Geometry Area Projects

Introduction :

In geometry,area is nothing but a region bounded by a closed curve. In differential geometry of surfaces, area is considered as an important invariant.

Doing projects on area of different shapes is an important and an interesting one. In this article of geometry area projects, the area of different shapes is calculated using the formulas.

Formulas for Geometry Area Projects:

Triangle: Area of Triangle = ½ b h

b —> base

h —> vertical height

Rectangle: Area of Rectangle = l w

l —-> length

w —-> width

Square: Area of a square = a2

a —-> side length

Parallelogram: Area of Parallelogram = b × h

b —-> breadth

h —-> height

Circle: Area of a Circle = `pi` r2

r —-> radius

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Example Problems for Geometry Area Projects:

Example 1:

Find the area of a triangle with base of 44 mm and a height of 14 mm.

Solution:

Area of a triangle = ½ b h

= ½ (44) (14)

= 308 mm2

Example 2:

Find the area of rectangle given the length is 23 cm and width is 17 cm.

Solution:

Area of a Rectangle = l * w

= 23 * 17

= 391 cm2

Example 3:

Find the area of a square of side length 10 cm

Solution:

Area of a square = a2

= 102

= 100 cm2

Example 4:

Find area of a parallelogram through base of 19 cm and a height of 6.5 cm.

Solution:

Area of a Parallelogram = b h

= (19) · (6.5)

= 123.5 cm2

Example 5:

The radius of a circle is 9 inches. Find its area.

Solution:

Area of Circle = `pi` r2

= 3.14 (9)2

= 3.14 (81)

= 254.34 in2
Practice Problems for Geometry Area Projects:

1) Find the area of rectangle given the length is 17 m and width is 8 m.

2) Find area of a square of side length 33 cm.

3) Find the area of triangle given base is 13 cm and height is 7 cm.

4) The radius of a circle is 11 inches. Find its area.

Answer key:

1) 136 m2 2) 1089 cm2 3) 25 cm2 4) 45.5 cm2 5) 379.94 in2

Derivative as Rate of Change Online Help

Let f be a given function and a be any point in its domain. Let h be any small but arbitrary positive or negative number. Then, the quantity `(f(a+h)-f(a))/h` is called the difference quotient of f at “a” with the increment h. It can be denoted by f(a,h). Clearly, it depends on both a and h.
Online help on derivative at a point and rate of change:

Actually this quotient represents the average rate of change of f in the interval [a, a+h] or in [a+h, a], according as h is positive or negative. Its limiting value as h tends to zero, i.e.,

`lim_(h->0) (f(a+h)-f(a))/h` represents the instantaneous rate of change of f at “a”.

Definition of derivative of a function at a point:

The derivative of a function f at a point x = a is defined by

f'(a) = `lim_(h->0)(f(a+h)-f(a))/h` (provided the limit exists)

In this article let us see how derivatives are helpful to find rate of change online.

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Determining Rate of Change Using Derivatives:

Ex 3: A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x co-ordinate.

Sol:

Step 1: Write down the given details.

The given curve is 6y = x^3 + 2.

`dy/dt = 8 (dx)/dt`

Step 2: Find the derivative of 6y = x3 + 2 with respect to the time t.

`6(dy)/dt = 3x^2 (dx)/dt + 0`

Step 3: Plug `(dy)/dt = 8(dx)/dt`

48`(dx)/dt = 3x^2(dx)/dt`

On simplification, we have 3x^2 = 48

x^2 = 16

So, x = ± 4

Step 4: Plug x = 4 and x = -4 in the equation of the curve 6y = x^3 + 2.

At x = 4, y = 11

At x = -4, y = -10.3

Step 5: Write the points in ordered pair.

The points on the curve at which the y coordinate is changing 8 times as fast as the x co-ordinate are (4, 11) and (-4, -10.3)
Problems to Find Rate of Change Using Derivatives:

An edge of a variable cube is increasing at the rate of 5 cm/second. How fast is the volume of the cube increasing when the edge is 7 cm long.

