RS Aggarwal Solutions: Congruence of Triangles and Inequalities in a Triangle- 2 | Mathematics (Maths) Class 9 PDF Download

 Page 1


                  
                                         
            
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
i
5 cm, 4 cm, 9 cm
ii
8 cm, 7 cm, 4 cm
iii
10 cm, 5 cm, 6 cm
iv
2.5 cm, 5 cm, 7 cm
v
3 cm, 4 cm, 8 cm
Solution:
i
No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9
ii
Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7
iii
Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6
iv
Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5
v
No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8
Question:29
In ?ABC, ?A = 50° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given: In ?ABC, ?A = 50° and ?B = 60°
In ?ABC,
?A + ?B + ?C = 180°           Anglesumpropertyofatriangle
?
50° + 60° + ?C = 180°
?
110° + ?C = 180°
?
?C = 180° -
 110°
?
?C = 70°
Hence, the longest side will be opposite to the largest angle ( ?C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle ( ?A = 50° ) i.e. BC.
Question:30
i
In ?ABC, ?A = 90°. Which is its longest side?
ii In ?ABC, ?A = ?B = 45°. Which is its longest side?
iii
In ?ABC, ?A = 100° and ?C = 50°. Which is its shortest side?
Solution:
i
Given: In ?ABC, ?A = 90°
So, sum of the other two angles in triangle ?B + ?C = 90°
i.e. ?B, ?C < 90°
Since, ?A is the greatest angle.
So, the longest side is BC.
ii
Given: ?A = ?B = 45°
Using angle sum property of triangle,
?C = 90°
Since, ?C is the greatest angle.
So, the longest side is AB.
iii
Given: ?A = 100° and ?C = 50°
Using angle sum property of triangle,
Page 2


                  
                                         
            
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
i
5 cm, 4 cm, 9 cm
ii
8 cm, 7 cm, 4 cm
iii
10 cm, 5 cm, 6 cm
iv
2.5 cm, 5 cm, 7 cm
v
3 cm, 4 cm, 8 cm
Solution:
i
No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9
ii
Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7
iii
Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6
iv
Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5
v
No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8
Question:29
In ?ABC, ?A = 50° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given: In ?ABC, ?A = 50° and ?B = 60°
In ?ABC,
?A + ?B + ?C = 180°           Anglesumpropertyofatriangle
?
50° + 60° + ?C = 180°
?
110° + ?C = 180°
?
?C = 180° -
 110°
?
?C = 70°
Hence, the longest side will be opposite to the largest angle ( ?C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle ( ?A = 50° ) i.e. BC.
Question:30
i
In ?ABC, ?A = 90°. Which is its longest side?
ii In ?ABC, ?A = ?B = 45°. Which is its longest side?
iii
In ?ABC, ?A = 100° and ?C = 50°. Which is its shortest side?
Solution:
i
Given: In ?ABC, ?A = 90°
So, sum of the other two angles in triangle ?B + ?C = 90°
i.e. ?B, ?C < 90°
Since, ?A is the greatest angle.
So, the longest side is BC.
ii
Given: ?A = ?B = 45°
Using angle sum property of triangle,
?C = 90°
Since, ?C is the greatest angle.
So, the longest side is AB.
iii
Given: ?A = 100° and ?C = 50°
Using angle sum property of triangle,
?B = 30°
Since, ?A is the greatest angle.
So, the shortest side is BC.
Question:31
In ?ABC, side AB is produced to D such that BD = BC. If ?B = 60° and ?A = 70°, prove that
i AD > CD and
ii AD > AC.
Solution:
In triangle CBA, CBD is an exterior angle.
i. e. , ?CBA + ?CBD = 180° ? 60°+ ?CBD = 180° ? ?CBD = 120°
Triangle BCD is isosceles and BC = BD.
Let ?BCD = ?BDC = x°
.
In ? CBD, we have: ? ?BCD + ?CBD + ?CDB = 180° ? x +120°+x = 180 ? 2x = 60° ? x = 30° ? ?BCD = ?BDC = 30°
In triangle ADC, ?C = ?ACB + ?BCD = 50°+30° = 80°
?A = 70°and ?D = 30°
? ?C > ?A ? AD > CD    . . . (1)
Also, ?C > ?D ? AD > AC   . . . (2)
Question:32
In the given figure, ?B < ?A and ?C < ?D. Show that AD > BC.
Solution:
Given: ?B < ?A and ?C < ?D
To prove: AD > BC
Proof: 
In ? AOB, ?B < ?A ? AO < BO       (Side opposite to the greater angle is longer)  . . . . . (1)
In ? COD, ?C < ?D ? OD < OC       (Side opposite to the greater angle is longer)  . . . . . (2)
Adding (1) and (2), we getAO+OD < BO+OC ? AD < BC
Question:33
AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ?A > ?C and
?B > ?D.
Solution:
Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.
To prove:
i
Page 3


