Triangles in Math

Introduction of Triangles in Math:

Triangles are closed loop that consists of three sides and three angles. The total interior angle of triangles is 180°. The triangles consist of vertex by joining the two sides of the triangles. The amplitude of the triangle is the distance between vertexes of triangles in to the opposite side of that same vertex of triangles. The length of the altitude is the height of the triangles .


Classification of Triangles in Math:


Classification of Triangles:

There are different types of triangles; they are acute angle triangles, obtuse angle triangles, right angle triangles, Isosceles triangles, equilateral triangles and scalene triangles.

Classification of triangles is based on sides and angles.

Classification based on the sides of the triangles:

Isosceles Triangles – Two sides are equal in this triangle.

Equilateral Triangles – All the three sides of the triangle is equal.

Scalene Triangles – All the three sides are different in length of the triangles.

Classification based on the angle of the triangles:

Acute angle Triangles – The three angles of the triangle is less than 90°.

Obtuse angle Triangles – One of the angles in triangle is greater than 90° (less than 180°).

Right angle Triangles – One of the sides of the triangle is 90°.


Formula and Example Problems – Triangles in Math:


Basic Formula – Triangles in Math:

Area of the triangles in math – ½ b(h)

Where,

b = Base length of the triangles.

h = Height length of the triangles.

Perimeter of the triangle – x + y + z

Where, x, y, and z are the sides of the triangles.

Example Problems – Triangles in Math:

Example 1:

Find the area and perimeter of the equilateral triangle, whose base is 5 cm and height, is 7 cm.

Solution:

Given: base (b) = 5 cm

Height (h) = 6 cm

Formula to find area and perimeter:

Area = 0.5 (b) (h)

Perimeter = a + b + c, where a, b, c are the sides of the equilateral triangle,

Here equilateral triangle have all three sides are equal. Therefore perimeter = 3 a

Area = 0.5 (5 (6)) = 15 cm^2

Perimeter = 3 (5) = 15 cm

Example 2:

Find the perimeter of the triangle whose sides are 4cm, 6cm and 7cm.

Solution:

Given: The sides of the triangle are, a = 4cm

b = 6cm and c = 7cm.

Formula:  Perimeter = a + b + c = 4 + 6 + 7 = 17cm.

Answer: perimeter of the triangle  = 17cm.

Grade 8 Math Inequalities

Introduction to grade 8 math inequalities:
In mathematics, an inequalities is a statement about the relative size or order of two objects or about whether they are the same or not.

The notation a < b means that a is less than b.
The notation a > b means that a is greater than b.
The notation a ? b means that a is not equal to b.
In this article we shall discuss about math grade 8 inequalities. (Source: Wikipedia)

Grade 8 math inequalities example problem


Here we are going to discuss some 8th grade inequalities problems with detailed solutions.

Example:

Solving the inequalities 5x – 12 > 4x +12

Solution:

The given inequalities is

5x – 12 > 4x +12

Adding the 12 on both side of equation

5x -12+12> 4x +12+12

5x>4x+24

Subtract 4x on both side of the inequality equation

x>24

Example:

Solving the following inequality equation -5< 4(x+2)-5<19 br="">
Solution:

The given inequality is

-5< 4(x+2)-5<19 br="">
Multiplying the factor for given inequality equation

-5<4x br="">
-5<4x br="">
Subtracting three on both sides of the inequality equation

-5-3<4x br="">
-8<4x br="">
Divide by 4 for all terms in inequality

-2
Conclusion:

The solution includes all real number value the interval is (-2, 4)

Example:

Solving the following inequality equation -4< 2(x+6)-2<12 br="">
Solution:

Given inequality is

-4< 2(x+6)-2<12 br="">
Multiplying factor values for given inequality equation

-4<2x br="">
-4<2x br="">
Subtracting 10 on both sides of equation

-4-10<2x br="">
-14<2x br="">
Divide by 2 for all terms in equation

-7
Conclusion:

The solution includes all real number value the interval is (-7, 1)

Example:

Solving the inequality 6x – 4 > 3x +14

Solution:

