RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

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 9A.

Other Exercises

Question 1.
Solution:
We know that sum of angles of a quadrilateral is 360°
Now, sum of three angles = 56° + 115° + 84° = 255°
Fourth angle = 360° – 255° = 105° Ans.

Question 2.
Solution:
Sum of angles of a quadrilateral = 360°
Their ratio = 2 : 4 : 5 : 7
Let first angle = 2x
then second angle = 4x
third angle = 5x
and fourth angle = 7x
∴ 2x + 4x + 5x + 7x = 360°
=> 18x = 360°
=> x = \(\frac { { 360 }^{ o } }{ 18 } \) = 20°
Hence, first angle = 2x = 2 x 20° = 40°
Second angle = 4x = 4 x 20° = 80°
Third angle = 5x = 5 x 20° = 100°
and fourth angle = 7x = 7 x 20° = 140°Ans.

Question 3.
Solution:
In the trapezium ABCD
DC || AB
∴ ∠ A + ∠ D = 180° (Co-intericr angles)
∴ 55°+ ∠D = 180°
∠D = 180° – 55°
∴ ∠D = 125°
Similarly, ∠B + ∠C = 180°
(Co-interior angles)
=> 70° + ∠C = 180°
=> ∠C = 180° – 70°
∠C = 110°
Hence ∠C = 110° and ∠D = 125° Ans.

Question 4.
Solution:
Given : In the figure, ABCD is a square and ∆ EDC is an equilateral triangles on DC. AE and BE are joined.
To Prove : (i) AE = BE
(ii) ∠DAE = 15°
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q4.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q4.2

Question 5.
Solution:
Given : In the figure,
BM ⊥ AC, DN ⊥ AC.
BM = DN
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q5.1

Question 6.
Solution:
Given : In quadrilateral ABCD,
AB = AD and BC = DC
AC is joined.
To Prove : (i) AC bisects ∠ A and ∠ C
(ii) BE = DE
(iii) ∠ABC = ∠ADC
Const. Join BD.
Proof : (i) In ∆ ABC and ∆ ADB
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q6.1

Question 7.
Solution:
Given : In square ABCD,
∠ PQR = 90°
PB = QC = DR
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q7.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q7.2

Question 8.
Solution:
Given : In quadrilateral ABCD, O is any point inside it. OA, OB, OC and OD are joined.
To Prove : OA + OB + OC + OD > AC + BD
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q8.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q8.2

Question 9.
Solution:
Given : In quadrilateral ABCD, AC is its one diagonal.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.2
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.3

Question 10.
Solution:
Given : A quadrilateral ABCD
To Prove : ∠A + ∠B + ∠C + ∠D = 360°
Const. Join AC.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q10.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q10.2

 

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

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.
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 1

Question 2.
Solution:
In the given equation.
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 2
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
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 2.1
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
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3.2

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
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.2
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.3
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.4
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.5

Question 5.
Solution:
In the given equation
2x – 3y = 5
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5.2

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
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 6
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 6.1

Question 7.
Solution:
In the given equation
3x + 2y = 6
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7.2

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RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A

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.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q2.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q2.2

Question 3.
Solution:
Sides of a triangle = 42cm, 34cm and 20cm
Let a = 42cm, b = 34cm and c = 20 cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q3.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q3.2

Question 4.
Solution:
Sides of the triangle = 18cm, 24cm and 30cm
Let a = 18 cm, b = 24 cm and c = 30cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q4.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q4.2

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
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q5.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q5.2

Question 6.
Solution:
Perimeter of triangle = 150m
Ratio in the sides = 5:12:13
Let sides be 5x, 12x and 13x
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q6.1

Question 7.
Solution:
Perimeter of a triangular field = 540m
Ratio is its sides = 25 : 17 : 12
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q7.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q7.2

Question 8.
Solution:
Perimeter of the triangular field = 324 m
Length of the sides are 85m and 154m
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q8.1

Question 9.
Solution:
Length of sides are
13 cm, 13 cm and 20cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q9.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q9.2

Question 10.
Solution:
Base of the isosceles triangle ABC = 80cm
Area = 360 cm²
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q10.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q10.2

Question 11.
Solution:
Perimeter of the triangle
ABC = 42 cm.
Let length of each equal sides = x
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q11.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q11.2
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q11.3

Question 12.
Solution:
Area of equilateral triangle = 36√3 cm².
Let length of each side = a
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q12.1

Question 13.
Solution:
Area of equilateral triangle = 81√3 cm²
Let length of each side = a
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q13.1

