Welcome to the Math Salamanders' Geometry Formulas Triangles area.
Here you will find information about the properties of triangles and
different triangle formulas and theorems.
The Math Salamanders have a large bank of free printable shape clipart.
Each of the printable shape sheets is available either in color or black and white.
Using this shape clipart will help your child understand to recognize shapes and learn
about the different properties that shapes have.
On the sheets with multiple shapes, we have shown the shapes in different sizes and orientations so that
your child will recognize variations of the same shape, and start noticing the properties the same shapes all have.
The sheets can be used as part of a Math display, as flashcards, or as printable coloring sheets.
Triangles have the following properties:


Equilateral Triangle  Isosceles Triangle  Scalene Triangle 
Equilateral Triangles
Equilateral triangles have the following properties (in addition to the properties above for all triangles):
Isosceles triangles have the following properties:
Please note: according to Wikipedia:
" In geometry, an isosceles triangle is a triangle that has two sides of equal length. Sometimes it is specified as having two and only two sides of equal length, and sometimes as having at least two sides of equal length, the latter version thus including the equilateral triangle as a special case. "
Source: https://en.wikipedia.org/wiki/Isosceles_triangle
This means that there is some dispute as to whether an equilateral triangle is a special case of an isosceles triangle or not!
Most modern textbooks include use the 'at least' definition for isosceles triangles.
Scalene Triangles
Scalene triangles have the following properties:
Acute Triangle  Right Triangle  Obtuse Triangle 
Acute triangles
Acute triangles have all acute angles (angles less than 90°).
It is possible to have an acute triangle which is also an isosceles triangle  these are called acute isosceles triangles.
Right triangles
Right Triangles (rightangled triangles) have one right angle (equal to 90°).
It is possible to have a right isosceles triangle  a triangle with a right angle and two equal sides.
Obtuse triangles
Obtuse triangles have one obtuse angle (angle which is greater than 90°).
It is possible to have a obtuse isosceles triangle  a triangle with an obtuse angle and two equal sides.
If you would like a printable version of the support page above on properties of triangles, use the link below.
Below are some of the triangle formulas and theorems for our Geometry Formulas Triangles page.
The angles in a triangle add up to 180°.
This can be shown using the following simple demonstration.
Step 1
Draw a triangle on a piece of paper. Mark the 3 angles a, b and c.
Step 2
Cut out (or tear out) the three angles.
Step 3
Put the angles a, b and c together. They should make a straight line when placed together, which is equivalent to 180°.
Example
Find the missing angle in the triangle below.
We know that all three angles in the triangle must add up to 180°.
So we know that ? + 128 + 23 = 180.
This means that ? = 180  128  23 = 29°.
So the missing angle is 29°.
The area of a triangle is equal to half of the base multiplied by the perpendicular height of the triangle.
A right triangle showing height and base.  A nonright triangle showing height and base. 
In a right triangle (left image) then we simply need to multiply the 2 sides together which are adjacent to the right angle, and then halve the answer.
In a triangle which is not a right triangle (right image), then we need to find the perpendicular height of the triangle.
We can find the perpendicular height by dropping a vertical line down from the highest point on the triangle to the base (with the longest side set horizontal).
Finally, we find the area by multiplying the base by this perpendicular height and halving the answer.
Example
The area of this triangle is \[{1 \over 2} bh = {1 \over 2} \times 6 \times 3 = 9cm^2 \]
Why does this work?
If we look at the rectangle below, we can see that it has twice the area of the blue triangle inside it because the two yellow triangles are exactly the same size as the two subdivided blue triangles.
The area of this rectangle is base x height, so the area of the triangle is half of this which is \[{1 \over 2} \times base \times height \]
Pythagoras' theorem states that in a right triangle (or rightangled triangle) the sum of the squares of the two smaller sides of the triangle is equal to the square of the hypotenuse.
In other words, \[a^2 + b^2 = c^2 \]
where c is the hypotenuse (the longest side) and a and b are the other sides of the right triangle.
What does this mean?
This means that for any right triangle, the orange square (which is the square made using the longest side) has the same area as the other two blue squares added together.
Other formulae
As a result of the formula \[a^2 + b^2 = c^2 \] we can also deduce that:
Square rooting the 3 equations above gives us:
In this example, we need to find the hypotenuse (longest side of a right triangle).
So using pythagoras, the sum of the two smaller squares is equal to the square of the hypotenuse.
This gives us: \[ 4^2 + 6^2 \; = \; ?^2 \]
Which means that \[ ?^2 = 16 + 36 = 52 \]
So \[ ? = \sqrt {52} = 7.21 \; to \; 2dp \]
In this example, we need to find the length of the base of the triangle, given the other two sides.
So using pythagoras, the sum of the two smaller squares is equal to the square of the hypotenuse.