The given inequality is

6x – 4 > 3x +14

Adding the value four on both side of equation

6x -4+4> 3x +14+4

6x>3x+18

Subtract value 3x on both side of the equation

3x>18

Simplifying the value x

x>18/3

x>6

Conclusion:

The solution includes all real number value the interval is (6, infinity)

Example:

Solving the inequality 8x – 2 > 3x +13

Solution:

The given inequality is

8x – 2 > 3x +13

Adding the value two on both side of equation

8x -2+2> 3x +13+2

8x>3x+15

Subtract value 3x on both side of the equation

5x>15

Simplifying the value x

x>15/5

x>3

Conclusion:

The solution includes all real number value the interval is (3, infinity)

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Grade 8 math inequalities practice problem


Problem:

Solving the inequality -4< 4(x+2)-4<16 br="">
Answer:

The solution includes the interval is (-2, 3)

Problem:

Solving the inequality equation -2< 4(x+6)-3<10 br="">
Answer:

The solution includes all real number value the interval is (-6, -2.75)

Multiplying Integers Practice

Introduction to multiplying integers practice:

Multiplying integers practice  is one of the important topics in mathematics. A multiplying is of the mathematical operation of scaling a number by another number. It is the basic operations in elementary arithmetic. Integer is the set of numbers in which positive whole numbers, negative whole numbers and zero. It has a complete unit or entity. But it has a no fractional part.

Positive whole number = {1, 2, 3, 4, …..}

Negative whole number = {-1, -2, -3, -4, …..}

Example for integers:

25, 1897, -665, 0, etc.,


Rules for multiplying integers practice:


Different rules for multiplying integers practice are,

Rule 1:

Positive number × Positive number = Positive number

Rule 2:

Positive number × - Negative number = - Negative number

Rule 3:

- Negative number × Positive number = - Negative number

Rule 4:

- Negative number × - Negative number = Positive number

Example problems for rules for multiplying integers practice:

Using rule 1:

12 × 12 = 144

Using rule 2:

15 × - 8 = - 120

Using rule 3:

-10 × 5 = - 50

Using rule 4:

- 14 × - 20 = 280


Example problems for multiplying integers practice:


Example 1:

Multiply the two integer numbers

12 × 3

Solution:

Given

12 × 3

Both are two positive integers so, the result is also a positive numbers

Here we add 12 into 3 times, then we get

12 + 12 + 12

36

It is the simplest method of multiplying integers.

Example 2:

15 × 10

Solution :

Given

15 × - 10

In the given integer numbers, positive and negative numbers so result of this given integer is also a negative numbers.

If we multiply by 10, we have add to zero to the result

- 150

Solution to the given two integer is - 150.

Example 3:

- 8 × - 7

Solution:

Given

- 8 × - 7

Both are negative numbers so the result is also a negative numbers

If we multiply the - 8 by -7 we get

56

Solution to the given integers is 56.

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Practice problems for multiplying integers:


Some practice problems for multiplying integers are

1). 15 × 20

Solution = 300

2). 8 × - 30

Solution = - 240

3). 2 × 3

Solution = 6

4). - 6 × 8

Solution = 48

Geometric Proofs Practice

Introduction for geometric proofs practice:

The geometric proofs practices are generally involved in solving the proof for some problems that involving the statements from different theorems that are earlierly solved. we have to analyze the question first and then we have to choose what kind of statements are used for the problems given. After those steps are taken, we have to match it for the statements with the problems. The proof represents the way that how to we prove the given problem or some postulates.

(Source from Wikipedia)

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Examples to explain "geometric proofs practice"


We have to prove the straight line's equation is 3x + 2y = 7 and which is passing through the point (1,2) and which making the intercepts on the axes of the co-ordinate which are in the ratio 2 : 3.
Geometry proof:

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

We know that the point (1, 2) lies on the straight line given above, 3 + 4 = 6k

i.e. 6k = 7

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

Thus we have proved the proof for finding the straight line's equation.

Practice:

In this problem as we are seen that it have specific points to prove the problems such that it having ratio on the concern and the point which the line passing.

