# RD Sharma Solutions for Class 9 Maths Chapter 12 Heron's Formula

## RD Sharma Solutions Class 9 Maths Chapter 12 – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 12 Heron’s Formula provides in-depth knowledge of concepts among students. Here, students will learn about the famous Heron’s formula, which is used for finding the area of a triangle in terms of its three sides. Class 9 students are already learnt about plane figures such as triangles, quadrilaterals, squares, rectangles, etc. This chapter also covered a brief look at plane figure definitions and their respective formulas. Regular practice of RD Sharma Solutions prepared by expert faculty helps students to grasp the covered concepts easily.

Students who are searching for the best reference guide can make use of RD Sharma Solutions, designed in a descriptive manner. These solutions provide a clear explanation for each step in accordance with the student’s intelligence levels. Students can now access RD Sharma Solutions Class 9 Maths for Chapter 12 in PDFs for free by clicking on the links given below. The solutions are created as per the latest CBSE syllabus for the 2023-24 exam.

## RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula

### Exercise 12.1 Page No: 12.8

Question 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Solution:

We know, Heron’s Formula

Here, a = 150 cm

b = 120 cm

c = 200 cm

Step 1: Find s

s = (a+b+c)/2

s = (150+200+120)/2

s = 235 cm

Step 2: Find the area of a triangle

= 8966.56

The area of a triangle is 8966.56 sq. cm.

Question 2: Find the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm.

Solution:

We know, Heron’s Formula

Here, a = 9 cm

b = 12 cm

c = 15 cm

Step 1: Find s

s = (a+b+c)/2

s = (9 + 12 + 15)/2

s = 18 cm

Step 2: Find the area of the triangle

= 54

The area of a triangle is 54 sq. cm.

Question 3: Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Solution:

Given:

a = 18 cm, b = 10 cm, and perimeter = 42 cm

Let c be the third side of the triangle.

Step 1: Find the third side of the triangle, that is c

We know, perimeter = 2s,

2s = 42

s = 21

Again, s = (a+b+c)/2

Put the value of s, and we get

21 = (18+10+c)/2

42 = 28 + c

c = 14 cm

Step 2: Find the area of the triangle

Question 4: In a triangle ABC, AB = 15cm, BC = 13cm and AC = 14cm. Find the area of triangle ABC and hence its altitude on AC.

Solution:

Let the sides of the given triangle be AB = a, BC = b, and AC = c, respectively.

Here, a = 15 cm

b = 13 cm

c = 14 cm

From Heron’s Formula;

= 84

Area = 84 cm2

Let BE is perpendicular on AC

Now, the area of the triangle = ½ x Base x Height

½ × BE × AC = 84

BE = 12cm

Hence, the altitude is 12 cm.

Question 5: The perimeter of a triangular field is 540 m, and its sides are in the ratio 25:17:12. Find the area of the triangle.

Solution:

Let the sides of a given triangle be a = 25x, b = 17x, c = 12x respectively,

Given, Perimeter of the triangle = 540 cm

2s = a + b + c

a + b + c = 540 cm

25x + 17x + 12x = 540 cm

54x = 540 cm

x = 10 cm

So, the sides of a triangle are

a = 250 cm

b = 170 cm

c = 120 cm

Semi perimeter, s = (a+b+c)/2

= 540/2

= 270

s = 270 cm

From Heron’s Formula;

= 9000

Hence, the area of the triangle is 9000 cm2.

### Exercise 12.2 Page No: 12.19

Question 1: Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

Area of the quadrilateral ABCD = Area of △ABC + Area of △ADC ….(1)

△ABC is a right-angled triangle, which is right-angled at B.

Area of △ABC = 1/2 x Base x Height

= 1/2×AB×BC

= 1/2×3×4

= 6

Area of △ABC = 6 cm2 ……(2)

Sides are given, apply Heron’s Formula.

