RD Sharma Solutions Class 9 Chapter 12 Herons Formula

Read RD Sharma Solutions Class 9 Chapter 12 Herons Formula below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

 

Exercise 12.1 
 
Question 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm. 
Solution:
It is given that,
a = 150 cm
b = 120 cm
c = 200 cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 2: Find the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm. 
Solution:
a = 9 cm
b = 12 cm
c = 15 cm
 
s = ((a+b+c))/2
s = ((9 + 12 + 15))/2
s = 18 cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
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:
It is given that,
a = 18 cm 
b = 10 cm
Perimeter = 42 cm
c be the third side of the triangle.
We know, perimeter = a + b + c
S = 42/2
S = 21
S = ((a+b+c))/2
Put the value of s, we get
21 = ((18 + 10 + c))/2
42 = 28 + c
c = 14 cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
 
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:
AB = a 
BC = b 
AC = c
a = 15 cm
b = 13 cm
c = 14 cm
s = ((a+b+c))/2
s = ((15 + 13 + 14))/2
s = 42/2
s = 21cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
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 us assumed that the sides of a triangle 
a = 25x 
b = 17x 
c = 12x
Perimeter of triangle = 540 cm
2s = a + b + c
a + b + c = 540cm
25x + 17x + 12x = 540cm
54x = 540cm
54x = 540/54cm
x = 10cm
The sides of a triangle are
a = 250 cm
b = 170 cm
c = 120 cm
S = ((a+b+c))/2
S = 540/2
S = 270cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
 
Question 6: The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle. 
Solution:
Let us assumed that the sides of a triangle
a = 3x
b = 5x 
c = 7x
Perimeter = a + b + c
300 = a + b + c
300 = 3x + 5x + 7x
300 = 15x
x = 20 m
So, the sides are
a = 60 m
b = 100 m
c = 140 m
S = ((a+b+c))/2
S = ((60+100+140))/2
S = 300/2
S = 150cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 7:  The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex. 
Solution:
It is given that, 
AB = a = 78 dm 
BC = b = 50 dm
Perimeter = 240 dm
AB + BC + AC = 240 dm
78 + 50 + AC = 240
AC = 240 - 78 - 50
AC = 112 dm 
AC = c = 112 dm
s = (a + b + c)/2
s = (78 + 50 + 112)/2
s = 240/2dm
s = 120dm
Area of a triangle = √(s(s-a)(s-b)(s-c))
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 8:  A triangle has sides 35 cm, 54 cm, 61 cm long. Find its area. Also, find the smallest of its altitudes? 
Solution:
It is given that given,
a = 35 cm
b = 54 cm
c = 61 cm
Perimeter 2s = a + b + c
s = (35 + 54 + 61)/2
s = 150/2
s = 75cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Longest side = 61 cm
Area of the triangle = 1/2 × H × B
939.14 = 1/2 × H × 61 
939.14 cm2 = 1/2 × H × 61
(939.14×2)/61   = H
H = 30.79 cm
Thus, the length of the smallest altitude is 30.79 cm
 
Question 9:  The lengths of the sides of a triangle are in a ratio of 3: 4: 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side? 
Solution:
Let us assumed that the sides of a triangle
a = 3x
b = 4x 
c = 5x
Perimeter = a + b + c
144 = 3x + 4x + 5x
144 = 12x
144/12 = x
x = 12 cm
So, the respective sides are
a = 3 × 12 = 36 cm
b = 4 × 12 = 48 cm
c = 5 × 12 = 60 cm
s = (a + b + c)/2
s = 144/2
s = 72cm
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Longest side = 60 cm
Area of the triangle = 1/2 × h × 60
864 cm2 = 1/2 × h × 60 
(864 × 2)/60 = h 
h = 28.8 cm
Thus the length of the smallest altitude is 28.8 cm 
 
