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CBSE Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 12  are very important to solve for your exam. Class 9 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  case study-based   questions for Class 9 Maths Chapter 12  Heron’s Formula

Case Study Questions Class 9 Maths Chapter 12

Case Study 1: A group of students is learning about Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Rohan and Kavya came across a triangular field in their village. They made the following observations:

• The lengths of the three sides of the triangular field are 8 meters, 12 meters, and 15 meters.
• The perimeter of the triangular field is 35 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular field. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The semiperimeter of the triangular field is: (a) 8 meters (b) 12 meters (c) 15 meters (d) 17.5 meters

Answer: (d) 17.5 meters

Q2. Using Heron’s Formula, the area of the triangular field is: (a) 24 square meters (b) 30 square meters (c) 36 square meters (d) 40 square meters

Answer: (b) 30 square meters

Q3. The type of triangle formed by the sides of the field is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Answer: (c) Scalene

Q4. The length of the altitude corresponding to the side of 15 meters is: (a) 2 meters (b) 4 meters (c) 6 meters (d) 8 meters

Answer: (c) 6 meters

Q5. The lengths of the altitudes corresponding to the sides of 8 meters and 12 meters are: (a) 4 meters and 6 meters (b) 6 meters and 8 meters (c) 8 meters and 10 meters (d) 10 meters and 12 meters

Answer: (a) 4 meters and 6 meters

Case Study 2: A group of students is studying Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Neha and Mohan went on a field trip to a riverbank. They noticed a triangular piece of land that they wanted to measure and calculate its area. They made the following observations:

• Neha measured the lengths of the three sides of the triangular piece of land as 7 meters, 9 meters, and 11 meters.
• Mohan measured the lengths of the three sides of the same triangular piece of land as 10 meters, 12 meters, and 15 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular piece of land. Let’s see if you can answer the questions correctly:

Q1. Using Neha’s measurements, the semiperimeter of the triangular piece of land is: (a) 13 meters (b) 16 meters (c) 19 meters (d) 23 meters

Answer: (c) 19 meters

Q2. Using Neha’s measurements, the area of the triangular piece of land is: (a) 24 square meters (b) 26 square meters (c) 28 square meters (d) 30 square meters

Answer: (a) 24 square meters

Q3. Using Mohan’s measurements, the semiperimeter of the triangular piece of land is: (a) 16 meters (b) 18 meters (c) 21 meters (d) 25 meters

Answer: (c) 21 meters

Q4. Using Mohan’s measurements, the area of the triangular piece of land is: (a) 40 square meters (b) 42 square meters (c) 45 square meters (d) 48 square meters

Answer: (b) 42 square meters

Q5. The measurements taken by Neha represent a triangle that is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula with Answers Pdf free download has been useful to an extent. If you have any other queries about Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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Test: Heron`s Formula- Case Based Type Questions - Class 9 MCQ

10 questions mcq test - test: heron`s formula- case based type questions, direction: isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. q. what is the length of equal sides.

So, x + x + 4 = 20

2x + 4 = 20

2x = 20 – 4

x = 16/2 = 8 cm.

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. Q. If the sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.

100√2 m 2

500√2 m 2

1500√3 m 2

200√3 m 2

Let the sides of a triangle are a = 3x, b = 5x, c = 7x

Then a + b + c = 300

3x + 5x + 7x = 300

So, a = 60, b = 100, c = 140

= 300/2 = 150

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. Q. What is the Heron's formula for the area of?

{Area} = area

s = semi-perimeter

a = length of side a

b = length of side b

c = length of side c

Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. What is the semi perimeter of the Isosceles triangle?

Required semi perimeter = Perimeter/2 = 20/2 = 10 m.

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm.

Q. What is the area of highlighted triangle ?

Thus, area of the triangle

Direction: Shakshi prepared a Rangoli in triangular shape on Diwali. She makes a small triangle under a big triangle as shown in figure.

Sides of big triangle are 25 cm, 26 cm and 28 cm. Also, ΔPQR is formed by joining mid points of sides of ΔABC.

