hypothesis formula class 9

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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• Monomial = \( {\rm{3}},2x,\frac{2}{3}y{\rm{etc}}{\rm{.}} \)
• Binomial = \( (2x + 3y),(3x – 2y){\rm{etc}}{\rm{.}} \)
• Trinomial = \( x^2 + 4x + 5{\rm{etc}}{\rm{.}} \)
• Linear Polynomial = \( x + 2,3x + 5{\rm{etc}}{\rm{.}} \)
• Quadratic Polynomial = \( ax^2 + bx + c{\rm{etc}}{\rm{.}} \)
• Cubic Polynomial = \( x^3 + 4x^2 + 5{\rm{etc}}{\rm{.}} \)
• Biquadratic Polynomial = \( x^4 + 5x^3 + 2x^2 + 3 \)
• Equation of a line = \( ax + by + c = 0 \)
• Equation of a circle = \( x^2 + y^2 = r^2 \) Here ‘r’ is the radius of the circle
• Equation of a parabola = \( y^2 = 4ax \)
• Equation of an ellipse = \( \frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} = 1 \)
• Equation of hyperbola = \( \frac{{x^2 }}{{a^2 }} – \frac{{y^2 }}{{b^2 }} = 1 \)
• Distance formula D = \( \sqrt {\begin{bmatrix}{\left( {x_2 – x_1 } \right)^2 } +\\ {\left( {y_2 – y_1 } \right)^2 }\end{bmatrix}} \)
• Angle between two lines = \( \tan ^{ – 1} \left( {\frac{{m_2 – m_1 }}{{1 + m_1 m_2 }}} \right) \)
• Area of Isoscele Triangle = \( \frac{1}{2}bh\)
• Altitude of an Isosceles Triangle= \( \sqrt{a^{2}-\frac{b^{2}}{4}}\)
• Perimeter of Isosceles Triangle = \( 2\,a+b\)<br/ >Where, b = Base , h = Height, a = length of the two equal sides
• Area of an Right Triangle = \( \frac{\sqrt{1}}{2}bh\)
• Perimeter of an Right Triangle = \( a+b+c\)
• semi Perimeter of an Right Triangle = \( \frac{a+b+c}{2}\) where:; b:Base, h:Hypotenuse a: Hight
• Area of Scalene Triangle = \( \sqrt{s(s-a)(s-b)(s-c)} \)
• Perimeter of Scalene Triangle = \( a+b+c \) Where: a, b, c are Side of Scalene Triangle
• Area of an Equilateral Triangle = \( \frac{\sqrt{3}}{4}a^{2}\)
• Perimeter of an Equilateral Triangle = \( 3a\)
• Semi Perimeter of an Equilateral Triangle = \( \frac{3a}{2}\)
• Height of an Equilateral Triangle = \( \frac{\sqrt{3}}{2}a\) Where, a:side, h: altitude
• Area of a Square = \( side^{2}\)
• Area of a Kite = \( \frac {1}{2} \times Diagonal_{1} \times Diagonal_{2} \)
• Perimeter of Kite= \( 2(a+b)\)
• Area of a Parallelogram = \( Base \times height \)
• Perimeter of Parallelogram= \( 2(Base + height)\)
• Area of a Rectangle = \( Length \times Breadth \)
• Perimeter of Rectangle= \( 2(Length + Breadth)\)
• Area of a Trapezoid = \( \frac {Base_{1} + Base_{2}}{2} \times hiehgt \)
• Perimeter of Trapezoid= \( Side_{1} + Side_{2} + Side_{3} + Side_{4}\)
• Area of a Parallelogram = \( b\times h\)
• Perimeter of Parallelogram = \( 2\left(b+h\right)\)
• Height of Parallelogram = \( \frac{Area}{Base}\) Where: b is the length of any base and h is the corresponding altitude or height.
• \( p^{2}+q^{2}=2(a^{2}+b^{2})\) Where: p,q are the diagonals, a,b are the parallel sides
• Area of a Triangle = \( \frac{1}{2}ah \) Where: a is the base of the triangle. h is the height of the triangle.
• Perimeter of a Triangle = \( a+b+c\) < br />Where: a, b, and c are three different sides of a Triangle.
• Area of a circle =\( \pi r^{2}=\frac{{\pi}d^{2}}{4}=\frac{{C} \times {r}}{2}\)
• Perimeter of Circle = \( 2 \pi r \)
• Area of a half circle = \( \frac{{\pi}r^{2}}{2}\)
• Area of a Quarter circle = \( \frac{{\pi}r^{2}}{4}\)
• Area of Sector of a circle = \( \frac{θ}{360}\pi r^{2}\)
• Length of an arc of a sector = \( \frac{θ}{360}{2 \pi r}\)
• Sector angle of circle = \( \frac{180 \times l}{\pi r}\)
• Area of the sector = \( \frac{θ}{2} \times r^{2}\)
• Area of the circular ring = \( \pi \times (R^{2} – r^{2})\) Where, r is the radius of the circle. d is the diameter of the circle. C is the circumference of the circle. θ is the Angle between two radius. R is the Radius of Outer Circle.
• Area of Triangle = \( \sqrt{s(s-a)(s-b)(s-c)} \)
• Perimeter of Triangle =\( a+b+c \)
• Semi perimeter of Triangle =\( \frac{a+b+c}{2} \) Where, a , b , c are Side of Triangle
• Surface area of Cube = \( 6a^{2}\)
• Volume of a cube = \( a^{3}\) Where, a is the side length of the cube.
• Surface area of Cuboid = \( 2(lb + bh + hl)\)
• Volume of a Cuboid = \( h \times l \times w\) Where, l: Height, h: Legth, w: Depth
• Diameter of a sphere = \( 2r\)
• Circumference of a sphere = \( 2\pi r\)
• Surface area of a sphere = \( 4\pi r^{2}\)
• Volume of a sphere = \( \frac{4}{3}\: \pi r^{3}\)
• Curved Surface area of a Hemisphere = \( 4\pi r^{2}\)
• Total Surface area of a Hemisphere = \( 3\pi r^{2}\)
• Volume of a Hemisphere = \( \frac{2}{3}\: \pi r^{3}\) Where, r: Radius
• Curved Surface area of a Cylinder = \( 2\pi rh\)
• Total Surface area of a Cylinder = \( 2\pi r(r+h)\)
• Volume of a Cylinder = \( \pi r^{2} h\)Where, r: Radius, h: Height
• Total Surface Area of cone = \( \pi r \left (s+r \right )\)
• Vomule of cone = \( \frac {1}{3}\pi r^{2}h\)
• Curved Surface Area of cone = \( \pi rs\) Where, r : radius of cone. h :height of cone. s : slant height of the cone.
• \(\ Mean\; \bar{x} = \frac{\sum x}{n}\) Where, x = Items given, n = Total number of items
• \(\ Range = Largest\; Value – Smallest\; Value \)
• If n is odd, \(\ Median = (\frac{n+1}{2})^{th}term\)
• If n is even, \(\ Median = \frac{(\frac{n}{2})^{th}term+(\frac{n}{2}+1)^{th}term}{2}\) where, n = Total number of items
• \(\ Variance = \sigma ^{2} = \frac{\sum (x- \bar{x})^{2}}{n}\) Where, x = Items given, x¯ = Mean, n = Total number of items
• Standard Deviation \(\ \sigma = \sqrt{\frac{\sum (x-\bar{x})^{2}}{n}}\) Where, x = Items given, x¯ = Mean, n = Total number of items
• Probability = \(\ \frac{No. \;of\; Favorable \;outcome}{No.\; of\; all\; possible\; outcome} \)
• \((a+b)^{2}=a^2+2ab+b^{2}\)
• \((a-b)^{2}=a^{2}-2ab+b^{2}\)
• \(\left (a + b \right ) \left (a – b \right ) = \\ a^{2} – b^{2}\)
• \(\left (x + a \right )\left (x + b \right ) = \\ x^{2} + \left (a + b \right )x + ab\)
• \(\left (x + a \right )\left (x – b \right ) = \\ x^{2} + \left (a – b \right )x – ab\)
• \(\left (x – a \right )\left (x + b \right ) = \\ x^{2} + \left (b – a \right )x – ab\)
• \(\left (x – a \right )\left (x – b \right ) = \\ x^{2} – \left (a + b \right )x + ab\)
• \(\left (a + b \right )^{3} = \\ a^{3} + b^{3} + 3ab\left (a + b \right )\)
• \(\left (a – b \right )^{3} = \\ a^{3} – b^{3} – 3ab\left (a – b \right )\)
• \( (x + y + z)^{2} = \\ x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2xz\)
• \( (x + y – z)^{2} = \\ x^{2} + y^{2} + z^{2} + 2xy – 2yz – 2xz\)
• \( (x – y + z)^{2} = \\ x^{2} + y^{2} + z^{2} – 2xy – 2yz + 2xz\)
• \( (x – y – z)^{2} =\\ x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2xz\)
• \( x^{3} + y^{3} + z^{3} – 3xyz = \\ (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz -xz\)
• \( x^{2} + y^{2} = \\ \frac{1}{2} \left [(x + y)^{2} + (x – y)^{2} \right ]\)
• \( (x + a) (x + b) (x + c) = \\ x^{3} + (a + b +c)x^{2} + (ab + bc + ca)x + abc\)
• \( x^{3} + y^{3} = \\(x + y) (x^{2} – xy + y^{2})\)
• \( x^{3} – y^{3} = \\ (x – y) (x^{2} + xy + y^{2})\)
• \( x^{2} + y^{2} + z^{2} -xy – yz – zx = \\ \frac{1}{2} [(x-y)^{2} + (y-z)^{2} + (z-x)^{2}]\)