                  
                                         
            
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
i
5 cm, 4 cm, 9 cm
ii
8 cm, 7 cm, 4 cm
iii
10 cm, 5 cm, 6 cm
iv
2.5 cm, 5 cm, 7 cm
v
3 cm, 4 cm, 8 cm
Solution:
i
No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9
ii
Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7
iii
Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6
iv
Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5
v
No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8
Question:29
In ?ABC, ?A = 50° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given: In ?ABC, ?A = 50° and ?B = 60°
In ?ABC,
?A + ?B + ?C = 180°           Anglesumpropertyofatriangle
?
50° + 60° + ?C = 180°
?
110° + ?C = 180°
?
?C = 180° -
 110°
?
?C = 70°
Hence, the longest side will be opposite to the largest angle ( ?C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle ( ?A = 50° ) i.e. BC.
Question:30
i
In ?ABC, ?A = 90°. Which is its longest side?
ii In ?ABC, ?A = ?B = 45°. Which is its longest side?
iii
In ?ABC, ?A = 100° and ?C = 50°. Which is its shortest side?
Solution:
i
Given: In ?ABC, ?A = 90°
So, sum of the other two angles in triangle ?B + ?C = 90°
i.e. ?B, ?C < 90°
Since, ?A is the greatest angle.
So, the longest side is BC.
ii
Given: ?A = ?B = 45°
Using angle sum property of triangle,
?C = 90°
Since, ?C is the greatest angle.
So, the longest side is AB.
iii
Given: ?A = 100° and ?C = 50°
Using angle sum property of triangle,
?B = 30°
Since, ?A is the greatest angle.
So, the shortest side is BC.
Question:31
In ?ABC, side AB is produced to D such that BD = BC. If ?B = 60° and ?A = 70°, prove that
i AD > CD and
ii AD > AC.
Solution:
In triangle CBA, CBD is an exterior angle.
i. e. , ?CBA + ?CBD = 180° ? 60°+ ?CBD = 180° ? ?CBD = 120°
Triangle BCD is isosceles and BC = BD.
Let ?BCD = ?BDC = x°
.
In ? CBD, we have: ? ?BCD + ?CBD + ?CDB = 180° ? x +120°+x = 180 ? 2x = 60° ? x = 30° ? ?BCD = ?BDC = 30°
In triangle ADC, ?C = ?ACB + ?BCD = 50°+30° = 80°
?A = 70°and ?D = 30°
? ?C > ?A ? AD > CD    . . . (1)
Also, ?C > ?D ? AD > AC   . . . (2)
Question:32
In the given figure, ?B < ?A and ?C < ?D. Show that AD > BC.
Solution:
Given: ?B < ?A and ?C < ?D
To prove: AD > BC
Proof: 
In ? AOB, ?B < ?A ? AO < BO       (Side opposite to the greater angle is longer)  . . . . . (1)
In ? COD, ?C < ?D ? OD < OC       (Side opposite to the greater angle is longer)  . . . . . (2)
Adding (1) and (2), we getAO+OD < BO+OC ? AD < BC
Question:33
AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ?A > ?C and
?B > ?D.
Solution:
Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.
To prove:
i
?A > ?C
ii
 ?B > ?D
Construction: Join AC.
Proof:
In ? ABC, ? BC > AB           (Given, AB is the smallest side) ? ?1 > ?2            . . . (1)
In ? ADC, ? CD > AD           (Given, CD is the largest side) ? ?3 > ?4            . . . (2)
Adding (1) and (2), we get ?1 + ?3 > ?2 + ?4 ? ?A > ?C
ii
Construction: Join BD.
Proof:
In ? ABD, ? AD > AB                  (Given, AB is the smallest side. ) ? ?5 > ?6                 . . . (3)
In ? CBD, ? CD > BC                  (Given, CD is the greatest side. ) ? ?7 > ?8                 . . . (4)
Adding (3) and (4), we get ?5 + ?7 > ?6 + ?8 ? ?B > ?D
Question:34
In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) > (AC + BD)
Proof:
In ? ABC, AB +BC > AC         . . . (i)In ? CAD, CD +AD > AC        . . . (ii)In ? BAD, AB +AD > BD        . . . (iii)In ? BCD, BC +CD > BD        . . . (iv)
Adding i
, ii
, iii
and iv
, we get
2(AB + BC + CD + DA) < 2( AC + BD)
Hence, (AB + BC + CD + DA) < (AC + BD).
Question:35
In a quadrilateral ABCD, show that
(AB +BC +CD +DA) < 2(BD +AC)
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
Page 4