Use the above data to help her in resolving below doubts.

Q. What is the semi-perimeter of ΔABC?

= (25 + 26 + 28) cm = 79 cm

Q. Area of ΔPQR =

where s is the semi-perimeter of ΔPQR.

Q. ½ of AB =

Q. What is the length of RQ?

Q. If colourful rope is to be placed along the sides of small ΔPQR. What is the length of the rope?

= (12.5 + 13 + 14) cm

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Important Questions for Heron`s Formula- Case Based Type Questions

Heron`s formula- case based type questions mcqs with answers, online tests for heron`s formula- case based type questions, welcome back, create your account for free.

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Heron’s Formula Exercises

Heron’s formula practice problems with answers.

There are six (6) practice problems here about Heron’s Formula. The more problems you solve, the more proficient you get. Please enjoy!

For your convenience, here’s the formula.

PRACTICE PROBLEMS

Note: drawings not to scale

Problem 1: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{SRQ}[/latex] is about [latex]13[/latex] square kilometers.

Problem 2: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{ABC}[/latex] is about [latex]52[/latex] square centimeters.

Problem 3: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{PRQ}[/latex] is about [latex]41[/latex] square yards.

Problem 4: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{XYZ}[/latex] is about [latex]34[/latex] square inches.

Problem 5: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{JKL}[/latex] is about [latex]62[/latex] square feet.

Problem 6: Find the area of the triangle below using Heron’s formula.

Therefore, the area of [latex]\triangle{VUT}[/latex] is about [latex]27[/latex] square feet.

You may also be interested in these related math lessons or tutorials:

Heron’s Formula

Heron's Formula Class 9

Heron’s formula class 9 is used to determine the area of a triangle when the length of all three sides is given. This formula does not involve the use of the angles of a triangle . Heron’s Formula class 9 is a fundamental math concept applied in many fields to calculate various dimensions of a triangle. Therefore, it is crucial for students to understand this formula along with its various applications.

List of Heron's Formula Class 9 

Heron’s Formula class 9 is used to find the area of triangles and quadrilaterals. The formula is used in various ways as shown below.

• Area of a triangle using Heron’s Formula = A = √{s(s - a)(s - b)(s - c)}, where a, b and c are the length of the three sides of a triangle and s is the semi-perimeter of the triangle which is calculated with the formula, s = (a + b + c)/2.
• The area of a quadrilateral whose sides and one diagonal are given can be calculated by dividing the quadrilateral into two triangles using Heron’s formula. 
• Area of quadrilateral ABCD = √(s(s - a)(s - d)(s - e)) + √(s'(s' - b)(s' - c)(s' - e)) , Where a, d and e represent the sides of one triangle and b, c and e represent the sides of the other triangle. It should be noted that ‘e’ is the length of the diagonal which is a common side for both triangles.

Applications of Heron’s Formula Class 9

Heron’s formula is used in various calculations. A few of them are given below.

• Heron’s Formula Class 9 is applied in finding the surface area of triangular plots or agricultural lands. Since not every plot is rectangular in shape, therefore, Heron's formula is of great use in such situations to estimate the cost of such plots. 
• Heron’s Formula can be used to determine the area of any irregular quadrilateral by dividing the quadrilateral into triangles. Hence, it can be used to determine the area of irregular plots, parks, farms, etc.

Tips to Memorize Heron’s Formulas class 9

Students should follow some creative ways to memorize Heron’s formula class 9 and the concepts related to it. It will enable them to use their knowledge of this formula in various situations. It is highly beneficial in real life and for competitive exams.

• Applying and solving problems based on Heron’s Formula to calculate the area of triangles and quadrilaterals is a smart way to memorize the formula and its steps. 
• Heron’s Formula is applied in various cases to determine the area of any irregular quadrilateral. Hence students should gain a proper understanding of the basic terms related to triangles and quadrilaterals used in these formulas. 
• Students should practice multiple problems and examples given in the textbook. It will provide optimal coverage of the usage of formulas in different contexts.

Heron’s Formula Class 9 Examples

Example 1: Find the area of a triangle with the side lengths given as 4 units, 6 units, and 8 units respectively.

