstly we find the value of the perimeter and then we divide it by 2.

P = a + a + a

P = 3a

$\displaystyle p=\frac{3a}{2}$

Then, we apply the formula

$\displaystyle A=\sqrt{p(p-a)(p-a)(p-a)}$

$\displaystyle A=\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$

$\displaystyle A=\sqrt{\frac{3a}{2}(\frac{3a}{2}-\frac{2a}{2})(\frac{3a}{2}-\frac{2a}{2})(\frac{3a}{2}-\frac{2a}{2})}$

$\displaystyle A=\sqrt{\frac{3a}{2}(\frac{a}{2})(\frac{a}{2})(\frac{a}{2})}$

$\displaystyle A=\sqrt{\frac{3{{a}^{4}}}{16}}$

$\displaystyle A=\frac{{{a}^{2}}\sqrt{3}}{4}$

As you can see we get the same result but you can use whichever method looks more easy to remember or just simply learn it by heart that in a equilateral triangle the area it’s always

$\displaystyle A=\frac{{{a}^{2}}\sqrt{3}}{4}$

The perimeter of the isosceles triangle.

**First method**: Firstly we draw the height to the side that is different from the two equal ones.

Then, we find the height by using the Pythagoras Theorem.

BC = b.

h² = $\displaystyle {{a}^{2}}-{{\left( \frac{b}{2} \right)}^{2}}$

h² = $\displaystyle {{a}^{2}}-\frac{{{b}^{2}}}{4}$

$\displaystyle {{h}^{2}}=\frac{4{{a}^{2}}}{4}-\frac{{{b}^{2}}}{4}$

h² = $\displaystyle \frac{4{{a}^{2}}-{{b}^{2}}}{4}$

h = $\displaystyle \sqrt{\frac{4{{a}^{2}}-{{b}^{2}}}{4}}$

h = $\displaystyle \frac{\sqrt{4{{a}^{2}}-{{b}^{2}}}}{2}$

Apply the formula A = $\displaystyle \frac{1}{2}\left( b\cdot h \right)$

$\displaystyle A=\frac{1}{2}(b\cdot \sqrt{\frac{4{{a}^{2}}-{{b}^{2}}}{4}})$

The perimeter of a scalene Triangle is the sum of all the side lengths. In a scalene triangle the side lengths are different, so:

P = a + b + c

First Method: Is the same as we did above by drawing the height and applying the Formula $\displaystyle A=\frac{1}{2}\left( b\cdot h \right)$.

Second method: using Heron formula

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

$\displaystyle A=\sqrt{p(p-a)(p-b)(p-c)}$

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Acute angled trianglearea of the equilateral triangleEquilateralHeron formulaHow to find the perimeter and area of an equilateral triangleHow to find the perimeter and area triangleIsoscelesObtuse angled triangleperimeter of the equilateral triangleProperties of TrianglesRight angle triangleTriangles