Perimeter = 2s = AC + CD + DA

2s = 5 cm + 4 cm + 5 cm

2s = 14 cm

s = 7 cm

Area of the △CAD = 9.16 cm2 …(3)

Using equations (2) and (3) in (1), we get

Area of quadrilateral ABCD = (6 + 9.16) cm2

= 15.16 cm2.

Question 2: The sides of a quadrilateral field, taken in order, are 26 m, 27 m, 7 m, and 24 m, respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

Here,

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

AC is the diagonal joined at A to C point.

From Pythagoras theorem,

AC2 = 142 + 72

AC = 25

Now, area of △ABC

All the sides are known, Apply Heron’s Formula.

Perimeter of △ABC= 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m

= 291.84

Area of a triangle ABC = 291.84 m2

Now, for the area of △ADC, (Right angle triangle)

Area = 1/2 x Base X Height

= 1/2 x 7 x 24

= 84

Thus, the area of a △ADC is 84 m2

Therefore, the area of rectangular field ABCD = Area of △ABC + Area of △ADC

= 291.84 m2 + 84 m2

= 375.8 m2

Question 3: The sides of a quadrilateral, taken in order as 5, 12, 14, and 15 meters, respectively, and the angle contained by the first two sides is a right angle. Find its area.

Solution:

Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m

Join the diagonal AC.

Now, the area of △ABC = 1/2 ×AB×BC

= 1/2×5×12 = 30

The area of △ABC is 30 m2

In △ABC, (right triangle).

From Pythagoras theorem,

AC2 = AB2 + BC2

AC2 = 52 + 122

AC2 = 25 + 144 = 169

or AC = 13

All sides are known, apply Heron’s Formula:

2s = 15 m +14 m +13 m

s = 21 m

= 84

Area of △ADC = 84 m2

Area of quadrilateral ABCD = Area of △ABC + Area of △ADC

= (30 + 84) m2

= 114 m2

Question 4: A park in the shape of a quadrilateral ABCD has ∠ C = 900, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?

Solution:

Here, AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.

And BD is a diagonal of ABCD.

In the right △BCD,

From Pythagoras theorem;

BD2 = BC2 + CD2

BD2 = 122 + 52 = 144 + 25 = 169

BD = 13 m

Area of △BCD = 1/2×BC×CD

= 1/2×12×5

= 30

Area of △BCD = 30 m2

Now, In △ABD,

All sides are known, Apply Heron’s Formula:

Perimeter of △ABD = 2s = 9 m + 8m + 13m

s = 15 m

= 35.49

Area of the △ABD = 35.49 m2

Area of quadrilateral ABCD = Area of △ABD + Area of △BCD

= (35.496 + 30) m2

= 65.5m2.

Question 5: Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium.

Solution:

Given: AB = 77 m , CD = 60 m, BC = 26 m and AD = 25m

AE and CF are diagonals.

DE and CF are two perpendiculars on AB.

Therefore, we get, DC = EF = 60 m

Let’s say, AE = x

Then BF = 77 – (60 + x)

BF = 17 – x …(1)

From Pythagoras theorem,

DE2 = 252 − x2 ….(2)

In right △BCF

From Pythagoras theorem,

CF2 = BC2 − BF2

CF2 = 262 − (17−x) 2

[Uisng (1)]

Here, DE = CF

So, DE2 = CF2

(2) ⇒ 252 − x2 = 262 − (17−x)2

625 − x2 = 676 – (289 −34x + x2)

625 − x2 = 676 – 289 +34x – x2

238 = 34x

x =7

(2) ⇒ DE2 = 252 – (7)2

DE2 = 625−49

DE = 24

Area of trapezium = 1/2×(60+77)×24 = 1644

Area of trapezium is 1644 m2 (Answer)

Question 6: Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Solution:

The perimeter of a rhombus = 80 m (given)

We know, Perimeter of a rhombus = 4×side

Let a be the side of a rhombus.

