How To Solve Geometry

Introduction :

Geometry is a branch of mathematics, which deals with lines, curves, solids, surfaces and points in space. In geometry, a point has a position only and is represented by a dot. A point has no length, width, or thickness. A line has length but no thickness or width. The position of a line with end points are called line segment.


How to solve Geometry Problems:


Geometry Problem 1:

Solve the equation of the straight line parallel to 6x + 4y = 12 and which passes through the point (3, − 3).

Solution:
The straight line parallel to 6x + 4y − 12 = 0 is of the form
6x + 4y + k = 0 … (1)
the point (3, − 3) satisfies the equation (1)
Hence 18 − 12 + k = 0 i.e. k = -6
3x + 2y - 6 = 0 is the equation of the required straight line.

Geometry Problem 2:

Solve the equation of the straight-line perpendicular to the straight line 3x + 4y + 28 = 0 and passing through the point (− 1, 4).

Solution:
The equation of any straight- line perpendicular to 3x + 4y + 28 = 0 is of the form4x − 3y + k = 0
the point (− 1, 4) lies on the straight line    4x − 3y + k = 0
− 4 − 12 + k = 0 ⇒ k = 16
the equation of the required straight line is 4x − 3y + 16 = 0

Geometry problem 3:

The lengths of two sides of right triangle are 7 cm and 24cm. Find its hypotenuses.

Solution:

AC = 7 cm
BC = 24 cm
AB  = ?
AB^2 = 7^2 + 24^2
= 49 + 576
AB^2  = 625
AB = √625 = 25

Thus, the hypotenuses are 25 cms in length.



Geometry Problems to practice:


1) Solve the equation of straight line passing through the points (1, 2) and (3, − 4).

Ans: 3x+y = 5

2) Solve the distance between the parallel lines 2x + 3y − 6=0 and 2x + 3y + 7 = 0.

Ans: √13 units

3) Find the point of intersection of the straight lines 5x + 4y − 13 = 0 and 3x + y − 5 = 0.

Ans: The point of intersection is (1, 2)

4) Solve the equation of the curve formed by the set of all those points the sum of whose distances from the points A (4, 0, 0) and B (-4, 0, 0) is 10 units.

Ans: 9x^2+25y^2+25z^2-225=0.

Geometry Expression

Introduction for geometry expression:
Geometry expression is one of the most important lesson in the geometry. Geometry gives the different geometrical shapes and diagrams in our daily life such as articles in the houses, wells, buildings, bridges etc. The word ‘Geometry’ means a learning of properties for diagrams and shapes. The basic shapes of geometry are point, line, square, rectangle, triangle, and circle. The geometry of plane figure is known as Euclidean geometry or plane geometry. Here we are going to learn about examples of geometry expression problems and practice problem. Understanding Definition for Trapezoid is always challenging for me but thanks to all math help websites to help me out.


Example problems for geometry expression:


Problem 1:

Find the equation of the line having slope 1/2 and y-intercept -3.

Given:

m = `1/2` , b = -3

y = mx +b

Solution:

Apply the slope-intercept formula, the equation of the line is

y = `1/2` x + (-3)

2y = x - 6

x - 2y - 6 = 0

Problem 2:

Solve of geometric expressions based on two angles of a triangle measure 35° and 75° and to find the measure of the third angle.

Solution:

Let the measure of third angle be X

We know that the sum, of the angles of a triangle is 180°

35° + 75° + x = 180°

Solving the expression we get,

110° + x = 180°

X  = 180° – 110°

= 70°

Problem 3:

Find the midpoint between the given points (1, 3) and (3, 7).

Solution:

Given: x1 = 1, y1 = 3 and x2 = 3, y2 = 7

Formula:

(xm, ym) = [`(x1 + x2) / 2 ` , `(y1 + y2) / 2` ].

=` (1 + 3) / 2` ,` (3 + 7) / 2`

= `4 / 2` , `10 / 2` .

= 2, 5

Answer:

The midpoint for the given points (2, 5)

Having problem with Arc Length Examples keep reading my upcoming posts, i will try to help you.

