Class 9 Mathematics - Heron's Formula Notes & Solutions

Chapter 12: Heron's Formula (NCERT Solutions)

Exercise 12.1

Q1. 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?

Part 1: Deriving Area using Heron's Formula
Let the sides of the equilateral triangle be a, b, c.
Here, a = a, b = a, c = a.
Semi-perimeter (s) = (a + b + c) / 2 = (a + a + a) / 2 = 3a / 2.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[(3a/2) × (3a/2 - a) × (3a/2 - a) × (3a/2 - a)]
Area = √[(3a/2) × (a/2) × (a/2) × (a/2)]
Area = √[3a&sup4; / 16]
Area = (√3/4)a²

Part 2: Calculating Specific Area
Perimeter = 180 cm
3a = 180 ⇒ a = 60 cm.
Area = (√3/4) × (60)²
Area = (√3/4) × 3600
Area = 900√3 cm²

Q2. 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 &rupee; 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Sides of the triangle: a = 122m, b = 120m, c = 22m.
Semi-perimeter (s) = (122 + 120 + 22) / 2 = 264 / 2 = 132 m.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[132 × (132-122) × (132-120) × (132-22)]
Area = √[132 × 10 × 12 × 110]
Area = √[ (12 × 11) × 10 × 12 × (11 × 10) ]
Area = √[ 12² × 11² × 10² ]
Area = 12 × 11 × 10 = 1320 m².

Calculating Rent:
Rate of rent = &rupee; 5000 per m² per year (12 months).
Rent for 1 month = &rupee; 5000 / 12 per m².
Rent for 3 months for 1320 m² = (5000 / 12) × 3 × 1320
Rent = 5000 × 1 × 110 × 3 = &rupee; 16,50,000.
The company paid &rupee; 16,50,000.

Q3. 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.

Sides of the triangle: a = 15m, b = 11m, c = 6m.
Semi-perimeter (s) = (15 + 11 + 6) / 2 = 32 / 2 = 16 m.
Area painted = √[s(s-a)(s-b)(s-c)]
Area = √[16 × (16-15) × (16-11) × (16-6)]
Area = √[16 × 1 × 5 × 10]
Area = √[ 16 × 5 × (5 × 2) ]
Area = √[ 4² × 5² × 2 ]
Area = 4 × 5 × √2
Area = 20√2 m².

Q4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Given sides: a = 18 cm, b = 10 cm.
Perimeter (2s) = 42 cm.
Let the third side be c.
a + b + c = 42
18 + 10 + c = 42
28 + c = 42 ⇒ c = 14 cm.

Semi-perimeter (s) = 42 / 2 = 21 cm.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[21 × (21-18) × (21-10) × (21-14)]
Area = √[21 × 3 × 11 × 7]
Area = √[ (7 × 3) × 3 × 11 × 7 ]
Area = √[ 7² × 3² × 11 ]
Area = 7 × 3 × √11 = 21√11 cm².

Q5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Let the sides be: a = 12x, b = 17x, c = 25x.
Perimeter = a + b + c = 540
12x + 17x + 25x = 540
54x = 540 ⇒ x = 10.

Therefore, the sides are:
a = 12(10) = 120 cm
b = 17(10) = 170 cm
c = 25(10) = 250 cm

Semi-perimeter (s) = 540 / 2 = 270 cm.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[270 × (270-120) × (270-170) × (270-250)]
Area = √[270 × 150 × 100 × 20]
Area = √[ 81000000 ]
Area = 9000 cm².

Q6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Given equal sides: a = 12 cm, b = 12 cm.
Perimeter (2s) = 30 cm.
Let the third side be c.
a + b + c = 30
12 + 12 + c = 30
24 + c = 30 ⇒ c = 6 cm.

Semi-perimeter (s) = 30 / 2 = 15 cm.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[15 × (15-12) × (15-12) × (15-6)]
Area = √[15 × 3 × 3 × 9]
Area = √[15 × 81]
Area = 9 × √15 = 9√15 cm².

Class 9 Mathematics - Heron's Formula Numericals & Practice

Chapter 12: Heron's Formula (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13 cm.

Using Pythagoras theorem:
Height² = Hypotenuse² - Base²
Height² = 13² - 12² = 169 - 144 = 25
Height = 5 cm.