Question 14.
Solution:
∆ ABC is a right angled triangle, right angle at B.
∴ BC 48cm and AC = 50cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q14.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q14.2

Question 15.
Solution:
Each side of equilateral triangle
(a) = 8cm.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q15.1

Question 16.
Solution:
Let a be the each side of
the equilateral triangle.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q16.1

Question 17.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q17.1
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.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q17.2
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q17.3

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
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q18.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q18.2
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q18.3

Question 19.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q19.1
Solution:
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q19.2
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q19.3

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°
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q20.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q20.2

Question 21.
Solution:
from the figure
∆DAB
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q21.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q21.2

Question 22.
Solution:
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q22.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q22.2

Question 23.
Solution:
from the figure,
We know that
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q23.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q23.2

Question 24.
Solution:
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q24.1
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q24.2

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

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.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q1.1

Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q2.1

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.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q5.1

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
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q6.1

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
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q7.1

Question 8.
Solution:
In the given equation,
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.1
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
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.2
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
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q9.1

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RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A

RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A.

Question 1.
Solution:
In ∆ ABC, ∠A = 70° and AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 1
=> 70° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B = 180°- 70° = 110°
∠B = \(\frac { { 110 }^{ o } }{ 2 } \) = 55° and
Hence ∠B = 55°,∠C = 55° Ans.

Question 2.
Solution:
In ∆ ABC, ∠A= 100°
It is an isosceles triangle
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 2
∴AB = AC
∠B = ∠C
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> 100° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B + 100° = 180°
=> 2∠B = 180°- 100° = 80°
∠B = \(\frac { { 80 }^{ o } }{ 2 } \) = 40°
and ∠C = 40°
∴ Base angles are 40°, 40° Ans.

Question 3.
Solution:
In ∆ ABC,
AB = AC
∴∠C = ∠B
(Angles opposite to equal sides)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 3
But ∠B = 65°
∴ ∠ C = 65°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 65° + 65° = 180°
=> ∠ A + 130° = 180°
=> ∠ A = 180° – 130°
=> ∠ A = 50°
Hence ∠ A = 50° and ∠ C = 65° Ans.

Question 4.
Solution:
In ∆ ABC
AB = AC
∴ ∠C = ∠B
(Angles opposite to equal sides)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 4
But ∠A = 2(∠B + ∠C)
=> ∠B + ∠C = \(\frac { 1 }{ 2 } \) ∠A 2
But ∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
=> ∠A+ \(\frac { 1 }{ 2 } \) ∠A = 180°
=> \(\frac { 3 }{ 2 } \) ∠A = 180°
=> ∠A = 180° x \(\frac { 2 }{ 3 } \) = 120°
and ∠B + ∠C = \(\frac { 1 }{ 2 } \) ∠A = \(\frac { 1 }{ 2 } \) x 120°
= 60°
∴ ∠ B = ∠ C
∴ ∠ B = ∠ C = \(\frac { { 60 }^{ o } }{ 2 } \) = 30°
Hence ∠ A = 120°, ∠B = 30°, ∠C = 30° Ans.

Question 5.
Solution:
In ∆ ABC,
AB = BC and ∠B = 90°
∴ AB = BC
∴ ∠ C = ∠ A
(Angles opposite to equal sides)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 5
Now ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠ A + 90° + ∠ A = 180°
(∴ ∠C = ∠A)
=> 2∠A + 90° – 180°
=> 2∠ A = 180° – 90° = 90°
∠ A = \(\frac { { 90 }^{ o } }{ 2 } \) = 45°
∴ ∠ C = 45° (∴ ∠ C = ∠ A)
Hence, each of the equal angles is 45° Ans.

Question 6.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 6
Base BC is produced to both sides upto D and E respectively forming exterior angles
∠ ABD and ∠ ACE
To Prove : ∠ABD = ∠ACE
Proof : In ∆ ABC
∴ AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
=> ∠ ABC = ∠ACB
But ∠ ABC + ∠ABD = 180° (Linear pair)
Similarly ∠ACB + ∠ACE = 180°
∠ ABC + ∠ABD = ∠ACB + ∠ACE
But ∠ ABC = ∠ ACB (proved)
∠ABD = ∠ACE
Hence proved.

