## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B.

Other Exercises

Question 1.
Solution:
We shall take the game along x-axis and number of students along y-axis.

Question 2.
Solution:
We shall take the time on x-axis and temperature (in °C) on y-axis.

Question 3.
Solution:
We shall take name of vehicle on x-axis and velocity (in km/hr) on y-axis

Question 4.
Solution:
We shall take sports on x-axis and number of students on y-axis.

Question 5.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 6.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 7.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 8.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 9.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 10.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 11.
Solution:
We shall take week on x-axis and Rate per 10g (in Rs.) on y-axis.

Question 12.
Solution:
We shall take mode of transport on x-axis and number of students on y-axis.

Question 13.
Solution:
We see from the graph that
(i) It shows the marks obtained by a student in various subjects.
(ii) The student is very well in mathematics.
(iii) The student is very’ poor, in Hindi.
(iv) Average marks
= $$\frac { 60+35+75+50+60 }{ 5 }$$ (Here x = 5)
= $$\frac { 280 }{ 5 }$$
= 56 marks

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B.

Other Exercises

Question 1.
Solution:
In parallelogram ABCD.
∠ A = 72°
But ∠ A = ∠ C (opposite angle of a ||gm)
∴ ∠ C = 72°
∴ ∠ A + ∠ B = 180° (co-interior angles)
=> 72° + ∠B = 180°
=> ∠B = 180° – 72°
=> ∠B = 108°
But ∠ B = ∠ D (opposite angles of a ||gm)
∴ ∠D = 108°
Hence ∠D = 108°, ∠ C = 72° and ∠ D = 108° Ans.

Question 2.
Solution:
In || gm ABCD, BD is its diagonal
and ∠DAB = 80° and ∠DBC = 60°
∴AB || DC
(co-interior angles)
=> 80° + ∠ADC = 180°
=> ∠ ADC = 180° – 80°
But ∠ ADB = ∠ DBC (Alternate angles)
But ∠ ADB + ∠ CDB = 100°
60° + ∠CDB = 100°
=> ∠CDB = 100° – 60° = 40°
Hence ∠CDB = 40° and ∠ ADB = 60° Ans.

Question 3.
Solution:
Given : In ||gm ABCD,
∠ A = 60° Bisectors of ∠ A and ∠ B meet DC at P.
To Prove : (i) ∠APB = 90°
(ii) AD = DP and PB = PC = BC
Proof: ∴ AD || B (opposite sides of a ||gm)
∴∠ A + ∠ B = 180° (co-interior angles)
But AP and BP are the bisectors of ∠A and ∠B
∴$$\frac { 1 }{ 2 }$$∠A + $$\frac { 1 }{ 2 }$$ ∠B = 90°

=> ∠PAB + ∠PBA = 90°
But in ∆ APB,
∠PAB + ∠PBA + ∠APB = 180° (angles of a triangle)
=> 90° + ∠APB = 180°
=> ∠APB = 180° – 90° = 90°
Hence ∠APB = 90°
(ii) ∠ A + ∠ D = 180° (co-interior angles)
and ∠ A – 60°
∴ ∠D = 180° – 60° = 120°
But ∠DAP = $$\frac { 1 }{ 2 }$$ ∠A = $$\frac { 1 }{ 2 }$$ x 60° = 30°
∴∠DPA = 180° – (∠DAP + ∠D)
= 180° – (30° + 120°)
= 180° – 150° = 30°
∠DAP = ∠DPA (each = 30°)
Hence AD = DP (sides opposite to equal angles)
In ∆ BCP,
∠ C = 60° (opposite to ∠ A)
∠CBP = $$\frac { 1 }{ 2 }$$ ∠ B = $$\frac { 1 }{ 2 }$$ x 120° = 60°
But ∠CPB + ∠CBP + ∠C = 180°
(Angles of a triangle)
=> ∠CPB + 60° + 60° = 180°
=> ∠CPB + 120° = 180°
=> ∠CPB = 180° – 120° = 60°
∆ CBP is an equilateral triangle and BC = CP = BP
=> PB – PC = BC
(iii) DC = DP + PC
(∴ DP = AD and PC = BC proved)
Hence proved.

Question 4.
Solution:
In ||gm ABCD,
AC and BD are joined
∠BAO = 35°, ∠ DAO = 40°
∠COD = 105°
∴ ∠AOB = ∠COD
(vertically opposite angles)
∴∠AOB = 105°
(i) Now in ∆ AOB,
∠ABO + ∠AOB + ∠OAB = 180°
(angles of a triangle)
=> ∠ABO + 105° + 35° = 180°
=> ∠ABO + 140° = 180°
=> ∠ABO = 180° – 140°
∠ ABO = 40°
(ii) ∴ AB || DC
∴ ∠ ABO = ∠ ODC (alternate angles)
∴ ∠ ODC = 40°
∴ ∠ ACB = ∠ DAO or ∠ DAC
(alternate angles)
= 40°
(iv) ∴ ∠ A + ∠ B = 180° (co-interior angles)
=> (40° + 35°) + ∠B = 180°
=> ∠B = 180° – 75° = 105°
=> ∠ CBD + ∠ABO = 105°
=> ∠CBD + 40° = 105°
=> ∠CBD = 105° – 40° = 65°
Hence ∠ CBD = 65° Ans.