Area = ½ × Base × Height
Area = ½ × 12 × 5
Area = 30 cm².

Q2. Find the area of an equilateral triangle having side length 8 cm. (Use √3 = 1.732)

Area of equilateral triangle = (√3 / 4) × a²
Area = (√3 / 4) × 8²
Area = (√3 / 4) × 64
Area = 16√3
Area = 16 × 1.732 = 27.712 cm².

Q3. The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.

Let sides be 6x, 7x, 8x.
Perimeter = 6x + 7x + 8x = 21x
21x = 420 ⇒ x = 20.

Sides: a=120m, b=140m, c=160m.
Semi-perimeter (s) = 420/2 = 210m.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[210(210-120)(210-140)(210-160)]
Area = √[210 × 90 × 70 × 50]
Area = √[ (21 × 10) × (9 × 10) × (7 × 10) × (5 × 10) ]
Area = 100 × √[ (7 × 3) × 3² × 7 × 5 ]
Area = 100 × 7 × 3 × √15 = 2100√15 m².

Q4. Find the cost of levelling a triangular ground of sides 50m, 65m, and 65m at the rate of &rupee; 8 per m².

Sides: a=50m, b=65m, c=65m. (Isosceles triangle).
s = (50+65+65)/2 = 180/2 = 90m.
Area = √[90(90-50)(90-65)(90-65)]
Area = √[90 × 40 × 25 × 25]
Area = 25 × √[3600]
Area = 25 × 60 = 1500 m².

Total cost = Area × Rate = 1500 × 8 = &rupee; 12,000.

Q5. If every side of a triangle is doubled, then find the percentage increase in its area.

Let original sides be a, b, c and semi-perimeter be s.
Original Area (A1) = √[s(s-a)(s-b)(s-c)]

New sides: 2a, 2b, 2c.
New semi-perimeter (S) = (2a+2b+2c)/2 = a+b+c = 2s.
New Area (A2) = √[2s(2s-2a)(2s-2b)(2s-2c)]
A2 = √[2s × 2(s-a) × 2(s-b) × 2(s-c)]
A2 = √[16 × s(s-a)(s-b)(s-c)]
A2 = 4 × √[s(s-a)(s-b)(s-c)] = 4 × A1.

Increase in Area = A2 - A1 = 4A1 - A1 = 3A1.
Percentage Increase = (Increase / Original) × 100 = (3A1 / A1) × 100 = 300%.

Q6. Find the area of an isosceles triangle whose base is 16 cm and one of its equal sides is 10 cm.

Sides: a = 10, b = 10, c = 16.
s = (10+10+16)/2 = 36/2 = 18.
Area = √[18(18-10)(18-10)(18-16)]
Area = √[18 × 8 × 8 × 2]
Area = 8 × √36 = 8 × 6 = 48 cm².

Q7. A parallelogram has sides 30 m and 14 m and one of its diagonals is 40 m long. Find its exact area.

A diagonal directly divides a parallelogram into two identically congruent triangles. We will find the area of one triangle with sides 30m, 14m, and 40m, then multiply it by 2.

Triangle Sides: a=30, b=14, c=40.
s = (30+14+40)/2 = 84/2 = 42.
Area of 1 triangle = √[42(42-30)(42-14)(42-40)]
Area = √[42 × 12 × 28 × 2]
Area = √[ (14 × 3) × (4 × 3) × (14 × 2) × 2 ]
Area = √[ 14² × 3² × 4² ]
Area = 14 × 3 × 4 = 168 m².

Total Area of Parallelogram = 2 × 168 = 336 m².

Q8. 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 rigorously required for the umbrella?

There are 5 pieces of each colour (Total 10 pieces). Let's find the area of 1 piece.

Sides: a = 20, b = 50, c = 50.
s = (20+50+50)/2 = 120/2 = 60.
Area of 1 piece = √[60(60-20)(60-50)(60-50)]
Area = √[60 × 40 × 10 × 10]
Area = 10 × √[2400] = 10 × 20√6 = 200√6 cm².

Cloth definitively required for EACH colour (5 pieces):
Area = 5 × 200√6 = 1000√6 cm².

Q9. An exactly square courtyard has a pathway forming a perimeter precisely measuring 200m. Using Heron's formula initially, prove its total area systematically.

