Perimeter of a polygon:
The perimeter of a Polygon is the sum of the lengths of all sides of the polygon.
Perimeter of a triangle:
-For a scalene Triangle of sides , and , the perimeter is:
-For Isoceles Triangle of Equal side and base :
-For an Equilateral Triangle of side :
-Perimeter of Quadrilaterals:
-For a quadrilateral of sides ,, and :
-For a Rectangle of Length and width
Or
-For a Square(or Rhombus) of side :
-For a Parallelogram of parallel equal sides and parallel equal sides :
Or
Areas of Circles and Polygons:
A triangular region consists of a triangle and its interior while a circular region is one of a circle and its interior.
The area of a triangle can be calculated using methods depending on information we have:
For a scalene triangle:
-Taking any altitude (as height, h) and its opposite side (as base, b).
The Area is:
We can also write:
For an isosceles triangle, given the three sides: a,a and b
We can calculate
Finally:
And the area of the triangle:
If we use the Heron formula, we find the same formula
In our situation and
That means
We get the following area:
Finally:
Special isosceles and right triangle:
In this situation we have :
Without going far, we can notice that each of the right triangle legs can be the base (a) or the height (a):
Using b:
This can be found from the previous formula as well.
Equilateral Triangle
We have the nomrmal formula:
Now for the area we have
finally:
Area Triangle using angles:
We can see that:
So:
The Area:
We can find all three relationships:
Circumference of a circle
The Circumference of a circle is as follows:
where is the radius of the circle.
Since the diameter , we can write:
Areas and circumferences of circular figures can be calculated without replacing with its value. That means in terms of .
Length of an arc:
The length of an arc of a circle is related to the central angle that intercepts it.
If is in , we have:
For an angle in , we use the proportion to the circumference:
The length of
This is the same as:
Area of a Circle of radius r:
If is the radius of a circle, the is:
This formula is widely used in geometry and other fields.
Area of a Sector AOB of a circle:
If is the radius of a circle, the central angle (or intercepted arc) in , the of is:
We simplify and get:
If the central angle (or intercepted arc) is in :
This is a simple conversion of radians above.
Area of a segment of a circle
This can be approached 2 ways:
-Area of the sector minus area of isosceles triangle
-By deriving the formula.
The central angle’s bisector bisects the chord AB at D.
We get
But
OD is the altitude:
But we know that:
That returns:
The Area of is:
We plug in:
Finally:
For sector AOB:
We take the difference to get the area of the segment:
Al-Kashi, Heron,Bretschneider’s, Brahmagupta’s and Coolidge formulas
The Bretscheiner’s formula helps find the area of a non-cyclic quadrilateral, that cannot be inscribed in a circle,
using only side lengths and possibly one angle measure or one diagonal length.
After the Bretschneider’s formula, we’ll simplify the quadrilateral to make it cyclic. We’ll get the Brahmagupta’s formula.
With the cyclic quadrilateral the product of the diagonals and is and opposite angles are supplementary.
Let be area.
From the figure, we notice
Squaring both sides:
Now let’s check 2 values of diagonal :
Taking the square:
Adding and :
Injected 2 in the last term.
To the same denominator and multiplying both sides by 4:
If we call the semi-perimeter
We have the final Bretscheiner’s formula:
Brahmagupta’s formula
The steps remain the same but this time since the quadrilateral is we know that the two opposite angles are supplementary.
We get from the Bretscheiner’s formula
Heron’s formula
The fourth side goes away and we get a triangle. From Brahmagupta’s formula:
In depth explanation:
From Al-Kashi to Bretshneider via Heron and Brahmagupta, we are going to go through some of their work:
Al-kashi is simply the Law of Cosines:
Law of cosines:
From what we know:
We have two right triangles: and
Using the Pythagorean Theorem:
Also:
But :
This means:
We know from above:
Finally we have the :
Heron Formula:
This one is widely used. We use it when we have only sides and we want to find the Area of the triangle
Let be area and the semi-perimeter.
.
We should remember this because we’ll need it in the following paragraphs.
But we know that:
This is a difference of squares, we can factor:
This is the Heron formula:
The Bretschneider formula:
We will demonstrate the following forms of the Area per BRETSCHNEIDER
We split the diagonal into and
The same way for : and
We express the area of the four triangles:
Now we add all four areas:
After simplication:
The first form:
Now the second form:
From the figure:
We expand and replace terms in by
Now we move to the
We expand and replace terms in by
Now we move to the
Let’s add as follows:
After simplification we get:
But:
We plug in
Second Form:
Third form:
From the sequences above:
But:
This is the actual BRETSCHNEIDER formula:
According to wiki, the following work is credited to COOLIDGE in 1939:
Let be area.
From the figure, we notice
Squaring both sides:
Now let’s check 2 values of diagonal :
Taking the square:
Adding and :
Injected 2 in the last term.
To the same denominator and multiplying both sides by 4:
If we call the semi-perimeter
We have the final COOLIDGE formula:
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