Question 5.
Solution:
In ||gm ABCD
( ∠A = (2x + 25)° and ∠ B = (3x – 5)°
∴AD || BC (opposite sides of parallelogram)
∴∠ A + ∠B = 180° (co-interior angles)
=> 2x + 25° + 3x – 5° = 180°
=> 5x + 20° = 180°
=> 5x = 180° – 20°

=> 5x = 160° => x = $$\frac { { 160 }^{ o } }{ 5 }$$ = 32°
∴x = 32°
Now ∠A = 2x + 25° = 2 x 32° + 25°
= 64° + 25° = 89°
∠B = 3x – 5 = 3 x 32° – 5°
= 96° – 5° = 91°
∠ C = ∠ A (∴ opposite angles of ||gm)
= 89°
Similarly ∠B = ∠D
∠D = 91°
Hence ∠ A = 89°, ∠ B = 91°, ∠ C = 89°, ∠D = 91° Ans.

Question 6.
Solution:
Let ∠A and ∠B of a ||gm ABCD are adjacent angles.
∠A + ∠B = 180°
Let ∠B = x

Then ∠ A = $$\frac { 4 }{ 5 }$$ x
∴ x + $$\frac { 4 }{ 5 }$$ x = 180°
$$\frac { 9 }{ 5 }$$ x = 180°
=>$$\frac { { 180 }^{ o }\quad X\quad 5 }{ 9 }$$ = 100°
∴ ∠A = $$\frac { 4 }{ 5 }$$ x 100° = 80°
and ∠B = 100°
But ∠ C = ∠ A and ∠ D = ∠ B
(opposite angles of a || gm)
∴∠C = 80°, and ∠D = 100°
Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠ D = 100° Ans.

Question 7.
Solution:
Let the smallest angle ∠ A and the other angle ∠ B
Let ∠ A = x
Then ∠ B = 2x – 30°
But ∠ A + ∠ B = 180° (co-interior angles)
∴x + 2x – 30° = 180°
=> 3x = 180° + 30° = 210°
=> x = $$\frac { { 210 }^{ o } }{ 3 }$$ = 70°
∴ ∠ A = 70°

and ∠ B = 2x – 30° = 2 x 70° – 30°
= 140° – 30° = 110°
But ∠C = ∠ A and ∠D = ∠B
(opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠A = 70°, ∠B = 110°, ∠C = 70° and ∠D = 110° Ans.

Question 8.
Solution:
In ||gm ABCD,
AB = 9.5 cm and perimeter = 30 cm
=> AB + BC + CD + DA = 30cm
=> AB + BC + AB + BC = 30 cm
( ∴ AB = CD and BC – DA opposite sides)

=> 2(AB + BC) = 30cm
=> AB + BC = 15cm
=> 9 5cm + BC = 15cm
∴BC = 15cm – 9.5cm = 5.5cm
Hence AB = 9.5cm, BC = 5.5cm,
CD = 9.5cm and DA = 5.5cm Ans.

Question 9.
Solution:
ABCD is a rhombus
AB = BC = CD = DA
(i)∴ AB || DC
∴ ∠ B + ∠ C = 180° (co-interior angles)
=> 110° + ∠C = 180°
=> ∠C = 180° – 110° = 70°

Question 10.
Solution:
In a rhombus,
Diagonals bisect each other at right angles
∴ AC and BC bisect each other at O at right angles.
But AC = 24 cm and BD = 18 cm

Question 11.
Solution:
Let ABCD he the rhombus whose diagonal are AC and BD which bisect each other at right angles at O.

Question 12.
Solution:
ABCD is a rectangle whose diagonals AC and BD bisect each other at O.

Question 13.
Solution:
ABCD is a square. A line CX cuts AB at X and diagonal BD at O such that
∠ COD = 80° and ∠ OXA = x°
∴∠ BOX = ∠ COD
(vertically opposite angles)
∴∠BOX = 80°
∴Diagonal BD bisects ∠ B and ∠ D
∴ ∠OBA or ∠OBX = 45°
Now in ∆ OBX,
Ext. ∠ OXA = ∠ BOX + ∠ OBX
=>x° = 80° + 45° = 125° Ans.

Question 14.
Solution:
Given : In ||gm ABCD, AC is joined. AL ⊥ BD and CM ⊥ BD
To prove :
(i) ∆ ALD ≅ ∆ CMB
(ii) AL = CM
Proof : In ∆ ALD and ∆ BMC
AD = BC (opposite sides of ||gm)
∠L = ∠M (each 90°)
∠ ADL = ∠ CBM (Alternate angles)
∴ ∆ ALD ≅ ∆ BMC. (AAS axiom)
∴ or A ALD ≅ A CMB.
AL = CM (c.p.c.t.) Hence proved.