Question 7.
Solution:
∆ ABC is an equilateral triangle
∴ AB = BC = CA
and ∠A = ∠B = ∠C = 60°
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 7
The sides of the ∆ ABC are produced in order to D,E and F forming exterior angles
∠ACP, ∠BAE and ∠CBF
∆ACD + ∠ACB = 180°
=> ∠ACD + 60° = 180°
=> ∠ACD = 180° – 60°
=> ∠ACD = 120°
Similarly, ∠BAE + ∠BAC = 180°
=> ∠BAE + 60° = 180°
=> ∠BAE = 180° – 60° = 120°
and ∠BAF + ∠ABC = 180°
=> ∠BAF + 60° = 180°
=> ∠BAF = 180° – 60° = 120°
Hence each exterior angle of an equilateral triangle is 120°.

Question 8.
Solution:
Given : In the figure,
O is mid-point of AB and CD.
i.e. OA = OB and OC = OD
To Prove : AC = BD and AC || BD.
Proof : In ∆ OAC and ∆ OBD,
OA = OB {given}
OC = OD {given}
∠AOC = ∠BOD
(Vertically opposite angles)
∴∆ OAC ≅ ∆ OBD (S.A.S. axiom)
∴AC = BD (c.p.c.t.)
and ∠ C = ∠ D
But these are alt. angles
∴AC || BD
Hence proved.

Question 9.
Solution:
Given : In the figure,
PA ⊥ AB, QB ⊥ AB and
PA = QB, PQ intersects AB at O.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 9
To Prove : O is mid-point of AB and PQ.
Proof : In ∆ AOP and ∆ BOQ,
∠ A = ∠ B
AP = BQ (given)
∠ AOP = ∠BOQ
(Vertically opposite angles)
∴ ∆AOP ≅ ∆ BOQ (A.A.S. axiom)
∴OA = OB (c.p.c.t)
and OP = OQ (c.p.c.t)
Hence, O is the mid-point of AB as well as PQ

Question 10.
Solution:
Given : Two line segments AB and CD intersect each other at O and OA = OB, OC = OD
AC and BD are joined.
To Prove : AC = BD
Proof : In ∆ AOC and ∆ BOD,
OA = OB {given}
OC = OD
∠AOC = ∠BOD
(vertically opposite angles)
∴ ∆ AOC ≅ ∆ BOD (S.A.S. axiom)
∴ AC = BD (c.p.c.t)
and ∠A = ∠D (c.p.c.t.)
Hence AC ≠ BD
Hence proved.

Question 11.
Solution:
Given : In the given figure,
l || m, m is mid-point of AB
CD is another line segment, which intersects AB at M.
To Prove : M is mid-point of CD
Proof : l || m
∴ ∠ CAM = ∠ MBD (Alternate angles)
Now, in ∆ AMC and ∆ BMD,
AM = MB (Given)
∠ CAM = ∠ MBD (proved)
∠ AMC = ∠BMD
(vertically opposite angles)
∆ AMC ≅ ∆ BMD (ASA axiom)
∴ CM = MD (c.p.c.t.)
Hence, M is mid-point of CD.
Similarly we can prove that M is mid point of any other line whose end points are on l and m.
Hence proved.

Question 12.
Solution:
Given : In ∆ ABC, AB = AC and in ∆ OBC,
OB = OC
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 12
To Prove : ∠ABO = ∠ACO
Construction. Join AO.
Proof : In ∆ ABO and ∆ ACO,
AB = AC (Given)
OB = OC (Given)
AO = AO (Common)
∴ ∆ ABO ≅ ∆ ACO (S.S.S. Axiom)
∴ ∠ABO = ∠ACO (c.p.c.t.)
Hence proved.

Question 13.
Solution:
Given : In ∆ ABC, AB = AC
D is a point on AB and a line DE || AB is drawn.
Which meets AC at E
To Prove : AD = AE
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 13
Proof : In ∆ ABC
AB = AC (given)
∴ ∠ C = ∠ B (Angles opposite to equal sides)
But DE || BC (given)
∴ ∠ D = ∠ B {corresponding angles}
and ∠ E = ∠ C
But ∠C = ∠B
∴ ∠E = ∠D
∴ AD = AE (Sides opposite to equal angles)
Hence proved.

Question 14.
Solution:
Given : In ∆ ABC,
AB = AC
X and Y are two points on AB and AC respectively such that AX = AY
To Prove : CX = BY
Proof : In ∆ AXC and ∆ AYB
AC = AB (given)
AX = AY (given)
∠ A = ∠ A (common)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 14
∴ ∆ AXC ≅ ∆ AYB (S.A.S. axiom)
∴ CX = BY (c.p.c.t.)
Hence proved.

