Tuesday, March 26, 2013

Solving 5th Grade Geometry Problems

Introduction to 5th Grade geometry problems:

Geometric is construction of object form given measurement.5th grade geometry contains the topic of  Understanding what is meant by point, line, and plane Identifying acute, right, obtuse, and straight angles as well as complementary angles, supplementary angles, and vertical angles Identifying parallel lines and perpendicular lines. 5th grade geometric contains the content of triangle, square, circle, rectangle, used to finding the area and volume of the given figure.


Solving Example problems in 5th grade geometry:


5th grade geometry Problems in angle:

Example 1:

1. Which of the following can be the angles of a triangle?

(a)90°, 45°, 55° (b)30°, 70°, 80°(c) 45°, 80°, 50°

Solution:

(a)Sum of the three angles = 90° + 45° + 55° = 190° > 180°

Hence, these cannot be the angles of a triangle.

(b)Sum of the three angles = 30° + 70° + 80° = 180°

Hence, these can be the angles of a triangle.

(c)Sum of the three angles = 45° + 80° + 50° = 175° < 180°

Hence, these cannot be the angles of a triangle.

5th grade geometry Example Problems :

Example 2:

2. find the area of the triangle, its base =4cm and height =6cm?

Solution:

Area of triangle=1/2(Base * height)

=1/2(4*6)

=2*6

=12cm

Problems in square:

A square is a closed figure made up of four line segments

Its side lengths are equal.

Area of square =(side * side)

Perimeter of square =4*side

Having problem with Volume of a Pyramid keep reading my upcoming posts, i will try to help you.

5th grade geometry Example Problems :


Example 3:

3.Find the area and perimeter of  square which side length is 5?

Solution

Side of the square=5m

The perimeter of the floor is given by P= 4×side

=4×5m = 20m

Thus, the perimeter is 20 m.

Area =side * side

=5*5

=25cm2

Problems in circle:

Example 4:

Length of the diameter = 2 × length of the radius

Find the diameter of the circles whose radii are:

(a)2 cm(b)5 cm (c)4.5 cm

Solution:

We have Diameter = 2 × radius

= 2×2cm

= 4cm

Diameter = 2 × radius

= 2×5cm

=10cm

Diameter = 2 × radius

= 2 × 4.5 cm

= 9 cm

Problems in rectangle:

Example 5:

Area of rectangle: Length * Width)

Perimeter of rectangle=2(length+ width

Find the area of rectangle which length is 4 and breath is 7?

Area of rectangle: Length * breath

=4*7

=28cm2

Monday, March 25, 2013

Introduction to Co-ordinate Geometry

Introduction to co-ordinate geometry:
Co-ordinate geometry is a branch of Mathematics that studies about points, lines and geometrical figures using co-ordinate systems. In geometry, we study the same using geometrical constructions and actual measurement but in co-ordinate geometry it is predominantly using co-ordinates of points.

Some of the topics covered in co-ordinate geometry are  finding distance between two points, slope of line, equations of lines, circles and geometrical figures etc

Let us solve some of co-ordinate geometry problems to get a feel of the subject

Understanding Vertex Form of Parabola is always challenging for me but thanks to all math help websites to help me out.

Solve Points Problems of Co-ordinate geometry:


Problem 1:

Find the slope of the line for these two points. ( 5 , 6 ) and ( -3 , 9 )

Solution:

The slope formula is given by

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

Given two points: (x1, y1) =  (5, 6) and (x2, y2) = (-3, 9)

Apply these two points into that formula for finding slope.

m = `(9-6)/(-3-5)`

m = `3/-8`

m = `- 3/8`

The slope of these two points is `-3/8` .

Problem 2:

Solve the distance for the given two points (5, 4) and (4,6)

Solution:

The distance formula is given by

d = `sqrt((x2-x1)^2+(y2-y1)^2)`

d = `sqrt((4-5)^2+(6-4)^2)`

d = `sqrt((-1)^2+(2)^2)`

d = `sqrt(1+ 4)`

d = `sqrt(5)`

The distance for these points is 2.23.


Solve Lines problem of Co-ordinate geometry:


Problem:

Solve the equation of a line between these two points (3, 8) and (-6, 4).

Solution:

Line equation form is y = mx + b

Solve m, slope between these two points.

The slope formula is given by

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

Given two points: (x1, y1) =  (3, 8) and (x2, y2) = (-6, 4)

Apply these two points into that formula for finding slope.

m = `(4-8)/(-6-8)`

m = `-4/-14`

m = `4/14` ---------------Simplify it.

m = `2/7`

Solve b, y intercept

For one point (x1, y1) = (3,8), the line equation becomes

y1 = mx1 + b

8  = `2/7` (3) + b

8  =  `6/7`   + b

b = 8 – `6/7`

b = `50/7`

Substitute m and b into line equation, we get

y = mx + b

y = `2/7` x + `50/7`

y = `1/7` (2x+50)

Multiply by 7 both on sides,

7y = 2x + 50

2x – 7y +50 = 0.

So the equation of line between these two points is 2x – 7y + 50 = 0.