Question 15.
Solution:
Given : In ∆ ABCD, bisectors of ∠A and ∠B meet each other at P.
To prove : ∠APB = 90°
∠A + ∠B = .180° (co-interior angles)
PA and PB are the bisectors of ∠ A and ∠B

Question 16.
Solution:
In ||gm ABCD,
P and Q are the points on AD and BC respectively such that AP = $$\frac { 1 }{ 3 }$$ AD and CQ = $$\frac { 1 }{ 3 }$$ BC

Question 17.
Solution:
Given : In ||gm ABCD, diagonals AC and BD bisect each other at O.
A line segment EOF is drawn, which meet AB at E and DC at F.
To -prove : OE = OF

Question 18.
Solution:
Given : ABCD is a ||gm.
AB is produced to E. Such that AB = BE
DE is joined which intersects BC in O.
To prove : ED bisects BC i.e. BO = OC

Question 19.
Solution:
Given : In ||gm ABCD, E is the midpoint BC
DE is joined and produced to meet AB on producing at F.

Question 20.
Solution:
Given : ∆ ABC and lines are drawn through A, B and C parallel to respectively BC, CA and AB forming ∆ PQR.
To prove : BC = $$\frac { 1 }{ 2 }$$ QR

Question 21.
Solution:
Given : In ∆ ABC, parallel lines are drawn through A, B and C respectively to the sides BC, CA and AB intersecting each other at P, Q and R.

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.

Question 1.
Solution:
Co-ordinates of
A are ( – 6, 5)
B are (5, 4)
C are ( – 3, 2)
D are (2, – 2)
E are ( – 1, – 4)
Ans.

Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:

Question 3.
Solution:
(i) (7, 0) lies on x-axis as its ordinate is (0)
(ii) (0, – 5) lies on y-axis as its abscissa is (0)
(iii) (0, 1) lies on y-axis as its abscissa is (0)
(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)
Ans.

Question 4.
Solution:
(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)
(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)
(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).

Question 5.
Solution:
In the given equation, .
y = x+1
Put x = 0, then y = 0 + 1 = 1
x = 1, then, y = 1 + 1=2
x = 2, then, y = 2 + 1 = 3
Now, plot the points as given in the table given below on the graph, and join them as shown.

Question 6.
Solution:
In the given equation
y = 3x + 2
Put x = 0,
then y = 3 x 0 + 2 = 0 + 2 = 2
x = 1, then
y = 3 x 1 + 2 = 3 + 2 = 5
and x = – 2, then
y = 3 x ( – 2) + 2 = – 6 + 2 = – 4

Question 7.
Solution:
In the given equation,
y = 5x – 3
Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2
x = 0 then,
y = 5 x 0 – 3 = 0 – 3 = – 3
and x = 2, then
y = 5 x 2 – 3 = 10 – 3 = 7

Question 8.
Solution:
In the given equation,

y = 3x,
Put x = 0,
then y = 3 x 0 = 0
Put x = 1, then
y = 3 x 1 = 3
Put x = – 1, then
y = 3 ( – 1) = – 3
Now, plot the points as given in the table below

and join them as shown.

Question 9.
Solution:
In the given equation, y = – x,
Put x = 1,
then y = – 1
Put x = 2, then
y = – 2
Put x = – 2, then
y = – ( – 2) = 2

Hope given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.

Other Exercises

Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.

Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.

Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.

Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : $$\frac { upper\quad limit+lower\quad limit }{ 2 }$$ is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.

Question 5.
Solution:
The given data can be represent in form of frequency table as given below:

Question 6.
Solution:
The frequency distribution table of the given data is given below :

Question 7.
Solution:
The frequency distribution table of the

Question 8.
Solution:
The frequency table is given below :

Question 9.
Solution:
The frequency table of given data is given below :

Question 10.
Solution:
The frequency distribution table of the given data in given below :

Question 11.
Solution:
The frequency table of the given data:

Question 12.
Solution:
The cumulative frequency of the given table is given below:

Question 13.
Solution:
The given table can be represented in a group frequency table in given below :

Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :

Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A.

Question 1.
Solution:
(i) x = 5 is the line AB parallel to the y-axis at a distance of 5 units.
(ii) y = – 2 is the line CD parallel to x-axis at a distance of – 2 units.
(iii) x + 6 = 0 => x = – 6 is the line EF parallel to y-axis at a distance of – 6 units.
(iv) x + 7 = 0 => x = – 7 is the line PQ parallel to y-axis at a distance of – 7 units.
(v) y = 0 is the equation of x-axis. The graph of y = 0 is the line X’OX
(vi) x = 0 is the equation of y-axis.The graph of x = 0 is the line YOY’
Ans.

Question 2.
Solution:
In the given equation.

y = 3x
Put x = 1, then y = 3 x 1 = 3
Put x = 2, then y = 3 x 2 = 6
Put x = – 1, then y = 3 ( – 1) = – 3
Now, plot the points (1, 3), (2, 6) and ( – 1, – 3) as given the following table

and join them to form a line of the given equation.
Now from x = – 2,
draw a line parallel to y-axis at a distance of x = – 2, meeting the given line at P. From P, draw, a line parallel to x-axis joining y-axis at M, which is y = – 6 Hence, y = – 6 Ans.