Once you are sure of the topics and the logic behind then it is always easy for you to learn the formulas and solve the most difficult problems too. In the next sections, we will discuss the chapter-wise marking scheme and the total number of chapters for study in class 9.

Moreover, once you learn Math Formulas, they are important for other subjects too like Physics, Economics, Chemistry etc. Learning online is the smartest way today with handy learning aid for students. When you learn online, the PDF formulas file is prepared carefully after deep research and observation.

At the same time, you will get hands-on practice material for exam preparation and complete their syllabus faster. For students, basic Mathematics formulas are the foundation for other subjects in class 9. Not only for students, but an online PDF guide and formulas are good for teachers as well and they can use it as a reference and brush up their mathematical skills in minutes.

NCERT Solutions For Class 9 Maths by Chapters

Recently, a new assessment scheme has been introduced for the class 9 th CBSE. Students will be assessed on the basis of final exams for each subject carrying 100 marks. It constitutes 80 marks for final exams and 20 marks for internal assessments. Students should know this assessment before they start preparing for the final exams for the current session. Also, students must-have a depth understanding of the basic concepts of each chapter and their respective formulas.

• Chapter 1 – Number Systems 
• Chapter 2 – Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 – Linear Equations in Two Variables
• Chapter 5 – Introduction to Euclids Geometry
• Chapter 6 – Lines and Angles
• Chapter 7 – Triangles
• Chapter 8 – Quadrilaterals
• Chapter 9 – Areas of Parallelograms and Triangles
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 12 – Heron’s Formula
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15 – Probability

In the previous section, we have already given chapters for class 9 th and their marking schemes. You should have a complete idea of each chapter before you start preparing for the final exams. These are the number system, Algebra, Geometry, Coordinate Geometry, Mensuration, Statistics, and Probability. These topics are common when you prepare for competitive exams too and they had plenty of real-life applications too. So, start your foundation today and be a front-runner in your future.