Solve Circle Problem of Co-ordinate geometry:


Problem:

Find the center and radius of (x – 5)^2 + (y – 7)^2 = 25 circle.

Solution:

The circle equation form is (x-h)^2 + (y-k)^2 = R2.

Here the center is (h, k) and Radius is R.

The given equation looks the same as circle equation form.

(x-5)^2 + (y-7)^2 = 52

From the given equation, the center is (5, 7) and Radius is 5.

Sunday, March 24, 2013

Solve Geometry Placement Test

Introduction to solve geometry placement test:

The division of math which deals with the measurement of lengths, angles, areas, perimeters and volumes of plane and solid figures is called geometry.we all knew placement is our dream in college life.In placement test, geometry plays an important role.Here solved geometry placement test papers with solutions given for your practice. Sample geometry placement test paper were given solve this test on your own without the help of a calculator, book, notes, or other people.


Solve geometry placement test:


Example 1:

A room inner space of diameter 150 cm has a wall around it. If the length of the outer edge of the wall is 60 cm, then find the width of the wall.

Solution:

Diameter of the room = 150 cm

Radius = 150 / 2 = 75 cm

Let width of wall = x cm then total radius = (75 + x) cm

Outer edge of the wall = 2  pi  (75 + x) = 44/7  (75 + x) cm

But outer edge of wall = 660 cm

44/7 (75 x) = 660

75 + x = 660  7 / 44

= 105

X = 105 - 75

X = 30 cm

Example 2:

Find number of times will the wheel of a car rotate in a journey of 76 km if the diameter of the wheel is 36 cm?

Solution:

Diameter of the wheel = 36 cm

The Circumference of the wheel of diameter  =  D = 22 / 7  36 = 113 cm

Length of the journey = 76 km = 76*1000*100 cm

Number of times the wheel will rotate in covering the above journey

= `7600000 / 113`

= 67,256.63 .

Please express your views of this topic Volume of Triangular Prism by commenting on blog

Solve geometry placement test:Examples


Example 3.

Find the area of a rectangle of length = 9 cm,breadth = 6 cm.

Solution:

Length and breadth is given here,

we need to find the area ,

Area = l × b sq.units

=` 9 * 6` sq.cm

= 54 sq.cm

Example 4.

Find the perimeter of a rectangle of length = 6 cm,breadth = 5 cm.

Perimeter = 2 (l + b) units

Perimeter  =  2 (6 + 5) units

= 22 cm.

Example 5:

A piece of thin wire which is circular, converted into a square of side 7cm. find the radius of circular wire.

Solution:

Side of a square = 7 cm

Its perimeter = `4* side` = `4 * 7` cm = 28 cm

Circumference of the circular wire = 28 cm. we know that c =`2*pi*r`

28 =` 2*(22/7)*r`

r = `(28*7)` /` (2* 22)`

= `196 / 44`

r = 4.454 cm

Thursday, March 21, 2013

Transformations Geometry

Introduction for transformations geometry:

The transformation Geometry is a copy of a geometric figure, where the copy holds some certain property. The original shape is called the pre-image the new picture is called the image of the transformation. A transformation is single in which the pre-image and the figure equally has the exact same dimension and shape. I like to share this Definition of Parabola with you all through my article.


Basic Transformation Geometry:


The two types of transformation geometry is given by

Rigid transformations

Non-rigid transformations.

This page will covenant with three rigid transformations known as translations, reflections and rotations.

About geometry transformations:

The main geometry transformations in the mathematics are given as,

Translations

Reflections

Rotations

Scaling

Shear

Translations:

The mainly basic transformation is the translation. The definition of a translation is the pre-image and then it can be moved to the equal distance in the same direction to form the image .The transformation is would be

T(x, y) = (x+7, y+4).

Reflections:

The reflection is a "flip" of an aim over a line.

The two very common reflections is given by

horizontal reflection

vertical reflection.

The line of reflection will be both red points, blue points, and green points. The line of reflection which is directly in the center of both points. Having problem with Surface Area of a Circle keep reading my upcoming posts, i will try to help you.


Other types of Transformations:


Rotations:

The transformations which are performed by spinning the object just about a point of the center rotation .You can able to change your object at some of the degree measure, but 90° and 180° are very important degrees.

Rotation 180° around the origin: T(x, y) = (-x, -y)

Scaling:

The scaling is a linear transformation which diminishes the objects and the scale factor is same for direction is called scaling. The resultant image of the uniform scaling is similar to the original transformations

Shear:

The Shear which transforms effectively to rotate one axis and that the axes are no longer at right angle. A rectangle becomes a parallelogram, and a round becomes an ellipse. Constant lines parallel to the axes continue the same length, others do not. As a plot of the plane, it deception in the class of equilateral mappings.

Tuesday, March 19, 2013

Geometry Definitions

That branch of mathematics which investigates the relationship, properties, and measurement of solids, surfaces, lines, and angles; the science which treats of the properties and relations of magnitudes; the science of the relations of space.


Line


In geometry a line:

·         is straight (no curves),

·          has no thickness, and

·         extends in both directions without end (infinitely)


Line segment:


If it does have ends it would be called a "Line Segment".

"Line" normally means straight, so say "curve" if it has a curve.