                  
                                         
            
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
i
5 cm, 4 cm, 9 cm
ii
8 cm, 7 cm, 4 cm
iii
10 cm, 5 cm, 6 cm
iv
2.5 cm, 5 cm, 7 cm
v
3 cm, 4 cm, 8 cm
Solution:
i
No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9
ii
Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7
iii
Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6
iv
Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5
v
No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8
Question:29
In ?ABC, ?A = 50° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given: In ?ABC, ?A = 50° and ?B = 60°
In ?ABC,
?A + ?B + ?C = 180°           Anglesumpropertyofatriangle
?
50° + 60° + ?C = 180°
?
110° + ?C = 180°
?
?C = 180° -
 110°
?
?C = 70°
Hence, the longest side will be opposite to the largest angle ( ?C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle ( ?A = 50° ) i.e. BC.
Question:30
i
In ?ABC, ?A = 90°. Which is its longest side?
ii In ?ABC, ?A = ?B = 45°. Which is its longest side?
iii
In ?ABC, ?A = 100° and ?C = 50°. Which is its shortest side?
Solution:
i
Given: In ?ABC, ?A = 90°
So, sum of the other two angles in triangle ?B + ?C = 90°
i.e. ?B, ?C < 90°
Since, ?A is the greatest angle.
So, the longest side is BC.
ii
Given: ?A = ?B = 45°
Using angle sum property of triangle,
?C = 90°
Since, ?C is the greatest angle.
So, the longest side is AB.
iii
Given: ?A = 100° and ?C = 50°
Using angle sum property of triangle,
?B = 30°
Since, ?A is the greatest angle.
So, the shortest side is BC.
Question:31
In ?ABC, side AB is produced to D such that BD = BC. If ?B = 60° and ?A = 70°, prove that
i AD > CD and
ii AD > AC.
Solution:
In triangle CBA, CBD is an exterior angle.
i. e. , ?CBA + ?CBD = 180° ? 60°+ ?CBD = 180° ? ?CBD = 120°
Triangle BCD is isosceles and BC = BD.
Let ?BCD = ?BDC = x°
.
In ? CBD, we have: ? ?BCD + ?CBD + ?CDB = 180° ? x +120°+x = 180 ? 2x = 60° ? x = 30° ? ?BCD = ?BDC = 30°
In triangle ADC, ?C = ?ACB + ?BCD = 50°+30° = 80°
?A = 70°and ?D = 30°
? ?C > ?A ? AD > CD    . . . (1)
Also, ?C > ?D ? AD > AC   . . . (2)
Question:32
In the given figure, ?B < ?A and ?C < ?D. Show that AD > BC.
Solution:
Given: ?B < ?A and ?C < ?D
To prove: AD > BC
Proof: 
In ? AOB, ?B < ?A ? AO < BO       (Side opposite to the greater angle is longer)  . . . . . (1)
In ? COD, ?C < ?D ? OD < OC       (Side opposite to the greater angle is longer)  . . . . . (2)
Adding (1) and (2), we getAO+OD < BO+OC ? AD < BC
Question:33
AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ?A > ?C and
?B > ?D.
Solution:
Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.
To prove:
i
?A > ?C
ii
 ?B > ?D
Construction: Join AC.
Proof:
In ? ABC, ? BC > AB           (Given, AB is the smallest side) ? ?1 > ?2            . . . (1)
In ? ADC, ? CD > AD           (Given, CD is the largest side) ? ?3 > ?4            . . . (2)
Adding (1) and (2), we get ?1 + ?3 > ?2 + ?4 ? ?A > ?C
ii
Construction: Join BD.
Proof:
In ? ABD, ? AD > AB                  (Given, AB is the smallest side. ) ? ?5 > ?6                 . . . (3)
In ? CBD, ? CD > BC                  (Given, CD is the greatest side. ) ? ?7 > ?8                 . . . (4)
Adding (3) and (4), we get ?5 + ?7 > ?6 + ?8 ? ?B > ?D
Question:34
In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) > (AC + BD)
Proof:
In ? ABC, AB +BC > AC         . . . (i)In ? CAD, CD +AD > AC        . . . (ii)In ? BAD, AB +AD > BD        . . . (iii)In ? BCD, BC +CD > BD        . . . (iv)
Adding i
, ii
, iii
and iv
, we get
2(AB + BC + CD + DA) < 2( AC + BD)
Hence, (AB + BC + CD + DA) < (AC + BD).
Question:35
In a quadrilateral ABCD, show that
(AB +BC +CD +DA) < 2(BD +AC)
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
In ?AOB, 
OA +OB > AB      . . . . . (i)
In ?BOC, 
OB +OC > BC      . . . . . (ii)
In ?COD, 
OC +OD > CD     . . . . . (iii)
In ?AOD, 
OD +OA > AD     . . . . . (iv)
Adding (i), (ii), (iii) and (iv)
, we get
     2(OA +OB +OC +OD) > (AB +BC +CD +DA) ? 2(OB +OD +OA +OC) > (AB +BC +CD +DA) ? 2(BD +AC) > (AB +BC +CD +DA)
Question:36
In ?ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in X. Arrange AX, BX and CX in descending order.
Solution:
Given: In ?ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in X.
In ? ABX, ? ?BAX > ?ABX ? BX > AX                 . . . (i)
Similarly, in ? ACX, ? ?ACX > ?XAC ? AX > CX                 . . . (ii)
From (i) and (ii), we getBX > AX > CX
Question:37
In the given figure, PQ > PR and QS and RS are the bisectors of ?Q and ?R respectively. Show that SQ > SR.
Solution:
Since the angle opposite to the longer side is greater, we have:
PQ > PR ? ?R > ?Q ?
1
2
?R >
1
2
?Q ? ?SRQ > ?RQS ? QS > SR
?SQ > SR
Question:38
D is any point on the side AC of ?ABC with AB = AC. Show that CD < BD.
Solution:
Given: In ?
ABC, AB = AC
To prove: CD < BD
Proof:
In ?
ABC,
Page 5