Here, are the benefits you should understand while studying important math formulas chapter-wise for class 9.

• You can score good marks in your final exams.
• You can complete the syllabus on time with confidence.
• You can revise the concepts and formulas quickly.
• This is easy to memorize formulas when they are available altogether.
• You would know your strengths and weakness. It would be great for spending more time in managing your weaknesses.
• Also, you will get a route to prepare for the competitive exams.
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• Direction Of A Vector Formula
• Probability Distribution Formula
• Quartile Formula
• Circumference of a Circle Formula
• Decay Formula
• Margin of Error Formula
• Population Mean Formula
• Infinite Geometric Series Formula
• Double Time Formula
• Linear Approximation Formula
• Cosine Formula
• Spherical Segment Formula
• Proportion Formula
• Rectangular Prism Formula
• R Squared Formula
• Triangular Prism Formula
• Statistical Significance Formula
• Difference of Squares Formula
• Vertex Formula
• Perfect Square Formula
• Vieta Formula
• Right Angle Formula
• U Substitution Formula
• Percentage Change Formula
• Regular Square Pyramid Formula
• Gross Profit Formulas
• Trajectory Formula
• Exponential Formulas
• Sample Mean Formula
• Quotient Formulas
• Real Number
• Box Formulas
• Simple Interest Formula
• X& Y Intercept Formulas
• Triangular Pyramid Formula
• Coefficient Formulas
• Distance Formula
• Arithmetic Sequence Formulas
• Weighted Mean Formula
• Parallelogram Formulas
• Multiplication Table
• Pentagon Formulas
• Volume Of Parallelepiped Formula
• Octagon Formulas
• Volume Of An Ellipsoid Formula
• Binary Formulas
• Volume Of A Square Pyramid Formula
• Binomial Formulas
• Charge Density Formula
• Isosceles Triangle Formulas
• Complex Number Power Formula
• Equilateral Triangle Formulas
• Diagonal Formula
• Triangle Formulas
• Division Formula
• Area of Cube Formulas
• Diameter Formula
• Sphere Formulas
• Coin Toss Probability Formula
• Trapezoid Formulas
• Circle Graph Formula
• Hexagon Formulas
• Chord Length Formula
• Cube Formulas
• Cofunction Formulas
• Discount Formulas
• Cofactor Formula
• Rhombus Formulas
• Cpk Cp Formula
• Cone Formulas
• Degree And Radian Measure Formula
• Equation Formulas
• Consecutive Integers Formula
• Cylinder Formulas
• Interpolation Formula
• Pythagoras Formulas
• Prime Number formula
• Matrix Formulas
• Euler Maclaurin formula
• Area of Circle Formulas
• Frequency Distribution formula
• Circle Formulas
• Hypergeometric Distribution formula
• Compound Interest Formulas
• Implicit Differentiation formula
• Completing the Square Formulas
• Inverse Function formula
• Square Formulas
• Inverse Hyperbolic Functions formula
• Factorial Formulas
• Pearson Correlation formula
• Statistics Formulas
• Confidence Interval formula
• Linear Formulas
• Lagrange Interpolation formula
• Derivative Formulas
• Hypothesis Testing formula
• Differentiation Formulas
• Kite formula
• Pyramid Formulas
• Degrees of Freedom formula
• Rectangle Formulas
• Interquartile Range formula
• Probability Formulas
• Function Notation formula
• Mean Median Mode Formulas
• Gaussian Distribution formula
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• Ellipse Formulas
• What Is Numbers?
• Permutation & Combination Formulas
• Area under the Curve Formula
• Complex Number Formulas
• Axis of Symmetry Formula
• Calculus Formulas
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• Trigonometry Formulas
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• Chain Rule Formula
• Geometry Formulas
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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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Hypothesis Testing Formula

Statistics is a discipline of applied mathematics that deals with gathering, describing, analyzing, and inferring conclusions from numerical data. Differential and integral calculus, linear algebra, and probability theory are all used substantially in statistics’ mathematical theories. Statisticians are especially interested in learning how to derive valid conclusions about big groups and general occurrences from the behavior and other observable features of small samples. These small samples reflect a subset of a larger group or a small number of occurrences of a common occurrence.

Table of Content

What is Hypothesis Testing in Statistics?

Hypothesis testing definition, steps in hypothesis testing, types of hypothesis testing, hypothesis testing z test, hypothesis testing t test, hypothesis testing chi square.

Hypothesis testing is a statistical procedure in which an analyst verifies a hypothesis about a population parameter. The analyst’s approach is determined by the type of the data and the purpose of the study. The use of sample data to assess the validity of a hypothesis is known as hypothesis testing . Such information might originate from a wider population or a data-gathering mechanism.

Hypothesis testing is a statistical method used to figure out if the results of an experiment actually mean something. It works by creating two different guesses: one is called the ‘null hypothesis’ and the other is the ‘alternative hypothesis.’ These two guesses never overlap. For example, we might use hypothesis testing to see if a new medicine is better at treating a disease. If the null hypothesis is true, then the alternative hypothesis must be false, and vice versa.

Step 1: Identifying the research questions and hypotheses is the first stage. Keep in mind that these are mutually incompatible options. If one theory asserts a truth, the other must contradict it.

Step 2: Consider statistical assumptions such as observation independence from one another, normality of data, random mistakes and their probability distribution , randomization during sampling, and so on.

Step 3: The third step involves deciding on the test that will be used to verify the hypothesis. At the same time, we need to figure out how we’ll test the null hypothesis with sample data.

