To 8th grade geometry problems

Introduction : 

Eighth grade Geometry contains the topics of  finding the area, volume and surface areas. In geometry the magnitude of the plane region is called area. The surface of these solids is generally curved surface areas(regions) of these solids are very useful because we come across cylinders, cones and spheres in our everyday life almost every step solid occupies some region in space and the magnitude of this region is called the volume of the solid.

Example of 8th Grade Geometry Problems:

Problems in CIRCLE:

CIRCLE:

Area of the circle=π*R*R (R=radius)

Circumference of circle=2*π*R

8^th grade geometry problem 1:

Diameter of a circular garden is 12.4 m. find its area.

Sol:

Diameter, d = 9.8 m. Therefore, radius r = 12.4 /2 = 6.2 m

Area of the circle =π*R*R

= 3.14*6.2*6.2

= 120.7 m²

TRIANGLE:

Area of triangle=`1/2` (Base * Height)

Perimeter of Triangle= (Sum of three sides)

8^th grade geometry problem 2:

Find the area and perimeter of triangle Base=5, Height=7, other two sides are 8, 9?

Area=`1/2` (5*7)

=`35/2`

=17.5 cm²

Perimeter= (5+8+9)

=22 cm

Other Examples of 8th Grade Geometry Problems:

CONE, CYLINDER, SPHERE, HEMISPHERE

Volume of a right circular cylinder:

V = (Area of the base) × (Height) = π*R*R*H 

Volume of a right circular cone: `1/3` *π*R3

 Volume of a sphere: `4/3` *π*R3

 Volume of a hemisphere: `2/3` *π*R3

8^th grade geometry problem 3:

1. Diameter of the base of a right circular cylinder is 7 cm. If its height is 40 cm, find its volume.

Sol:

Since the diameter of the base is 7 cm, its radius r = 3.5cm. Also, h = 40 cm,

Volume of a right circular cylinder:

V = (Area of the base) × (Height) = π*R*R*H

=3.14*3.5*3.5*40

= 1538.6 cm³

8^th grade geometry problem 4:

Find curved surface area of a right circular cylinder of base radius 3 cm

and height 5 cm. (Take P= 3.14.)

Sol:

 Curved surface area of a cylinder = 2 Ph

= 2 × 3.14 × 3 × 5 cm²

= 94.2 cm²

PARALLELLOGRAM:

Area of parallelogram = Area of rectangle = length (l) × breadth (b) = l × b

8^th grade geometry problem 5:

1. One sides and the corresponding height of a parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram

Sol:

Given that length of base (b) = 4 cm, height (h) = 3 cm

Area of the parallelogram = b × h

= 4 cm × 3 cm

= 12 cm²

 

I like to share this math problems 7th grade with you all through my article

 

To practice algebra linear equation

Introduction :

Algebra linear equation is a mathematical expression that contains an equal sign and the linear expressions. Algebra is a Arabic word al–jabr.  In mathematics, we use letters like a, b, x and y to denote numbers. We performs addition, subtraction, multiplication, division or extraction of roots  and real numbers, we obtain what are called algebraic expressions. Symbols in an algebraic expression are called variables . An  Algebraic expression of the form axⁿ is called a monomial in x,

Practice Algebra Linear Equation – Example

A set of common solution set of practice algebraic expression equation are said to be linear equations. A linear equations are set of variables in equation form

For example, this linear equation:  x + 1 = 4 means that when we add 1 to the unknown value, ‘x’, the answer is equal to 4.

Two variable equation: 2x+3y=6 where x and y are two variables.

Practice Algebra Linear Equation – Solved Examples

 Example 1:

Solve for x

           X+ 2 =   -3

solution:

1. Subtract 1 from both sides:

x + 2 – 1 =   -3 – 2

2. Simplify both sides:

x   =   -5

Example 2:

Solve for x

-3x   =   12

solution:

1. Divide both sides by -3:

`(-3x)/-3` =`12/-3`

2. Simplify both sides:

          x   =   -4

Example 3:

Solve for x

`x/4` =-2

solution:

1. Multiply both sides by 4:

X*4/4=-2*4

2. Simplify both sides:

x   =   -8

Example 4:

Solve for x

2 x + 2 = -18

solution:

1. Subtract 1 from both sides:

2x + 2 -2 = -18 – 2

2. Simplify both sides:

2x = -20

3. Divide both sides by 2:

2x/2=-20/2

4. Simplify both sides:

x   =   -10

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Example 5:

Solve the given Equation: 8z + 50 = 4z + 62

Solution:

Given equation:    8z + 50 = 4z + 62

Subtracting both sides from 50:    8z + 50- 50 = 4z + 62 – 50

Simplification gives:        8z = 4z + 12

Subtracting both sides 4z:  8z – 4z = 4z – 4z + 12

Simplification gives:      4z = 12

Answer:    z = 3

Example 6:

         Given question is, Y = 3x +4

Solution: General equation is y = mx + b

Where m = Slope of the line

b = y- intercept

(Slope) m= 3

Y-intercept = 1

Here the values of slope (3) and intercept (1) are the prepared values from the given linear equation.

Practice Algebra Linear Equation – Practice Problems:

Problem 1:

       solve for x in (x+5)=7

Solution:

       x=-2

 

Problem 2:

      solve `x/5` +3=5

Solution:

      x= -10

Problem 3:

      solve x+3x-5=4

Solution:

       x= `9/4`

Problem 4:

      solve 2(17x+5)=2

Solution:

      x = -`5/17` .