 
Question 10: The perimeter of an isosceles triangle is 42 cm and its base is 3/2 times each of the equal side. Find the length of each of the triangle, area of the triangle and the height of the triangle. 
Solution:
Let the two sides of an isosceles triangle be x and x.
Third side ‘base’ = 3/2x
Perimeter = a + b + c
42 = x + x + 3/2x
42 = 2x + 3/2x
42 = (4x + 3x)/2
42 = 7x/2
(42 × 2)/7 = x
6×2 = x
12 = x
So, the sides are
a = x = 12 cm
b = x = 12 cm
c = 3/2×12 = 18 cm
s = (a + b + c)/2
s = (12 + 12 + 18)/2
s = 42/2
s = 21cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 11: Find the area of the shaded region in fig. below. 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Solution:
Area of the shaded region = Area of ΔABC − Area of ΔADB
Now in triangle ADB
AB2 = AD2 + BD
It is given that,
AD = 12 cm 
BD =16 cm
AB2 = 122 + 162
AB2 = 400cm2
AB2 = √(400cm2 )
AB = 20 cm
In ∆ADB
Area of a triangle = 1/2 × AD × BD
Area of a triangle = 1/2 × 12 × 16
Area of a triangle = 96 cm2
In ∆ABC,
S = (AB + BC + CA)/2
S = (52 + 48 + 20)/2
S = 120/2
S = 60 cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Exercise 12.2 
 
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: 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
∆ABC is a right-angled triangle
Area of ∆ABC = 1/2  x Base x Height
Area of ∆ABC = 1/2 × AB × BC
Area of ∆ABC = 1/2×3 × 4 
Area of ∆ABC = 6 cm2
From Pythagoras theorem
AC2 = AB2 + BC2
AC2 = (3)2 + (4)2
AC2 = 9 + 16
AC2 = 25
AC = √25
AC = 5cm
In ∆CAD,
s = (a + b + c)/2
s = (5 + 4 + 5)/2
s = 14/2
s = 7cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 2:  The sides of a quadrilateral field, taken in order are 26m, 27m, 7m, and 24m respectively. The angle contained by the last two sides is a right angle. Find its area. 
Solution: 
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
It is given that,
AB = 26m 
BC = 27m 
CD = 7m 
DA = 24m
AC is the diagonal joined at A to C point.
Now, in ∆ADC,
From Pythagoras theorem;
AC2 = AD2 + CD2
AC2 = (24)2 + (7)2
AC2 = 576 + 49
AC2 = 625
AC = √625
AC = 25cm
s = (a + b + c)/2
s = (25 + 26 + 27)/2
s = 78/2
s = 39cm
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of right triangles △ADC = 1/2 × Base × Height
Area of right triangles △ADC = 1/2 x 7 x 24
Area of right triangles △ADC = 84 m2
 
Area of rectangular field ABCD = Area of △ABC + Area of △ADC
Area of rectangular field ABCD = 291.84 m2 + 84 m2
Area of rectangular field ABCD = 375.8 m2
 
 
Question 3:  The sides of a quadrilateral, taken in order as 5, 12, 14, 15 meters respectively, and the angle contained by first two sides is a right angle. Find its area. 
Solution:  
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
It is given that, 
AB = 5 m 
BC = 12 m 
CD =14 m 
DA = 15 m
Join the diagonal AC.
Area of △ABC = 1/2 × AB × BC
Area of △ABC = 1/2 × 5 × 12 
Area of △ABC = 30m2
In △ABC
From Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (5)2 + (12)2
AC2 = 25 + 144 
AC2 = 169
AC = 13
In △ADC,
s = (a + b + c)/2
s = (13 + 14 + 15)/2
s = 42/2
s = 21cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of quadrilateral ABCD = Area of △ABC + Area of △ADC
Area of quadrilateral ABCD = 30 + 84 m2
Area of quadrilateral ABCD = 114 m2
 
Question 4:  A park in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy? 
Solution:  
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
It is given that, 
AB = 9 m
BC = 12 m 
CD = 5 m 
DA = 8 m.
BD is a diagonal of ABCD.
In right ∆BCD,
From Pythagoras theorem;
BD2 = BC2 + CD2
BD2 = 122 + 5
BD2 = 144 + 25 
BD2 = 169
BD = 13m
Area of ∆BCD = 1/2 × BC × CD
Area of ∆BCD = 1/2 × 12 × 5
Area of ∆BCD = 30m2
In △ABD,
s = (a + b + c)/2
s = (9 + 8 + 13)/2
s = 30/2
s = 15m
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
 
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:  
RD Sharma Solutions Class 9 Chapter 12 Herons Formula

It is given that,

AB = 77m  

CD = 60 m 

BC = 26m 

AD = 25m

AE and CF are diagonals.

DE and CF are two perpendiculars on AB.