                  
                                         
            
Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
i
5 cm, 4 cm, 9 cm
ii
8 cm, 7 cm, 4 cm
iii
10 cm, 5 cm, 6 cm
iv
2.5 cm, 5 cm, 7 cm
v
3 cm, 4 cm, 8 cm
Solution:
i
No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9
ii
Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7
iii
Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6
iv
Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5
v
No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8
Question:29
In ?ABC, ?A = 50° and ?B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given: In ?ABC, ?A = 50° and ?B = 60°
In ?ABC,
?A + ?B + ?C = 180°           Anglesumpropertyofatriangle
?
50° + 60° + ?C = 180°
?
110° + ?C = 180°
?
?C = 180° -
 110°
?
?C = 70°
Hence, the longest side will be opposite to the largest angle ( ?C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle ( ?A = 50° ) i.e. BC.
Question:30
i
In ?ABC, ?A = 90°. Which is its longest side?
ii In ?ABC, ?A = ?B = 45°. Which is its longest side?
iii
In ?ABC, ?A = 100° and ?C = 50°. Which is its shortest side?
Solution:
i
Given: In ?ABC, ?A = 90°
So, sum of the other two angles in triangle ?B + ?C = 90°
i.e. ?B, ?C < 90°
Since, ?A is the greatest angle.
So, the longest side is BC.
ii
Given: ?A = ?B = 45°
Using angle sum property of triangle,
?C = 90°
Since, ?C is the greatest angle.
So, the longest side is AB.
iii
Given: ?A = 100° and ?C = 50°
Using angle sum property of triangle,
?B = 30°
Since, ?A is the greatest angle.
So, the shortest side is BC.
Question:31
In ?ABC, side AB is produced to D such that BD = BC. If ?B = 60° and ?A = 70°, prove that
i AD > CD and
ii AD > AC.
Solution:
In triangle CBA, CBD is an exterior angle.
i. e. , ?CBA + ?CBD = 180° ? 60°+ ?CBD = 180° ? ?CBD = 120°
Triangle BCD is isosceles and BC = BD.
Let ?BCD = ?BDC = x°
.
In ? CBD, we have: ? ?BCD + ?CBD + ?CDB = 180° ? x +120°+x = 180 ? 2x = 60° ? x = 30° ? ?BCD = ?BDC = 30°
In triangle ADC, ?C = ?ACB + ?BCD = 50°+30° = 80°
?A = 70°and ?D = 30°
? ?C > ?A ? AD > CD    . . . (1)
Also, ?C > ?D ? AD > AC   . . . (2)
Question:32
In the given figure, ?B < ?A and ?C < ?D. Show that AD > BC.
Solution:
Given: ?B < ?A and ?C < ?D
To prove: AD > BC
Proof: 
In ? AOB, ?B < ?A ? AO < BO       (Side opposite to the greater angle is longer)  . . . . . (1)
In ? COD, ?C < ?D ? OD < OC       (Side opposite to the greater angle is longer)  . . . . . (2)
Adding (1) and (2), we getAO+OD < BO+OC ? AD < BC
Question:33
AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ?A > ?C and
?B > ?D.
Solution:
Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.
To prove:
i
?A > ?C
ii
 ?B > ?D
Construction: Join AC.
Proof:
In ? ABC, ? BC > AB           (Given, AB is the smallest side) ? ?1 > ?2            . . . (1)
In ? ADC, ? CD > AD           (Given, CD is the largest side) ? ?3 > ?4            . . . (2)
