Isosceles Triangle Proof

In an isosceles triangle, two sides are equal in length. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem. Some mathematicians define isosceles triangles to have only two equal sides, whereas others define that an isosceles triangle is one with at least two equal sides. The latter definition would make all equilateral triangles isosceles triangles. (Source : WIKIPEDIA)

Things need to remember for Proofs of isosceles triangle:

For proving isosceles triangle we need to know the following information about  isosceles triangle,
There are two types of isosceles triangles,
1.Normal isosceles triangle
2.Right isosceles triangle.

Properties:
Properties of parts of  isosceles triangle:
1.The two sides of the isosceles tringles are equal.
2.Two base angles has same measure.
3. Angle ratio of the right isosceless triangle is   45:90:45.
4.The side ratio of the isosceles triangle is  1:1:`sqrt(2)`

Problems on isosceles triangle proof:

Problem 1:
Prove that the following triangle is isosceles triangle.

Proof:
Given , The angle We know that the sum of the angles are 180.
So 100 + x +x = 180
      100 + 2x = 180
Subtract 100 0n both sides.
    2x =180 -100
    2x = 80
Divide by x on both sides,
 x = 40
The base anles are equal two 40 .
According to the properties of isosceles triangle,we can determine that the given triangle is isosceles triangle.
Hence the proof.

Problem 2:
Prove that the triangle with the sides 5 : 5 : 5`sqrt(2)` is an isosceles right triangle triangle.
Proof:
Given,The sides of the triangle is 5 , 5 ,5`sqrt(2)`
We know that in a right Angle triangle,the hypotenuse is greater then the legs and it satisfies the pythagorean theorem,
5`sqrt(2)` > 5 , 5
5²+5² = (5`sqrt(2)` )²
25+25 = 25 *2
50 = 50
So the given sides satisfies the pythagoren theorem.So we can say that it is a right triangle.
The given sides are in the ratio of 1:1:`sqrt(2)`
so the given triangle is isosceles triangle.
Hence the proof.

Pyramid with a Trapezoid Base

A pyramid is a building where the outer surfaces are triangular and converge at a point. The base of a pyramid can be trilateral, quadrilateral, or any polygon shape, meaning that a pyramid has at least three outer surfaces (at least four faces including the base). The square pyramid, with square base and four triangular outer surfaces, is a common version. (Source: Wikipedia)

I like to share this Trapezoid Shape with you all through my article.

Pyramid with trapezoid base:
      A pyramid with a trapezoid base is called as trapezoidal pyramid.
     Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height

Example problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 12 cm, 14 cm and its height of the pyramid is 6 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 12 cm, b₂ = 14 cm, and h = 6 cm.
Formula:
   Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (12 + 14) * 6 cm³
= (1 / 3) * 78 cm³
= 26 cm³

Answer:
 Volume of the trapezoidal pyramid is 26 cm³

Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 15 cm, 24 cm and its height of the pyramid is 9 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 15 cm, b₂ = 24 cm, and h = 9 cm.
Formula:
Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
 Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (15 + 24) * 9 cm³
 = (1 / 3) * 175.5 cm³
                                               = 58.5 cm³
Answer:
 Volume of the trapezoidal pyramid is 58.5 cm³

Trapezoid pyramid problem 3:
        A base length of the trapezoid pyramid is 6 cm, 11 cm and its height of the pyramid is 5 cm. Find the area of the trapezoid.
Solution:
     Given base lengths are b₁ = 6 cm, b₂ = 11 cm, and h = 5 cm.
Formula:
Area of trapezoid = ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
                                               = (1 / 2) * (6 + 11) * 5 cm³
                                               = (1 / 2) * 85 cm³
                                               = 42.5 cm²
Answer:
 Volume of the trapezoidal pyramid is 42.5 cm²

Practice problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 7 cm, 9 cm and its height of the pyramid is 12 cm. Find the volume of the trapezoidal pyramid.
Answer:
 Volume of the trapezoid pyramid is 32 cm³
Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 10 cm, 22 cm and its height of the pyramid is 7 cm. Find the volume of the trapezoidal pyramid.
Answer:

 Volume of the trapezoid pyramid is 37.33 cm³

Inverting Matrices

The inverse of a square matrix A, sometimes called a reciprocal matrix, is a matrix   such that         

Where I is the identity matrix.

A square matrix A has an inverse if the determinant . A matrix possessing an inverse is called nonsingular, or invertible.

The matrix inverse of a square matrix m may be taken in mathematics using the function inverse[m].

Inverse of 2x2 Matrix

For a 2X2  matrix

A= `[[a,b],[c,d]]`

The matrix inverse is

                  

             =  

                   =    

Inverse of 3x3 Matrix

For a 3X3 matrix

          

The matrix inverse is

A general matrix can be inverted using methods such as the gauss Jordan elimination, Gaussian elimination.