This gives us \[ ?^2 \; + \; 5^2 \; = \; 8^2 \]
So \[ ?^2 \; = \; 8^2 \;  \; 5^2 \; = \; 64  25 \; = \; 39 \]
So \[ ? \; = \; \sqrt {39} \; = \; 6.25 \; to \; 2dp \]
In basic triangle trigonometry, we look at three trigonometric functions: sine, cosine and tangent (sin, cos and tan) with right triangles.
Let us start by naming the parts of this triangle.
θ is one of the angles of the right triangle (but not the right angle!)
h is the hypotenuse  the longest side of the right triangle
o is opposite to the angle
a is adjacent to the angle
The three formulas to rememeber can be summed up in one acronym:
where s stands for sine, c for cosine and t for tangent.
The three formulas are written below:
\[ \sin (θ) \; = \; {o \over h} \] \[ \cos (θ) \; = \; {a \over h} \] \[ \tan (θ) \; = \; {o \over a} \]
In the triangle below, we have been given the angle which is 35°
We have also been told the length of the adjacent side a, which is 8cm.
We need to find out the length of the hypotenuse h.
So as the letters a and h are the two letters involved, we need the cosine operation (CAH).
So the formula we need is \[ \cos (θ) \; = \; {a \over h} \]
So \[ \cos (35°) = {8cm \over ?} \]
This means that \[ ? \; = \; {8cm \over \cos (35° ) } \]
So \[ ? \; = \; 9.77cm \; to \; 2dp.\]
In the triangle below, we have been given the angle, which is 48°.
We have also been told the adjacent side a, which is 11m.
We need to find out the length of the opposite side o.
So as the letters o and a are the two letters involved, we need the tangent operation (TOA).
The formula we need is: \[ \tan (θ) \; = \; {o \over a} \]
So \[ \tan (48°) \; = \; {? \over 11} \]
So \[ ? \; = \; 11 \times \tan (48°) \; = \; 12.22m \; to \; 2dp \]
Example 3) Find the length of the missing side.
In this triangle, we need to find the length of the opposite side of the triangle.
We have been given the angle and the hypotenuse.
So as the letters o and h are used, we need the sine operation (SOH).
The formula we need is: \[ \sin (θ) \; = \; {o \over h} \]
So \[ \sin (42° ) \; = \; {? \over 4.2} \]
So \[ ? \; = \; 4.2 \times \sin (42° ) \; = \; 2.81cm \; to \; 2dp \]
The basic triangle trigonometry formulas above will only work for right triangles  they will not work for triangles without right angles.
The following formulas below will work for any triangle.
Again, if we start with the naming of the sides and angles of the triangle:
a is the side opposite angle A;
b is the side opposite angle B;
c is the side opposite angle C.
The sine rule states that: \[ { a \over \sin A} \; = \; {b \over \sin B} \; = \; {c \over \sin C} \]
The Cosine rule states that:
\[ a^2 \; = \; b^2 \; + \; c^2 \;  \; 2bc \; \cos A \]
OR
\[ \cos A \; = {b^2 \; + \; c^2 \;  \; a^2 \over 2bc} \]
Example 1) Find the missing side.
In this example, we have been given 2 of the angles, but only one side measurement.
The rule we need is the sine rule, as 2 angles are involved.
So using the sine rule gives us:
\[ {12 \over \sin (123° )} \; = \; { ? \over \sin (27° )} \]
So this means that:
\[ ? \; = \; { 12 \times \sin (27° ) \over \sin (123° )} \]
So \[ ? \; = \; 6.50cm \; to \; 2dp. \]
Example 2)
Find the missing angle in the triangle below. Give your answer to 1 decimal place.
In this example, we have been given three sides and we need to find the angle.
The rule we need is the cosine rule which involves 3 sides and one angle.
It is easier to use the cos A formula, as it is the angle we want to find.
\[ \cos (A) \; = \; {b^2 \; + \; c^2 \;  \; a^2 \over 2bc} \]
Using ? as our angle A, this gives us: \[ \cos (?) \; = \; {10^2 \; + \; 11^2 \;  \; 6^2 \over 2 \times 10 \times 11} \]
So \[ \cos (?) \; = \; {185 \over 220} \; = \; {37 \over 44}\]
This means that: \[ ? \; = \; \cos^{1} {37 \over 44} \]
This gives us: \[ ? \; = \; 32.8° \; to \; 1dp \]
Take a look at some more of our worksheets similar to these.
Here you will find a support page packed with a range of geometric formula.
Included in this page are formula for:
This page will provide a useful reference for anyone needing a geometric formula.
Here you will find a range of geometry cheat sheets to help you answer a range of geometry questions.
The sheets contain information about angles, types and properties of 2d and 3d shapes, and also common formulas associated with 2d and 3d shapes.
Included in this page are:
Using the sheets in this section will help you understand and answer a range of geometry questions.
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