SELINA Solutions for Class 9 Maths Chapter 11 - Inequalities
Chapter 11 - Inequalities Exercise Ex. 11
In ABC,
AB = AC[Given]
ACB =
B[angles opposite to equal sides are equal]
B = 700[Given]
ACB = 700 ……….(i)
Now,
ACB +
ACD = 1800[ BCD is a straight line]
700 +
ACD = 1800
ACD = 1100 …………(ii)
In ACD,
CAD +
ACD +
D = 1800
CAD + 1100 +
D = 1800 [From (ii)]
CAD +
D = 700
But D = 400 [Given]
CAD + 400= 700
CAD = 300 ………………(iii)
In ACD,
ACD = 1100[From (ii)]
CAD = 300[From (iii)]
D = 400 [Given]
[Greater angle has greater side opposite to it]
Also,
AB = AC[Given]
Therefore, AB > CD.
In PQR,
QR = PR[Given]
P =
Q[angles opposite to equal sides are equal]
P = 360[Given]
Q = 360
In PQR,
P +
Q +
R = 1800
360 + 360 +
R = 1800
R + 720 = 1800
R = 1080
Now,
R = 1080
P = 360
Q = 360
Since R is the greatest, therefore, PQ is the largest side.
The sum of any two sides of the triangle is always greater than third side of the triangle.
Third side < 13+
8 =
21 cm.
The difference between any two sides of the triangle is always less than the third side of the triangle.
Third side > 13-
8 =
5 cm.
Therefore, the length of the third side is between 5 cm and 9 cm, respectively.
The value of a =5 cm and b
= 21
cm.
In BEC,
B +
BEC +
BCE = 1800
B = 650 [Given]
BEC = 900[CE is perpendicular to AB]
650 + 900 +
BCE = 1800
BCE = 1800 - 1550
BCE = 250 =
DCF …………(i)
In CDF,
DCF +
FDC +
CFD = 1800
DCF = 250 [From (i)]
FDC = 900[AD is perpendicular to BC]
250 + 900 +
CFD = 1800
CFD = 1800 - 1150
CFD = 650 …………(ii)
Now, AFC +
CFD = 1800[AFD is a straight line]
AFC + 650 = 1800
AFC = 1150 ………(iii)
In ACE,
ACE +
CEA +
BAC = 1800
BAC = 600 [Given]
CEA = 900[CE is perpendicular to AB]
ACE + 900 + 600 = 1800
ACE = 1800 - 1500
ACE = 300 …………(iv)
In AFC,
AFC +
ACF +
FAC = 1800
AFC = 1150 [From (iii)]
ACF = 300[From (iv)]
1150 + 300 +
FAC = 1800
FAC = 1800 - 1450
FAC = 350 …………(v)
In AFC,
FAC = 350[From (v)]
ACF = 300[From (iv)]
In CDF,
DCF = 250[From (i)]
CFD = 650[From (ii)]
ACB = 740 …..(i)[Given]
ACB +
ACD = 1800[BCD is a straight line]
740 +
ACD = 1800
ACD = 1060 ……..(ii)
In ACD,
ACD +
ADC+
CAD = 1800
Given that AC = CD
ADC=
CAD
1060 +
CAD +
CAD = 1800[From (ii)]
2
CAD = 740
CAD = 370 =
ADC………..(iii)
Now,
BAD = 1100[Given]
BAC +
CAD = 1100
BAC + 370 = 1100
BAC = 730 ……..(iv)
In ABC,
B +
BAC+
ACB = 1800
B + 730 + 740 = 1800[From (i) and (iv)]
B + 1470 = 1800
B = 330 ………..(v)
(i) ADC +
ADB = 1800[BDC is a straight line]
ADC = 900[Given]
900 + ADB = 1800
ADB = 900 …………(i)
In ADB,
ADB = 900[From (i)]
B +
BAD = 900
Therefore, B and
BAD are both acute, that is less than 900.
AB > BD …….(ii)[Side opposite 900 angle is greater than
side opposite acute angle]
(ii) In ADC,
ADB = 900
C +
DAC = 900
Therefore, C and
DAC are both acute, that is less than 900.
AC > CD ……..(iii)[Side opposite 900 angle is greater than
side opposite acute angle]
Adding (ii) and (iii)
AB + AC > BD + CD
AB + AC > BC
Const: Join AC and BD.