Question 15.
Solution:
Given : In the figure,
C is mid point of AB
∠DCA = ∠ECB and
∠DBC = ∠EAC
To Prove : DC = EC
Proof : ∠ DCA = ∠ FCB (given)
Adding ∠ DCE both sides,
∠DCA +∠DCE = ∠DCE + ∠ECB
=> ∠ACE = ∠ BCD
Now, in ∆ ACE and ∆ BCD,
AC = BC (C is mid-point of AB)
∠EAC = ∠DBC (given)
∠ACE =∠ BCD (proved)
∴ ∆ ACE ≅ ∆ BCD (ASA axiom)
∴ CE = CD (c.p.c.t.)
=> EC = DC
or DC = EC
Hence proved.

Question 16.
Solution:
Given : In figure,
BA ⊥ AC, DE ⊥ EF .
BA = DE and BF = DC
To Prove : AC = EF
Proof : BF = DC (given)
Adding FC both sides,
BF + FC = FC + CD
=> BC = FD.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 16
Now, in right-angled ∆ ABC and ∆ DEF,
Hyp. BC = Hyp. FD (proved)
Side AB = side DE (given)
∴ ∆ ABC ≅ ∆DEF (RHS axiom)
∴ AC = EF (c.p.c.t.)
Hence proved.

Question 17.
Solution:
To prove : AE = CD
Proof: x° + ∠ BDC = 180° (Linear pair)
Similarly y°+ ∠AEB = 180°
∴ x° + ∠BDC = y° + AEB
But x° = y° (given)
∠ BDC = ∠ AEB
Now, in ∆ AEB and ∆ BCD,
AB = CB (given)
∠B = ∠B (common)
∠ AEB = ∠ BDC (proved)
∴ ∆ AEB ≅ ∆ BCD (AAS axiom)
∴ AE = CD. (c.p.c.t.)
Hence proved.

Question 18.
Solution:
Given : In ∆ ABC,
AB = AC.
Bisectors of ∠ B and ∠ C meet AC and AB in D and E respectively
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 18
To Prove : BD = CE
Proof : In ∆ ABC,
AB = AC
∠ B = ∠ C (Angles opposite to equal sides)
Now, in ∆ ABD and ∆ ACE,
AB = AC (given)
∠ A = ∠ A (common)
∠ ABD = ∠ACE
(Half of equal angles)
∴ ∆ A ABD ≅ ∆ ACE (ASA axiom)
∴ BD = CE (c.p.c.t)
Hence proved.

Question 19.
Solution:
Given : In ∆ ABC,
AD is median. BL and CM are perpendiculars on AD and AD is produced to E
To prove : BL = CM.
Proof : In ∆ BLD and ∆ CMD,
BD = DC (D is mid-point of BC)
∠LDB = ∠CDM
(Vertically opposite angles)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 19
∠L = ∠M (each 90°)
∴ ∆ BLD ≅ ∆ CMD (A.A.S. axiom)
Hence, BL = CM (c.p.c.t)
Hence proved.

Question 20.
Solution:
Given : In ∆ ABC, D is mid-point of BC. DL ⊥ AB and DM ⊥ AC and DL = DM
To prove : AB = AC
Proof: In right angled ∆ BLD and ∆ CMD
Hyp. DL = DM (given)
Side BD = DC (D is mid-point of BC)
∴ ∆ BLD ≅ ∆ CMD (R.H.S. axiom)
∴ ∠B = ∠C (c.p.c.t.)
∴ AC = AB
(sides opposite to equal angles)
Hence AB = AC
Hence proved.

Question 21.
Solution:
Given : In ∆ AB = AC and bisectors of ∠B and ∠C meet at a point O. OA is joined.
To Prove : BO = CO and Ray AO is the bisector of ∠ A
Proof : AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
=> \(\frac { 1 }{ 2 } \) ∠C = \(\frac { 1 }{ 2 } \) ∠B
=>∠OBC = ∠OCB
∴ in ∆OBC,
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 21
OB = OC (Sides opposite to equal angles)
Now in ∆ OAB and ∆ OCA
OB = OC (proved)
OA = OA (common)
AB = AC (given)
∴ ∆ OAB ≅ ∆ OCA (S.S.S. axiom)
∴ ∠OAB = ∠OAC (c.p.c.t.)
Hence OA is the bisector of ∠ A.
Hence proved.