Question 3.
Solution:
In the given equation x + 2y – 3 = 0
=> 2y = 3 – x
y = $$\frac { 3-x }{ 2 }$$
put x = 1,then

Question 4.
Solution:
(i) In the equation y = x
When x = 1, then y = 1
when x = 2, then y = 2
and when x = 3, then y = 3

Question 5.
Solution:
In the given equation
2x – 3y = 5

Question 6.
Solution:
In the given equation
2x + y = 6
=> y = 6 – 2x
Put x = 1, then y – 6 – 2 x 1 = 6 – 2 = 4
Put x = 2, then y = 6 – 2 x 2 = 6 – 4 = 2
Put x = 3, then y = 6 – 2 x 3 = 6 – 6 = 0

Question 7.
Solution:
In the given equation
3x + 2y = 6

Hope given RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A.

Question 1.
Solution:
Given : In the figure, ABCD is a quadrilateral and
AB = CD = 5cm

Question 2.
Solution:
In ||gm ABCD,
AB = 10cm, altitude DL = 6cm
and BM is altitude on AD, and BM = 8 cm.

Question 3.
Solution:
Diagonals of rhombus are 16cm and 24 cm.
Area = $$\frac { 1 }{ 2 }$$ x product of diagonals
= $$\frac { 1 }{ 2 }$$ x 1st diagonal x 2nd diagonal
= $$\frac { 1 }{ 2 }$$ x 16 x 24
= 192 cm² Ans.

Question 4.
Solution:
Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm

Question 5.
Solution:
from the figure
(i) In ∆ BCD, ∠ DBC = 90°

Question 6.
Solution:
In the fig, ABCD is a trapezium. AB || DC
AB = 7cm, AD = BC = 5cm.
Distance between AB and DC = 4 cm.
i.e. ⊥AL = ⊥BM = 4cm.

Question 7.
Solution:
Given : In quad. ABCD. AL⊥BD and CM⊥BD.

Question 8.
Solution:
In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD

Question 9.
Solution:
Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : ar(∆ AOD) = ar(∆ BOC)

Question 10.
Solution:
Given : In the figure,
DE || BC.

Question 11.
Solution:
Given : In ∆ ABC, D and E are the points on AB and AC such that
ar( ∆ BCE) = ar( ∆ BCD)
To prove : DE || BC.
Proof : (∆ BCE) = ar(∆ BCD)
But these are on the same base BC.
Their altitudes are equal.
Hence DE || BC
Hence proved.

Question 12.
Solution:
Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.

Question 13.
Solution:
A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.
To prove : ar( ∆ ABP) = ar(quad. ABCD).
Const. Join AC

Question 14.
Solution:
Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and
ar( ∆ ABC) = ar( ∆ DBC).

Question 15.
Solution:
Given : In ∆ ABC, AD is the median and P is a point on AD
BP and CP are joined
To prove : (i) ar(∆BDP) = ar(∆CDP)
(ii) ar( ∆ ABP) = ar( ∆ ACP)

Question 16.
Solution:
Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD
To prove : ar(∆ ABC) = ar(∆ ADC)
Proof : In ∆ ABD,

Question 17.
Solution:
In ∆ ABC,D is mid point of BC
and E is midpoint of AD and BE is joined.

Question 18.
Solution:
Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.
To prove : ar( ∆ BEC) = $$\frac { 1 }{ 2 }$$ ar( ∆ ABC)
Proof : In ∆ ABD,
BE is its median
ar(∆ EBD) = ar(∆ ABE)

Question 19.
Solution:
Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.
To prove that ar( ∆ BOE) = $$\frac { 1 }{ 8 }$$ ar(∆ ABC).

Question 20.
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC.
To prove : ar( ∆ AOB) = ar( ∆ AOD)
Const. Join BD which intersects AC at P
Proof : In ∆ OBD,
P is midpoint of BD
(Diagonals of ||gm bisect each other)

Question 21.
Solution:
Given : ABCD is a ||gm.
P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.
PQ, QR, RS and SP are joined.

Question 22.
Solution:
Given : In pentagon ABCDE,
EG || DA meets BA produced and
CF || DB, meets AB produced.
To prove : ar(pentagon ABCDE) = ar(∆ DGF)

Question 23.
Solution:
Given ; A ∆ ABC in which AD is the median.
To prove ; ar( ∆ ABD) = ar( ∆ ACD)
Const : Draw AE⊥BC.
Proof : Area of ∆ ABD

Question 24.
Solution:
Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.

Question 25.
Solution:
Given : In ∆ ABC,
D is a point on BC such that
BD = $$\frac { 1 }{ 2 }$$ DC

Question 26.
Solution:
Given : In ∆ ABC, D is a point on BC such that
BD : DC = m : n

Hope given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C.

Other Exercises

Question 1.
Solution:
It is clear that the given frequency distribution is in exclusive form, we represent the daily wages (in rupees) along x-axis and no. of workers along y-axis. Then we construct rectangles with class intervals as bases and corresponding frequencies as heights as shown given below:

Question 2.
Solution:
We shall take daily earnings along x-axis and number of stores along y-axis. Then we construct rectangles with the given class intervals as bases and corresponding frequency as height as shown in the graph.