DC = EF = 60 m

Let us assumed that, AE = x

BF = 77 – (60 + x)

BF = 17 – x

In ∆ADE,

RD Sharma Solutions Class 9 Chapter 12 Herons Formula

 

Question 6: Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m. 
Solution:   
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
It is given that perimeter of a rhombus = 80m
Let the sides will be ‘a’
Perimeter of a rhombus = 4 × side
80 = 4 × a
80/4 = a
20 = a 
Diagonal AC = 24m
OA = 1/2 × AC
OA = 1/2 × 24
OA = 12
In △AOB,
OB2 = AB2 − OA
OB2  = 202 −12
OB2  = 400 – 144 
OB2  = 256
OB2  = √256
OB = 16
Since diagonal of rhombus bisect each other at 90 degrees.
And OB = OD
BD = 2 × OB 
BD = 2 x 16m 
BD = 32m
Area of rhombus = 1/2 × BD × AC 
Area of rhombus = 1/2  × 32 × 24 
Area of rhombus = 384m2
 
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:    
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
It is given that, Perimeter of a rhombus = 32 m
Let the side will be a.
Perimeter of a rhombus = 4 × side
32 = 4 × Side
32/4 = Side
8m = Side
So the side of a rhombus is 8.
AC = 10m 
OA = 1/2 × AC
OA = 1/2 × 10
OA = 5m
In ∆AOB,
From Pythagoras theorem;
OB2 = AB2 − OA2  
OB2  = 82 – 52 
OB2  = 64 – 25 
OB2  = 39
OB = √39m
BD = 2 x OB
BD = 2√39 m
Area of the sheet = 1/2 × BD × AC 
Area of the sheet = 1/2 x (2√39 × 10) 
Area of the sheet = 10√39 
Area of the sheet is 10√39 m2
Cost of printing on both sides of the sheet, Rs. 5 per m2
Cost of printing = Rs. 2 x (10√39 x 5) 
Cost of printing = Rs. 625 
 
Question 8:  Find the area of the quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. [Take √3 = 1.73] 
Solution:    
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
It is given that, ABCD is a quadrilateral 
AD = 24cm
∠BAD = 90°
BCD is an equilateral triangle 
BC = CD = BD = 26 cm
In ∆BAD
BA2 = BD2 − AD2
BA2 = 262 - 242
BA2 = 676 - 576
BA = √100
BA = 10 cm
 
Area of the ∆BAD = 1/2 × BA × AD
Area of the ∆BAD = 1/2 × 10 × 24
Area of the ∆BAD = 120cm
 
Area of the equilateral triangle = √3/4 × side
Area of the equilateral triangle BCD = √3/4 × 26 
Area of the equilateral triangle BCD = 292.37 cm2
 
The area of quadrilateral ABCD = Area of triangle BAD + Area of the triangle BCD
The area of quadrilateral ABCD = 120 + 292.37
The area of quadrilateral ABCD = 412.37 cm2
 
Question 9:  Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and the diagonal BD = 20 cm. 
Solution:

RD Sharma Solutions Class 9 Chapter 12 Herons Formula

 
It is given that,
AB = 42 cm 
BC = 21 cm 
CD = 29 cm 
DA = 34 cm
The diagonal BD = 20cm
In ∆ABD
s = (a + b + c)/2
s = (34 + 42 + 20)/2
s = 96/2
s = 48 cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
Area of quadrilateral ABCD = 336 + 210
Area of quadrilateral ABCD = 546 cm2
 
Question 10: Find the perimeter and the area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB = 90°. 
Solution: 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
It is given that, 
AB = 17 cm 
AD = 9 cm 
CD = 12 cm 
AC = 15 cm 
∠ACB = 90°
In Quadrilateral ABCD
BC2 = AB2 − AC2
BC2 = 172 − 152
BC2 = 289 − 225
BC = √64
BC = 8cm
In ∆ABC
Area of ∆ABC = 1/2 × AC × BC
Area of ∆ABC = 1/2 × 8 × 15
Area of ∆ABC = 60 cm2
 
In ∆ACD
s = (a + b + c)/2
s = (15 + 12 + 9)/2
s = 36/2
s = 18 cm
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 60 cm2 + 54 cm2
Area of quadrilateral ABCD = 114 cm2
 
Question 11:  The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of parallelogram. 
Solution: 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Area of triangle ABC = Area of triangle ADC
Area of the parallelogram = Area of triangle ADC + Area of triangle ABC
Area of the parallelogram = 2 × (Area of triangle ABC)
Area of triangle ABC
s = (34 + 20 + 42 )/2
s = 96/2
s = 48 cm
 
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of parallelogram ABCD = 2 × (Area of triangle ABC)
Area of parallelogram ABCD = 2 × 336 cm2
Area of parallelogram ABCD = 672 cm2
 