Sol: 735 cm3/s

The radius of the circle is increasing uniformly at the rate of 4 cm per second. Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Sol: 64πcm2/s.

Ray that Completes the Angle

Definition of a ray
End point of a ray
Definition of an angle
Vertex of an angle

Definition of a ray – A ray in a plane is a part of a line which has one end point and it can be extended endlessly in one direction.

End point of a ray – The end point of the ray is a terminal point from where the ray can not be extended.

Definition of an angle – When two rays are joined together at a common end point then an angle is formed.

Vertex of an angle – The common end point where the two rays meet to form an angle is known as the vertex.
How a Ray Completes the Angle:

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Angle – An angle is formed by two rays with a common vertex. One of the ray is base and the other ray is known as revolving ray.It is the revolving ray completes the angle

Steps To draw an angle –

Take a point O .
From O as starting point draw a ray OA.
Again,from O draw another ray OB .
Then Angle AOB is formed.

Here OA is the base Ray and OB is the revolving ray. Both these rays are called the arms of the angle. The rays OB is that ray which completes the angle AOB.O is referred as the vertex of the angle.
Few Points about Angles Completed by Rays

Few facts about the ray that completes the angle-

For an angle measuring ZERO Degree the base ray and the revolving ray are completely overlapped.
For an angle measuring 180 Degrees the base ray & revolving ray are exactly opposite to each other at the vertex.

Pictures of 0 and 180 degree angle.

Triangles Obtuse Acute Meaning

Triangles are the one of the most important geometric shapes. Comparing with angle we consider the triangle as acute angle and obtuse angle. acute angle is one, which is less than 90o. The obtuse angle is one, which is greater than 90o. In this article, we will discuss about triangle obtuse acute meaning. Also we shall solve problems regarding triangles obtuse acute meaning.
Triangles Obtuse Acute Meaning:

The following are the some of the triangles – obtuse acute angles

Acute angle triangle:

acute angle triangle

It is the angle, which is less than 90o.

Obtuse angle triangle:

obtuse angle triangle

It is the angle, which is more than 90o and less than 180o.
Example Problems – Triangles Obtuse Acute Meaning:

I like to share this Acute Isosceles Triangle with you all through my article.

Problem 1:

Identify the pairs of acute and obtuse angles of 50o,30o, and 40o.

Solution:

Given the pairs of acute and obtuse angles of 50o,30o, and 40o.

Step 1: Here the angle 50o is the acute angle because the acute angle, which is less than 90o.

Step 2: Here the angle 30o, is the also acute angle because the acute angle, which is less than 90o

Step 3: Here the angle 40o, is the also acute angle because the acute angle, which is less than 90o.

Therefore the given three angles are the acute angles because the acute angle, which is less than 90o. This is an acute triangles.

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Problem 2:

Identify the pairs of acute and obtuse angles of 100o,130o, and 140o.

Solution:

Given the pairs of acute and obtuse angles of 100o,30o, and 140o.

Step 1: Here the angle 100o is the obtuse angle because the obtuse angle, which is greater than 90o less than 180o.

Step 2: Here the angle 30o, is the also acute angle because the acute angle, which is less than 90o

Step 3: Here the angle 140o, is the obtuse angle because the obtuse angle, which is greater than 90o less than 180o.

Therefore the given triangles is an obtuse angled triangles.

Five Sided Polygon

In geometry, the polygon is the shapes. The number of shapes is available in polygon. The names are triangle, quadrilateral, heptagon, hexagon, pentagon, octagon, nonagon, decagon etc. A pentagon is a model of a self-intersecting polygon. The vertex normally a corner or a point where lines get together. Here we will see about the five sided polygon, the name is called as pentagon.