Step 4: The data from a sample is evaluated in the fourth stage. It’s when we look for scores such as mean values, normal distributions , t distributions , and z scores , among other things.

Step 5: The final stage entails deciding whether there shall be a rejection of the null hypothesis in favour of the alternative or not to reject it.

We use a hypothesis test to see if the evidence in a sample data set is sufficient to establish that research conditions are true or untrue for the full population. A Z-test is used to determine the assumption of a given sample. Normally, we compare two sets in hypothesis testing by comparing them to a synthesized data set and an idealized model.

[Tex]z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] where, [Tex]\overline{x} [/Tex]  is the sample mean, μ represents the population mean,  σ is the standard deviation and  n is the size of the sample.

Check this Hypothesis Testing Short Lesson: Click Here ✅

When you want to test a hypothesis, you might feel lost about which test to choose. These tests help us figure out if our idea about something is likely true or not. Here are some important tests we use for this:

This is for big groups (more than 30 people). We use it to see if there’s a difference between what we think the whole group is like and what we found in a smaller part of the group. We can also compare two smaller groups. To do this test, we use some formulas.

This one’s for smaller groups (less than 30 people). Like the Z test, we use it to compare what we found in a small group with what we think the whole group is like. But here, we don’t know the exact numbers for the whole group, so we use what we found in the small group instead. Again, we can compare two small groups.

This test helps us see if things in a big group are connected or if they happen randomly. We use it when the numbers we get from our test follow a special pattern.

Check: Hypothesis in Machine Learning

Sample Problems

Question 1. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 600, 533, 6 and 140.

Given:  [Tex]\overline{x} [/Tex] = 600 μ = 533 σ = 6 n = 140 As per the formula for hypothetical testing,  [Tex]z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{600-533 }{\frac{6 }{\sqrt{140}}} [/Tex] ⇒ z = 132.125

Question 2. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 600, 585, 100 and 150.

Given:  [Tex]\overline{x}  [/Tex]  = 600 μ = 585 σ = 100 n = 150 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{600-585 }{\frac{100 }{\sqrt{150}}} [/Tex] ⇒ z = 1.837

Question 3. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 600, 577, 77 and 140.

Given:  [Tex]\overline{x}  [/Tex]  = 600 μ = 577 σ = 77 n = 140 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{600-577 }{\frac{77 }{\sqrt{140}}} [/Tex] ⇒ z = 2.765

Question 4. Conduct the z test if the sample means, the population mean , standard deviation and sample size are given to be 600, 456, 77 and 140.

Given: \overline{x} = 600 μ = 456 σ = 77 n = 140 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{600-456 }{\frac{77 }{\sqrt{140}}} [/Tex] ⇒ z = 2.987

Question 5. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 600, 533, 45 and 120.

Given: \overline{x} = 410 μ = 256 σ = 45 n = 120 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{410-256 }{\frac{45 }{\sqrt{120}}} [/Tex] ⇒ z = 6.879

Question 6. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 322, 125, 6 and 140.

Given: \overline{x} = 322 μ = 125 σ = 6 n = 15 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{322-125 }{\frac{6 }{\sqrt{15}}} [/Tex] ⇒ z = 4.9765

Question 7. Conduct the z test if the sample means, the population mean, standard deviation and sample size are given to be 600, 533, 6 and 120.

Given: \overline{x} = 600 μ = 533 σ = 6 n = 120 As per the formula for hypothetical testing, z =  [Tex]\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} [/Tex] z =  [Tex]\frac{600-533 }{\frac{6 }{\sqrt{120}}} [/Tex] ⇒ z = 142.15

Check: Understanding Hypothesis Testing

Hypothesis Testing – FAQs

What is hypothesis testing in statistics.

Hypothesis testing is a statistical method that uses sample data to evaluate a hypothesis about a population parameter.

How do I formulate a null hypothesis?

The null hypothesis (H0) is a statement of no effect or no difference and is usually the hypothesis that sample observations result purely from chance.

What’s the difference between type I and type II errors?

A type I error occurs when the null hypothesis is true but is incorrectly rejected. A type II error happens when the null hypothesis is false but erroneously fails to be rejected.

How do I choose the right significance level for a test?

The significance level, often denoted as alpha, is typically set at 0.05 (5%), but it can vary based on the context of the study and the consequences of decision errors.

What are p-values and how are they used in hypothesis testing?

A p-value is the probability of observing test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis.

How do I perform a hypothesis test for a mean?

To test a hypothesis about a mean, collect your sample data, calculate the sample mean, and use a t-test or z-test to compare it to the hypothesized population mean, depending on the sample size and variance known.

Can I use hypothesis testing for proportions?

Yes, hypothesis testing can be applied to proportions. Use a z-test for proportions to determine if the observed proportion significantly differs from the expected proportion under the null hypothesis.

What is the power of a hypothesis test?

The power of a hypothesis test is the probability that the test correctly rejects a false null hypothesis (1 minus the probability of a Type II error). Increasing sample size or significance level can increase power.

When should I use a one-tailed test vs. a two-tailed test?

Use a one-tailed test when the research hypothesis specifies a direction of the effect. Use a two-tailed test when the direction of the effect is not specified, as it checks for any significant difference, regardless of direction.

What are common software tools for performing hypothesis testing?

Common software tools include R, Python (with libraries like SciPy and StatsModels), SPSS, and SAS, which provide extensive capabilities for conducting various statistical tests.

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Notes of Ch 9 Force and Laws of Motion| Class 9th Science

Study material and notes of ch 9 force and laws of motion class 9th science.