Question 12:  Find the area of the blades of the magnetic compass shown in figure given below: 
Solution: 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB
In ∆ADB
s = (5 + 1 + 5 )/2
s = 11/2
s = 5.5 cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Area of triangle ADB = Area of triangle CDB
Area of the blades of the magnetic compass = 2 × area of triangle ADB
Area of the blades of the magnetic compass = 2 × 2.49
Area of the blades of the magnetic compass = 4.98 cm2
 
Question 13:  A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.
 RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Solution:
It is given that,
The sides of AOB
AO = 25 cm
OB = 25 cm
BA = 14 cm
Area of each strip = Area of triangle AOB
In ∆AOB
s = (25 + 14 + 25 )/2
s = 64/2
s = 32 cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
Area of each type of paper needed to make a fan = 5 × Area of triangle AOB
Area of each type of paper needed to make a fan = 5 × 168 cm2
Area of each type of paper needed to make a fan = 840 cm2
 
Question 14:  A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of a parallelogram. 
 
Solution:
The sides of the triangle DCE are
DC = 15 cm,
CE = 13 cm,
ED = 14 cm
 
s = (15 + 13 + 14 )/2
s = 42/2
s = 21cm
 
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
 
Exercise VSAQs 
 
Question 1: Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively. 
Solution:
It is given that the Base of a triangle = 5cm and altitude = 4 cm
Area of triangle = 1/2 x base x altitude
Area of triangle = 1/2 x 5 x 4
Area of triangle = 10cm2
 
Question 2:  Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively. 
Solution:
s = (3+4+5)/2 
s = 12/2
s = 6cm
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
 
Question 3: Find the area of an isosceles triangle having the base x cm and one side y cm. 
Solution:
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
 
Question 4:  Find the area of an equilateral triangle having each side 4 cm. 
Solution: 
Side of an equilateral triangle = 4 cm
Let the side will be ‘a’
Area of an equilateral triangle = √3/4 × side²
Area of an equilateral triangle =  √3/4 × 4²
Area of an equilateral triangle = 4√3cm2
 
Question 5: Find the area of an equilateral triangle having each side x cm. 
Solution: 
Side of an equilateral triangle = x cm
Area of an equilateral triangle = √3/4 × side²
Area of an equilateral triangle =  √3/4 × x²
Area of an equilateral triangle = (√3 x)/4cm2

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RS Aggarwal Class 9 Mathematics Solutions Chapter 1 Real Numbers
RS Aggarwal Class 9 Mathematics Solutions Chapter 2 Polynomials
RS Aggarwal Class 9 Mathematics Solutions Chapter 3 Introduction to Euclids Geometry
RS Aggarwal Class 9 Mathematics Solutions Chapter 4 Lines and Triangles
RS Aggarwal Class 9 Mathematics Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle
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RS Aggarwal Class 9 Mathematics Solutions Chapter 9 Quadrilaterals and Parallelograms
RS Aggarwal Class 9 Mathematics Solutions Chapter 10 Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 11 Circle
RS Aggarwal Class 9 Mathematics Solutions Chapter 12 Geometrical Constructions
RS Aggarwal Class 9 Mathematics Solutions Chapter 13 Volume and Surface Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 14 Statistics
RS Aggarwal Class 9 Mathematics Solutions Chapter 15 Probability
RD Sharma Solutions Class 9 Maths
RD Sharma Solutions Class 9 Chapter 1 Number System
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 3 Rationalisation
RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
RD Sharma Solutions Class 9 Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions Class 9 Chapter 6 Factorization of Polynomials
RD Sharma Solutions Class 9 Chapter 7 Introduction To Euclids Geometry
RD Sharma Solutions Class 9 Chapter 8 Lines and Angles
RD Sharma Solutions Class 9 Chapter 9 Triangle and its Angles
RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
RD Sharma Solutions Class 9 Chapter 15 Area of Parallelograms and Triangles
RD Sharma Solutions Class 9 Chapter 16 Circles
RD Sharma Solutions Class 9 Chapter 17 Construction
RD Sharma Solutions Class 9 Chapter 18 Surface Area and Volume of Cuboid and Cube
RD Sharma Solutions Class 9 Chapter 19 Surface Area and Volume of A Right Circular Cylinder
RD Sharma Solutions Class 9 Chapter 20 Surface Area and Volume of A Right Circular Cone
RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere
RD Sharma Solutions Class 9 Chapter 22 Tabular Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 23 Graphical Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 24 Measure of Central Tendency
RD Sharma Solutions Class 9 Chapter 25 Probability