Question 22.
Solution:
Given : ∆ PQR is an equilateral triangle and QRST is a square. PT and PS are joined.
To Prove : (i) PT = PS
(ii) ∠PSR = 15°
Proof : In ∆ PQT and ∆ PRS,
PQ = PR (Sides of equilateral ∆ PQR)
QT = RS (sides of in square PRST) .
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 22
∠PQT = ∠PRS
(each angle = 90° + 60° = 150°)
∴ ∆ PQT ≅ ∆ PRS (S.A.S. axiom)
∴ PT = PS (c.p.c.t)
In ∆ PRS, ∠PRS = 60° + 90° = 150°
∠ RPS + ∠ PSR = 180° – 150° = 30°
But ∠RPS = ∠PSR ( ∴PR = RS)
∴∠PSR + ∠PSR = 30°
=> 2∠PSR = 30°
∴ ∠PSR = \(\frac { { 30 }^{ o } }{ 2 } \) = 15°
Hence proved.

Question 23.
Solution:
Given : In right angle ∆ ABC, ∠B is right angle. BCDE is square on side BC and ACFG is also a square on AC.
AD and BF are joined.
To Prove : AD = BF
Proof : ∠ACF = ∠BCD (Each 90°)
Adding ∠ ACB both sides,
∠ ACF + ∠ ACB = ∠ BCD + ∠ ACB
=> ∠ BCF = ∠ ACD
Now in ∆ ACD and ∆ BCF,
AC = CF (sides of a squares)
CD = BC (sides of a square)
∴∠ ACD = ∠ BCF (proved)
∴ ∆ ACD ≅ ∆ BCF (S.A.S. axiom)
AD = BF (c.p.c.t)
Hence proved.

Question 24.
Solution:
Given : ∆ ABC is an isosceles in which AB = AC. and AD is the median which meets BC at D.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 24
To Prove : AD is the bisector of ∠ A
Proof : In ∆ ABD and ∆ ACD,
AD = AD (Common)
AB = AC (Given)
BD = CD (D is mid-point of BC)
∴ ∆ ABD ≅ ∆ ACD (S.S.S. axiom)
∠ BAD = ∠ CAD (c.p.c.t.)
Hence, AD in the bisector of ∠ A.

Question 25.
Solution:
Given : ABCD is a quadrilateral in which AB || DC. P is mid-point of BC. AP and DC are produced to meet at Q.
To Prove : (i) AB = CQ.
(ii) DQ = DC + AB
Proof:
(i) In ∆ ABP and ∆ CPQ,
BP = PC (P is mid-point of BC)
∠ BAP = ∠ PQC (Alternate angles)
∠ APB = ∠ CPQ
(Vertically opposite angles)
∴ ∆ ABP ≅ ∆ CPQ (A.A.S. axiom)
∴ AB = CQ. (c.p.c.t.)
(ii) Now DQ = DC + CQ
=> DQ = DC + AB ( CQ = AB proved)
Hence proved.

Question 26.
Solution:
Given : In figure,
OA = OB and OP = OQ.
To Prove : (i) PX = QX (ii) AX = BX
Proof : In ∆OAQ and ∆OBP,
OA = OB. (Given)
OQ = OP (Given)
∠O = ∠O (Common)
∴ ∆ OAQ ≅ ∆ OBP (SAS axiom)
∴ ∠ A = ∠ B (c.p.c.t)
Now OA = OB and OP = OQ
Subtracting
OA – OP = OB – OQ
=>PA = QB
Now, in ∆ XPA and ∆ XQB,
PA = QB (Proved)
∠AXP = ∠BXQ
(Vertically opposite angles)
∠ A = ∠ B (Proved)
∴ ∆ XPA ≅ ∆ XQB (A.A.S. axiom)
∴ AX = BX (c.p.c.t.) and PX = QX (c.p.c.t.)
Hence proved.

Question 27.
Solution:
Given : ABCD is a square in which a point P is inside it. Such that PB = PD.
To prove : CPA is a straight line.
Proof : In ∆ APB and ∆ ADP,
AB = AD (Sides of a square)
AP = AP (Common)
PB = PD (Given)
∴ ∆ APB ≅ ∆ ADP (S.S.S. axiom)
∠APD = ∠ APB (c.p.c.t) …(i)
Similarly, in ∆ CBP and ∆ CPD,
CB = CD (Sides of a square)
CP – CP (Common)
PB = PD (Given)
∴ ∆ CBP ≅ ∆ CPD (S.S.S. axiom)
∴ ∠BPC = ∠CPD (c.p.c.t.) …(i)
Adding (i) and (ii),
∴ ∠ APD + ∠ CPD = ∠ APB + ∠ BPC
∠APC = ∠APC
∠APC = 180°
(v sum of angles at a point is 360°)
APC or CPA is a straight line.
Hence proved.