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NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion – Here are all the NCERT solutions for Class 9 Science Chapter 9. This solution contains questions, answers, images, step by step explanations of the complete Chapter 9 titled Force and Laws of Motion of Science taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 9 Force and Laws of Motion. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion in one place. For a better understanding of this chapter, you should also see Chapter 9 Force and Laws of Motion Class 9 notes , Science.

Topics and Sub Topics in Class 9 Science Chapter 9 Force and Laws of Motion:

• Force and Laws of Motion
• Balanced and Unbalanced Forces
• First Law of Motion
• Inertia and Mass
• Second Law of Motion
• Third Law of Motion
• Conservation of Momentum

These solutions are part of NCERT Solutions for Class 9 Science . Here we have given Class 9 NCERT Science Text book Solutions for Chapter 9 Force and Laws of Motion.

In-Text Questions Solved

NCERT Textbook for Class 9 Science – Page 118 Question 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin? Answer: (a) A stone of the same size (b) a train (c) a five-rupees coin As the mass of an object is a measure of its inertia, objects with more mass have more inertia.

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Formulae Handbook for Class 9 Maths and Science Educational Loans in India

Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Answer:  When the tree’s branch is shaken vigorously the branch attain motion but the leaves stay at rest. Due to the inertia of rest, the leaves tend to remain in its position and hence detaches from the tree to fall down.

Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Answer:  When a moving bus brakes-to a stop: When the bus is moving, our body is also in motion, but due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion. When the bus accelerates from rest we fall backwards: When the bus’ is stationary our body is at rest but when the bus accelerates, the lower part of our body being in contact with the floor of the bus comes in motion, but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction.

Class 9 Science NCERT Textbook – Page 126-127 Question 1. If action is always equal to the reaction, explain how a horse can pud a cart? Answer:  The third law of motion states that action is always equal to the reaction but they act on two different bodies. In this case the horse exerts a force on the ground with its feet while walking, the ground exerts an equal and opposite force on the feet of the horse, which enables the horse to move forward and the cart is pulled by the horse.

Question 2. Explain, why is it difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity. Answer:  The water that is ejected out from the hose in the forward direction comes out with a large momentum and equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward. It becomes difficult for a fireman to hold a hose which experiences this large momentum.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (Hindi Medium)

Extra Questions from NCERT Textbook for Class 9 Science

Question 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. Answer:  When an object experiences a net zero external unbalanced force, in accordance with second law of motion its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in same direction with same speed. It means that the object may be travelling with a non-zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain. Answer:  The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest the dust particles retain their position of rest and falls down due to gravity.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Answer:  In moving vehicle like bus, the motion is not uniform, the speed of vehicle varies and it may apply brake suddenly or takes sudden turn. The luggage will resist any change in its state of rest or motion, due to inertia and this luggage has the tendency to fall sideways, forward or backward. To avoid the fall of the luggage, it is tied with the rope.

Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. Answer:  (c) there is a force 6n the ball opposing the motion.

Question 9.What is the momentum of an object of mass m, moving with a velocity v? . (a) (mv) 2           (b) mv 2                     (c) 1/2 mv 2                   (d) mv Answer:  (d) mv

Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. Answer:  The mass of truck is too large and hence its inertia is too high. The small force exerted on the truck cannot move it and the truck remains at rest. For the truck to attain motion, an external large amount of unbalanced force need to be exerted on it.

Question 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. Answer:  Rahul gave the correct reasoning and explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum. When 2 bodies collide: Initial momentum before collision = Final momentum after collision m 1  u 1 + m 2  u 2  = m 1  v 1 + m 2  v 2 The equal force is exerted on both the bodies but, because the mass of insect is very small it will suffer greater change in velocity.

Additional Exercises

Question 5. A large truck and a car, both moving with a velocity of magnitude v, have a head- on collision and both of them come to a halt after that. If the collision lasts for 1 s: (a) Which vehicle experiences the greater force of impact? (b) Which vehicle experiences the greater change in momentum? (c) Which vehicle experiences the greater acceleration? (d) Why is the car likely to suffer more damage than the truck? Answer: (a) During head on collision forces applied by truck and car are action-reaction forces. Hence both vehicles experience same (equal) force of impact. (b) Here initial velocity of both car and truck is same equal to v and final velocity of both is zero. But mass of truck is much more than that of car, hence change in momentum of truck is more than change in momentum of car. (c) For same force of impact, the acceleration of car will have greater magnitude because its mass is less. (d) Car suffers more damage than the truck, as acceleration of car is more, its velocity falls to zero in a shorter time and consequently, its momentum changes in a shorter time.

More Questions Solved

NCERT Solutions for Class 9 Science Chapter 9 Multiple Choice Questions

NCERT Solutions for Class 9 Science Chapter 9 Very Short Answer Type Questions

Question 1. Define force. Answer:  It is a push or pull on an object that produces acceleration in the body on which it acts. 4

Question 2. What is S.I. unit of force? Answer:  S.I. unit of force is Newton.

Question 3. Define one Newton. Answer:  A force of one Newton produces an acceleration of 1 m/s 2 on an object of mass 1 kg. . 1 N = 1 kg m/s 2

Question 4. What is balanced force? Answer:  When forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.

Question 5. What is frictional force? Answer:  The force that always opposes the motion of object is called force of friction.

Question 6. What is inertia? Answer:  The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia.

Question 7. State Newton’s first law of motion. Answer:  An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Question 8. State Newton’s second law of motion. Answer:  The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Question 9. What is momentum? Answer:  The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. The S. I. unit is kg m/s. (p = mv)

Question 10. State Newton’s III law of motion. Answer:  To every action, there is an equal and opposite reaction and they act on two different bodies.

Question 11. Which will have more inertia a body of mass 10 kg or a body of mass 20 kg? Answer:  A body of mass 20 kg will have more inertia.