Question 3.
Solution:
We shall take heights along x-axis and number of students along y-axis. Then we shall complete the rectangles with the given class intervals as bases and frequency as height and complete the histogram as shown below :

Question 4.
Solution:
We shall take class intervals along x-axis and frequency along y-axis Then we shall complete the rectangles with given class interval as bases and frequency as heights and complete the histogram as shown below.

Question 5.
Solution:
The given frequency distribution is in inclusive form first we convert it into exclusive form at given below.

Now, we shall take class intervals along x-axis and frequency along y-axis and draw rectangles with class intervals as bases and frequency as heights and complete the histogram as shown.

Question 6.
Solution:

Question 7.
Solution:
In the given distribution class intervals are different size. So, we shall calculate the adjusted frequency for each class. Here minimum class size is 4. We know that adjusted
frequency of the class is $$\frac { max\quad class\quad size }{ class\quad size\quad of\quad this\quad class }$$  x its frequency

Question 8.
Solution:
We shall take two imagined classes one 0-10 at the beginning with zero frequency a id other 70-80 at the end with zero frequency.

Now, plot the points” (5, 0), (15, 2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4), (75, 0) on the graph and join them its order to get a polygon as shown below.

Question 9.
Solution:
We take Age (in years) along x-axis and number of patients along y-axis. First we complete the histogram and them join the midpoints of their tops in order

We also take two imagined classes. One 0-10 at the beginning and other 70-80 at the end and also join their midpoints to complete the polygon as shown.

Question 10.
Solution:
We take class intervals along x-axis and frequency along y-axis.
First we complete the histogram and then join the midpoints of the tops of adjacent rectangles in order. We shall take two imagined classes 15-20 at the beginning and 50-55 at the end.
We also join their midpoints to complete the polygon as shown

Question 11.
Solution:
We represent class interval on x-axis and frequency on y-axis. First we construct the histogram and then join the midpoints of the tops of each rectangle by line segments. We shall take two imagined class i.e. 560-600 at the beginning and 840-900 at the end. Join their midpoints also to complete

Question 12.
Solution:
We will take the imagined class 11-0 at the beginning and 61-70 at the end. Each with frequency zero. Thus, we have,

Now we will mark the points A (-5.5, 0), B(5.5, 8), C(15.5, 3) D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0) on the graph and join them in order to get the required frequency polygon.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D.

Other Exercises

Question 1.
Solution:
In ∆ABC,
∠B = 76° and ∠C = 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 76° + 48° = 180°
=> ∠ A + 124° = 180°
=> ∠A= 180° – 124° = 56°

Question 2.
Solution:
Angles of a triangle are in the ratio = 2:3:4
Let first angle = 2x
then second angle = 3x
and third angle = 4x
2x + 3x + 4x = 180°
(Sum of angles of a triangle)
=> 9x = 180°
=> x = $$\frac { { 180 }^{ o } }{ 9 }$$ = 20°
First angle = 2x = 2 x 20° = 40°
Second angle = 3x = 3 x 20° = 60°
and third angle = 4x = 4 x 20° = 80° Ans.

Question 3.
Solution:
In ∆ABC,
3∠A = 4∠B = 6∠C = x (Suppose)

Question 4.
Solution:
In ∆ABC,
∠ A + ∠B = 108° …(i)
∠B + ∠C – 130° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangle)
Subtracting (i) from (iii),
∠C = 180° – 108° = 72°
Subtracting (ii) from (iii),
∠A = 180°- 130° = 50°
But ∠ A + ∠B = 108° (from i)
50° + ∠B = 108°
=> ∠B = 108° – 50° = 58°
Hence ∠A = 50°, ∠B = 58° and ∠C = 72° Ans.

Question 5.
Solution:
In ∆ABC,
∠A+∠B = 125° …(i)
∠A + ∠C = 113° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangles) Subtracting, (i), from (iii),
∠C = 180°- 125° = 55°
Subtracting (ii) from (iii),
∠B = 180°- 113° – 67°
∠A + ∠B = 125°
∠ A + 67° = 125°
=> ∠ A = 125° – 67°
∠A = 58°
Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.

Question 6.
Solution:
In ∆ PQR,
∠ P – ∠ Q = 42°
=> ∠P = 42°+∠Q …(i)
∠Q – ∠R = 21°
∠Q – 21°=∠R …(ii)
But ∠P + ∠Q + ∠R = 180°
(Sum of angles of a triangles)
42° + ∠Q + ∠Q + ∠Q – 21°= 180°
=> 21° + 3∠Q = 180°
=> 3∠Q = 180°- 21° = 159°
from ∠Q = $$\frac { { 159 }^{ o } }{ 3 }$$ = 53°
(i)∠P = 42° + ∠Q = 42° + 53° = 95°
and from (ii) ∠R = ∠Q – 21°
= 53° – 25° = 32°
Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.

Question 7.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of A ABC.
and ∠A + ∠B = 116° …(i)

Question 8.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC
Let ∠ A = ∠ B = x
then ∠C = x + 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
x + x + x + 18° = 180°
=> 3x + 18° = 180°
=> 3x = 180° – 18° = 162°
x = $$\frac { { 162 }^{ o } }{ 3 }$$ = 54°
∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72°
Hence angles are 54°, 54 and 72° Ans.