Question 28.
Solution:
Given :∆ ABC is an equilateral triangle PQ || AC and AC is produced to R such that CR = BP
To prove : QR bisects PC
Proof : ∴ PQ || AC
∴ ∠BPQ = ∠BCA
(Corresponding angles)
But ∠ BCA = 60°
(Each angle of the equilateral triangle)
and ∠ ABC or ∠QBP = 60°
∴ ∆ BPQ is also an equilateral triangle.
∴ BP = PQ
But BP = CR (Given)
∴ PQ = CR
Now in ∆ PQM and ∆ RMC,
PQ = CR (proved)
∠QMP = ∠RMC
(vertically opposite angles)
∠ PQM = ∠ MRC (alternate angles)
∴ ∆ PQM ≅ ∆ RMC (AAS axiom)
∴ PM = MC (c.p.c.t.)
Hence, QR bisects PC.
Hence proved

Question 29.
Solution:
Given : In quadrilateral ABCD,
AB = AD and BC = DC
AC and BD are joined.
To prove : (i) AC bisects ∠ A and ∠ C
(ii) AC is perpendicular bisector of BD.
Proof : In ∆ ABC and ∆ ADC,
AB = AD (given)
BC = DC (given)
AC = AC (common)
∴ ∆ ABC ≅ ∆ ADC (S.S.S. axiom)
∴ ∠BCA = ∠DCA (c.p.c.t.)
and ∠BCA = ∠DAC (c.p.c.t.)
Hence AC bisects ∠ A and ∠ C
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 29
(ii) In ∆ ABO and ∆ ADO,
AB = AD (given)
AO = AO (common)
∠BAO = ∠DAO
(proved that AC bisects ∠ A)
∴ ∆ ABO ≅ ∆ ADO (SAS axiom)
∴ BO = OD and ∠AOB = ∠AOD (c.p.c.t.)
But ∠ AOB + ∠ AOD = 180° (linear pair)
∠AOB = ∠AOD = 90°
Hence AC is perpendicular bisector of BD. Hence proved.

Question 30.
Solution:
Given : In ∆ ABC,
Bisectors of ∠ B and ∠ C meet at I
From I, IP ⊥ BC. IQ ⊥ AC and IR ⊥ AB.
IA is joined.
To prove : (i) IP = IQ = IR
(ii) IA bisects ∠ A.
Proof : (i) In ∆ BIP and ∆ BIR
BI = BI (Common)
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 30
and ∠P = ∠R (Each 90°)
∴ ∠ IBP ≅ ∠ IBR
( ∴ IB is bisector of ∠ B)
∴ ∆ BIP = ∆ BIR (A.A.S. axiom)
∴ IP = IR (c.p.c.t) …(i)
Similarly, in ∆ CIP and ∆ CIQ,
CI = CI (Common)
∠P = ∠Q (each = 90°) and ∠ICP = ∠ ICR
( IC is bisector of ∠ C)
∴ ∆CIP ≅ ∆CIQ (A.A.S. axiom)
∴ IP = IQ (c.p.c.t.) …(ii)
From (i) and (ii),
IP = IQ = IR.
(i) In right angled ∆ IRA and ∆ IQA,
Hyp. IA = IA (Common)
side IR = IQ (Proved)
∴ ∆ IRA ≅ ∆ IQA (R.H.S. axiom)
∴ ∠IAR = ∠ IRQ (c.p.c.t.)
Hence IA is the bisector of ∠ A
Hence proved.

Question 31.
Solution:
Given. P is a point in the interior of ∠ AOB
PL ⊥ OA and PM ⊥ OC are drawn and PL = PM.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 31
To prove : OP is the bisector of ∠ AOB
Proof: In right angled ∆ OPL and ∆ OPM,
Hyp. OP OP (Common)
Side PL = PM (Given)
∴ ∆ OPL ≅ ∆ OPM (R.H.S. axiom)
∴ ∠POL = ∠POM (c.p.c.t.)
Hence, OP is the bisector of ∠AOB.
[ Hence proved.