Question 12. Name the factor on which the inertia of the body depends. Answer:  Inertia of a body depends upon the mass of the body.

Question  13. Name two factors which determine the momentum of a body. Answer:  Two factors on which momentum of a body depend is mass and velocity. Momentum is directly proportional to the mass and velocity of the body.

Question 14. What decides the rate of change of momentum of an object? Answer:  The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

NCERT Solutions for Class 9 Science Chapter 9 Short Answer Type Questions

Question 2. What change will force bring in a body? Answer:  Force can bring following changes in the body:

• It can change the speed of a body.
• It can change the direction of motion of a body,
• It can change the shape of the body.

Question 3. When a motorcar makes a sharp turn at a high speed, we tend to get thrown to one side. Explain why? Answer:  It is due to law of inertia. When we are sitting in car moving in straight line, we tend to continue in our straight-line motion. But when an unbalanced force is applied by the engine to change the direction of motion of the motorcar. We slip to one side of the seat due to the inertia of our body.

Question 4. Explain why it is dangerous to jump out of a moving bus. Answer:  While moving in a bus our body is in motion. On jumping out of a moving bus our feet touches the ground and come to rest. While the upper part of our body stays in motion and moves forward due to inertia of motion and hence we can fall in forward direction. Hence, to avoid this we need to run forward in the direction of bus.

Question 5. Why do fielders pull their hand gradually with the moving ball while holding a catch? Answer:  While catching a. fast moving cricket ball, a fielder on the ground gradually pulls his hands backwards with the moving ball. This is done so that the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact of catching the fast moving ball is reduced.

Question 6. In a high jump athletic event, why are athletes made to fall either on a cushioned bed or on a sand bed? Answer:  In a high jump athletic event, athletes are made to fall either on a cushioned bed or on a sand bed so as to increase the time of the athlete’s fall to stop after making the jump. This decreases the rate of change of momentum and hence the force.

Question 7. How does a karate player breaks a slab of ice with a single blow? Answer:  A karate player applied the blow with large velocity in a very short interval of time on the ice slab which therefore exerts large amount of force on it and suddenly breaks the ice slab.

Question 8. What is law of conservation of momentum? Answer:  Momentum of two bodies before collision is equal to the momentum after collision. In an isolated system, the total momentum remain conserved.

Question 9. Why are roads on mountains inclined inwards at turns? Answer:  A vehicle moving on mountains is in the inertia of motion. At a sudden turn there is a tendency of vehicle to fall off the road due to sudden change in the line of motion hence the roads are inclined inwards so that the vehicle does not fall down the mountain.

Question 10. For an athletic races why do athletes have a special posture with their right foot resting on a solid supporter? Answer:  Athletes have to run the heats and they rest their foot on a solid supports before start so that during the start of the race the athlete pushes the support with lot of force and this support gives him equal and opposite push to start the race and get a good start to compete for the race.

Question 11.Why do you think it is necessary to fasten your seat belts while travelling in your vehicle? Or How are safety belts helpful in preventing any accidents? Answer:  While we are travelling in a moving car, our body remains in the state of rest with respect to the seat. But when driver applies sudden breaks or stops the car our body tends to continue in the same state of motion because of its inertia. Therefore, this sudden break may cause injury to us by impact or collision. Hence, safety belt exerts a force on our body to make the forward motion slower.

Question 13. When you kick a football it flies away but when you kick a stone you get huh why? Answer:  This is because stone is heavier than football and heavier objects offer larger inertia. When we kick a football its mass is less and inertia is also less so force applied by our kick acts on it and hence it shows larger displacement but in case of stone, it has larger mass and offers larger inertia. When we kick (action) the stone it exerts an equal and opposite force (reaction) and hence it hurts the foot.

Question 14. If a person jumps from a height on a concrete surface he gets hurt. Explain. Answer:  When a person jumps from a height he is in state of inertia of motion. When he suddenly touches the ground he comes to rest in a very short time and hence the force exerted by the hard concrete surface on his body is very high, and the person gets hurt.

Question 15. What is the relation between Newton’s three laws of motion? Answer:  Newton’s first law explains about the unbalanced force required to bring change in the position of the body. Second law states/explains about the amount of force required to produce a given acceleration. And Newton’s third law explains how these forces acting on a body are interrelated.

Question 16. Give any three examples in daily life which are based on Newton’s third law of motion. Answer:  Three examples based on Newton’s third law are :

• Swimming: We push the water backward to move forward. action – water is pushed behind reaction – water pushes the swimmer ahead
• Firing gun: A bullet fired from a gun and the gun recoils. action – gun exerts force on the bullet reaction – bullet exerts an equal and opposite force on the gun
• Launching of rocket action – hot gases from the rocket are released reaction – the gases exert upward push to the rocket

NCERT Solutions for Class 9 Science Chapter 9 Long Answer Type Questions

Question 2. State all 3 Newton’s law of motion. Explain inertia and momentum. Answer: Newton’s I law of motion: An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force. Newton’s II law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the-force. Newton’s III law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies. Inertia: The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. Momentum: The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its S.I. unit is kgm/s. p = m x v

Question 3. Define force. Give its unit and define it. What are different types forces? Answer:   Force: It is a push or pull on an object that produces acceleration in the body on which it acts. A force can do 3 things on a body (a) It can change the speed of a body. (b) It can change the direction of motion of a body. (c) It can change the shape of the body. The S.I. unit of force is Newton. Newton: A force of one Newton produces an acceleration of 1 m/s 2 on an object of mass 1 kg. 1N = 1kg m/s 2 Types of forces:

• Balanced force: When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
• Unbalanced force: When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced force.
• Frictional force: The force that always opposes the motion of object is called force of friction.