Prism Solve for h

Prism is one of the three dimensional shape. Normally the prisms faces are flat surface. Here we have two bases, one is at the top and another is at the bottom. In prisms we have two kinds of prisms; they are rectangular prism and triangular prism. When a prism bottom is rectangular shape then it’s denoted as rectangular prism. Cube and cuboids are some examples are rectangular prism. When a prism bottom is triangular shape then its said to be triangular prism. 

The diagram given below the structure of prism:

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Description to prism solve for h:

The following types are used in the prism shape they are

Volume of the regular prism is = BH cubic units,

Where B represents the base area of shape, and H represents height.

The Lateral surface area of the regular prism is given by,

              L.S.A = PH square units.

Where P represents the perimeter of shape and H represents the height of prism

Prism based on rectangular:

The area of the base is calculated by l*w

The perimeter of the base can be calculated by 2(l+w)

Therefore the total surface area is calculated by 2lw+2(l+w)H

Volume of the rectangular prism is V= L * W * H

Triangular based prism:

The area is measured as 1/2 bh

The perimeter of the base is measured by a+b+c.

So the surface are is measured by  bh+ (a+b+c) H

Circular based prism:

The area of the base is calculated as `pi` r²

The perimeter of the base is calculated by 2`pi` r

Therefore the surface area of the prism is 2`pi` r²+2`pi` rH

Let we see some basic problems regarding prisms.

Example 1:

 The dimensions of the rectangular prism is given below ,they are length is 10cm, height is 4 cm, and width is 14 cm.

Solution:

Given dimensions length = 10 cm, height = 4 cm, width = 14 cm.

Formula:

  Surface area = 2lw+2(l+w)H

apply  the values in the above formula, we get

                       = 2 *10*14+2(10+14)4

                       = 240 + 2(24)4

                      = 240 + 48*4

                     = 240 +192

                     = 432 cm²

      Volume = l * w * h  

                      = 10 * 4 * 14 cm³

                      = 560 cm³

Answer:

Surface area = 432 cm²

Volume = 560 cm³

Examples problems to solve h in prism

Example 1:

solve the h value whose prism volume is 1210 cubic units and base is 21 cm.

Solution:

The volume of the prism is V = 1210cubic units.

Base of the prism is = 21cm

                                   V= BH

apply the values in the formula we get

                              1210 = 121*H

                                     h = 1210 /121

                                     h = 10cm.

Example2:

The volume of the  rectangular prism is 360 cm³ and its length and widths are 10 ,4 cm , solve the h value

Solution:

The volume of the rectangular prism is V = 360cm³

Length of the rectangular prism is l = 10cm

Width of the rectangular prism is W = 4 cm

Apply these values in formula we get

                                     360 = 10*4*h

                                      360 = 40 *h

                                     h = 360 / 40

                                    h = 9 cm

Sample problems:

1) The volume of a regular prism is  120cm³.The base of the regular prism is 10 cm , solve its height h?

solution : height h = 12 cm.

2) The volume of the rectangular prism is 420cm3 and its length = 7 cm and width =6 cm and solve its height h?

solution: height h = 10 cm.

Conic Tutorial

The sections of a cone are  ellipses, Circles, parabolas and hyperbolas. Because these diagrams can be obtained by passing a plain through vertex of the double-napped cone. In this conic tutorial we are going to see the sections of the cone.

Circle - conic tutorial:
The circle in the section of a cone is like ellipse. When two foci are coincided to each other the circle is formed. This conic section can be formed when a circular cone is intersected with a perpendicular plane.

Sections of a cone - Parabolas - conic tutorial:

One of the section of the cone is the parabola. The points will not be on the same line but in the plain are equal from a fixed line and in a fixed point.

Solve Maxima and Minima:
The local minimum and the local maximum value must be known for finding the maxima and minima values.

Ellipse - conic tutorial:
The section of a cone is an ellipse. It is from a constant two fixed point with the sum of its focus of all the points. The concept of an ellipse makes different terminologies like focus of ellipse, loci of ellipse, etc.

Equation of ellipse:
Equation of ellipse is `(x - h)^2 / a^2 ` + `(y - k)^2 / b^2` = 1 . here real numbers are h, k, a and b.

Sections of a cone-General Equation of an Ellipse

The points of locus will be F ( p, q ) and the equation of directrix ax + by + c = 0. This is the general equation of ellipse

Theorem on Line contact with an Ellipse:
The condition for the line y = m x + c to touch the ellipse `(x^2/a^2)+(y^2/b^2) ` = 1 is that c = ±  `sqrt(a^2 . m^2 + b^2)` .