The word "segment" is significant, because a line normally extends in both directions without end.


angle-angle-angle (AAA) similarity


The amount of turn between two straight lines that have a common end point (the vertex). An acute triangle is a triangle with all angles lesser than 90 degrees.

The angle-angle-angle (AAA) relationship test says that if two triangles have corresponding angles that are congruent, then the triangles are similar. Because the sum of the angles in a triangle must be 180°, we really only need to know that two pairs of corresponding angles are congruent to know the triangles are similar.

The centroid of a triangle is the point where the three medians meet. This point is the center of mass for the triangle. If you cut a triangle out of a piece of paper and put your pencil point at the centroid, you could balance the triangle.Having problem with Surface Area Sphere keep reading my upcoming posts, i will try to help you.


Congruent


Two figures are congruent if all corresponding lengths are the equal, and if all corresponding angles have the same measure. Colloquially, we say they "are the same size and shape," though they may have different orientation. (One might be rotated or flipped compared to the other.)

Sat Geometry Problems

Introduction of sat geometry:
The Abbreviation of SAT is Scholastic Aptitude Test. This test is used for admission to college in United States. The total time has given for SAT test is 70 minutes.There are two sections , one section is given 50 minutes, another section is given 20 minutes.

Topics of sat geometry:

Area and perimeter of a polygon in sat geometry

Area and circumference of a circle in sat geometry

Volume of a box, cube and cylinder in sat geometry

Pythagorean Theorem in sat geometry

Coordinate geometry in sat geometry

Slope

Triangles


SAT Geometry Problems


Problem1:

Which one of these have the largest volume?

Square based prism with sides 7

Rectangular prism of dimensions 5x5x4

Rectangular prism of dimensions 5x6x8

Cylinder of base radius 3 and height 6

Cylinder of base radius 3 and height 8

Problem 2:

My triangular prism has a triangle base with base 5 and height 6, and the prism has a height of 7. What is the volume of the triangular prism?

96

24

100

84

105

Problem 3:

My shape has 4 sides. One of which are parallel. The sides are not all equatl. What shape do I have?

Parallelogram

Trapezoids

Rhombus

Prism

Problem 4:

I have a polygon with 6 equal sides that has 6 equal angles. What is the size of each angle?

180

120

90

154.3

720

Problem 5:

What is the equation of a line with slope is -2 and y intercept is -1 is ?

2x-y+2=0

2x+y+1=0

2x+y-1=0

Is this topic Area of Ellipse hard for you? Watch out for my coming posts.

Problem 6:


The slope of the line 3x+4y+5 = 0 is

3/4

4/3

-3/4

Problem 7:

The straight line x+2y+7=0 passes through (3,k) then valuee of k=?

5

-5

0

Problem 8:

Equation of line parallel to y-axis and passing through the point (3,2) is ?

X = 3

x = -3

y = -3

Problem 9:

A line passing through (0,3) and (4,5) is ?

x – 2y +6 = 0

2x – y + 6 = 0

x – 2y - 6 = 0

Problem 10:

What is the perimeter of my triangle with given three vertices (2,3), (6,2) and (4,2)?

9.7

9.0

8.9

7.9

Monday, March 18, 2013

Solve Geometry Exam

Introduction to solve geometry exam:

Geometry is a branch of mathematics that can be deals with the size, shape, position of shapes, and the properties of space. The geometry is also deals with the applications such as surveying, measurements, areas, and volumes. In Theoretical geometry or pure geometry, we give proofs for theorems on the properties of geometrical figures by applying axioms and reasoning. In practical geometry, we do not construct exactly the geometrical figures but draw rough sketches of the figures to give support to our logical reasoning. I like to share this Quadrilateral Formula with you all through my article.


Example problems to solve geometry exam:


Example problems to solve geometry exam are as follows:

1) The side length of cube is 10 cm. Find the volume of the cube.

Solution:

Formula for volume of the cube = a^3.
a= side length of the cube.

a=10 cm.

= (10)^3.

Volume of the cube =1000 cm^3.
This is the solution for the given geometry problem.


2)A triangle has a perimeter of 56. If 2 of its sides are equal and the third side is 8 more than the equal sides, what is the length of the third side?

Solution:

Let y = length of the equal side


Perimeter = sum of three sides.
Plug in the values from the question.
56 = y + y + y + 8

Combine like terms
56 = 3y + 8

3y = 56 – 8 (by equating the given equation)
3y = 48
y =16

Note: the third side is 5 more than the equal sides.

So, the length of third side = 16 + 8 =24

Answer: The length of third side is 24



Additional problems to solve geometry exam:


Additional problems to solve geometry exam are as follows:

1)The ratio of two supplementary angles is 12 to 6. Find the measure of each angle.

Solution:

Let measure of smaller angle = 12x, measure of larger angle = 6x.
12x + 6x = 180° (The sum of supplementary angles is 180°.)

18x = 180°

x = 10°
Then, 12x = 12(10°) and 6x = 6(10°).

So, 12x = 120° and 6x = 60° (by equating the given equation)

The angles have measures of 120° and 60°.
This is the solution for the given geometry example problem.