Example of Congruence

Introduction to congruence:

Two objects are congruent if they consist of the similar shape with size. The given two triangles are congruent if their equivalent sides are equal within length also their equivalent angles are the same in size. Assume the triangles DEF and RST is congruent. These can be written as ? DEF ? ? RST. In geometry two congruent triangles contains the equal corresponding angles.

Example of Congruence

The followings are the important congruence test

ASA congruence

The two angles with the integrated faces of one triangle are equal to the corresponding two angles with the integrated faces of another triangle.

SAS congruence

The two faces with the included angle of triangle are the same to two faces with the included angle of another triangle.

AAS congruence

The two angles along throughout a non integrated side of one triangle are congruent to the identical measurements of a different triangle.

SSS congruence

Three sides of one triangle are identical to corresponding three sides of another triangle.

Example

Consider the following two triangles.




The triangle IJK is congruent to the triangle LMN

Angle I = Angle L

Angle J = Angle M

Angle K = Angle N

Length IJ = Length LM

Length JK = Length MN

Length KI = Length LN

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Examples for Congruence

Example 1 for congruence

In the following triangles are congruent then find the length of sides a, b, c.



Solution

The given triangles are congruent. Therefore the lengths of the sides of the triangles are equal.

Length EG = 52

Therefore the length VT = a = 52

Length FG = 48

Therefore the length UT = b = 48

Length EF = 50

Therefore the length UV = c = 50

Thus the a = 52, b = 48, c = 50.

Example 2 for congruence

Prove that triangle LMN is congruent to triangle PQR.



Solution

Given figure the angle L and angle P are the same.

Angle L = Angle P = 75 degree

Given figure the angle N and angle Q are the same.

Angle N = Angle Q = 65 degree.

Line segment LM is equal to the line segment PR.

Line LM = Line PR = 40 cm.

Therefore ? LMN and ? PQR are congruent through AAS congruence.

Unit Circle Equation

Introduction :

A unit circle is defined as a circle with the radius value is one. Particularly in the trigonometry the unit circle with radius one is pointed at (0, 0) that is the origin in Euclidean plane of the Cartesian coordinate system. The unit circle is represented as S1. The higher dimension of the unit circle is called as the unit sphere.

Formula for Unit Circle Equation:

If the point (x, y) is on the first quadrant of the unit circle, then the point x and y are the lengths of the right triangle and the hypotenuse length is one. By using the Pythagorean Theorem, the equation of the unit circle is,

` x^2 +y^2=1`

consider` x^2=(-x)^2` for all the value of x ,it gives the positive x value and the reflection of any point of x and y axis of the unit circle is provides the unit circle equation that is ` x^2 +y^2=1` and this is not only for the first quadrant of the unit circle.

The unit circle coordinates:

The unit circle having the angle theta and also having the radius one for the unit circle. The unit circle coordinates are,(x,y) that is,

`x=cos theta or cos theta=x/1=x`

`y=sin theta or sin theta=y/1=y`

By using the Pythagorean Theorem, the equation of the unit circle is,

`cos^2 theta + sin^2 theta=1`

Example 1 for Unit Circle Equation:

To check whether the following points are on the unit circle equation or not.

i) ` (1/ 2, sqrt3/2)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

` x=1/2 and y=sqrt 3/2`

put x and y values in the unit circle equation

`(1/2)^2+(sqrt 3/2)^2 =1/4 +3/4 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =1/2`

`theta =cos^(-1) (1/2) =pi/3 =60^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=60^@.`

Example 2 for unit circle equation:

To check whether the following points are on the unit circle equation or not.

i) ` (0, 1)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

here x=0 and y=1

put x and y values in the unit circle equation

`(0)^2+(1)^2 =0 +1 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =0`

`theta =cos^(-1) (0) =pi/2 =90^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=90^@.`

Three Dimensional Pyramid

Introduction:

In geometry, we have two-dimensional shape and three dimensional shape. Pyramid is a thee dimensional shape. It is also called as Polyhedron. The base of pyramid may be square, rectangle, triangle or any other polygon shape. The sides of pyramid look triangular faces. In Pyramid, all vertices of base are connecting a point to above, it is called apex of pyramid.

We have various kind of three dimensional pyramids. They are following.

For example, we have Square based pyramid, Rectangular pyramid, Triangular Pyramid, Pentagonal pyramid, Hexagonal pyramid and etc.

In this article, we will see about volume of all kinds of three dimensional pyramid.

Three Dimensional Pyramid: Square Pyramid, Rectangle Pyramid

Square based pyramid:

It is a three dimensional pyramid shape having a base is square. In square pyramid, we have five vertices, four triangular faces.

Formula:

Volume of square pyramid = `1/3*base^2*height`

Example 1:

Find the volume of square pyramid for the given base side is 7 meter and height is 10 meter.

Solution:

Given:

Base side = 7 m

Height = 10 m

Volume of square pyramid           = `1/3*base^2*height`

= `1/3*7^2*10`

= 0.333 * 49 * 10

= 163.17

Therefore, Volume of square pyramid is 163.2 cubic meter.

Rectangular pyramid:

It is a three dimensional pyramid shape having a base is Rectangle. In Rectangle pyramid, we have five vertices, four triangular faces.

Formula:

Volume of Rectangular pyramid = `1/3*LenGth*width*height`

Example 2:

Find the volume of Rectangular pyramid for the given length is 10 meter , width is 8 meter and height is 12 meter.

Solution:

Given:

Length = 10 m

Width = 8 m

Height = 12 m

Volume of rectangular pyramid          = `1/3*LenGth*width*height`

= `1/3*10*8*12`

= 0.333 * 960

= 319.68

Therefore, Volume of Rectangular pyramid is 319.7 cubic meter.

Three Dimensional Pyramid: Triangular Pyramid, Hexagonal Pyramid

Triangular pyramid:

It is a three dimensional pyramid shape having a base is Triangle. In triangular pyramid, we have four vertices, four triangle faces.

Formula:

Volume of triangular pyramid = `1/6*base*height*HEIGHT`

Example 3:

Find the volume of triangular pyramid for the given base is 6 meter , height is 10 meter and HEIGHT is 12 meter.

Solution:

Given:

Base = 6 m

Height = 10 m

HEIGHT = 12 m

Volume of Triangular pyramid      = `1/6*base*height*HEIGHT`

= `1/6*6*10*12`

= 0.167 * 720

= 120.24

Therefore, Volume of triangular pyramid is 120.2 cubic meter.

Hexagonal pyramid:

It is a three dimensional pyramid shape having a base is Hexagon. In Hexagonal pyramid, we have seven vertices, six triangle faces.

Formula:

Volume of Hexagonal pyramid = `Apothem*Side*Height`

Example 3:

Find the volume of Hexagonal pyramid for the given side is 6 meter , height is 10 meter and Apothem is 7 meter.

Solution:

Given:

Side = 6 m

Height = 10 m

Apothem = 7 m

Volume of Hexagonal pyramid     = `Apothem*Side*Height`

= `7*6*10`

= 420

Therefore, Volume of Hexagonal pyramid is 420 cubic meter.