Perimeter of square = 4a = 200 ⇒ a = 50 m.
A square can be cleanly divided into 2 right-angled triangles with sides 50m, 50m and a diagonal.
Diagonal(c) = √(50² + 50²) = 50√2 m.

Applying Heron's Formula for one triangle:
Sides: a=50, b=50, c=50√2.
s = (50+50+50√2)/2 = 50 + 25√2.
Area = √[s(s-a)(s-b)(s-c)]
Area = √[ (50+25√2) × (25√2) × (25√2) × (50 - 25√2) ]
Area = 25√2 × √[ (50)² - (25√2)² ]
Area = 25√2 × √[ 2500 - 1250 ] = 25√2 × √1250
Area = 25√2 × 25√2 = 625 × 2 = 1250 m².

Total Area of Square = 2 × 1250 = 2500 m². (Verifies standard formula a²).

Q10. Calculate the exact area of the shaded triangular region whose perfectly measured sides are 40cm, 51cm, and 85cm. Validate if Heron's formula accommodates these dimensions successfully.

Before proceeding, we thoroughly check the triangle inequality theorem: Sum of any two sides explicitly must be strictly greater than the third side.
Let a=40, b=51, c=85.
Check: a + b = 40 + 51 = 91 > 85. (True).

s = (40+51+85)/2 = 176/2 = 88.
Area = √[88(88-40)(88-51)(88-85)]
Area = √[88 × 48 × 37 × 3]
Area = √[ (11 × 8) × (16 × 3) × 37 × 3 ]
Area = √[ 11 × 8 × 16 × 9 × 37 ]
Area = 4 × 3 × √[ 8 × 11 × 37 ] = 12√3256 = 24√814 cm².
Yes, it flawlessly accommodates irrational square root solutions accurately portraying specific distinct regions.

Q11. The edges of a triangular board are 6 cm, 8 cm, and 10 cm. Find the structural cost of precisely painting it at the strict rate of 9 paise per cm².

Checking if it's a precisely verified Right Angled Triangle:
6² + 8² = 36 + 64 = 100 = 10². Yes, it absolutely is. (Pythagorean triplet).
Base = 8, Height = 6.
Area = ½ × Base × Height = ½ × 8 × 6 = 24 cm².

Total exact cost = Area × Rate
Cost = 24 × 0.09 = &rupee; 2.16.

Q12. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal objectively identical measured side to its strict base is 3:2. Find the total internal area.

Let equal sides be exactly 3x. The base is exactly 2x.
Perimeter = 3x + 3x + 2x = 8x.
8x = 32 ⇒ x = 4.

Sides physically are: 12 cm, 12 cm, 8 cm.
s = 32 / 2 = 16.
Area = √[16(16-12)(16-12)(16-8)]
Area = √[16 × 4 × 4 × 8]
Area = 4 × 4 × √8 = 16 × 2√2 = 32√2 cm².

Class 9 Mathematics - Heron's Formula (Concepts & Formulas)

Chapter 12: Heron's Formula (Summary & Concepts)

1. Standard Area Formula

The standard, general area of a triangle when its base and corresponding altitude (height) are known is given by:

Area of a Triangle = ½ × Base × Height

2. Heron's Formula

If the lengths of all three sides of a triangle are known (say a, b, and c), its area can be calculated using Heron's Formula perfectly without needing the height.

Here's the two-step process:

Step 1: Find the Semi-perimeter (s)
The semi-perimeter is exactly half of the triangle's total perimeter.
s = (a + b + c) / 2

Step 2: Apply the Formula
Area = √[ s(s - a)(s - b)(s - c) ]
Where a, b, c are unequivocally the precisely measured side lengths.

3. Special Triangle Cases

Equilateral Triangle: If all three sides are equal (a = b = c), the area definitively simplifies to:
Area = (√3 / 4)a²
Right-Angled Triangle: If the triangle has a 90° angle, you can just use the legs as the base and height:
Area = ½ × (Leg 1) × (Leg 2)

4. Application in Quadrilaterals

To find the area of an irregular quadrilateral using Heron's logically proven formula:

Divide the quadrilateral cleanly into two distinct triangles by systematically drawing one of its diagonals.
Determine the strict lengths of all sides for both separated triangles.
Calculate the exact area of each distinct triangle firmly using Heron's Formula individually.
Add the two separated areas logically to rigidly get the total composite area of the entire quadrilateral structure.

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