Question 9.
Solution:
Let the smallest angle of a triangle = x°
their second angle = 2x°
and third angle = 3x°
But sum of angle of a triangle = 180°
x + 2x + 3x = 180°
=> 6x = 180°
=> x – $$\frac { { 180 }^{ o } }{ 6 }$$ = 30°
Hence smallest angle = 30°
Second angle = 2 x 30° = 60°
and third angle = 3 x 30° = 90° Ans.

Question 10.
Solution:
In a right angled triangle.
one angle is = 90°
Sum of other two acute angles = 90°
But one acute angle = 53°
Second acute angle = 90° – 53° = 37°
Hence angle of the triangle with be 90°, 53°, 37° Ans.

Question 11.
Solution:

Given : In ∆ ABC,
∠ A = ∠B + ∠C
To Prove : ∆ABC is a right-angled
Proof : We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
(angles of a triangle)
But ∠ A = ∠ B + C given
∠A + (∠B + ∠C) = 180°
=> ∠A + ∠A = 180°
=> 2∠A = 180°
=> ∠ A = $$\frac { { 180 }^{ o } }{ 2 }$$ = 90°
∠ A = 90°
Hence ∆ ABC is a right-angled Hence proved.

Question 12.
Solution:
Given. In ∆ ABC, ∠A = 90°
AL ⊥ BC.

To Prove : ∠BAL = ∠ACB
Proof : In ∆ ABC, AL ⊥ BC
In right angled ∆ALC,
∠ ACB + ∠ CAL = 90° …(i)
( ∴∠L = 90°)
But ∠ A = 90° ‘
=> ∠ BAL + ∠ CAL = 90° …(ii)
From (i) and (ii),
∠BAL + ∠ CAL= ∠ ACB+ ∠CAL
=> ∠ BAL = ∠ ACB Hence proved.

Question 13.
Solution:
Given. In ∆ABC,
Each angle is less than the sum of the other two angles
∠A< ∠B + ∠C
∠B < ∠C + ∠A
and ∠C< ∠A + ∠C
Proof : ∠ A < ∠B + ∠C
∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180°
(∴ ∠A+∠B+∠C=180°)
∠A < $$\frac { { 180 }^{ o } }{ 2 }$$ => ∠A< 90
Similarly, we can prove that,
∠B < 90° and ∠C < 90°
∴ each angle is less than 90°
Hence, triangle is an acute angled triangle. Hence proved.

Question 14.
Solution:
Given. In ∆ABC,
∠B > ∠A + ∠C

Question 15.
Solution:
In ∆ABC
∠ ABC = 43° and Ext. ∠ ACD = 128°

Question 16.
Solution:
∠ ABC + ∠ ABD = 180°
(Linear pair)

Question 17.
Solution:
(i)In the figure, ∠BAE =110° and ∠ACD = 120°.

Question 18.
Solution:
In the figure,
∠A = 55°, ∠B = 45°, ∠C = 30° Join AD and produce it to E

Question 19.
Solution:
In the figure,
∠EAC = 108°,
AD divides ∠ BAC in the ratio 1 : 3
∠EAC + ∠ BAC = 180°
(Linear pair)

Question 20.

Solution:
Sides BC, CA and AB
are produced in order forming exterior
angles ∠ ACD,

Question 21.
Solution:
Given : Two ∆ s DFB and ACF intersect each other as shown in the figure.
To Prove : ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Proof : In ∆ DFB,
∠D + ∠F + ∠B = 180°
(sum of angles of a triangle)
Similarly,in ∆ ACE
∠A + ∠C + ∠E = 180° …(ii)
Adding (i) and (ii), we get :
∠D + ∠F + ∠B+ ∠A+ ∠C + ∠E = 180° + 180°
=> ∠A+∠B+∠C+∠D+∠E + ∠ F = 360°
Hence proved.

Question 22.
Solution:
In the figure,
ABC is a triangle
and OB and OC are the angle
bisectors of ∠ B and ∠ C meeting each other at O.
∠ A = 70°
In ∆ ABC,
∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)

Question 23.
Solution:
In ∆ABC, ∠ A = 40°
Sides AB and AC are produced forming exterior angles ∠ CBD and ∠ BCE

Question 24.
Solution:
In the figure, ∆ABC is triangle and ∠A : ∠B : ∠C = 3 : 2 : 1
AC ⊥ CD.
∠ A + ∠B + ∠C = 180°
(sum of angles of a triangle)
But ∠A : ∠B : ∠C = 3 : 2 : 1

Question 25.
Solution:
In ∆ ABC
AN is the bisector of ∠ A
∠NAB =$$\frac { 1 }{ 2 }$$ ∠A.
Now in right angled ∆ AMB,
∠B + ∠MAB = 90° (∠M = 90°)

Question 26.
Solution:
(i) False: As a triangle has only one right angle
(ii) True: If two angles will be obtuse, then the third angle will not exist.
(iii) False: As an acute-angled triangle all the three angles are acute.
(iv) False: As if each angle will be less than 60°, then their sum will be less than 60° x 3 = 180°, which is not true.
(v) True: As the sum of three angles will be 60° x 3 = 180°, which is true.
(vi) True: A triangle can be possible if the sum of its angles is 180°
But the given triangle having angles 10° + 80° + 100° = 190° is not possible.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C.