Practice problems for geometry expression:


1. Find the area of the rectangle with the length is 13 cm and breadth is 10 cm

Ans: 130

2. Solve the geometric expression based on the triangle ratio. The triangle ratios are 3: 2: 4. Find the angles of a triangle.

Ans: 60°, 40°, 80°

3. Find the slope and y-intercept of the line equation is 3x + 4y + 5 = 0.

Ans: Slope(m) = -3/4, y - intercept(c) = -5/4

Definitions of Geometry

Introduction to definitions of geometry:

"Earth-measuring" is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow. A mathematician who works in the field of geometry is called a geometer. Source – wikipedia.


Importants definitions of Geometry :

There are various terms and definitions involved in geometry. Some of them are listed below:

Lines:

In geometry if A and B are the two points in the plane. There is only one line AB containing the points. The region where two points connects via the shortest path and continues indefinitely in both the directions is referred as a line.

Line segments:

Line segment is a part of line between two points. The line Segments that intersect at an angle of 90° is called Perpendicular line segments and the line segments that never intersect are known as parallel line segments.

Angles:

An angle is an inclination between two rays with the same initial point.

Right angle:

Angle that measures 90° is referred as right angle

Acute angle:

Angle that measures less than 90° is referred as Acute angle

Obtuse angle:

An angle that measures more than 90° is referred as Obtuse angle.

Scalene triangle:

A triangle in which all three sides has different lengths is known as Scalene Triangles.

Isosceles triangle:

A triangle with two equal length sides and also with two equal internal angles is referred as an isosceles triangle.

Equilateral triangle:

In geometry if a triangle has the equal length on all three sides, then it is referred as Equilateral triangle.

Axioms:

Certain statements are assumed as being true without proof apart from the theorems. Such assumptions are called axioms.

Complementary angles:

Two angles are said to be complementary if their sum is 90° and each is called the complement of the other.

Supplementary angles:

Two angles are said to be supplementary if their sum is 180° and each is called the supplement of the other.


Important definitions of geometry: Circles


The followings are some of the important definitions of geometry in circles.

Circles:

A and B are two concentric circles with radii r and R respectively and O is the center of the circle.

Circumference:

The distance around a circle is called the circumference of a circle.

Radius:

It is the distance from center of a circle to any point on that circle's circumference.

Chord:

Chord is a line segment joining two points on a curve.

Arc:

Part of a curve is referred as an arc.

Concentric circles:

Circles having the same center but different radii are called concentric circles.

Intersecting circles:

Two circles are said to be intersecting when they cut at two different points.

Touching circles:

In geometry two circles are said to touch one another if they meet only at one point. The point at which they touch one another is called the point of contact.

6th Grade Geometry Problems

Introduction:

Sixth grade geometric contains the basic of geometricals .It includes the topic of geometric in Points, Lines ,Line segment, Triangles, Types of triangles, circles, Angles, Types of Angles, Quadrilaterals

Geometric Definitions:

Point: A point   determines the location of particular area.

Line:   A line through two points A and B is written as AB.. It extends

Indefinitely in both directions. So it contains countless number of points. Two points are enough to fix a line

Types of lines:

Intersecting lines
Parallel lines
Perpendicular lines

Triangles in Geometry:


Triangles:

A triangle is a three-sided polygon. In fact, it is the polygon with the least number of sides

Types of Triangle:

Equilateral Triangle
When all the three sides of a triangle are equal to each other, it is called an Equilateral triangle. Each angle measures to 60 degrees. It is a type of regular polygon.

Isosceles Triangle
When two sides of a triangle are equal it is called an Isosceles triangle. It also have two equal angles.

Scalene Triangle
When no two sides of a triangle are equal the triangle is called Scalene triangle. It has three unequal sides.