Solution: As we know, a = 4 units, b = 6 units and c = 8 units

Thus, Semi-perimeter, s = (a + b + c)/2 = (4 + 6 + 8)/2 = 9 units

Area of triangle = √[s(s - a)(s - b)(s - c)] = √[9(9 - 4)(9 - 6)(9 - 8)]

⇒ Area of triangle = √(9 × 5 × 3 × 1) = √135 = 11.61 unit 2

∴ The area of the triangle is 11.61 unit 2

Example 2: If the sides of a triangular field are 50 m, 52 m and 34 m find the area of the triangular field.

Solution Given, sides of the triangular field are 50 m, 52 m and 34 m 

By Heron's formula:

Area of triangle =√(s(s - a)(s - b)(s - c)); where, a, b, c are sides of triangle and semi perimeter (s) = (a + b + c)/2

S = {50 + 52 + 34} / 2 = {136} / 2= 68 m

Area of triangular field = √{68(68 - 50)(68 - 52)(68 - 34) = √68 × 18 × 16 × 34 

Students can download the printable Maths Formulas Class 9  sheet from below.

FAQs On Heron’s Formula Class 9

What is heron’s formula class 9.

Heron’s formula class 9 is used to obtain the area of a triangle when the length of its three sides is given. It states that the area of a triangle of sides a, b, c  is √s(s - a)(s - b)(s - c)  where ‘s’ is the semi perimeter = (a + b + c) /2 .

It can also be used as a way of checking whether a triangle with three given sides can be actually drawn. If the area obtained by Heron’s formula is zero or imaginary (that is, if  (s - a)(s - b)(s - c) ≤ 0), then a triangle with sides a, b, c  cannot be physically illustrated.

What are the Basic Formulas Covered Under Heron’s Formula Class 9?

Here is a list of some basic formulas covered under heron's formula class 9 that are used to find the area of a triangle and the area of a quadrilateral:

• Heron’s Formula for Area of scalene triangle  = √{s(s - a)(s - b)(s - c); where a, b, and c are the 3 sides of a triangle.
• Area of equilateral triangle = (√3 × a 2 ) / 4; where ‘a’ is the side of the triangle.
• Area of Isosceles Triangle = (b/4)√(4a 2 - b 2 ); where a is the different side and ‘b’ is the length of the two equal sides of the triangle.
• Area of quadrilateral ABCD = √(s(s - a)(s - d)(s - e)) + √(s'(s' - b)(s' - c)(s' - e)), where a, d and e represent the sides of one triangle and b, c and e represent the sides of the other triangle. It should be noted that ‘e’ is the length of the diagonal which is a common side for both triangles. 

What are the Practical Applications of Heron’s Formula Class 9?

There are many practical applications of Heron’s formula class 9. For example, the area of any irregular quadrilateral can be easily calculated by splitting the quadrilateral into triangles. Using Heron's formula, the area of these triangles can be calculated and added together to find the area of the irregular quadrilateral. By using Heron’s formula we can determine the area of irregular plots, parks, farms, etc. Heron's formula is of great use in such situations to estimate the cost of plots. 

Why is it Important to Practice Problems Based on Heron’s Formula Class 9?

Practicing problems based on Heron’s formula will benefit students to form a deep understanding of each and every term used in this topic. It will also help students to attain the problem-solving skills required for higher-level studies and competitive exams.

How to Memorize Heron’s Formula?

Memorizing Heron’s formula requires practice and perseverance. With the practice of a wide range of questions based on this formula, students will gain deep knowledge of the core concepts. They will also learn some creative ways to memorize formulas and important theorems. It will enable them to employ their knowledge of formulas in various situations, which is highly useful for facing competitive exams.

