Draw a perpendicular bisector

You need 5 isosceles triangles with a base of 30 mm. To get the height of the triangles (perpendicular to the 30 mm bases) you need to calculate the apothem of the pentagon (assumed to be regular).

If you cut the pyramid by a plane passing through its apex, the center of the base-pentagon,and the midpoint of one side, the plane figure created by the three points (apex, center, and midpoint) is a right triangle. The legs are the apothem, and the altitude of the pyramid. The hypotenuse is the slant height of the pyramid, and is thus the height of the triangles in the development. pyramid (60 mm) form.
Use the Pythagorean Theorem to find that slant height.
s^2=a^2+h^2.

Since you have the coordinates of the three vertices, the most straightforward method is to calculate the length of the sides using the distance formula
d(P_1,P_2)=SQRT((X_1-X_2)^2+(Y_1-Y_2)^2)
where SQRT is the square root function, X_1, Y_1) are the coordinates of point P_1, etc.
With the three lengths available, use Heron's (sometimes called Hero's) to find the area.
Here is Heron's formula.
Let's call the lengths a, b, and c
Let p be the semi-perimeter p= (a+b+c)/2
Then
Area= SQRT [ p(p-a)(p-b)(p-c) ]
Make sure that there is a matching ) parenthesis to the one in the SQRT.

Alternatively,
You can choose the base as the side opposite the vertex (0,0)
Calculate the equation of the line that supports the base.
Calculate the equation of the line issuing from (0,0) and perpendicular t the base.
Calculate the coordinates of the intersection point , call it H, of the base and its perpendicular line (coming from (0,0)).
Calculate the distance OH, that is the height relative to the chosen base.
Use the formula Area= base*height/2

Now it is up to you to choose one of the two methods and calculate the area of that triangle. The second method involves more calculations than the first, and more possibilities of errors. Good Luck

Since you have the coordinates of the three vertices, the most straightforward method is to calculate the length of the sides using the distance formula
d(P_1,P_2)=SQRT((X_1-X_2)^2+(Y_1-Y_2)^2)
where SQRT is the square root function, X_1, Y_1) are the coordinates of point P_1, etc.
With the three lengths available, use Heron's (sometimes called Hero's) to find the area.
Here is Heron's formula.
Let's call the lengths a, b, and c

Let p be the semi-perimeter p= (a+b+c)/2
Then
Area= SQRT [ p(p-a)(p-b)(p-c) ]
Make sure that there is a matching ) parenthesis to the one in the SQRT.

Alternatively,
You can choose the base as the side opposite the vertex (0,0)
Calculate the equation of the line that supports the base.
Calculate the equation of the line issuing from (0,0) and perpendicular t the base.
Calculate the coordinates of the intersection point , call it H, of the base and its perpendicular line (coming from (0,0)).
Calculate the distance OH, that is the height relative to the chosen base.
Use the formula Area= base*height/2

Now it is up to you to choose one of the two methods and calculate the area of that triangle. The second method involves more calculations than the first, and more possibilities of errors. Good Luck

Draw two chords of the circle (any two chords will do, as long as they aren't parallel). Construct perpendicular bisectors of each chord. The center of the circle will be where the two bisectors intersect.

Draw two chords of the circle (any two chords will do, as long as they aren't parallel). One way to avoid parallel chords is to have them share an endpoint. Construct perpendicular bisectors of each chord. The center of the circle will be where the two bisectors intersect.

o find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that aparallelogram can be divided into 2 triangles. For example, in the diagram to the left, the area of each triangle is equal to one-half the area of the parallelogram.
Since the area of a parallelogram is , the area of a triangle must be one-half the area of a parallelogram. Thus, the formula for the area of a triangle is: or where is the base, is the height and · means multiply. The base and height of a triangle must be perpendicular to each other. In each of the examples below, the base is a side of the triangle. However, depending on the triangle, the height may or may not be a side of the triangle. For example, in the right triangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. In the triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dotted line is drawn to represent the height.

Formulas relating to right angled triangles are:
Sine = perpendicular divided by hypotenuse
Cosine = base divided by hypotenuse
Tangent = perpendicular divided by base
Sine Tangent and Cosine functions are all available in Excel

In right triangle we are making 90 degree angle triangle, we can have problem for finding hypotenuse or finding sin or cos values of the side of the triangle. For ex,Find out the hypotenuse,sin and cos value of the right triangle with base 4 cm and perpendicular 3 cm Solution:Hypotenuse = SQRT(4^2 + 3^2) =SQRT(4*4 + 3*3) =SQRT(16+9)=SQRT(25)=5 cm For right triangle, sin(x)=3/5=0.6 cos(x)=4/5 =0.8

Assuming M is the intersection of MN with AB, and N is the intersection of MN and AC:

Angle ACP = angle BCP (by definition)
Angle NCP = angle BCP (intersection of line with parallel lines produces equal angles)
Triangle CPN is isoceles (two equal angles), and line NP = CN

Same argument for line MP = BM

Therefore NP + MP (i.e, MN) = CN + BM

To start points and lines are not shapes, and a triangle is not representable with a single function, therefore you need a function to draw shapes , given a number of points as parameters, and one to draw math functions (lines etc.).

Here is an example: Draw Line, Ellipse, Polygon ...

From website above, here is script in JS:

"<script type="text/javascript">
<!--
function myDrawFunction()
{
jg_doc.setColor("#00ff00"); // green
jg_doc.fillEllipse(100, 200, 100, 180); // co-ordinates related to the document
jg_doc.setColor("maroon");
jg_doc.drawPolyline(new Array(50, 10, 120), new Array(10, 50, 70));
jg_doc.paint(); // draws, in this case, directly into the document

jg.setColor("#ff0000"); // red
jg.drawLine(10, 113, 220, 55); // co-ordinates related to "myCanvas"
jg.setColor("#0000ff"); // blue
jg.fillRect(110, 120, 30, 60);
jg.paint();

jg2.setColor("#0000ff"); // blue
jg2.drawEllipse(10, 50, 30, 100);
jg2.drawRect(400, 10, 100, 50);
jg2.paint();
}

var jg_doc = new jsGraphics(); // draw directly into document
var jg = new jsGraphics("myCanvas");
var jg2 = new jsGraphics("anotherCanvas");

myDrawFunction();

//-->
</script>"