Area of triangle: 1/2(Base*Height)

Perimeter of Triangle: (Sum of three sides)

Example problem:

1.Find the area of triangle base is 4cm,height is 2cm

Solution:

Area=1/2(4*2)

=8/2

=4cm2

2.Find perimeter of Triangle side lengths are 5cm,5cm,8cm

Solution:

Perimeter=(A+B+C)(Sum of three side lengths)

A=5, B=5, C=8

=(A+B+C)

=5+5+8

=18cm


Angles and Circle in Geometry


Angle:

Right Triangle
. Right angle is equal to 90 degrees. It obeys Pythagoras theorem.

Acute angle
. Acute angle is an angle which is less than 90 degrees.

Obtuse angle
An Obtuse angle is an angle which is greater than 90 degrees but less than 180 degrees.

Acute angle:
Acute and Obtuse triangles are also called as Oblique triangles because they don’t have any angle measuring 90 degrees.

Quadrilateral:
A four sided polygon is a quadrilateral. It has sides and 4 angles

Circle:

Are of circle=Pi*r*r

Circumference of Triangle=2*Pi*r

Diameter=2*Radius

Example:

Find the area  and circumference of the circle when the radius is 4cm?

Solution:

1.     Area=Pi*r*r (r=4) (Pi=3.14 constant)

=3.14*4*4

=50.24cm2

2. Circumference =2*pi*r

=2*3.14*4

=25.12cm

Answers to Geometry Homework

Introduction to answers to geometry homework:

Learning geometry has traditionally been regarded as important in the secondary schools, at least partly because it has been the primary means of teaching the art of reasoning.

Geometry is a theoretical subject, but easy to understand, and it has many real practical applications. Eventually, geometry has evolved into a skillfully arranged and sensibly organized body of knowledge. I like to share this Triangular Prism Net with you all through my article.


Part 1 -answers to geometry homework:


Geometry homework example 1:

If the perimeter of a cube is 52.5 ft, find its surface area.

Geometry homework solution:

Perimeter of a cube P=12a

52.5=12a

a=52.5/12

a=4.375 ft.

So, the value of a=4.375 ft.

Surface area of a cube SA=6a2

=6(4.375) 2

=6(19.14) ft2

=114.84 ft2

Answer of example 1: Surface area of cube   = 114.84 ft2

Geometry homework example 2:

A barrier of length 15 m was to be built across an open ground. The height (h) of the wall is 5 m and thickness of the barrier is 32 cm. If this barrier is to be built up with bricks whose dimensions are 25 cm × 18 cm × 12 cm, how many bricks would be required?

Geometry homework solution:

1 m=100 cm

Here, Length = 15 m = 1500 cm

Thickness = 32 cm

Height = 5 m = 500 cm

Therefore, Volume of the barrier = length × thickness × height

= 1500 × 32 × 500 cm3

Now, each brick is a cuboid with length = 25 cm, breadth = 18 cm and height = 12 cm

So, volume of each brick = length × breadth × height

= 25 × 18 × 12 cm3

So, number of bricks required =volume of the barrier divided by volume of each brick.

Substituting the values,then we get the final answer.

= (1500 × 32 × 500)/ (25 ×18 × 12)

= (24000000)/5400

=4444.44

Answer: The barrier requires 6416 bricks.

I have recently faced lot of problem while learning math word problems for 7th grade, But thank to online resources of math which helped me to learn myself easily on net.


answers to geometry homework:


Geometry homework example 3:

A line passes through (–3, 4) with a slope of -1/5. If another point on this line has coordinates (x, 2), find x.

Geometry homework solution:

Slope m= (y2-y1)/(x2-x1)

-1/5= (2-4)/ (x-(-3))

-1/5= (-2)/(x+3)

We can take cross multiplication.

-1(x+3) =5(-2)

-x-3= -10

-x=-10+3

-x=-7

In both sides cancel for the negative sign and then we get the final answer.

x =7

Answer: The x value is 7

Positive and Negative Angles

Introduction to Positive and Negative Angles:

An ANGLE is strong-minded by rotating a ray about it's endpoint. The initial location of the ray is the INITIAL SIDE of the angle, and the ending position of the ray is its TERMINAL SIDE. The endpoint of ray called the VERTEX.