Finding the point of contact for Ellipse: 
The tangent y = m x + `sqrt(a^2 . m^2 + h^2)`  touches the ellipse `(x^2/a^2)+(y^2/b^2) `= 1, at the point `[(-a^2m/sqrt(a^2m^2+b^2) , b^2/sqrt(a^2m^2+b^2)]`

Hyperbola - conic tutorial:

A hyperbola is the section of a cone. A conic which an eccentricity greater than 1. It has two branches extending up to infinity, two foci, two directions, two vertices and two axes. ( i . e . )  transverse and conjugate and has a center which is the midpoint of the line of two foci and perpendicular conjugate axis is the line through the center and to the transverse axis.

Two Non Intersecting Lines

The two lines are concurrent means then the pencil is formed by the set of lines passing in their common point: if they are parallel, by the set of parallels to both, in the same direction means, they are non-intersecting lines. In fact, if the lines meet in the ordinary sense of the term or are parallel, the point can be at once obtained. These are called non-intersecting lines.

Example problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
     Y = x – 5 and y = x + 5.
     Two non intersecting lines graph for the following equation:
     Y = x - 5
Solution:
   Now plug x values for 0, 1, 2, 3, 4
   When x = 0                                                 
   Y = (0) -5
   Y = 0-5
   Y = -5

   The co ordinate’s point is (0,-5).
   When we plug x = 1
   We get
   Y = 1 -5
   Y = -4

   The co ordinate’s point is (1, -4).
   When we plug x = 2
   We get
   Y = 2 -5
   Y = 2-5
   Y = -3

   The co ordinate’s point is (2,-3).
    When we plug x = 3
    We get
    Y = 3 -5
    Y = -2

   The co ordinate’s point is (3, -2).
    When we plug x = 4
    We get
    Y = 4-5
    Y = -1
The co ordinate’s point is (4, -1).

X	0	1	2	3	4
Y	-5	-4	-3	-2	-1

Two non intersecting lines graph for the following another equation:
   Y = x + 5
Solution:
    Now plug x values for 0, 1, 2, 3, 4
    When x = 0                                                 
    Y = 0+5
    Y = 0+5
    Y = 5

    The co ordinate’s point is (0, 5).
    When we plug x = 1
    We get
     Y = 1 +5
     Y = 6

    The co ordinate’s point is (1, 6).
     When we plug x = 2
     We get
     Y = 2 +5
     Y = 7

     The co ordinate’s point is (2,7).
     When we plug x = 3
     We get
     Y = 3 +5
     Y = 8

     The co ordinate’s point is (3, 8).
     When we plug x = 4
     We get
     Y = 4+5
     Y = 9

     The co ordinate’s point is (4, 9).

X	0	1	2	3	4
Y	5	6	7	8	9

The non intersecting lines represent in graph is,

Practice problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
    Y = x – 4 and y = x + 4.
2. Find the equation for two non intersecting lines that passes through the two points:
    Y = x + 8 and y = x - 8.
3. Find the equation for two non intersecting lines that passes through the two points:

    Y = x + 2 and y = x - 2.

Trinomial Notes

Trinomial notes is a polynomial with three monomial terms. In trinomial notes, sum of three monomials is said to be  trinomial. For example, 2x5 – 3x2 + 3 is a trinomial . An Algebraic expression of the form axn is called a monomial in x, The sum of two monomials is called a binomial and the sum of three monomials is called a trinomial.  sum of the finite number of a monomials in term x is called a polynomial of x.

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How To Factor trinomial Notes:

Example 1:
Factor   trinomial notes x² + 3x − 10.

Solution.
The binomial factors will have this form:
(x   a)(x   b)
What are the factors of 10?  Let us take that they are 2 and 5:

x² + 3x − 10 = (x   2)(x   5).

We have to know how to choose the signs so that,  coefficient for middle term  will be sum of the outers plus the inners -- will be +3.                                  Choose factors , +5 and −2.

=  x² + 3x − 10

= (x − 2)(x + 5)
When 1 is the coefficient for x², the order of  factors it does not matter.

(x − 2)(x + 5) = (x + 5) (x − 2) are same factors.

Example Problems In trinomial notes:

Example 1:
Find factors for trinomial notes x² − x − 12.

Solution:
We must finding factors for 12 whose algebraic sum will be  coefficient of x is −1.
Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3). are facors

Example 2:

Find factors for trinomial notes    x2 + 5xy + 4y2
Solution:
x2+5xy+4y2

=x2+xy+4xy+4y2

=x(x+y)+4y(x+y)

=(x+y)(x+4y)

Example 3:

Find factors for trinomial notes    2a2 - 7ab + 6b2
Solution:
2a2-7ab+6b2
=2a2-3ab-4ab+6b2

=a(2a-3b)-2b(2a+3b)

=(2a-3b)(a-2b)

Example 4:

Find factors for trinomial notes 4x2 - 13xy + 3y2
Solution:
4x2-13xy+3y2
=4x2-xy-12xy+3y2

=x(4x-y)-3y(4x-y)

=(4x-y)(x-3y)