Hypotenuse of a Right Triangle in brief

There are different types of triangle which we have learnt; one of them is the right angled triangle in other words a right triangle. A triangle in which one of the angles is a right angle that is 90 degrees is called a right triangle. The longest side of a right triangle is called the Hypotenuse of a Right Triangle and the other two sides are called the legs of the right triangle. We use the Pythagorean Theorem in finding the hypotenuse of a Right Triangle. The Pythagorean Theorem states that ‘the sum of the squares of the two sides (legs) of a right triangle is equal to the square of the hypotenuse’. Let us assume the lengths of the legs of a right triangle to be ‘a’ and ‘b’ units and the hypotenuse length to be ‘c’. By using Pythagorean Theorem, we can calculate the hypotenuse of a right triangle,
(Hypotenuse) ^2 = (sum of the squares of the sides (legs)^2
  c^2 = (a^2 +b^2)

This gives us the Hypotenuse of a Right Triangle formula, c = sqrt(a^2 + b^2)
Given the lengths of the sides or legs of a right triangle as 3 cm and 4 cm respectively, find the hypotenuse of the right triangle.  Here we are given the lengths of the two sides a = 3cm and b = 4cm, we need to find c. let us apply the Pythagorean Theorem, we get, c = sqrt(a^2+b^2). a^2= (3)^2 = 9 and b^2 = (4)^2 = 16, that gives us a^2+b^2 = 9 +16 = 25. So  c = sqrt(25) = +/- 5 , as length cannot be negative, the hypotenuse of the given right triangle is 5 cm.

Let us now learn how to calculate the Hypotenuse of a Right Triangle using the Pythagorean theorem given by hypotenuse = sqrt[(sum of the squares of the sides or legs)^2]. A ladder is placed against a wall of height 12 ft. The distance between the base of the ladder and the wall is 5ft, find the length of the ladder. In this problem, the triangle formed by the wall, the floor and the ladder is a right triangle and hence the length of the ladder would be the hypotenuse which we need to find. We are given the two lengths of the sides of the triangle which are 12ft and 5 ft respectively. We know c= sqrt(a^2 +b^2); here a = 12ft, b = 5 ft which gives us a^2 = 144 and b^2 = 25. So, c = sqrt(144 +25) = sqrt(169) = +/-13. The length of the ladder is 13 ft

Solving fourth grade math homework

In this article we are going to discuss about the mathematical concepts for fourth grade students. The students of fourth grade learn the different areas of mathematics, like place values of six digits numbers, expanded notations, place value chart, addition, subtraction, multiplication and division of three and four digits numbers, multiples and factors, HCF and LCM. Grade fourth students also learn unitary method, fractions, decimal numbers and measurement of time, length, mass and capacity.

Here we are going to discuss about some of them. This article will be helpful in solving fourth grade math home work.

Topics of Solving Fourth Grade Math

Some of fourth grade mathematical topics are as follows:

Place value: To read and write large number easily, the Indian place value chart is divided into periods as shown below:

Practice Questions: 

(1)Write the number name of these numbers:

(a)    2, 50,946 = Two Lakh fifty thousand nine hundred forty six.

(b)    6, 92,438 = Six lakh ninety two thousand four hundred thirty eight.

(c)    20, 10,101 = Twenty lakh ten thousand one hundred one.

 (1)   Write the numeral of these number names:

(a)    One lakh fifty thousand two hundred eighteen = 1,50,218

(b)   Nine lakh ninety five thousand sixty three = 9,95,063

(c)    Seventeen lakh fifteen = 17,00,015

Simplification involving four fundamental operations: 

In this lesson we will learn to use all the operations together. The fourth grade learners are able to learn the order of operations through this section.

Step 1 ---- Of

Step 2------Division

Step 3 ------ Multiplication

Step 4 ------ Addition

Step 5 ------ Subtraction

 So the order of operation is ODMAS.Now we will do some simplification using ODMAS rule:

 Practice Questions:

(a)    Simplify 36÷ 6 x 4 + 2 – 8

                   = 6x 4 + 2 – 8

                   =   24 +2 – 8

                    =    26 – 8 =18 Ans.

