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Class 9 Mathematics Herons Formula Assignments

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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.

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Chapter 10 Class 9 Herons Formula

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Updated for new NCERT - 2023-24 Curriculmn

Get NCERT Solutions of all exercise questions and examples of Chapter 10 Class 9 Herons Formula. Answers to all question have been solved in a step-by-step manner, with videos of all questions available.

We have studied that

Area of triangle  = 1/2 × Base × Height

In questions where Height and Base is given, we can find the area of triangle easily.

But, in cases where all 3 sides are given, how will we find the area?

If all 3 sides are given, we find Area of Triangle using Herons (or Hero's) Formula

By Hero's Formula

Area of triangle = Square root (s (s-a) (s-b) (s-c))

where a,b, c are sides of the triangle

and s = Semi-Perimeter of Triangle

i.e. s = (a+b+c)/2

In this chapter, we will find Area of Triangle using Herons formula

We will also find Area of Quadrilateral by dividing it into two triangles, and then finding Area of triangle using Hero's Formula

Click on an exercise link or a topic link below to start doing the chapter.

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• Math Article
• Herons Formula Class 9

Heron’s Formula Class 9 Notes: Chapter 12

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

CBSE Class 9 Maths Heron’s Formula Notes –  Download PDF Here

In geometry, a triangle is a closed three-dimensional figure. In this article, you are going to learn Heron’s formula for class 9, which is used to find the area of triangles. You will also learn how Heron’s formula is used to find the area of other polygons in detail.

  The plane closed figure, with three sides and three angles is called as a triangle.

Types of triangles: Based on sides –  a) Equilateral b) Isosceles c) Scalene Based on angles – a) Acute-angled triangle b) Right-angled triangle c) Obtuse-angled triangle

For more information on Triangles, watch the below video

To know more about Triangles and Its Type, visit here .

Area of a Triangle

A r e a  =  (1/2) ×  b a s e  ×  h e i g h t

In the case of equilateral and isosceles triangles, if the lengths of the sides of triangles are given, we use Pythagoras’ theorem in order to find the height of a triangle.

To know more about the Area of a triangle, visit here .

Area of an Equilateral Triangle

Consider an equilateral Δ A B C , with each side as a unit. Let AO be the perpendicular bisector of BC. In order to derive the formula for the area of an equilateral triangle, we need to find height AO.

Using Pythagoras’ theorem, A C 2  =  O A 2  +  O C 2 O A 2  =  A C 2  −  O C 2 Substitute A C  =  a , O C = a/2 in the above equation. O A 2  =  a 2  −  a 2 /4 O A  =  √3a/2

We know that the area of the triangle is: A  =  (1/2) ×  b a s e  ×  h e i g h t A  =  (1/2) ×  a  × (√3a/2)

∴  A r e a   o f   E q u i l a t e r a l   t r i a n g l e  =  √3a 2 /4

Area of an Isosceles Triangle

Consider an isosceles Δ A B C with equal sides as a units and base as b units.

Isosceles triangle ABC

The height of the triangle can be found by Pythagoras’ Theorem : C D 2  =  A C 2  −  A D 2 ⇒ h 2  =  a 2 − (b 2 /4) = (4 a 2  –  b 2 )/4

⇒ h  = (1/2) √(4 a 2  –  b 2 ) Area of triangle is A  =  (1/2)b h ∴  A  =  (1/2) ×  b  ×  (1/2) √(4 a 2  –  b 2 )

∴  A  =  (1/4) ×  b  ×  √(4 a 2  –  b 2 )

Area of a Triangle – By Heron’s formula

Area of a Δ A B C , given sides a, b, c  by Heron’s formula  (also known as Hero’s Formula) is:

Triangle ABC

Find semi perimeter (s) =  (a + b + c)/2

A r e a  =  √[s(s – a)(s – b)(s – c)]

This formula is helpful in finding the area of a scalene triangle, given the lengths of all its sides.

To know more about Heron’s Formula, visit here .

Area of Any Polygon – By Heron’s formula

For a quadrilateral, when one of its diagonal values and the sides are given, the area can be calculated by splitting the given quadrilateral into two triangles and using Heron’s formula.

