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

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.

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