4×a = 80

or a = 20

One of the diagonal, AC = 24 m (given)

Therefore OA = 1/2×AC

OA = 12

In △AOB,

Using Pythagoras theorem:

OB2 = AB2 − OA2 = 202 −122 = 400 – 144 = 256

or OB = 16

Since the diagonal of the rhombus bisect each other at 90 degrees.

And OB = OD

Therefore, BD = 2 OB = 2 x 16 = 32 m

Area of rhombus = 1/2×BD×AC = 1/2×32×24 = 384

Area of rhombus = 384 m2.

Question 7: A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per m2. Find the cost of painting.

Solution:

The perimeter of a rhombus = 32 m

We know, Perimeter of a rhombus = 4×side

⇒ 4×side = 32

side = a = 8 m

Each side of the rhombus is 8 m

AC = 10 m (Given)

Then, OA = 1/2×AC

OA = 1/2×10

OA = 5 m

In right triangle AOB,

From Pythagoras theorem;

OB2 = AB2–OA2 = 82 – 52 = 64 – 25 = 39

OB = √39 m

And, BD = 2 x OB

BD = 2√39 m

Area of the sheet = 1/2×BD×AC = 1/2 x (2√39 × 10 ) = 10√39

The area of the sheet is 10√39 m2

Therefore, the cost of printing on both sides of the sheet, at the rate of Rs. 5 per m2

= Rs. 2 x (10√39 x 5) = Rs. 625.

### Exercise VSAQs Page No: 12.23

Question 1: Find the area of a triangle whose base and altitude are 5 cm and 4 cm, respectively.

Solution:

Given: Base of a triangle = 5 cm and altitude = 4 cm

Area of triangle = 1/2 x base x altitude

= 1/2 x 5 x 4

= 10

The area of the triangle is 10 cm2.

Question 2: Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm, respectively.

Solution:

Given: Sides of a triangle are 3 cm, 4 cm and 5 cm, respectively

Apply Heron’s Formula:

S = (3+4+5)/2 = 6

Semi perimeter is 6 cm

Now,

= 6

The area of the given triangle is 6 cm2.

Question 3: Find the area of an isosceles triangle having the base x cm and one side y cm.

Solution:

In right triangle APC,

Using Pythagoras theorem,

AC2 = AP2 + PC2

y2 = h2 + (x/2)2

or h2 = y2 – (x/2)2

Question 4: Find the area of an equilateral triangle having each side 4 cm.

Solution: Each side of an equilateral triangle = a = 4 cm

Formula for Area of an equilateral triangle = ( √3/4 ) × a²

= ( √3/4 ) × 4²

= 4√3

The area of an equilateral triangle is 4√3 cm2.

Question 5: Find the area of an equilateral triangle having each side x cm.

Solution:

Each side of an equilateral triangle = a = x cm

Formula for Area of an equilateral triangle = ( √3/4 ) × a²

= ( √3/4 ) × x²

= x2 √3/4

The area of an equilateral triangle is √3x2/4 cm2.

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

In Chapter 12 of Class 9 RD Sharma Solutions, students will study important concepts as listed below:

• Heron’s Formula Introduction
• Applications of Heron’s Formula

Students can download the chapter-wise solutions in PDFs for the questions provided in the RD Sharma Class 9 textbook and prepare for examinations to score good marks.

## Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 12

Q1

### What is the main objective of RD Sharma Solutions for Class 9 Maths Chapter 12?

Students who want to boost their problem-solving skills and score good marks in the annual exam are encouraged to follow RD Sharma Solutions, created by expert faculty. These solutions are created in a clear manner, which aims to help students to grasp the concepts in detail and clear their doubts instantly. Students can also cross-check their answers while revising the problems of the textbook and know their level of preparation.

Q2

### Mention the key benefits of RD Sharma Solutions for Class 9 Maths Chapter 12.

Following are the key benefits of RD Sharma Solutions for Class 9 Maths Chapter 12:
1. Students can easily download the solutions for each exercise for a better understanding of concepts.
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