Problems to explain "geometric proofs practice"


We have to prove that the point's co-ordinates are (8, 9) and (- 42, - 41) and given that it is reside on the straight line y = x + 1 which are at a distance of 5 units from the straight line 4x - 3y + 20 = 0
Geometry proof:

Let (x1, y1) be a point on y = x + 1

? y1 = x1 + 1 … (1)

The length of the perpendicular from (x1, y1) to the straight line

4x - 3y + 20 = 0 is `|(4x1 - 3y1 + 20)/sqrt(4^2 + (-3)^2)|`   = `+-`    `((4x1 - 3y1 + 20)/5)`

But the length of the perpendicular is given as 5.

?  `+-`    `((4x1 - 3y1 + 20)/5)`  = 5

4x1- 3y1 + 20 = `+-`  25

Considering the positive sign, 4x1- 3y1 + 20 = `+` 25

?                                   4x1 - 3y1 = 5 … (2)

Considering the positive sign, 4x1- 3y1 + 20 = `-` 25

?                                   4x1 - 3y1 =`-` 45 … (3)

Solving (1) and (2), we get x1 = 8, y1 = 9

Solving (1) and (3), we get x1 = - 42, y1 = - 41.

? The required points co-ordinates are (8, 9) and (- 42, - 41).

Thus we have proved the proof with the given points.

Practice:

In this problem as we are seen that it have specific points to prove the problems such that it having the perpendicular distance from the two straight lines.

Practice problems on geometric proof:

We have to prove the straight line's equation is x - `sqrt(3)` y + 12 = 0, if the perpendicular comes from the origin, which makes an 120° angle with x-axis and the distance from perpendicular comes from the origin is 6 units.
We have to show that the straight lines 132x +13 y - 9 = 0 and 132x + 13y - 10 = 0 are parallel.

Division Practice problems

Introduction to division practice problems:

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

The following division methods are all based on the form Q = N / D where

• Q = Quotient

• N = Numerator (dividend)

• D = Denominator (divisor).

Specifically, if c times b equals a, written:

c x b = a

Where b is not zero, then a divided by b equals c, written:

a / b = c

In the above expression, a is called the dividend, b the divisor and c the quotient. (Source: Wikipedia)

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Example problems on division practice


Ex:1  Jack built a tower of blocks forty-five inches high. Each block in the tower is five inches tall. How many blocks were used to build the tower?

Sol: Jack built a tower of blocks forty-five inches high.

Each block in the tower is five inches tall.

So, total blocks = 45 / 5

Therefore total blocks used to build the tower = 9 blocks

Ex:2 The school's Internet connection transferred thirty six megabytes of data in six seconds. How many megabytes can it transfer in just one second?

Sol:

The school's Internet connection transferred thirty six megabytes Data in six seconds.

So, therefore total megabytes can it transfer in just one second

= 36 / 6 = 6 megabytes.

Ex:3 There are eight soft drink machines in the university. They hold sixty - four cases of soda altogether. How many cases does each machine hold?

Sol:

There are eight soft drink machines in the university.

They hold sixty - four cases of soda altogether.

Total cases of soda machine hold = 64 / 8

= 8 Cases of soda

Ex:4 Zachary used five thousand, five hundred forty-four chips to make a big batch of giant chocolate chip cookies. Each cookie got about eighteen chips. How many cookies did Zachary make?

Sol:

Zachary used five thousand, five hundred forty-four chips to make a big batch of giant chocolate chip cookies

Each cookie got about eighteen chips

So, total cookies Zachary make = 5544 / 18

= 308 Cookies

Ex:5 Twenty-five busses brought a total of seven hundred fifty passengers to the city. About how many passengers were on each bus?

Sol:

Twenty-five busses brought a total of seven hundred fifty passengers to the city.

So, total passengers were on each bus = 750 / 25

= 30 Passengers 

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Division practice problems:


Ex:1 There are thirty apartments in that building. The building has five stories. How many apartments does each story of the building have?

Ans: 6 apartments

Ex:2 Jenna will mail out seven copies of her resume on special paper. She needs twenty-eight sheets of the paper. How many pages long is her resume?