The unit circle of the angle is said to be in STANDARD POSITION because its vertex is the origin, and its initial side lies on the x-axis. This is also called positive angle, meaning it's created by a COUNTERCLOCKWISE rotation.
The unit circle of angle is in standard position, but it's called a NEGATIVE ANGLE, since it is created by a CLOCKWISE rotation. I like to share this Inscribed Angles with you all through my article.


Rules of Positive and Negative angles:

Positive Angle:

An angle formed by anti-clockwise rotation is a positive angle. In the figure initial side is OX. When these side is rotated by an angle θ in counter clockwise direction then angle is generated is called positive angle.

Negative Angle:

An angle generated by clockwise rotation is a positive angle. In these diagram let the initial side is OX. When these side is rotated by an angle θ in clockwise direction then angle called as negative angle.

Rule I:

Sign of an angle is always positive when measured in anti-clockwise direction.

Rule II:

Sign of an angle is always negative when measured in clockwise direction. Understanding Volume of Right Prism is always challenging for me but thanks to all math help websites to help me out.


Example of Positive and Negative Angles:

Positive and Negative Angles:

Positive Angles start from 0 degrees and turn around counterclockwise.

Negative Angles start from 0 degrees and turn around clockwise.

You can translate your negative angle to its equivalent positive angle by adding 360 degrees to it until it turns positive.

Once it is positive, you can pleasure it the same as you would any other positive angle in the quadrant that it is in.

Example 1:

Angle is -135 degrees.

sum  360 degrees to it until it turns positive.

It turn positive then first time we add 360 degrees to it.

The equivalent is positive angle is 225 degrees.

It is in the quadrant of 3.

Example 2:

Is 300 not same as -300?

Solution:

The answer to this question is NO. Here why the angle have  two attributes attached to it: Degree of rotation (or magnitude of rotation) and Direction of rotation (clockwise or anticlockwise). While those wo angles have same degree of rotation, direction of rotation is just opposite as signified by there opposite signs. Therefore those two angles are different.

Angle of Rotational Symmetry

Introduction to angle of rotational symmetry:

Rotational symmetry is definite as angle, while we rotate as alternate a shape in its center point; you may notice that at a certain angle, the shape coincides with its not rotated itself. When this happens, the shape is said that it have rotational symmetry. A shape has rotational symmetry if it fits on to itself two or more times in one turn. The number of times the rotational symmetry is the shape fits on to itself in one turn. Is this topic Straight Angle Definition hard for you? Watch out for my coming posts.


Type of symmetry:

Symmetry has,

Line symmetry
2D rotational symmetry
3D rotational symmetry
A 2D shape has a line of symmetry if the lines separate the shape into two share equally – one being the mirror image of the other.

Rotation: whatever rotate the shape on it around, every rotation has depending on center point and an angle.

Translation:  Translation is to be in motion without rotating or reflecting, every translation has depending on distance and

a direction.

Reflection: reflection is seemed mirror image. Every reflection has a mirror line.

Glide Reflection: With the direction of the reflection line, glide reflection is the symmetry of its collection of reflection and  translation.

Angle of rotation:


When you can turn a figure around a center point by less than 270° and the figure appear that there is no changed, and then the figure has rotation symmetry. The middle of rotation, and then the smallest angle require turning and the point around which you rotate is called the angle of rotation.

For instance take any figure on the left can be turned by 160 degree the same way you would turn an hourglass and it look like the same. Her take 2nd figure one can be turned by 120 degree and other one is 72 degree. The 2nd figure comes from 72 degree that it has five points, just rotate it until it looks the same; we need to make 1 / 5 of a completed 360 degree. So 1 / 5 * 360 degree = 72 degree.

Geometry Book Answers

Introduction:

Geometry is nothing but the act of determining the dimensions or volume of an object. Buildings, cars, planes, maps are of great examples of geometry. The plane geometry, normally used in finding the area, perimeter, circumference of  Two-dimensional figures like triangle, circle, rectangle, rhombus, trapezoid, quadrilateral etc.  Let us solve some sample problems from plane geometry. Understanding Geometry Lines is always challenging for me but thanks to all math help websites to help me out.