(b)   Simplify 6 + 8 ÷2 -2 x 1 + 5 of  2÷ 5

                    = 6 + 8 ÷ 2 – 2 x1 + 10 ÷5

                    = 6 + 4 – 2 + 2

                    = 12 - 2 = 10 Ans.

Now simplify these questions:

(a)    23741 -  3826 ÷  2 x 6 + 221

(b)   529 x 71 – 630 of 6 ÷ 3 + 4

(c)    6000 +9000 ÷  500 of 6 – 2000

(d)   6699 ÷ 33 +3075 ÷ 25 – 203

Answers: (a) 12484 (b) 36303 (c) 4003 (d) 123

Solving Fourth Grade Math Homework-unitary Method

Some basic knowledge is very important for fourth grade learners like:

• We have to add when we find the total cost of things tat we purchase.
• We have to subtract when we take back the balance.
• We have to multiply to calculate the cost of more articles.
• We have to divide when we find the cost of one.

Practice Question: 

If the cost of 25 Milton jugs is Rs. 8,250.  What is the cost of 48 such Milton jugs?

Solution: The cost of 25 Milton jugs = Rs.8, 250

                The cost of 1 Milton jug is Rs.8, 250 ÷ 25 = Rs.330

                The cost of 48 Milton jugs = Rs. 330 x 48 = Rs.15840

 Solve these problems: 

 28 digital diaries cost Rs.70, 560. What is the cost of 15 such diaries?

 Answer: 37,800

Statistics homework answers

Statistics is the branch of applied mathematics which deals with scientific analysis of data. The subject statistics had been started in early days as arithmetic to aid a ruler who needed to know the wealth of his subjects to levy new taxes. Now a days statistics  plays an important role in all organizations in their decision-making and planning. Statistics homework deals with mean,median and mode.

Classification of Statistics Homework Answers:

In statistics homework answers has following topics are

• Arithmetic mean
• Median
• Mode

The arithmetic mean;

In Statistics the arithmetic mean (A.M) or simply the mean or average of n observations x1, x2, …, xn is defined to be number x such that the sum of the deviations of the observations from x is 0.

         x1 +x2 +x3.......xn
x =  --------------------
                     n

the symbol Σ, called sigma notation is used to represent summation.

x=Σ xi
   -------
      n

Homework Example:

Calculate  mean of the data 9, 11, 13, 15, 17, 19.

Sol :

X =   Σxi/n = `[9+11+13+15+17+19]/[6] ` =` [14]/[6]`

Median :

Median is defined as central most or middle value for given series of data it should be arranged in ascending or descending order.

Homework Example:

Find the median of 23, 25, 29, 30, 39.

Sol:

The given values are already in  ascending order. No. of observations N = 5.

So the median = ` (N+1)/(2)`  =`(5+1)/(2)`
= 3 rd term =29
∴ Median = 29.

Mode:

In Statistics answers,Mode is also a measure of central tendency.

In a set of each observations, mode is defined as value which occurs most often.

If the data are arranged in the form of a frequency table, the class corresponding to  greatest frequency is called  modal class.

Homework Example:

Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7.

Sol:

Arranging the data in  ascending order, we get 1, 3, 4, 4, 5, 6, 7, 7, 7.

In above data 7 occurs several times. Hence mode = 7.