Ans: 4 pages

Ex:3 Jeremy has twenty-four balloons. He wants to give each of his four friends an equal number. How many balloons should each friend be given?

Ans: 6 balloons

Practice Act Question

Act Test preparation:

Act test is one of test which is used to get admissions in mid east countries. Depending on the test marks students get admissions in mid east colleges. Many students around the globe are writing this test hence this test mark is now accepted in US countries also. Act test is some what different from sat test. Act test consists of science reasoning questions but sat test does not. The grammatical talent of the student is checked in the act test while it is not in the sat test. Both the sat and act test are skill oriented test.  The act test is conducted for nearly 4 hrs and optional of 30 minutes.

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Act practice problems:

Act practice question 1:

Anand's new job comes with a salary increase of 8%. If he currently makes $74,000 per year, which of the following is the amount he will earn per year at his new job?

A. $79,000
B. $72,000
C. $79,200
D. $75,550
E. $79,920

Solution:

Increase percent in salary = 8 %

8% of $74,000                     = (8 x 74,000) / 100

= 8 x 740

= $5920

The total amount her earned is = $74,000 + $5920

= $79,920

The amount he earned is $79,920

Answer: E



Act practice question 2:

If Suresh biked 10 miles in 4 hours and arun biked three times as much in half the time, what was arun's average rate of speed?



A. 10 mph
B. 12 mph
C. 15 mph
D. 23 mph
E. 27 mph

Solution:

Let the speed of suresh per hour = x

Speed of arun                              = 3x

Arun biked suresh distance in ½ hour.

Total distance covered by arun   = 3*10

= 30 miles

Arun covered the distance in half the time of suresh,

= 30/2

Average speed of arun                  = 15 miles/hr

The answer is 15 miles/hr

Answer: C


Practice act questions:


Practice question 1:

In an urn there are 15 balls: 8 balls are black, 4 are red and 3 are orange. Then 1 black and 1 red ball are taken from the urn and put away. What is the probability that a red ball is selected at random from the urn?

A) 3/13
B) 5/15
C) 6/15
D) 7/13
E) 4/13

Answer: A

Practice question 2:

A group of 7 friends are having dinner together. Each person eats at least 3/4 of a cake. What is the smallest number of whole cakes needed for dinner?

A) 7
B) 5
C) 6
D) 28
E) 21

Answer: C

Reasoning Practice Test

Introduction:

Let us learn about the reasoning practice test. Reasoning test is a test which is conducted to get placed in international colleges, universities and in companies. Reasoning test  practice consists of quantitative aptitude, verbal reasoning and analytical writing. Quantitative aptitude involves math problem solving questions on various topics. Practice of reasoning test questions helps the students to get skilled in aptitude questions. It also helps the students to get prepared for SAT and ACT test.

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Reasoning practice test:


Example 1:

A trader mixes 30 kg of rice at Rs. 18 per kg with 28 kg of rice of other variety at Rs. 34 per kg and sells the mixture at Rs. 32 per kg. What is the profit percentage?

Solution:

Total price of 58kg mixture rice = ( 30 x 18 + 28 x 34)

= 540 + 952

= Rs 1492.

Sell price of 58 kg rice = (58 x 32)

= Rs 1852

Total gain amount        = Rs 1852 - Rs 1492

= Rs 360

Total gain percentage = (360/1492)*100

= 0.24 * 100

= 24%

The profit gain is 24%



Example 2:

Vinay has 100 kg of wheat, part of wheat which he sells at 8% profit and the rest of wheat at 18% profit. He gains 14% on the whole. What is the quantity sold at 18% profit?

Solution:

From the given data, form two equations,

Part of wheat sold at 8%    = X

Part of wheat sold at 18%  = Y

X + Y = 100  equation 1

0.08X + 0.18Y = 100(0.14)

0.08X + 0.18Y = 14 equation 2

Multiply equation 1 with 0.18

0.18X + 0.18Y = 18  equation 1

0.08X + 0.18Y = 14  equation 2     (subtract)

0.10X                = 4

Divide 0.10 on both sides,

0.10X/0.10 = 4/0.10

X = 40 kg

Total quantity            = 100

Remaining quantity = 100 - 40

Y  = 60 kg

The quantity which is sold at  18% profit is 60 kg.