Question 4. What is inertia? Explain different types of inertia. Give 3 examples in daily life which shows inertia. Answer: Inertia: The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg. Types of inertia: Inertia of rest: The object at rest will continue to remain at rest unless acted upon by an external unbalanced force. Inertia of motion: The object in the state of uniform motion will continue to remain in motion with same speed and direction unless it is acted upon by an external unbalanced force. . Three examples of inertia in daily life are:

• When we are travelling in a vehicle and sudden brakes are .applied we tend to fall forward.
• When we shake the branch of a tree vigorously, leaves fall down.
• If we want to remove the dust from carpet we beat the carpet so that dust fall down.

NCERT Solutions for Class 9 Science Chapter 9 Activity-Based Questions

Question 1.

• Make a pile of similar carom coins on a table, as shown in the figure.

Question 2.

• Set a five-rupees coin on a stiff card covering an empty glass tumbler standing on a table as shown in the figure.
• Give the card a sharp horizontal flick with a finger. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.

Answer:  The force applied on the card due to flicking changes the inertia of the card but the coin resist a change and stay at the rest i.e. inertia of rest and due to gravity falls down in the tumbler.

Question 3.

• Place a water-filled tumbler on a tray.
• Hold the tray and turn around as fast as you can.
• We observe that the water spills. Why?

Answer:  The water-filled in tumbler and tray are at rest. On moving/turning around the tray at faster speed the water spills because the tray and the tumbler comes into motion while the water in the tumbler remain at inertia of rest.

Question 4.

• Request two children to stand on two separate carts as shown on the next page.
• Give them a bag full of sand or some other heavy object. Ask them to play a game of catch with the bag.
• Does each of them receive an instantaneous reaction as a result of throwing the sand bag (action)?
• You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Question 5.

• Take a big rubber balloon and inflate it fully. Tie its neck using a thread. Also using adhesive tape, fix a straw on the surface of this balloon.
• Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.
• Ask your friend to hold the other end of the thread or fix it on a wall at some distance. The arrangement is shown in the figure below.
• Now remove the thread tied on the neck of balloon. Let the air escape from the mouth of the balloon.

Question 6.

• Take a test tube of good quality glass material and put a small amount of water in it. Place a stop cork at the mouth of it.
• Now suspend the test tube horizontally by two strings or wires as shown in the figure on next page.
• Heat the test tube with a burner until water vaporises and the cork blows out.

NCERT Solutions for Class 9 Science Chapter 9 Value-Based Questions

Question 1. Class V students were playing cricket with the cork hall in the school campus. Charu a senior student told them about the accidents that can occur due to cork ball in the campus and also advised them to bring soft cosco ball to play the game. (a) Why it was safe to play with soft ball and not with hard cork ball? (b) A player pulls his hands backwards after holding the ball shot at high speed. Why? (c) What value of Charu is seen in this act? Answer: (a) The soft ball will have less inertia as compared to the heavy ball and it would not hurt the players. (b) By pulling the hand backwards it reduces the force exerted by the ball on hands. (c) Charu showed the value of being responsible and helpful by nature.

Question 2. Saksham saw his karate expert friend breaking a slate. He tried to break the slate but Saksham’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing so. (a) How can a karate expert break the slate without any injury to his hand? (b) What is Newton’s third law of motion? (c) What value of Saksham’s friend, is seen in the above case? Answer: (a) A karate player applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks. (b) To every action there is an equal and opposite reaction, both act on different bodies. Saksham’s friend showed the value of being responsible and caring friend.

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NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula

Ncert solutions class 9 maths chapter 12 – download free pdf.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula is provided here. Heron’s formula is a fundamental concept that finds significance in countless areas and is included in the CBSE Syllabus of Class 9 Maths. Therefore, it is imperative to have a clear grasp of the concept. And one of the best ways to do just that is by referring to the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula. These solutions are designed by knowledgeable teachers with years of experience in accordance with the latest update on the CBSE syllabus 2023-24. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.

Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 12 Heron’s Formula

Download most important questions for class 9 maths chapter – 12 heron’s formula.

Hence, one of the best guides you could adapt to your study needs is NCERT Solutions . Relevant topics are presented in an easy-to-understand format, barring the use of any complicated jargon. Furthermore, its content is updated as per the last CBSE syllabus and its guidelines.

• Chapter 1 Number System
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclids Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Introduction to Probability

Download PDF of NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula

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List of Exercises in NCERTclass 9 Maths Chapter 12:

• Exercise 12.1 Solutions – 6 Questions
• Exercise 12.2 Solutions – 9 Questions

Access Answers of NCERT Class 9 Maths Chapter 12 – Heron’s Formula

Exercise: 12.1 (Page No: 202)

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Side of the signal board = a

Perimeter of the signal board = 3a = 180 cm

∴ a = 60 cm

Semi perimeter of the signal board (s) = 3a/2

By using Heron’s formula,

Area of the triangular signal board will be =

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m 2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.

Now, the perimeter will be (122+22+120) = 264 m

Also, the semi perimeter (s) = 264/2 = 132 m

Using Heron’s formula,

Area of the triangle =

We know that the rent of advertising per year = ₹ 5000 per m 2

∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

It is given that the sides of the wall as 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Area of the message =

= √[16(16-15)(16-11) (16-6)] m 2

= √[16×1×5×10] m 2 = √800 m 2

4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.

Assume the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42cm

So, x = 42-(18+10) cm = 14 cm

∴ The semi perimeter of triangle = 42/2 = 21 cm

Area of the triangle,

= √[21(21-18)(21-10)(21-14)] cm 2

= √[21×3×11×7] m 2

= 21√11 cm 2

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

The ratio of the sides of the triangle are given as 12 : 17 : 25

Now, let the common ratio between the sides of the triangle be “x”

∴ The sides are 12x, 17x and 25x

It is also given that the perimeter of the triangle = 540 cm

12x+17x+25x = 540 cm

54x = 540cm

Now, the sides of triangle are 120 cm, 170 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 540/2 = 270 cm

Area of the triangle

= 9000 cm 2

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

First, let the third side be x.