Homework Problems and Answers for Statistics:

Calculate mean :    
Problem 1:
Calculate the mean of the data 7, 12, 18, 14, 19, 20.
Problem 2:
Calculate the mean of the data   16,18,14,15,21,30,26,44
Problem 3: 
Calculate the mean of the data   22,77,55,11,22,26,38,72
Answers:
1) 15
2) 23
3) 40.35

Calculate median:
Problem 1: 
Find the median of 3, 4, 10, 12, 27, 32, 41, 49, 50, 55, 60, 63, 71, 75, 80.
Problem 2:
Find the median of 29, 23, 25, 29, 30, 25, 28.
Problem 3: 
Find the median of 26, 25, 29, 23, 25, 29, 30, 25, 28, 30.
Answers:
1) 49
2) 28
3) 27

Calculate mode:
Problem 1:
Find the mode of 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15
Problem 2:
Find the mode of 3,5,8,3,9,3
Problem 3: 
Find the mode of 3,5,8,5,6,7
Answers:
1) 12 and 15
2) 3
3) 5
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An Introduction to Calculating the Area of a Right Triangle

The Area of a Right Triangle
A Right Triangle or a Right Angled Triangle is a triangle which has right angle or 90 degrees as one of its interior angles. We have two types of Right Triangles namely, scalene right triangle in which one angle is right angle and the other two are unequal angles and the corresponding sides are also unequal. The other type is the Isosceles Right Triangle in which one angle is right angle and the other two angles are equal and also their corresponding sides are of equal length.  A right triangle is very useful in trigonometry, a branch of mathematics. Area of a triangle is given by half times the base times the height of the triangle. We can deduce the formula for Area of a Right Triangle using the following method. Let us consider a rectangle of length l cm and width w cm. Let us now cut the rectangle into two equal halves diagonally as shown in the figure below.

 
The triangles ABC and ADC are congruent triangles which means when placed one over the other they overlap exactly. Also they are right triangles with 90 degrees as one of their angles. They are of the same size and hence we can say that they have the equal area. From this it is clear that the area of a right triangle will be half the area of the rectangle.
The Area of the right triangle = ½ [Area of the Rectangle]
               = ½ [length x width]
             = ½ [l w]

From the figure, we can see that the length of the rectangle is one of the sides of the triangle which is called the base of the triangle. So, we have base of the triangle equals length of the rectangle.  Also the distance from the vertex A of the triangle ABC to the vertex B is the height of the triangle ABC which is the width of the rectangle. So, we have height of the triangle equals width of the rectangle. Substituting these in the above equation, Formula for Area of a Right Triangle = ½ [base x height]. So, to calculate area of a right triangle we use the formula ½ times [base x height]
Calculate the Area of a Right Triangle with base and height 12 cm and 17 cm respectively
Solution:  Base of the right triangle   = 12 cm
         Height of the right triangle = 17 cm
       Area of the right triangle   = ½ x base x height
                      = ½ x 12 x 17 = 6 x 17 = 102 cm^2

Characteristics of Right Angle Triangle

Triangle is a basic shape in geometry. A right triangle means 90 - degree angle that is one angle is right angle. It also called as right angled. It shows the relations between the sides and angles of a right triangle and it is the basic concept of trigonometry. The opposite side of the right angle is known as hypotenuse. The adjacent side to the right angle is known as legs or sometime known as catheters. This phenomenon is the basis of Pythagorean triangle theorem. This theorem states that the length of all three sides of a right triangle are integer and the side length are known as Pythagorean triple.

The basic principle properties of a right triangle are area, altitude, Pythagorean Theorem, inradius and circumradius. First we discuss about area of a right triangle. Area of a right triangle is equal to the one half of the base multiplied by the corresponding height. It is true for all right angled triangles. In a right triangle from the two legs if one leg is taken as a base then other leg is taken as height. So in other way we can say that area of a right triangle is one half of the product of the two legs means one half products of base and height. Mathematically suppose in a right triangle two legs are (a) and (b) and hypotenuse is (c) then formula for the area of a right triangle is expressed as [A=(1/2 ab)].

For finding area of a right triangle we have to calculate perimeter also. The formula we can see above. Suppose in a right triangle two legs (a=5) and (b=12) where a is base and b is height. Hypotenuse (c=13) is given. We know the formula area of a right triangle so we substitute all the values in formula. Finally we get area= (1/2 *5*12) = (1/2*60) =30 and unit of right angle triangle is square meter or square centimeters.

