Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 8 - Triangles
Triangles Exercise MCQ
Solution 1
Solution 2
Correct option: (c)
∠A - ∠B = 42°
⇒ ∠A = ∠B + 42°
∠B - ∠C = 21°
⇒ ∠C = ∠B - 21°
In ΔABC,
∠A + ∠B + ∠C = 180°
⇒ ∠B + 42° + ∠B + ∠B - 21° = 180°
⇒ 3∠B = 159
⇒ ∠B = 53°
Solution 3
Correct option: (b)
∠ACD = ∠B + ∠A (Exterior angle property)
⇒ 110° = 50° + ∠A
⇒ ∠A = 60°
Solution 4
Correct option: (d)
Solution 5
Correct option: (a)
Solution 6
Solution 7
Correct option: (b)
∠EAF = ∠CAD (vertically opposite angles)
⇒ ∠CAD = 30°
In ΔABD, by angle sum property
∠A + ∠B + ∠D = 180°
⇒ (x + 30)° + (x + 10)° + 90° = 180°
⇒ 2x + 130° = 180°
⇒ 2x = 50°
⇒ x = 25°
Solution 8
Correct option: (a)
∠ABF + ∠ABC = 180° (linear pair)
⇒ x + ∠ABC = 180°
⇒ ∠ABC = 180° - x
∠ACG + ∠ACB = 180° (linear pair)
⇒ y + ∠ACB = 180°
⇒ ∠ACB = 180° - y
In ΔABC, by angle sum property
∠ABC + ∠ACB + ∠BAC = 180°
⇒ (180° - x) + (180° - y) + ∠BAC = 180°
⇒ ∠BAC - x - y + 180° = 0
⇒ ∠BAC = x + y - 180°
Now, ∠EAD = ∠BAC (vertically opposite angles)
⇒ z = x + y - 180°
Solution 9
Correct option: (b)
In ΔOAC, by angle sum property
∠OCA + ∠COA + ∠CAO = 180°
⇒ 80° + 40° + ∠CAO = 180°
⇒ ∠CAO = 60°
∠CAO + ∠OAE = 180° (linear pair)
⇒ 60° + x = 180°
⇒ x = 120°
∠COA = ∠BOD (vertically opposite angles)
⇒ ∠BOD = 40°
In ΔOBD, by angle sum property
∠OBD + ∠BOD + ∠ODB = 180°
⇒ ∠OBD + 40° + 70° = 180°
⇒ ∠OBD = 70°
∠OBD + ∠DBF = 180° (linear pair)
⇒ 70° + y = 180°
⇒ y = 110°
∴ x + y = 120° + 110° = 230°
Solution 10
Solution 11
Solution 12
Correct option: (a)
∠ACB + ∠ACD = 180° (linear pair)
⇒ 5y + 7y = 180°
⇒ 12y = 180°
⇒ y = 15°
Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle property)
⇒ 7y = x + 3y
⇒ 7(15°) = x + 3(15°)
⇒ 105° = x + 45°
⇒ x = 60°
Triangles Exercise Ex. 8
Solution 1
Since, sum of the angles of a triangle is 180o
A + B + C = 180o
A + 76o + 48o = 180o
A = 180o - 124o = 56o
A = 56o
Solution 2
Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o.
Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180o ]
9x = 180
The measures of the required angles are:
2x = (2 20)o = 40o
3x = (3 20)o = 60o
4x = (4 20)o = 80o
Solution 3
Let 3A = 4B = 6C = x (say)
Then, 3A = x
A =
4B = x
and 6C = x
C =
As A + B + C = 180o
A =
B =
C =
Solution 4
A + B = 108o [Given]
But as A, B and C are the angles of a triangle,
A + B + C = 180o
108o + C = 180o
C = 180o - 108o = 72o
Also, B + C = 130o [Given]
B + 72o = 130o
B = 130o - 72o = 58o
Now as, A + B = 108o
A + 58o = 108o
A = 108o - 58o = 50o
A = 50o, B = 58o and C = 72o.
Solution 5
Since. A , B and C are the angles of a triangle .
So, A + B + C = 180o
Now, A + B = 125o [Given]
125o + C = 180o
C = 180o - 125o = 55o
Also, A + C = 113o [Given]
A + 55o = 113o
A = 113o - 55o = 58o
Now as A + B = 125o
58o + B = 125o
B = 125o - 58o = 67o
A = 58o, B = 67o and C = 55o.
Solution 6
Since, P, Q and R are the angles of a triangle.
So,P + Q + R = 180o(i)
Now,P - Q = 42o[Given]
P = 42o + Q(ii)
andQ - R = 21o[Given]
R = Q - 21o(iii)
Substituting the value of P and R from (ii) and (iii) in (i), we get,
42o + Q + Q + Q - 21o = 180o
3Q + 21o = 180o
3Q = 180o - 21o = 159o
Q =
P = 42o + Q
= 42o + 53o = 95o
R = Q - 21o
= 53o - 21o = 32o
P = 95o, Q = 53o and R = 32o.