Sample Geometry Problems:


Example 1:
The radius of a right circular cylinder is 7 cm and its height is 20 cm. Find its curved surface area and total surface area.

Solution:
Radius(r) = 7cm, Height(h) = 20 cm.


Curved surface area = (2`pi`r h )= 2 * 22 / 7 * 7 * 20 cm^2

= 880 cm^2

Area of the base and the top    = (2`pi`r2)

= 2 * (22 / 7) * 7 * 7 cm^2

= 308 cm^2

Total surface area = (2`pi`r h) + (2`pi`r2) = 880 cm^2 + 308 cm^2

= 1188 cm^2

Example 2:

Surface area of a right circular cylinder of height 35 cm is 121 cm^2. find the radius of the base.

Solution:

Curved surface area = (2`pi`r h) = 121 cm^2

2*(22/7)* r *35 = 121

r = (7*121)/(2*22*35)

= 11/20

= 0.55 cm

Hence, radius of the base = 0.55 cm.

Example 3:

Curved surface area of a right circular cylinder is 6.6 m^2. if the radius of the base of the cylinder is 1.4 m , find the height.

Solution:

Curved surface area = (2`pi`r h) = 6.6

= 2 * (22/7) * 1.4 * h

= h = (7*6.6) / (2*22*1.4)

= (7*66) / (2*22*14)

= 3/4 = 0.75

Required height of the cylinder = 0.75 m.

Please express your views of this topic Surface Area of Cylinder by commenting on blog.

Geometry Practice Exam Problems:


1. the circumference of the base of a right circular cylinder is 220cm. if the height of the cylinder is 2m, find the lateral surface area of the cylinder.

2. A closed circular cylinder has diameter 20cm and height 30 cm. find the total surface area of the cylinder.

3. the radius of the base of a closed right circular cylider is 35 cm and its height is 0.5m. find the total surface area of the cylinder.

Answers:

1. 4.4 m^2

2. 2512 cm^2

3. 18700 cm^2

Construction of Triangles

Introduction to construction of triangles:

The triangles can be constructed if the following requirements are given such as follows,

The measurement of three sides should be given (or)

The measurement of the two sides and the included angle should be given (or)

The measurement of a side and any two angles should be given.

Now we are going to see about the construction of triangles. Is this topic Scalene Triangles hard for you? Watch out for my coming posts.


Construction of triangles:


Construction of triangles if three sides are given:

Construct a triangle if three sides are given with x, y and z measurements.

Steps of construction:

First a line segment QR of x cm length should be drawn.

With Q as center and radius of y cm be drawn and it equals to PQ and draw an arc of a circle.

With R as center and radius of PR = z cm and draw an arc and it will intersects at the first arc of point P.

Now join the points of line segments PQ and PR.

Thus, PQR is a required triangle. I have recently faced lot of problem while learning Geometry Definition, But thank to online resources of math which helped me to learn myself easily on net.

Other constructions of triangles:

Construction of triangles if two sides and angle are given:

Construct a triangle if two sides and an angle are given.

Steps of Construction:

First we have to draw a ray of QX of some length.

With the help of protractor measure the given angle and draw the line to meet Q.

The ray QY which may cut line segment QR of x cm.

The ray QY which may cut the line segment QP of y cm.

Now we can join the two points P and R.

Thus, PQR is the required triangle.

Construction of triangles if two angles and Side are given:

Construct a triangle if two angles and a side are given.


Steps of Construction:

First we should draw the line segment of QR of given length.

With the help of the protractor measure the given angle at RQX

Then, draw QRY for the given angle such that XY lie on the same side of the PQ.

Then, label the point where it intersects at QX and QY as P.

Thus, the PQR is the required triangle.

Base Pentagon Prism

Introduction:-
In geometry, the pentagonal prism is a prism with a pentagonal base.

Pentagonal prism is a type of heptahedron.