(i) In ABC,
AB + BC > AC….(i)[Sum of two sides is greater than the
third side]
In ACD,
AC + CD > DA….(ii)[ Sum of two sides is greater than the
third side]
Adding (i) and (ii)
AB + BC + AC + CD > AC + DA
AB + BC + CD > AC + DA - AC
AB + BC + CD > DA …….(iii)
(ii)In ACD,
CD + DA > AC….(iv)[Sum of two sides is greater than the
third side]
Adding (i) and (iv)
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(iii) In ABD,
AB + DA > BD….(v)[Sum of two sides is greater than the
third side]
In BCD,
BC + CD > BD….(vi)[Sum of two sides is greater than the
third side]
Adding (v) and (vi)
AB + DA + BC + CD > BD + BD
AB + DA + BC + CD > 2BD
(i) In ABC,
AB = BC = CA[ABC is an equilateral triangle]
A =
B =
C
In ABP,
A = 600
ABP< 600
[Side opposite to greater side is greater]
(ii) In BPC,
C = 600
CBP< 600
[Side opposite to greater side is greater]
Let PBC = x and
PCB = y
then,
BPC = 1800 - (x + y) ………(i)
Let ABP = a and
ACP = b
then,
BAC = 1800 - (x + a) - (y + b)
BAC = 1800 - (x + y) - (a + b)
BAC =
BPC - (a + b)
BPC =
BAC + (a + b)
BPC >
BAC
We know that exterior angle of a triangle is always greater than each of the interior opposite angles.
In
ABD,
ADC >
B ……..(i)
In ABC,
AB = AC
B =
C …..(ii)
From (i) and (ii)
ADC >
C
(i) In ADC,
ADC >
C
AC > AD ………(iii) [side opposite to greater angle is greater]
(ii) In ABC,
AB = AC
AB > AD[ From (iii)]
Const: Join ED.
In AOB and
AOD,
AB = AD[Given]
AO = AO[Common]
BAO =
DAO[AO is bisector of
A]
[SAS criterion]
Hence,
BO = OD………(i)[cpct]
AOB =
AOD .……(ii)[cpct]
ABO =
ADO
ABD =
ADB ………(iii)[cpct]
Now,
AOB =
DOE[Vertically opposite angles]
AOD =
BOE[Vertically opposite angles]
BOE =
DOE ……(iv)[From (ii)]
(i) In BOE and
DOE,
BO = CD[From (i)]
OE = OE[Common]
BOE =
DOE[From (iv)]
[SAS criterion]
Hence, BE = DE[cpct]
(ii) In BCD,
ADB =
C +
CBD[Ext. angle = sum of opp. int. angles]
ADB >
C
ABD >
C[From (iii)]
In ABC,
AB > AC,
ABC <
ACB
1800 -
ABC > 1800 -
ACB
Since AB is the largest side and BC is the smallest side of the triangle ABC
In the quad. ABCD,
Since AB is the longest side and DC is the shortest side.
(i) 1 >
2[AB > BC]
7 >
4[AD > DC]
1 +
7 >
2 +
4
C >
A
(ii) 5 >
6[AB > AD]
3 >
8[BC > CD]
5 +
3 >
6 +
8
D >
B
In ADC,
ADB =
1 +
C.............(i)
In ADB,
ADC =
2 +
B.................(ii)
But AC > AB[Given]
B >
C
Also given, 2 =
1[AD is bisector of
A]
2 +
B >
1 +
C …….(iii)
From (i), (ii) and (iii)
ADC >
ADB
We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at right angle.
Using Pythagoras theorem in AFB,
AB2 = AF2 + BF2…………..(i)
In AFD,
AD2 = AF2 + DF2…………..(ii)
We know ABC is isosceles triangle and AB = AC
AC2 = AF2 + BF2 ……..(iii)[ From (i)]
Subtracting (ii) from (iii)
AC2 - AD2 = AF2 + BF2 - AF2 - DF2
AC2 - AD2 = BF2 - DF2
Let 2DF = BF
AC2 - AD2 = (2DF)2 - DF2
AC2 - AD2 = 4DF2 - DF2
AC2 = AD2 + 3DF2
AC2 > AD2
AC > AD
Similarly, AE > AC and AE > AD.
The sum of any two sides of the triangle is always greater than the third side of the triangle.
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