Other Exercises

Question 1.
Solution:
Given : In trapezium ABCD,
AB || DC and E is the midpoint of AD.
A line EF ||AB is drawn meeting BC at F.
To prove : F is midpoint of BC

Question 2.
Solution:
Given : In ||gm ABCD, E and F are the mid points of AB and CD respectively. A line segment GH is drawn which intersects AD, EF and BC at G, P and H respectively.
To prove : GP = PH

Question 3.
Solution:
Given : In trapezium ABCD, AB || DC
P, Q are the midpoints of sides AD and BC respectively
DQ is joined and produced to meet AB produced at E
Join AC which intersects PQ at R.

Question 4.
Solution:
Given : In ∆ ABC,
AD is the mid point of BC
DE || AB is drawn. BE is joined.
To prove : BE is the median of ∆ ABC.
Proof : In ∆ ABC

Question 5.
Solution:
Given : In ∆ ABC, AD and BE are the medians. DF || BE is drawn meeting AC at F.
To prove : CF = $$\frac { 1 }{ 4 }$$ BC.

Question 6.
Solution:
Given : In ||gm ABCD, E is mid point of DC.
EB is joined and through D, DEG || EB is drawn which meets CB produced at G and cuts AB at F.
To prove : (i)AD = $$\frac { 1 }{ 2 }$$ GC

Question 7.
Solution:

Given : In ∆ ABC,
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and FD are joined.

Question 8.
Solution:
Given : In ∆ ABC, D, E and F are the mid points of sides BC, CA and AB respectively

Question 9.
Solution:
Given : In rectangle ABCD, P, Q, R and S are the midpoints of its sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
To prove : PQRS is a rhombus.

Question 10.
Solution:
Given : In rhombus ABCD, P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.

Question 11.
Solution:
Given : In square ABCD, P,Q,R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

Question 12.
Solution:
Given : In quadrilateral ABCD, P, Q, R and S are the midpoints of PQ, QR, RS and SP respectively PR and QS are joined.
To prove : PR and QS bisect each other
Const. Join PQ, QR, RS and SP and AC
Proof : In ∆ ABC,

But diagonals of a ||gm bisect each other PR and QS bisect each other.
∴ PR and QS bisect each other

Question 13.
Solution:
Given : ABCD is a quadrilateral. Whose diagonals AC and BD intersect each other at O at right angles.
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, QS and SP are joined.

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A.

Question 1.
Solution:
Base of the triangle (b) = 24cm and height (h) = 14.5 cm
∴ Area = $$\frac { 1 }{ 2 }$$ x b x h = $$\frac { 1 }{ 2 }$$ x 24 x 14.5 cm²
= 174 cm² Ans.

Question 2.
Solution:
Let the length of altitude of the triangular field = x then its base = 3x.

Question 3.
Solution:
Sides of a triangle = 42cm, 34cm and 20cm
Let a = 42cm, b = 34cm and c = 20 cm

Question 4.
Solution:
Sides of the triangle = 18cm, 24cm and 30cm
Let a = 18 cm, b = 24 cm and c = 30cm

Question 5.
Solution:
Sides of triangular field ABC arc 91m, 98m and 105m
Let AC be the longest side
∴ BD⊥AC
Here a = 98m, b = 105m and c = 91m

Question 6.
Solution:
Perimeter of triangle = 150m
Ratio in the sides = 5:12:13
Let sides be 5x, 12x and 13x

Question 7.
Solution:
Perimeter of a triangular field = 540m
Ratio is its sides = 25 : 17 : 12

Question 8.
Solution:
Perimeter of the triangular field = 324 m
Length of the sides are 85m and 154m

Question 9.
Solution:
Length of sides are
13 cm, 13 cm and 20cm

Question 10.
Solution:
Base of the isosceles triangle ABC = 80cm
Area = 360 cm²

Question 11.
Solution:
Perimeter of the triangle
ABC = 42 cm.
Let length of each equal sides = x

Question 12.
Solution:
Area of equilateral triangle = 36√3 cm².
Let length of each side = a

Question 13.
Solution:
Area of equilateral triangle = 81√3 cm²
Let length of each side = a

Question 14.
Solution:
∆ ABC is a right angled triangle, right angle at B.
∴ BC 48cm and AC = 50cm

Question 15.
Solution:
Each side of equilateral triangle
(a) = 8cm.

Question 16.
Solution:
Let a be the each side of
the equilateral triangle.

Question 17.

Solution:
The given umbrella has 12 triangular pieces of the size 50cm x 20cm x 50cm. We see that each piece is of an isosceles triangle shape and we have to find firstly area of one such triangle.

Question 18.
Solution:
The given floral design is made of 16 tiles
The size of each tile is 16cm 12cm, 20cm
Now we have to find the area of firstly one tile

Question 19.

Solution:

Question 20.
Solution:
In the figure, ABCD is a quadrilateral
AB = 42 cm, BC = 21 cm, CD = 29 cm,
DA = 34 cm and ∠CBD = 90°

Question 21.
Solution:
from the figure
∆DAB

Question 22.
Solution:

Question 23.
Solution:
from the figure,
We know that

Question 24.
Solution:

Hope given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A.