Example : A park, in the shape of a quadrilateral ABCD, has ∠ C = 90 ∘ , AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm. How much area does it occupy?

⇒ We draw the figure according to the information given.

The figure can be split into 2 triangles Δ B C D   a n d   Δ A B D From Δ B C D , we can find BD (Using Pythagoras’ Theorem) B D 2  =  12 2  +  5 2  =  169 B D  =  13 c m Semi-perimeter for Δ B C D   S 1  =  (12 + 5 + 13)/2 = 15

Semi-perimeter Δ A B D   S 2  =  (9 + 8 + 13)/2 = 15

Using Heron’s formula A 1   a n d   A 2  will be:

A 1  =  √[15(15 – 12)(15 – 5)(15 – 13)]

A 1  = √(15 × 3 × 10 × 2 ) A 1  = √900 =  30  c m 2 Similarly,

A 2 will be 35.49  c m 2 . The area of the quadrilateral A B C D  =  A 1  +  A 2  =  65.49   c m 2

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Assignment - Herons Formula, Class 9 Mathematics PDF Download

VERY SHORT ANSWER TYPE QUESTIONS : 1. Write the area of a triangle having 5 cm base and height 6 cm.

2. Write the area of an equilateral triangle whose side is 6 cm.

3. State Heron's Formula for area of a triangle.

4. In ΔABC, BC = a, CA = b and AB = c. Write the semiperimeter s.

5. Find the area of isosceles triangle ABC in which AB = AC = 5 cm and BC = 8 cm.

6. Find the area of an equilateral triangle having each side of length a cm.

7. Find the area of the triangle having three sides given as 5 cm, 6 cm and 7 cm.

SHORT ANSWER TYPE QUESTIONS : 1. A triangular park in a city has dimensions 100 m × 90 m × 110 m. A contract is given to a company for planting grass in the park at the rate of Rs. 4000 per hectare. Find the amount to be paid to the company. (Take √2 = 1.414) (one hectare = 10,000 m 2 )

2. There is a slide in a children park. The front side of the slide has ben painted and a message "ONLY FOR CHILDREN" is written on it as shown in figure. If the sides of the triangular front wall of the slide are 9 m, 8 m and 3 m, then find the area which is painted in colour.

3. The perimeter of a triangular park is 180 m and its sides are in the ratio 5 : 6 : 7. Find the area of the park.

4. A triangle has sides 35 mm, 54 mm and 61 mm long. What is its area. Find also the smallest altitude of the triangle.

5. The perimeter of a right triangle is 12 cm and its hypotenuse is of length 5 cm. Find the other two sides and calculate its area. Verify the result using Heron's Formula.

6. Using Heron's Formula, find the area of an isosceles triangle, the measure of one of its equal sides being a units and the third side 2b units.

7. The sides of a triangle are 39 cm, 42 cm and 45 cm. A parallelogram stands on the greatest side of the triangle and has the same area as that of the triangle. Find the height of the parallelogram.

8. From a point in the interior of an equilateral triangle perpendiculars drawn to the three sides are 8 cm, 10 cm and 11 cm respectively. Find the area of the triangle to the nearest cm. (use √3 = 1.73)

9. A municipal corporation wall on road side has dimensions as shown in fig. The wall is to be used for advertisements and it yields an earning of Rs. 400 per m 2 in a year. Find the total amount of revenue earned in a year.

10. ABCD is a quadrilateral such that AB = 5 cm, BC = 4 cm, CD = 7cm, AD = 6 cm and diagonal BD = 5 cm. Prove that the area of the quadrilateral ABCD is 4(3 + √6 cm 2  ).

11. Find the area of the quadrilateral ABCD in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm. (Take √110 = 10.5 approx.)

12. A rhombus has perimeter 64 m and one of the diagonals is 22 m. Prove that the area of the rhombus is 66 √15 m 2

13. ABCD is a trapezium in which AB║CD ; BC and AD are non-parallel sides. It is given that AB = 75 cm, BC = 42 cm, CD = 30 cm and AD = 39 cm. Find the area of the trapezium.

14. OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. If the radius of the circle is 10 cm, find the area of the rhombus.

15. The cross-section of a canal is in the shape of a trapezium. If the canal is 12 m wide at the top and 8 m wide at the bottom and the area of its cross-section is 84 m 2 , determine its depth.

16. Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA ; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m. CD = 15 m, DA = 28 m and ∠B = 90°, which group cleaned more area and by how much? Find the total area cleaned by the students.

17. Find the perimeter of a square, the sum of lengths of whose diagonals is 144 cm.

18. Find the area of a quadrilateral piece of ground one of whose diagonals is 60 metres long and the perpendiculars from the other two vertices are 38 and 22 metres respectively.

LONG ANSWER TYPE QUESTIONS :  

1. In figure, AB = 28 m, AC = 24 m, BC = 20 m, CG = 32 m, AG = 40 m and D is mid-point of AG. Find the area of the quadrilateral ABCD.

2. White and grey coloured triangular plastic sheets are used to make a toy as shown in fig. Find the total areas of white and grey coloured sheets used for making the toy.

3. Suman made an arrangement with white and black coloured paper sheets as showin in fig. Find the total areas of the white and black paper sheets used in making the arrangement.

4. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangular tiles are 26 cm, 20 cm and 10 cm. The tiles are polished at the rate of 20 p per cm2. Find the cost of polishing the tiles. (Take √14 = 3.74)

5. Suman made a picture with some white paper and a single coloured paper as shown in figure. White paper is available at her home and free of cost. The cost of coloured paper used is at the rate of 10 p per cm 2 . Find the total cost of the coloured paper used. (Take √3 = 1.732 and √11 = 3.31)

6. In figure, P and Q are two lamp posts. If the area of the ΔPDC is same as that of the rectangle ABCD, find the distance between the two lamp posts.

7. A triangle and a parallelogram have same base and same area. If the sides of the triangle are 20 cm, 25 cm and 35 cm, and the base side is 25 cm for the triangle as well as the parallelogram, find the vertical height of the parallelogram.

8. A triangle and a parallelogram have a common side and are of equal areas. The triangle having sides 26 cm, 28 cm and 30 cm stands on the parallelogram. The common side of the triangle and the parallelogram is 28 cm. Find the vertical height of the triangle and that of the parallelogram.

9. A farmer has two triangular fields in the form of ΔABC and ΔACD in which the side AC is common as shown in figure. AB = 840 m, BC = 600 m, AC = 480 m, AD = 800 m and CD = 640 m. He has marked midpoints E and F on the sides AB and AD respectively. By joining CE and CF, he has made a field in the shape of quadrilateral AECF. He grew wheat in the quadrilateral plot AECF, potatoes in ΔCFD and onions in ΔBEC. How much area has been used for each crop? (Take √6 = 2.45; one hectare = 10000 m 2 )

10. A field in the form of a quadrilateral ABCD whose sides taken in order are respectively equal to 192, 576, 288 and 480 dm has the diagonal equal to 672 dm. Find its area to the nearest square metre.

11. A trapezium with its parallel sides in the ratio 16 : 15 is cut from a rectangle whose sides measure 63 m and 5 m respectively. The area of the trapezium is 4/15 of the area of the rectangle. Find the lengths of the parallel sides of the trapezium

12. Find the cost, at Rs. 25 per 10 square metres, of turfing a plot of land in the form of a parallelogram whose adjacent sides and one of the diagonals measure 39 m, 25 m and 56 m respectively.

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• Heron's Formula

Class 9 NCERT (CBSE and ICSE) Heron's Formula

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• Class 9 NCERT Solutions- Chapter 12 Heron's Formula - Exercise 12.1
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• Class 9 RD Sharma Solutions - Chapter 12 Heron’s Formula- Exercise 12.1
• Class 9 RD Sharma Solutions - Chapter 12 Heron’s Formula- Exercise 12.2
• Class 8 NCERT Solutions - Chapter 14 Factorization - Exercise 14.1
• NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula
• Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.3
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• Class 9 NCERT Solutions - Chapter 10 Circles - Exercise 10.3
• Class 9 NCERT Solutions- Chapter 10 Circles - Exercise 10.5
• Class 9 NCERT Solutions - Chapter 10 Circles - Exercise 10.2

NCERT Solutions Class 9 – Chapter 10 Heron’s Formula – Exercise 10.1

Question 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.