• CBSE- Herons Formula
• Sample Questions

Herons Formula-Sample Questions

• STUDY MATERIAL FOR CBSE CLASS 9 MATH
• Chapter 1 - Area of Parallelograms and Triangles
• Chapter 2 - Circles
• Chapter 3 - Constructions
• Chapter 4 - Coordinate Geometry
• Chapter 5 - Herons Formula
• Chapter 6 - Introduction to Euclids Geometry
• Chapter 7 - Linear Equations in two variables
• Chapter 8 - Lines and Angles
• Chapter 9 - Number Systems
• Chapter 10 - Polynomials
• Chapter 11 - Probability
• Chapter 12 - Quadrilaterals
• Chapter 13 - Statistics
• Chapter 14 - Surface Areas and Volumes
• Chapter 15 - Triangles

• Class 9 Maths MCQs
• Chapter 12 Herons Formula Mcq

Class 9 Maths Chapter 12 Heron’s Formula MCQs

Class 9 Maths Chapter 12 Heron’s Formula MCQs are available online here, with solved answers. The MCQs are prepared as per the latest exam pattern for Class 9. The objective questions are prepared chapter-wise, as per the CBSE syllabus (2022-2023) and NCERT curriculum. These questions are provided with detailed explanations. Get Chapter-wise Class 9 Maths MCQs at BYJU’S.

Download the below PDF to get more MCQs on Class 9 Maths Chapter 12 Heron’s Formula.

Class 9 Maths Chapter 12 Heron’s Formula MCQs – Download PDF

MCQs on Class 9 Maths Chapter 12 Heron’s Formula

Choose the correct answer and solve the MCQs on Heron’s formula.

1) Area of a triangle is equal to:

a. Base x Height

b. 2(Base x Height)

c. ½(Base x Height)

d. ½ (Base + Height)

2) If the perimeter of an equilateral triangle is 180 cm. Then its area will be:

a. 900 cm 2

b. 900√3 cm 2

c. 300√3 cm 2

d. 600√3 cm 2

Explanation: Given, Perimeter = 180 cm

3a = 180 (Equilateral triangle)

Semi-perimeter = 180/2 = 90 cm

Now as per Heron’s formula,

In the case of an equilateral triangle, a = b = c = 60 cm

Substituting these values in the Heron’s formula, we get the area of the triangle as:

= √(90× 30 × 30 × 30)

A = 900√3 cm 2

3) The sides of a triangle are 122 m, 22 m and 120 m respectively. The area of the triangle is:

a. 1320 sq.m

b. 1300 sq.m

c. 1400 sq.m

d. 1420 sq.m

Explanation: Given,

Semi-perimeter, s = (122 + 22 + 120)/2 = 132 m

Using heron’s formula:

= √(132 × 10 × 110 × 12)

= 1320 sq.m

4) The area of a triangle with given two sides 18 cm and 10 cm, respectively and a perimeter equal to 42 cm is:

a. 20√11 cm 2

b. 19√11 cm 2

c. 22√11 cm 2

d. 21√11 cm 2

Explanation: Perimeter = 42

a + b + c = 42

18 + 10 + c = 42

c = 42 – 28 = 14 cm

Semi perimeter, s = 42/2 = 21 cm

Using Heron’s formula:

= √(21 × 3 × 11 × 7)

= 21√11 cm 2

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

a. 1000 sq.cm

b. 5000 sq.cm

c. 9000 sq.cm

d. 8000 sq.cm

Explanation: The ratio of the sides is 12: 17: 25

Perimeter = 540 cm

Let the sides of the triangle be 12x, 17x and 25x.

12x + 17x + 25x = 540 cm

54x = 540 cm

a = 12x = 12 × 10 = 120

b = 17x = 17 × 10 = 170

c = 25x = 25 × 10 = 250

Semi-perimeter, s = 540/2 = 270 cm

= √(270 × 150 × 100 × 20)

= 9000 sq.cm

6) The equal sides of the isosceles triangle are 12 cm, and the perimeter is 30 cm. The area of this triangle is:

a. 9√15 sq.cm

b. 6√15 sq.cm

c. 3√15 sq.cm

d. √15 sq.cm

Perimeter = 30 cm

Semiperimeter, s = 30/2 = 15 cm

a = b = 12 cm

a + b + c = 30

12 + 12 + c = 30

c = 30 – 24 = 6 cm

= √(15 × 3 × 3 × 9)

= 9√15 sq.cm

7) A quadrilateral whose sides are 3 cm, 4 cm, 4 cm, 5 cm and one of the diagonal is equal to 5 cm as per the below figure. The area of the quadrilateral is:

a. 19.17 sq.cm

b. 15.17 sq.cm

c. 20.17 sq.cm

d. 22.17 sq.cm.