Practice questions:


Practice problem 1:

The ratio between the speeds of two trains is 6 : 8. If the second train runs 300 km in 3 hours, then the speed of the first train is:

A) 60 km/hr

B) 65 km/hr

C) 75km/hr

D) 73 km/hr

Answer: Option C



Practice problem 2:

544, 509, 474, 439, ... Find the next number in the given series.

A) 420

B) 404

C) 454

D) 528

Answer: Option B



Practice problem 3:

201, 202, 204, 207, ... Find the next number in the given series.

A) 211

B) 212

C) 225

D) 230

Answer: Option A

Practice 9th Grade Algebra Midterm

Introduction to practice algebra problem for 9th grade midterm:

Algebra which is fetched by a finite number which is used to define the following operations like exponents, adding the values, subtracting the values, multiplying and dividing as well.
Basically algebra midterm defines about the number systems , and here in 9th grade we can see about the following methods. They are,
Numbers

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In number systems, we can see the operations of fractions,exponents,basic arithmetic operations and place values.
Measurement:

Measurement is a process of measuring the object or shapes values in the required units.
Geometry:

Geometry is a process,in which we can declare the different shapes of polygons like regular shapes and irregular shapes.
Algebra / pattern and Probability.

Algebra pattern is a method of arranging the algebraic values in the proper order and it performs a prominent role in algebra terms.

practice problem for 9th grade algebra midterm


Example 1:

Solve the  practice algebraic problem 5(-5x - 2) - (-x - 6) = -6(4x - 5) -36

Solution:

Given the equation

5(-5x - 2) - (-x - 6) = -6(4x - 5) - 36

Multiply factors.
-25x - 10 + x + 6 = -24x + 30-36

Grouping the terms.

-24x - 4 = -24x - 4

Add 24x +4 to both sides, the above equation becomes

0 = 0

All real values are solution to this equation.

Example 2:

Simplify the  practice algebraic expression    3(a -4) + 6b - 3(a -b -2) + 8

Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

= 9b + 2.

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Practice algebra problem for 9th grade midterm:


Try to solve this practice 9th grade algebra midterm problem:  a+ b = 3, b + c = –5, c + a = 2.

(solution: a = 5, b = -2, c = -3)

Trigonometry Practice Questions

Introduction to trigonometry practice questions:

The branch of mathematics which deals with the relations of the sides and angles of triangles, which the methods of deducing from certain given parts other required parts, and also of the general relations which exist between the trigonometrical functions of arcs or angles with the trigonometric functions.


Solved problems on trigonometry practice questions


1. Solve the trigonometric equation and find the (6 tan^2 x - 2) (2 tan^2 x - 6) = 0

2. Solve the trigonometric equation and find the interval [0 , 2Pi).

-2 sec^2 x + 4 = -2sec x

3. Solve the trigonometric equation and find the interval [0 , 2Pi).

2sin (x) cos (-x) = 2 sin (-x) sin (x)

4. Two tennis players stand on opposite sides of a net. When the ball is directly above the net, at a height of 7m above the ground, the angle of elevation from player A's position to the ball is 41±  and the angle of elevation from player B's position to the ball is 53±. What is the distance between player A and player B?

5. A pilot flying at an altitude of 1200 feet’s is starting his approach. He is 2400 feet along his flight path from the runway. What is his horizontal distance x from the runway and what is his angle of approach µ?

6. The distance between two points on the map is 5 cm correct to the nearest centimeter.

(a) Write down the

(i) Least upper bound of the measurement

(ii) Greatest lower bound of the measurements

(b) The scale of the map is 1 to 20 000. Work out the actual distance in real life (in kilometers) between the upper and lower bounds.

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7. The temperature from the factory furnace a varies inversely as the square of the distance from the furnace.

The temperature 2 meters from the furnace is 50 degrees Celsius.

Calculate the temperature 3.5 meters from the furnace. Give your answer to 2 decimal places.