Question 1.
Solution:
Steps of Constructions :
(i) Draw a line segment AB = 5cm.
(ii) With A as centre and a radius equal to more than half of AB, drawn two arcs one above and other below of AB.

(iii) With centre B, and with same radius, draw two arcs intersecting the previously arcs at C and D respectively.
(iv) Join CD, intersecting AB at P.
Then CD is the perpendicular bisector of AB at the point P.

Question 2.
Solution:
Steps of constructions.
(i) Draw a line segment AB.
(ii) With A as centre and with small radius drawn arc cutting AB at P.
(iii) With P as centre and same radius draw another arc cutting the previous arc at Q and then R.

(iv) Bisect arc QR at S.
(v) Join AS and produce it to X such that ∠ BAX = 90°.
(vi) Now with centres P and S and with a suitable radius, draw two arcs intersecting each other at T.
(vii) Join AT and produced it to C Then ∠BAC = 45°.
(viii) Again with centres P and T and suitable radius draw two arcs intersecting each at D.
AD is the bisector of ∠ BAC

Question 3.
Solution:
Steps of construction.
(i) Draw a line segment AB.
(ii) With centre A and same radius draw an arc which meets AB at P.
(iii) With centre P and same radius, draw arcs first at Q and then at R.
(iv) With centres Q and R, draw arcs intersecting each other at C intersecting the first arc at T.

(v) Join AC
Then ∠BAC = 90°
(vi) Now with centres P and T and with some suitable radius, draw two arcs intersecting each other at L.
(vii) Join AL and produce it to D.
Then AD is the bisector of ∠ BAC.

Question 4.
Solution:
Steps of construction.
(i) Draw a line segment BC = 5cm.
(ii) With centres B and C and radius
5cm, draw two arcs intersecting each other at A.

(iii) Join AB and AC.
Then ∆ ABC is the required equilateral triangle.

Question 5.
Solution:
We know that altitudes of equilateral triangle are equal and each angle is 60°.
Steps of construction.
(i) Draw a line XY and take a point D on it.
(ii) At D, draw a perpendicular and cut off DA = 5.4cm.

(iii) At A draw angles of 30° on each side of AD which meet XY at B and C respectively.
Then ∆ ABC is the required triangle.

Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5cm
(ii) With centre B and radius 3.8 cm draw an arc.

(iii) With centre C and radius 2.6 cm draw another arc intersecting the first arc at A.
(iv) Join AB and AC.
Then ∆ ABC is the required triangle.

Question 7.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.7cm.
(ii) At B, draw a ray BX making an angle of 60° with BC.

(iii) At C, draw another ray, CY making an angle of 30° which intersects the ray BX at A ,
Then ∆ ABC is the required triangle On measuring ∠ A, it is 90°.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment QR * 5cm.
(ii) With centres Q and R and radius equal to 4.5cm, draw arcs intersecting eachother at P.

(iii) Join PQ and PR.
Then ∆ PQR is the required triangle.

Question 9.
Solution:
We know that in an isosceles triangle, two sides are equal and so their opposite angles are also equal.

Question 10.
Solution:
Steps of constructions :
(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 90° with BC.

(iii) With centre C and radius 5.3 cm, draw an arc intersecting BX at A.
(iv) Join AC.
Then ∆ ABC is the required right angled triangle.

Question 11.
Solution:
Steps of constructions :
(i) Draw a line XY.
(ii) Take a point D on XY.

(iii) Draw a perpendicular at D and cut off DA = 4.8 cm
(iv) At A, draw a line LM parallel to XY.
(v) At A, draw an angle of 30° with LM on one side and an angle of 60° with LM on other side meeting XY at B and C respectively
Then ∆ ABC is the required triangle.

Question 12.
Solution:
Steps of constructions :
(i) Draw a line segment EF = 12cm.
(ii) At E, draw a ray EX making an acute angle with EF.

(iii)From EX,cut off 3+2+4=9 equal parts.
(iv) Join E9 F.
(v) From E5 and E3, draw lines parallel to E9 F meeting EF at C and B respectively.
(vi) With centre B and radius EB and with centre C and radius CF, draw arcs intersecting eachother at A.
(vii) Join AB and AC.
Then ∆ ABC is the required triangle.

Question 13.
Solution:
Steps of constructions :

(i) Draw a line segment BC = 4.5cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BD = 8cm.
(iii) Join DC.
(iv) Draw the perpendicular bisector of BD which intersects BX at A.
(v) Join AC.
Then ∆ ABC is the required triangle.

Question 14.
Solution:
Steps of Constructions :
(i) Draw a line segment BC = 5.2 cm.
(ii) At B draw a ray BX making an angle of 30°.

(iii) From BX, cut off BD = 3.5cm.
(iv) Join DC.
(v) Draw perpendicular bisector of DC which intersects BX at A.
(vi) Join AC.
Then ∆ ABC is the required triangle.

Hope given RS Aggarwal Class 9 Solutions Chapter 12 Geometrical Constructions Ex 12A are helpful to complete your math homework.

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