Explanation: Using Pythagoras theorem, in ΔABC,

AC 2 = AB 2 + BC 2

⇒ 5 2 = 3 2 + 4 2

Hence, ABC is a right triangle.

Area of ΔABC = ½ x 3 x 4 = 6 sq.cm

Semiperimeter of ΔACD = s = (5+5+4)/2 = 14/2 = 7 cm

Area of ΔACD can be determined by using Heron’s formula.

= √(7 × 2 × 2 × 3)

= 9.17 sq.cm

Therefore, the area of quad.ABCD = Area of ΔABC + Area of ΔACD = (6 + 9.17) sq.cm = 15.17 sq.cm

8) The area of an equilateral triangle having side length equal to √3/4 cm (using Heron’s formula) is:

a. 2/27 sq.cm

b. 2/15 sq.cm

c. 3√3/64 sq.cm

d. 3/14 sq.cm

Explanation: Here, a = b = c = √3/4

Semiperimeter = (a + b + c)/2 = 3a/2 = 3√3/8 cm

Using Heron’s formula,

= 3√3/64 sq.cm

9) The sides of a parallelogram are 100 m each and the length of the longest diagonal is 160 m. The area of a parallelogram is:

a. 9600 sq.m

b. 9000 sq.m

c. 9200 sq.m

d. 8800 sq.m

Explanation: The diagonal divides the parallelogram into two equivalent triangles. Hence, its area will be equal to the sum of the area of the two triangles.

Thus, the sides of one triangle will be 100 m, 160 m, and 100 m.

So, a = 100 m, b = 160 m, c = 100 m

Semiperimeter (s) = (a + b + c)/2 = (100 + 160 + 100)/2 = 360/2 = 180 m

= √(180 × 80 × 20 × 80)

= 4800 sq.m

Thus, the area of the parallelogram = 2 × 4800 sq.m = 9600 sq.m

10) The sides of a triangle are in the ratio of 3: 5: 7 and its perimeter is 300 cm. Its area will be:

a. 1000√3 sq.cm

b. 1500√3 sq.cm

c. 1700√3 sq.cm

d. 1900√3 sq.cm

Explanation:

The ratio of the sides is 3: 5: 7

Perimeter = 300 cm

Let the sides of the triangle be 3x, 5x and 7x.

3x + 5x + 7x = 300 cm

15x = 300 cm

a = 3x = 3 × 20 = 60

b = 5x = 5 × 20 = 100

c = 7x = 7 × 20 = 140

Semi-perimeter, s = 300/2 = 150 cm

= √(150 × 90 × 50 × 10)

= 1500√3 sq.cm

11) The base of a right triangle is 8 cm and the hypotenuse is 10 cm. Its area will be

(a) 24 cm 2

(b) 40 cm 2

(c) 48 cm 2

(d) 80 cm 2

Given: Base = 8 cm and Hypotenuse = 10 cm

Hence, height = √[(10 2 – 8 2 ) = √36 = 6 cm

Therefore, area = (½)×b×h = (½)×8×6 = 24 cm 2 .

12) The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm 2 is

(a) Rs 2.00

(b) Rs 2.16

(c) Rs 2.48

(d) Rs 3.00

Given: a = 6 cm, b = 8 cm, c = 10 cm.

s = (6 + 8 + 10)/2 = 12 cm

Hence, by using Heron’s formula, we can write:

A = √[12(12 – 6)(12 – 8)(12 – 10)] = √[(12)(6)(4)(2)] = √576 = 24 cm 2

Therefore, the cost of painting at a rate of 9 paise per cm 2 = 24 × 9 paise = Rs. 2.16

13) An isosceles right triangle has an area of 8 cm 2 . The length of its hypotenuse is

Given that area of the isosceles triangle = 8 cm 2 .