Question 32.
Solution:
Given : ABCD is a square, M is midpoint of AB.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 32
PQ is perpendicular to MC which meets CB produced at Q. CP is joined.
To prove : (i) PA = BQ
(ii) CP = AB + PA
Proof : (i) In ∆ PAM and ∆ QBM,
AM = MB (M is midpoint of AB)
∠AMP = ∠BMQ
(vertically opposite angles)
∠ PAM = ∠ QBM
(Each = 90° angles of a square)
∴ ∆ PAM ≅ ∆ QBM (A.S.A. axiom)
∴ AP = BQ
=> PA = BQ (c.p.c.t.)
and PM = QM (c.p.c.t.)
Now, in ∆ CPM and ∆ CQM,
CM = CM (Common)
PM = QM (Proved)
∠CMP =∠CMQ (Each 90° as PQ ⊥ MC)
∴ ∆ CPM ≅ ∆ CQM (SAS axiom)
∴ CP = CQ (c.p.c.t.)
= CB + BQ
= AB + PA
( CB = AB sides of squares and BQ = PA proved)
Hence, CP = AB + PA.

Question 33.
Solution:
Construction. Let AB be the breadth of the river. Mark any point M on the bank on which B is situated. Let O be the midpoint of BM. From M move along the path MN perpendicular to BM to a point N such that A, O, N are on the same straight line. Then MN is the required breadth of the river.
Proof : In ∆ ABO and ∆ MNO,
BO = OM (const.)
∠ AOB = ∠ MON (vertically opposite angle)
∠B = ∠M (each 90°)
∴ ∆ ABO ≅ ∆ MNO (A.S.A. axiom)
∴ AB = MN. (c.p.c.t.)
Hence, MN is the required breadth of the river.

Question 34.
Solution:
In ∆ ABC,
∠A = 36°, ∠B = 64°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> 36° + 64° + ∠C = 180°
=> 100° + ∠C = 180°
=> ∠C = 180° – 100° = 80°
∴ ∠ C = 80° which is the greatest angle.
∴ The side AB, opposite to it is the longest side.
∴∠ A is the shortest angle
∴ BC is the shortest side.

Question 35.
Solution:
In ∆ ABC,
∴ ∠A = 90°
∴ ∠B + ∠C 180° – 90° = 90°
Hence, ∠ A is the greatest angle of the triangle.
Side BC, opposite to this angle be the longest side.

Question 36.
Solution:
In ∆ ABC,
∠ A = ∠ B = 45°
∴ ∠A + ∠B = 45° + 45° = 90°
and ∠C= 180°-90° = 90°
∴ ∠ C is the greatest angle.
∴ Side AB, opposite to ∠ C will be the longest side of the triangle.

Question 37.
Solution:
Given : In ∆ ABC, side AB is produced to D such that BD = BC.
∠B = 60°, ∠A = 70°
To Prove : (i) AD > CD (ii) AD > AC
Proof : In ∆ BCD,
Ext. ∠ B = 60°
∴ ∠ CBD = 180° – 60° = 120°
and ∠ BCD + ∠ BDC = Ext. ∠ CBA = 60°
But ∠BCD = ∠ BDC ( BC = BD)
∴ ∠BCD = ∠BDC = \(\frac { { 60 }^{ o } }{ 2 } \) = 30°.
∴ ∠ ACD = 50° + 30° = 80°
(i) Now in ∆ ACD,
∴ ∠ ACD > ∠ CAD ( 80° > 70°)
AD > CD .
(ii) and ACD > ∠D (80° > 30°)
∴ AD > AC
Hence proved.

Question 38.
Solution:
Given : In ∆ ABC, ∠B = 35°and ∠C = 65°
AX is the bisector of ∠ BA C meeting BC in X
∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠ A + 35° + 65° = 180°
=> ∠A + 100° = 180°
=> ∠A = 180° – 100° = 80°
∴ AX is the bisector of ∠ BAC
∴ ∠ BAX = ∠ CAX = 40°
In ∆ ABX,
∠BAX = 40° and ∠B = 35°
∴ ∠BAX > ∠B
∴ BX > AX …(i)
and in ∆ AXC,
∠C = 65° and ∠CAX = 40°
∴ ∠C > ∠CAX
∴ AX > XC …(ii)
From (i) and (ii),
BX > AX > XC
or BX > AX > CX