If faces are all regular, the pentagonal prism is a semi regular polyhedron and the third in an infinite set of prisms formed by square sides and two regular polygon caps.

A pentagonal prism has                                                                                                                                                                         Source:- Wikipedia.

7 faces
10 vertices
15 edges.
The pentagonal prism looks like .


Formulas For Pentagonal Prism:-


`Base area = 5/ 2 * a * s`
`Base perimeter = 5s`
`Prisms Surface area = 5as + 5sh.`
`Volume of Prism = (5/2)ash`
`here`
`a = apothem Leng th,`
`s = side,`

`h = height.`

Please express your views of this topic Congruence of Triangles by commenting on blog.

Problems on Prism:-


Problem 1:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 4 and the side is 2.

Solution:-

Given

The apothem, length is 4

The side is 2.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2` * 4 * 2.

By crossing 2 and 2 in the above equation we get

Area of the base = 5 * 4 = 20.square units.

The formula that is used to find the base perimeter is  `5s` .

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 2 = 10.

Problem 2:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 6 and the side is 4.

Solution:-

Given

The apothem, length is 6

The side is 4.

The formula that is used to calculate the area of the base is` 5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 6 * 4.`

By crossing 2 and 4 in the above equation we get

Area of the base = 5 * 12 = 60.square units.

The formula that is used to find the base perimeter is  `5s.`

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 4 = 20.

Problem 3:-

Find the base area and base perimeter of the pentagonal prism if the apothem length is 5 and the side is .

Solution:-

Given

The apothem, length is 5

The side is 7.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 5 * 7.`

Area of the base = `175/ 2` square units.

The formula that is used to find the base perimeter is  5s.

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 7 = 35.

angles at a point

Introduction (Angles at point):

In geometry an angle is the figure produced by two ray’s distribution a common endpoint, called the vertex of angle. The degree of the angle is the quantity of revolution that separates the two waves, and deliberate by considering the length of circular curve is out when one ray is rotate regarding the vertex to correspond with the other. The angle along with a line and a curve or along with two intersecting curve.


Positive and negative angles at a point:

In mathematical script is that angles specified a sign are positive angles if considered anticlockwise and negative angles ? is efficiently the same to a positive angle of one full rotation less ?. if considered clockwise, from a known line. If no line is specified, that can be understood to be the x-axis in the Cartesian plane. In many geometrical situations a pessimistic angle of ?? is efficiently the same to a positive angle of one full rotation less ?.

Example, a clockwise rotation of 45° (angle of ?45°) is efficiently the same to an anticlockwise rotation of 360° ? 45° (angle of 315°).

Types of Angles:

Right angle
Acute angle
obtuse angle
reflex angle
Vertical opposite angles
Co-responding angles and Alternative angles
Interior angle
Identifying angles:

Angles may be recognized by the labels involved to the three points to identify them. Example, the angle by vertex A with this by the rays AB and AC.

Potentially, an angle denoted,  ?BAC may refer to any of four angles: the clockwise angle from B to C, the anticlockwise angle from B to C, the clockwise angle as of C to B, or the anticlockwise angle as of C to B, wherever the way in that the angle is deliberate determines its sign.


Examples for angles at a point:


Example 1:

Find the value of x.

Solution:

x + 80° + 2x + x = 180° (contiguous angles on a straight line)

4x = 180° - 80°
= 100°

x = 100°
4
The answer of x = 25°

Example 2:

Find the value of x.

Solution:

48° + 90° + 120° + x = 360° ( Angles at a point )
x = 360° - (48° + 90° + 120° )
= 360° - 258°
The answer of x= 102°

Solve Allied Angles Axiom

Introduction to solve allied angles axiom:

The allied angles are nothing but the co-interior angles where they are transversely cuts the two parallel lines and the allied angles will be formed. The supplementary angles present in the geometry figures are also called as one of the types of allied angles. The allied angles total measurements are about 180 degrees. The tutors will describe the concepts to students with some example problems. Now we see how to solve allied angles axiom with the help of the tutor.