It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.

So, 30 = 12+12+x

∴ The length of the third side = 6 cm

Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm

= √[15(15-12)(15-12)(15-6)] cm 2

= √[15×3×3×9] cm 2

= 9√15 cm 2

Exercise: 12.2 (Page No: 206)

1. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

First, construct a quadrilateral ABCD and join BD.

We know that

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

BD 2 = BC 2 +CD 2

⇒ BD 2 = 12 2 +5 2

⇒ BD 2 = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (½ ×12×5) = 30 m 2

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Area of ΔABD

= 6√35 m 2 = 35.5 m 2 (approximately)

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m 2 +35.5m 2 = 65.5 m 2

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC 2 = AB 2 +BC 2

⇒ 5 2 = 3 2 +4 2

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (½ ×3×4) = 6 cm 2

The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Now, using Heron’s formula,

Area of ΔACD

= 2√21 cm 2 = 9.17 cm 2 (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 6 cm 2 +9.17 cm 2 = 15.17 cm 2

3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

For the triangle I section:

It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5+5+1 = 11 cm

So, semi perimeter = 11/2 cm = 5.5 cm

= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm 2

= √[5.5×0.5×0.5×4.5] cm 2

= 0.75√11 cm 2

= 0.75 × 3.317cm 2

= 2.488cm 2 (approx)

For the quadrilateral II section:

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

∴ Area = 6.5×1 cm 2 =6.5 cm 2

For the quadrilateral III section:

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

And, the area of the equilateral triangle will be (√3/4×a 2 ) = 0.43

∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm 2 (approximately).

For triangle IV and V:

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2×(½×6×1.5) cm 2 = 9 cm 2

So, the total area of the paper used = (2.488+6.5+1.3+9) cm 2 = 19.3 cm 2

4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

It is given that the parallelogram and triangle have equal areas.

The sides of the triangle are given as 26 cm, 28 cm and 30 cm.

So, the perimeter = 26+28+30 = 84 cm

And its semi perimeter = 84/2 cm = 42 cm

Now, by using Heron’s formula, area of the triangle =

= √[42(42-26)(42-28)(42-30)] cm 2

= √[42×16×14×12] cm 2

Now, let the height of parallelogram be h.

As the area of parallelogram = area of the triangle,

28 cm× h = 336 cm 2

∴ h = 336/28 cm

So, the height of the parallelogram is 12 cm.

5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Area of the ΔBCD =

∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m 2 = 864 m 2

Thus, the area of the grass field that each cow will be getting = (864/18) m 2 = 48 m 2

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

For each triangular piece, The semi perimeter will be

s = (50+50+20)/2 cm = 120/2 cm = 60cm

Area of the triangular piece

= √[60(60-50)(60-50)(60-20)] cm 2

= √[60×10×10×40] cm 2

= 200√6 cm 2

∴ The area of all the triangular pieces = 5 × 200√6 cm 2 = 1000√6 cm 2

7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

As the kite is in the shape of a square, its area will be

A = (½)×(diagonal) 2

Area of the kite = (½)×32×32 = 512 cm 2.

The area of shade I = Area of shade II

512/2 cm 2 = 256 cm 2

So, the total area of the paper that is required in each shade = 256 cm 2

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm

By using Heron’s formula, the area of the III triangular piece will be

= √[10(10-6)(10-6)(10-8)] cm 2

= √(10×4 ×4×2) cm 2

= 8√5 cm 2 = 17.92 cm 2 (approx.)

8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm 2 .

The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm

The area of each triangular shape will be

= 36√6 cm 2 = 88.2 cm 2

Now, the total area of 16 tiles = 16×88.2 cm 2 = 1411.2 cm 2

It is given that the polishing cost of tiles = 50 paise/cm 2

∴ The total polishing cost of the tiles = Rs. (1411.2×0.5) = Rs. 705.6

9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25-ED = 25-10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+14+15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

We also know that the area of ΔBEC = (½)×CE×BF

84 cm 2 = (½)×15×BF

BF = (168/15) cm = 11.2 cm

So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m 2

∴ Area of the field = 84+112 = 196 m 2

Chapter 12 Heron’s Formula belongs to Unit 5: Mensuration. This unit carries a total of 13 marks out of 100. Therefore, this is an important chapter and should be studied thoroughly. The important topics that are covered in this chapter are:

• Area of a Triangle – by Heron’s Formula
• Application of Heron’s Formula in Finding Areas of Quadrilaterals

Heron’s formula helps us to find the area of a triangle with 3 side lengths. Besides the formula, Heron also contributed in other ways – the most notable one being the inventor of the very first steam engine called the Aeolipile. However, Heron couldn’t find any practical applications for it; instead, it ended up being used as a toy and an object of curiosity for the ancient Greeks.

Explore more about Heron’s Formula and learn how to solve various kinds of problems only on NCERT Solutions For Class 9 Maths . It is also one of the best academic resources to revise for your exams.

Key Features of NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

• Well-structured content
• Relevant formulas are highlighted
• Easy-to-understand language and jargon-free explanations
• Designed by qualified teachers
• Latest questions with solutions from the updated syllabus
• A thorough analysis of previous year question papers
• Access to other learning resources, such as sample papers and more

The expert faculty team of members have formulated the solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Heron’s Formula, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 9 Maths Chapter 12 – Heron’s Formula

Disclaimer:

Dropped Topics –  12.1 Introduction and 12.3 Application of Heron’s formula in finding areas of quadrilaterals.

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