As the given triangle is isosceles triangle, let base = height = h

(½)×h×h = 8

Since it is isosceles right triangle, Hypotenuse 2 = Base 2 +Height 2

Hypotenuse 2  = 4 2  + 4 2

Hypotenuse 2  = 32

Hypotenuse = √32 cm

14) The area of an isosceles triangle having a base 2 cm and the length of one of the equal sides 4 cm, is

(a) √15 cm 2

(b) √(15/2) cm 2

(c) 2√15 cm 2

(d) 4√15 cm 2

Given that a = 2 cm, b= c = 4 cm 

s = (2 + 4 + 4)/2 = 10/2 = 5 cm

By using Heron’s formula, we get:

A =√[5(5 – 2)(5 – 4)(5 – 4)] = √[(5)(3)(1)(1)] = √15 cm 2 .

15) The perimeter of an equilateral triangle is 60 m. The area is

(a) 10√3 m 2

(b) 15√3 m 2

(c) 20√3 m 2

(d) 100√3 m 2

Given: Perimeter of an equilateral triangle = 60 m

3a = 60 m (As the perimeter of an equilateral triangle is 3a units)

We know that area of equilateral triangle = (√3/4)a 2  square units

A = (√3/4)20 2

A = (√3/4)(400) = 100√3 m 2 .

16) The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude

(a) 16√5 cm

(b) 10√5 cm

(c)  24√5 cm

Given: a = 35 cm, b = 54cm, c = 61cm

s = (35 + 54 + 61)/2 = 150/2 = 75 cm.

Hence, by using Heron’s formula, A = √[75(75 – 35)(75 – 54)(75 – 61)] = √(882000) = 420√5 cm 2

The area of triangle with longest altitude “h” is given as”

(½)×a×h = 420√5 {a is less than b and c}

(½)×35×h = 420√5

h = (840√5)/35 = 24√5 cm.

17) The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

(a) 1322 cm 2

(b) 1311 cm 2

(c) 1344 cm 2

(d) 1392 cm 2

Since, all the sides of a triangle are given, we can find the area of a triangle using Heron’s formula.

Let a = 56 cm, b= 60 cm, c = 52 cm

s = (56+60+52)/2 = 84 cm.

Area of triangle using Heron’s formula, A = √[s(s-a)(s-b)(s-c)] square units

A = √[84(84-56)(84-60)(84-52)] = √(1806336) =1344 cm 2 .

18) If the area of an equilateral triangle is 16√3 cm 2 , then the perimeter of the triangle is

Explanation: 

Given: Area of equilateral triangle = 16√3 cm 2

(√3/4)a 2  = 16√3

a 2  = [(16√3)(4)]/ √ 3

a = 8cm Therefore, perimeter = 3(8) = 24 cm.

19) The area of an equilateral triangle with sides 2√3 cm is

(a) 5.196 cm 2

(b) 0.866 cm 2

(c) 3.496 cm 2

(d) 1.732 cm 2

Given: Side = 2√3 cm

We know that, area of equilateral triangle = (√3/4)a 2  square units

A = (√3/4)(2√3) 2  = (√3/4)(12) = 3√3 = 3(1.732) = 5.196 cm 2  .

20) The length of each side of an equilateral triangle having an area of 9√3 cm 2 is

Given: Area of equilateral triangle = 9√3 cm 2

Hence, (√3/4)a 2  = 9√3

a 2  = [(9√3)(4)]/√3

Related Articles

• Heron’s Formula
• Maths Formulas for Class 9
• Important Questions for Class 9 Maths
• Heron’s Formula Class 9 Notes: Chapter 12
• Important Questions Class 9 Maths Chapter 12-Heron’s Formula
• CBSE Class 9 Maths Examination 2018: Important 3 marks questions

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Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

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Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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14 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

MATHS PAAGAL HAI

All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

Where is search ? bar

maths is love

Can I have more questions without downloading the app.

I love math

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