About How to Solve Allied Angles Axiom:

Now we see about how to solve for the allied angles axiom and its concepts with the help of the tutor. The allied angles are nothing but the angles where they are formed when two parallel lines are cut by a transverse line and the interior angles forms are called as the allied angles. The allied angles measurements are be about 180 degrees.

From the above figure it is given that as there are two parallel lines such as L1 and L2. These two lines can be cut by the traversal line T and the allied angles forms in the figure are given as a, b, c, d.

`angle a` +`angle d` = 180 degrees.

`angle b`  +`angle c` = 180 degrees.

The angle measures are given and we can determine the other angle with the total measurement of the angles are about 180 degrees.

Now we see some of the problems on allied angles axiom with the help of the tutor. Understanding Alternate Interior Angle is always challenging for me but thanks to all math help websites to help me out.

Problems to Solve Allied Angles Axiom:
Example:

Calculate the allied angles from the given figure?

Solution:

Now we see how to solve the allied angles measurement as follows,

The allied angles measurements are about 180 degrees.

From the given diagram, S1and S2 are the two parallel lines and cut by the transverse line T.

The allied angle are thus formed between the parallel lines.

For calculating the allied angle we have to do as follows,

Y + 70 = 180

Y = 180 - 70

Y = 110

Thus, the allied angles for 70 is about 110 degrees.

11th Grade Geometry

Introduction of 11th grade geometry :-

In grade (11) means eleventh grade is a year of education in all over the world. The eleventh grade is the final year of the secondary school. Students are usually 16 - 17 years old. Geometry is concerned with size, shape, relative figures etc. In 11th grade geometry lessons study about the angle, circle , quadrilateral etc.

Example Problems for 11th Grade Geometry :-

Problem 1:-

Determine the equation of the straight line passing through the points (1, 2) and (3, − 4).

Solution:

The equation of a straight line passing through two points is

`(y - y1)/ (y1 - y2)` = `(x - x1)/( x1 - x2)`


Here (x1, y1) = (1, 2) and (x2, y2) = (3, − 4).

Substituting the above, the required line is

`(y - 2)/(2 + 4)` = `(x - 1)/(1 - 3)`

`(y - 2)/6` = `(x - 1)/(- 2)`

`(y-2)/3` = `(x-1)/(-1)`

y − 2 = − 3 (x − 1)
y − 2 = − 3x + 3

3x + y = 5 is the required equation of the straight line.Is this topic AAA Postulate hard for you? Watch out for my coming posts.

Problem 2:-

Find the equation of the straight line passing through the point (1,2) and making intercepts on the co-ordinate axes which are in the ratio 2 : 3.

Solution:-

The intercept form is

`x/a +y/b` = 1 … (1)

The intercepts are in the ratio 2 : 3  a = 2k, b = 3k.

(1) becomes

`x/(2k) +y/(3k)` = 1     i.e. 3x + 2y = 6k

Since (1, 2) lies on the above straight line, 3 + 4 = 6k i.e. 6k = 7

Hence the required equation of the straight line is 3x + 2y = 7


Problem 3:-

Find the distance between the parallel lines 2x + 3y − 6=0 and 2x + 3y + 7 = 0.

Solution:-

The distance between the parallel lines is

`|(c_1 - c_2)/sqrt(a^2 + b^2)|` .

Here `c_1` = − 6, `c_2` = 7, a = 2, b = 3

The required distance is

`|(- 6 -7)/sqrt(2^2 + 3^2)|` = `| (-13)/sqrt(13)|`

= `sqrt(13)` units.


Practice Problems for 11th Grade Geometry :-

Problem 1:-

Find the equation of the straight line, if the perpendicular from the origin makes an angle of 120° with x-axis and the length of theperpendicular from the origin is 6 units.

Answer: The required equation of the straight line is x − `sqrt(3)` y + 12 = 0


Problem 2:-

Find the points on y-axis whose perpendicular distance from the straight line 4x − 3y − 12 = 0 is 3.