Similarly we can find surface area of a right triangle.  Formula for surface area of right angle triangle is [Area (A) =bc/2] where (a=√b^2+c^2), (b=√a^2-c^2) and (c=√a^2-b^2). Suppose for above problem we have to find surface area. We substitute all the values in formula that is (A=12*13/2).
From the right angle triangles we also find the Pythagorean Theorem. It is very useful in all parts of mathematics. It states that in any right angle square of hypotenuse is equal to sum of square of base and square of height that is (c^2=a^2+b^2).

Trigonometric equations

Trigonometric equations
In this article we shall study some basics of how to solve trigonometric equations. A trigonometric equation is an equation that involves trigonometric ratios in addition to algebraic terms. As trigonometric  functions are many to one type of functions, it is quite possible that such trigonometric- equations may have many roots (solutions).

This can be made clear by solving trigonometric equations examples:
For instance, sin ( pi/6) = ½ but the equation sin x = ½ has not only the solution x =  pi/6, but also x = 5 pi/6, x = 2 pi +  pi/6, x = 3 pi -  pi/6 etc. Thus, we can say that x =  pi/6 is a solution of sin x = ½ but it is not the general solution of the equation.

A general solution gives all the roots of an equation. When attempting to solve trigonometric equations online it is very imperative that the equations fed in are valid. For example, and equation like sin x = 2 would not have any solution. That is because we know that the range of the sine function is [-1,1]. So for any value of angle x, the value of sin x can never be = 2. Therefore the solution of such an equation would be : no solution.

The objective should be to develop methods to find general solutions of trigonometric-equations.
We know that sine, cosine and tangent functions are all periodic. Sine and cosine functions have a period of 2 pi and the tangent function has a period of  pi. Therefore the general solutions of the equations, sin t = 0, Cos t = 0 and tan t = 0 can be found as follows:
Sin t = 0 ? t = k pi, k belongs to the set of integers ------------- (1)
Cos t = 0 ? t = (2k+1)/2, k belongs to set of integers ----------- (2)
Tan t = 0 ? t = k pi, k belongs to the set of integers -------------- (3)

These results can be used to solve the general trigonometric-equations which are as follows:
Cos t = a, |a| = 1
Sin t = a, |a| = 1
Tan t = a, a’ belongs to set of all real numbers.
By solving a trigonometric equation we intend to find a set of solutions for the given set of trigonometric-equations such that each member of the solution set satisfies the set of equations.

Learn more about how to solve Trigonometry Problems.

Limit of a function

For understanding the limit of a function we must know the fundamental concepts of calculus and analysis which belonging to that particular function near a define input. In another way we can simplify by taking a simple example. Suppose function is f(x) and limit is x tends to c where c is a constant, then it means that function f(x) is get closer to limit as x get closer to c. more accurately when function is applied to each input , the result in an output value that is close to limit.

Limits of a sequence
As we know about the limits. Limits of a sequence means a value in term of sequences. If limit value exist then such sequence occurs. Limits of a sequence of any function can be understood by applying the function on a real line. This is one way to know the limits of function in term of sequence.

Limits of sequences
For knowing the limits of sequences we choose such type of example where limits value exist then the sequences will convergence. Limits are first apply for the real numbers and then for others such as metric spaces and topological spaces.

Limits of functions
In limits of functions we take any calculus function like additive or subtractive functions, because for simplify purpose these functions are easy. Functions are with limits where in limit x is approaches to any real number. When we solve such function in last we must put the limits. So in limits of functions output result are moves according to limits.

Limit of sequence
In limit of sequence, first take a sequence such as (Xn) with limit where x is tens to a and a is a constant. When we apply limit to this function if and only if for all sequences(Xn) means with (Xn) with all value of n but not equal to constant a.

Limits of a function
Limits of function means limits should apply to a function. Function may be different types. In limits of function we explain that what it means for any function which tends to real number, infinity or minus infinity. Limits of function can be both right handed and left handed.
Suppose in any case if both right hand and left hand limits of  function as x approaches to constant exist and are equal in value, their common value evidently will be the limits of a function. If however either or both of these limits do not exist then the limits of a function does not exist.