Adding (1) and (2), we get ?1 + ?3 > ?2 + ?4 ? ?A > ?C
ii
Construction: Join BD.
Proof:
In ? ABD, ? AD > AB                  (Given, AB is the smallest side. ) ? ?5 > ?6                 . . . (3)
In ? CBD, ? CD > BC                  (Given, CD is the greatest side. ) ? ?7 > ?8                 . . . (4)
Adding (3) and (4), we get ?5 + ?7 > ?6 + ?8 ? ?B > ?D
Question:34
In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) > (AC + BD)
Proof:
In ? ABC, AB +BC > AC         . . . (i)In ? CAD, CD +AD > AC        . . . (ii)In ? BAD, AB +AD > BD        . . . (iii)In ? BCD, BC +CD > BD        . . . (iv)
Adding i
, ii
, iii
and iv
, we get
2(AB + BC + CD + DA) < 2( AC + BD)
Hence, (AB + BC + CD + DA) < (AC + BD).
Question:35
In a quadrilateral ABCD, show that
(AB +BC +CD +DA) < 2(BD +AC)
Solution:
Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:
In ?AOB, 
OA +OB > AB      . . . . . (i)
In ?BOC, 
OB +OC > BC      . . . . . (ii)
In ?COD, 
OC +OD > CD     . . . . . (iii)
In ?AOD, 
OD +OA > AD     . . . . . (iv)
Adding (i), (ii), (iii) and (iv)
, we get
     2(OA +OB +OC +OD) > (AB +BC +CD +DA) ? 2(OB +OD +OA +OC) > (AB +BC +CD +DA) ? 2(BD +AC) > (AB +BC +CD +DA)
Question:36
In ?ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in X. Arrange AX, BX and CX in descending order.
Solution:
Given: In ?ABC, ?B = 35°, ?C = 65° and the bisector of ?BAC meets BC in X.
In ? ABX, ? ?BAX > ?ABX ? BX > AX                 . . . (i)
Similarly, in ? ACX, ? ?ACX > ?XAC ? AX > CX                 . . . (ii)
From (i) and (ii), we getBX > AX > CX
Question:37
In the given figure, PQ > PR and QS and RS are the bisectors of ?Q and ?R respectively. Show that SQ > SR.
Solution:
Since the angle opposite to the longer side is greater, we have:
PQ > PR ? ?R > ?Q ?
1
2
?R >
1
2
?Q ? ?SRQ > ?RQS ? QS > SR
?SQ > SR
Question:38
D is any point on the side AC of ?ABC with AB = AC. Show that CD < BD.
Solution:
Given: In ?
ABC, AB = AC
To prove: CD < BD
Proof:
In ?
ABC,
Since, AB = AC       Given
So, ?ABC = ?ACB
      ...i
In ? ABC and ? DBC, ? ABC > ?DBC ? ?ACB > ? DBC         [From (i)] ? BD > CD               (Side opposite to greater angle is longer. ) ? CD < BD
Question:39
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 
2
3
of a right angle.
Solution:
Given: In ?
ABC, BC is the longest side.
To prove: ?
BAC > 
2
3
 of a right angle, i.e., ?
BAC > 60°
Construct: Mark a point D on side AC such that AD = AB = BD.
Proof:
In ?
ABD,
?
 AD = AB = BD    Byconstruction
? ?1 = ?3 = ?4 = 60°
Now, ?BAC = ?1 + ?2 = 60°+ ?2but 60° =
2
3
 of a right angleSo, ?BAC =
2
3
 of a right angle + ?2
Hence, ?
BAC > 
2
3
 of a right angle.
Question:40
In the given figure, prove that
i
CD + DA + AB > BC
ii
CD + DA + AB + BC > 2AC.
Solution:
Given: Quadrilateral ABCD
To prove:
i
 CD + DA + AB > BC
ii
 CD + DA + AB + BC > 2AC
Proof:
i
In ? ACD, CD +DA > CA        . . . (1)In ? ABC, AB +CA > BC         . . . (2)