As it is mentioned here that, ΔABC is an equilateral triangle having side length = a. So, here AB =BC = AC = a Perimeter of Equilateral triangle = 3× (Length of a side)  = 3×a = 3a and perimeter = 180 cm (given) So, 3a = 180 a = 60 cm. Hence, length of each side is 60 cm. Now, Area of △ABC can be calculated by Heron’s Formula, where AB = a = 60 cm BC = b = 60 cm AC = c = 60 cm Semi Perimeter (s) = (a+b+c)/2 s = 180/2 s = 90 cm ar(△ABC) = √s(s-a)(s-b)(s-c) = √90(90-60)(90-60)(90-60) = √90×(30)×(30)×(30) = 900√3 cm 2 Hence, the area of the signal board = 900√3 cm 2

Question 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.). 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?

Here, Area of △ABC can be calculated by Heron’s Formula , where AB = a = 120 m BC = b = 22 m AC = c = 122 m Semi Perimeter (s) = (a+b+c)/2 s = (120+22+122)/2 s = 132 m ar(△ABC) = √s(s-a)(s-b)(s-c) = √132(132-120)(132-22)(132-122) = √132×(12)×(110)×(10) = 1320 m 2 As it is given that, For 1 year we cost 1 m 2 = ₹ 5000  So, for 3 months, 1 m 2 = ₹ 5000 × (1/4) For area of walls 1320 m 2 = 5000×(1/4)×(1320) = ₹ 16,50,000 Hence, ₹ 16,50,000 much rent company will be pay for 3 months.

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

Here, Area of △ABC can be calculated by Heron’s Formula , where AB = a = 11 m BC = b = 6 m AC = c = 15 m Semi Perimeter (s) = (a+b+c)/2 s = (11+6+15)/2 s = 16 m ar(△ABC) = √s(s-a)(s-b)(s-c) = √16(16-11)(16-6)(16-15) = √16×(5)×(10)×(1) = 20√2 m 2 Hence, the area painted in colour is 20√2 m 2

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

Here, length of two sides are given as 18 cm and 10 cm respectively. and, perimeter = 42 cm. Hence, length of third side = (Perimeter)-(length of two side) = 42-(18+10) AC = 14 cm Here, Area of △ABC can be calculated by Heron’s Formula , where AB = a = 18 cm BC = b = 10 cm AC = c = 14 cm Semi Perimeter (s) = (a+b+c)/2 s = (18+10+14)/2 s = 21 cm ar(△ABC) = √s(s-a)(s-b)(s-c) = √21(21-18)(21-10)(21-14) = √21×(3)×(11)×(7) = 21√11 cm 2 Hence, the area of triangle is 21√11 cm 2

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

The ratio of the sides of the triangle are given as 12 : 17 : 25 Lets consider the common ratio between the sides of the triangle be p Then, the sides are 12p, 17p and 25p The perimeter of the triangle = 540 cm  ( Given ) 12p+17p+25p = 540 cm 54p = 540cm So, p = 10 Hence, the sides of triangle are 120 cm, 170 cm, 250 cm. Here, Area of △ABC can be calculated by Heron’s Formula , where AB = a = 250 cm BC = b = 120 cm AC = c = 170 cm Semi Perimeter (s) = (a+b+c)/2 s = (250+120+170)/2 s = 270 cm ar(△ABC) = √s(s-a)(s-b)(s-c) = √270(270-250)(270-120)(270-170) = √270×(20)×(150)×(100) = 9000 cm 2 Hence, the area of triangle is 9000 cm 2 .

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

Here, length of two equal sides of isosceles triangle are given as 12 cm. and, perimeter = 30 cm. Hence, length of third side = (Perimeter)-(length of two side) = 30-(12+12) AC = 6 cm Here, Area of △ABC can be calculated by Heron’s Formula , where AB = a = 12 cm BC = b = 12 cm AC = c = 6 cm Semi Perimeter (s) = (a+b+c)/2 s = (12+12+6)/2 s = 15 cm ar(△ABC) = √s(s-a)(s-b)(s-c) = √15(15-12)(15-12)(15-6) = √15×(3)×(3)×(9) = 9√15 cm 2 Hence, the area of triangle is 9√15 cm 2

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