Answer: The required points are (0, 1) and (0, − 9).

Sum of Two Squares

Introduction to sum of two squares

In algebra, we have some formulae to expand squares.

They are:

`( a + b ) ^2 = a^2 + 2ab + b^2`
`( a ** b ) ^2` = `a^2 ** 2ab + b^2`
`( a + b ) ^2 + ( a ** b ) ^2` = `2 ( a^2 + b^2 )`
`( a + (1/a) ) ^2` = `a ^2 + (1/a^2) + 2`
`( a ** 1/a ) ^2` = `a ^2 + 1/a ^2 ** 2`
`( a + (1/a) ) ^2` + `( a ** 1/a ) ^2`   = `2 ( a ^2 + (1/a ^2))`
`a + b = sqrt (( a ** b ) ^2 + 4 ab)`
`a ** b ` = `sqrt (( a + b ) ^2 ** 4 ab)`


Keeping these formulae in mind, we can break up the square values to get the final answer. Now let us see few problems on sum of two squares. I like to share this Example of Obtuse Angle with you all through my article.

Example Problems on Sum of Two Squares

Ex 1: Find the value of a ^2 + b^2, If  a + b = 7 and ab = 7.

Soln: By using the above formulae, `a^2 + b ^2 = (1/2) [ ( a + b ) ^2 + ( a ** b ^2]`

Therefore  a – b = `sqrt (( a + b ) ^2 **4 ab)`

a – b = `sqrt (7 ^2 ** 4 ( 7 ))` = `sqrt (49 **28)` = `sqrt 21`

Therefore `a ^2 + b ^2` = `(1/2)[ ( a+ b ) ^2 + ( a ** b) ^2]`

= `(1/2) [ 7^2 + ( sqrt21)^2]`   =  `(1/2) [ 49 + 21 ]`

Therefore  `a^2 + b^2 = 35`

Ex 2: Find the value of A^2 + b^2, if a – b = 7 and ab = 18.

Soln: By using the above formula, `a + b = sqrt (( a ** b) ^2 + 4 ab)`

= `sqrt ((7) ^2 + 4 (18)) = sqrt (49 + 72)`

= `sqrt 121`    = 11

Therefore `a^2 + b^2` = `(1/2) [( a + b ) ^2 + ( a ** b ) ^2]`

= `(1/2) [ 11^2 + 7 ^2 ] = (1/2)[ 121 + 49 ]`

= `(1/2)` [ 170 ]  =  85

Therefore `a^2 + b^2 = 85`

Ex 3: If `a + (1/a) = 6` , find the value of `a ^2 + (1/a^2)` .

Soln: `a ^2 + (1/a^2) = (a + (1/a) ) ^2 **2 = ( 6 ^2) ** 2 = 34`

Therefore `a^2 + (1/a^2) = 34` [By using formula in 4]

Ex 4: If `a ** (1/a) = 8` , find the value of `a^2 + (1/a ^2)`

Soln: Therefore `a^2 + 1/a^2` = `(a ** (1/a)) ^2` + 2

= `8^2 + 2`

= 64 + 2 = 66.

Therefore `a^2 + (1/a^2)` = 66.   [By using formula in 5]

Ex 5: If a^2 – 5a – 1 = 0, find the value of `a^2 + (1/a^2)`

Soln: Given:  a2 – 5a – 1 = 0

`rArr` a – 5 – (1/ a) = 0    [Divide throughout by]

`rArr` `a **(1/a)`  = 5

Therefore `a^2 + (1/a^2)`  = `(a ** (1/a) ) ^2` + 2 =  `5^2` + 2 = 27. Understanding Area of Hexagon is always challenging for me but thanks to all math help websites to help me out.

Practice Problems on Sum of Two Squares

1. If a + 1/a  = 2, find a^2 + 1/a^2

Ans: 2

2. If a + b = 9 and ab = -22, find the values of a ^2 + b^2.

[And: 125]

3. If a^2 – 3a + 1 = 0, find the value of a^2 + 1/a ^2.

[Ans: 7]