Question 39.
Solution:
Given : In ∆ ABC,
AD is the bisector of ∠ A
To prove : (i) AB > BD and
(ii) AC > DC
Proof : (i) In ∆ ADC,
Ext. ∠ ADB > ∠ CAD
=>∠ ADB > ∠BAD ,
( AD is bisector of ∠ A)
In ∆ ABD,
AB > BD.
(ii) Again, in ∆ ADB,
Ext. ∠ADC > ∠BAD
=> ∠ADC > ∠CAD
( ∠ CAD = ∠BAD)
Now in ∆ ACD.
AC > DC.
Hence proved

Question 40.
Solution:
Given : In ∆ ABC, AB = AC
BC is produced to D and AD is joined.
To Prove : AD > AC
Proof : In ∆ ABC,
AB = AC (given)
∴ ∠B = ∠C (Angles opposite to equal sides)
Ext. ∠ ACD > ∠ ABC
∠ACB = ∠ABC
∴ ∠ABC > ∠ADC
Now, in ∆ ABD,
∴ ∠ABC > ∠ADC or ∠ADB
∴ AD > AC
Hence proved.

Question 41.
Solution:
Given : In ∆ ABC,
AC > AB and AD is the bisector of ∠ A which meets BC in D.
To Prove : ∠ADC > ∠ADB
Proof : In ∆ ABC,
AC > AB
∴∠B > ∠C
∴∠ 1 = ∠ 2 (AD is the bisector of ∠ A)
∴ ∠B + ∠2 = ∠C + ∠1
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 41
But in ∆ ADB
Ext. ∠ADC = ∠B + ∠2
and in ∆ ADC,
Ext. ∠ADB > ∠C + ∠1
∴ ∠B + ∠2 > ∠C + ∠1 (Proved)
∴ ∠ ADC > ∠ ADB Hence proved.

Question 42.
Solution:
Given : In ∆ PQR,
S is any point on QR and PS is joined
To Prove : PQ + QR + RP > 2PS
Proof : In ∆ PQS,
PQ + QS > PS
(Sum of two sides of a triangle is greater than the third side) …(i)
Similarly, in ∆ PRS.
PQ + SR > PS …(ii)
Adding (i), and (ii),
PQ + QS + PR + SR > PS + PS
=> PQ + QS + SR + PR > 2PS
=> PQ + QR + RP > 2PS
Hence proved.

Question 43.
Solution:
Given : O is the centre of the circle and XOY is its diameter.
XZ is the chord of this circle
To Prove : XY > XZ
Const. Join OZ
Proof : OX, OZ and OY are the radii of the circle
∴ OX = OZ = OY
In ∆ XOZ,
OX + OZ > XZ (Sum of two sides of a triangle is greater than its third side)
=> OX + OY > XZ (∴ OZ = OY)
=> OXY > XZ
Hence proved.

Question 44.
Solution:
Given : In ∆ ABC, O is any point within it OA, OB and OC are joined.
To Prove : (i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 } \)
(AB + BC + CA)
Const. Produce BO to meet AC in D.
RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A 44
Proof : (i) In ∆ ABD,
AB + AD > BD …(i)
(Sum of two sides of a triangle is greater than its third side)
=> AB + AD > BO + OD …(i)
Similarly in ∆ ODC
OD + DC > OC. …(ii)
Adding (i) and (ii)
AB + AD + OD + DC > OB + OD + OC
=> AB + AD + DC > OB + OC
AB + AC > OB + OC.
(ii) In (i) we have proved that
AB + AC > OB + OC
Similarly, we can prove that
AC + BC > OC + OA
and BC + AB > OA + OB
Adding, we get:
AB + AC + AC + BC + B + AB > OB + OC + OC + OA + OA + OB
=> 2(AB + BC + CA) > 2(OA + OB + OC)
=> AB + BC + CA > OA + OB + OC.
(iii) In ∆ AOB,
OA + OB > AB
Similarly, in ∆ BOC
OB + OC > BC
and in ∆ COA
OC + OA > CA
adding we get:
OA + OB + OB + OC + OC + OA > AB + BC + CA
=> 2(OA + OB + OC) > (AB + BC + CA)
=> OA + OB + OC > \(\frac { 1 }{ 2 } \) (AB + BC + CA)
Hence proved.

Question 45.
Solution:
Sides of ∆ ABC are AB = 3cm, BC = 3.5cm and CA = 6.5cm
We know that sum of any two sides of a triangle is greater than its third side.
Here, AB = 3 cm and BC = 3.5 cm
∴ AB + BC = 3cm + 3.5 cm = 6.5 cm and CA = 6.5 cm
∴ AB + BC = CA
Which is not possible to draw the triangle.
Hence, we cannot draw the triangle with the given data.

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