Types of Pentagon

There are two types of pentagon available in polygon. The names are,

Regular Pentagon
Irregular Pentagon
Regular Pentagon

The regular pentagon has the five sides with the equal sided length. The following is the regular pentagon picture,

Irregular Pentagon

The irregular pentagon also has the five sides but the sides are varying from one side to another side. The irregular pentagon figure is given below,

Properties

Properties of Regular Pentagon

The interior angle is 108 degree. We can find the interior angle by the formula of (180n-360)/n.
The exterior angle of regular pentagon is 72 degree. The formula for finding interior angle is 180-interior angle.
The area is calculated 1.72 s2 approximately. Here S is the length of a sided. We can find the exact area of a regular pentagon by using different methods.
Properties of All Pentagons

All pentagons have the five sided with equal sides. The number of different diagonals feasible from all vertices. In general ½ n (n-3).
This pentagon’s triangle is three. Many triangles are created by drawing the diagonals from a given highest point. In general n-2.
The total interior angles are 540 degree.
Examples

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1) Find the regular pentagon shape from the following figures.

Solution

Shape 2 is the regular pentagon. Because it is the five sided polygon.

2) Which one of the following is the irregular pentagon shape?

Solution

Shape 2 is the irregular pentagon. Since it has five sides but with the different size of the sides.

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Online Circumcentre Tutoring

Definition of Terms

Line – A line in a plane joins any two points and is extensible on both the sides, so it does not have any end points. In the diagram  below PQ represents a line.

Line segment –  A line segment is a part of a line. It has two end points and its also has a fixed measure. It can not be extended on either side. In the diagram below AB represents a line segment.

Perpendicular bisector – The perpendicular bisector of a line segment divides it into two equal length at right angle. All the points on a perpendicular bisectors are equidistant from the end points of the line segment.
Circumcenter – The circumcenter of a triangle is the point of intersection of  perpendicular bisectors of its sides. It is the center of the circumcircle drawn around the triangle.

Constructions Of Perpendicular Bisector Of A Line Segment – To draw  perpendicular bisector of a line segment AB we need to perform following steps :

Draw a line segment AB .
Take more than half of AB and with A as center draw arcs on both the sides of AB.
Now, with the same measure and B as center draw arcs on both the sides of AB to intersect the arcs of step 2 at P & Q.
Join PQ with a line to meet AB at O.
Thus  PQ is the required perpendicular bisector of line segment AB.

perpendicular bisector
Construction of Circumcenter of a Triangle

Following steps are performed to get the circumcenter of a triangle :

Assume that we are given a triangle with some specified measurement.
Take one of its side AB & draw its perpendicular bisector.
Now, take another side AC of the triangle and draw its perpendicular bisector.
These two perpendicular bisectors intersect each other at a point say O, this point is known as circumcenter.
With O as center and OA ( here OA = OB = OC )as radius draw circumcircle.
Thus O is the desired circumcenter.

circumcenter
Geometrical Proof for Basic Properties of Circumcenter

Aim –

The circumcenter of a triangle is the point of concurrence of the perpendicular bisectors of its sides.

The circumcenter is equidistant from all the three vertices of the triangle.

Diagram –

Proof –

Let the perpendicular bisectors BE & CD of AB and AC respectively meet at point O.
Now, extend AO to BC such that OF is perpendicular to BC .

In right angled Δs ADO and BDO

AD  = BD               ( CD is the perpendicular bisector of AB)
L ADO = L BDO   ( each of 90° )
OD = OD               ( common side )

Thus Δs ADO and BDO are congruent ( by SAS )   and hence OA = OB

Similarly Δs AOE and COE are congruent ( by SAS )  hence OA = OC

So OA = OB = OC

Δs BOF and COF are congruent (by RHS)  showing that OF is the perpendicular bisector of side BC.

 

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Now from the above discussion it is clear that –

CD , BE and AF are the perpendicular bisectors which are concurrent at O.
OA = OB = OC hence any circle with O as center and passing through one of the vertices will pass thrtough all of them.