Solution 7
Given that the sum of the angles A and B of a ABC is 116o, i.e., A + B = 116o.
Since, A + B + C = 180o
So, 116o + C = 180o
C = 180o - 116o = 64o
Also, it is given that:
A - B = 24o
A = 24o + B
Putting, A = 24o + B in A + B = 116o, we get,
24o + B + B = 116o
2B + 24o = 116o
2B = 116o - 24o = 92o
B =
Therefore, A = 24o + 46o = 70o
A = 70o, B = 46o and C = 64o.
Solution 8
Let the two equal angles, A and B, of the triangle be xo each.
We know,
A + B + C = 180o
xo + xo + C = 180o
2xo + C = 180o(i)
Also, it is given that,
C = xo + 18o(ii)
Substituting C from (ii) in (i), we get,
2xo + xo + 18o = 180o
3xo = 180o - 18o = 162o
x =
Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.
Solution 9
Let C be the smallest angle of ABC.
Then, A = 2C and B = 3C
Also, A + B + C = 180o
2C + 3C + C = 180o
6C = 180o
C = 30o
So, A = 2C = 2 30o = 60o
B = 3C = 3 30o = 90o
The required angles of the triangle are 60o, 90o, 30o.
Solution 10
Let ABC be a right angled triangle and C = 90o
Since, A + B + C = 180o
A + B = 180o - C = 180o - 90o = 90o
Suppose A = 53o
Then, 53o + B = 90o
B = 90o - 53o = 37o
The required angles are 53o, 37o and 90o.
Solution 11
Let ABC be a right angled triangle and C = 90o
Since, A + B + C = 180o
A + B = 180o - C = 180o - 90o = 90o
Suppose A = 53o
Then, 53o + B = 90o
B = 90o - 53o = 37o
The required angles are 53o, 37o and 90o.
Solution 12
Let ABC be a triangle.
So, A < B + C
Adding A to both sides of the inequality,
2 A < A + B + C
2 A < 180o [Since A + B + C = 180o]
Similarly, B <A + C
B < 90o
and C < A + B
C < 90o
ABC is an acute angled triangle.
Solution 13
Let ABC be a triangle and B > A + C
Since, A + B + C = 180o
A + C = 180o - B
Therefore, we get,
B > 180o - B
Adding B on both sides of the inequality, we get,
B + B > 180o - B + B
2B > 180o
B >
i.e., B > 90o which means B is an obtuse angle.
ABC is an obtuse angled triangle.
Solution 14
Since ACB and ACD form a linear pair.
So, ACB + ACD = 180o
ACB + 128o = 180o
ACB = 180o - 128 = 52o
Also, ABC + ACB + BAC = 180o
43o + 52o + BAC = 180o
95o + BAC = 180o
BAC = 180o - 95o = 85o
ACB = 52o and BAC = 85o.
Solution 15
As DBA and ABC form a linear pair.
So,DBA + ABC = 180o
106o + ABC = 180o
ABC = 180o - 106o = 74o
Also, ACB and ACE form a linear pair.
So,ACB + ACE = 180o
ACB + 118o = 180o
ACB = 180o - 118o = 62o
In ABC, we have,
ABC + ACB + BAC = 180o
74o + 62o + BAC = 180o
136o + BAC = 180o
BAC = 180o - 136o = 44o
In triangle ABC, A = 44o, B = 74o and C = 62o
Solution 16
(i) EAB + BAC = 180o [Linear pair angles]
110o + BAC = 180o
BAC = 180o - 110o = 70o
Again, BCA + ACD = 180o [Linear pair angles]
BCA + 120o = 180o
BCA = 180o - 120o = 60o
Now, in ABC,
ABC + BAC + ACB = 180o
xo + 70o + 60o = 180o
x + 130o = 180o
x = 180o - 130o = 50o
x = 50
(ii)
In ABC,
A + B + C = 180o
30o + 40o + C = 180o
70o + C = 180o
C = 180o - 70o = 110o
Now BCA + ACD = 180o [Linear pair]
110o + ACD = 180o
ACD = 180o - 110o = 70o
In ECD,
ECD + CDE + CED = 180o
70o + 50o + CED = 180o
120o + CED = 180o
CED = 180o - 120o = 60o
Since AED and CED from a linear pair
So, AED + CED = 180o
xo + 60o = 180o
xo = 180o - 60o = 120o
x = 120
(iii)
EAF = BAC [Vertically opposite angles]
BAC = 60o
In ABC, exterior ACD is equal to the sum of two opposite interior angles.
So, ACD = BAC + ABC
115o = 60o + xo
xo = 115o - 60o = 55o
x = 55
(iv)
Since AB || CD and AD is a transversal.
So, BAD = ADC
ADC = 60o
In ECD, we have,
E + C + D = 180o
xo + 45o + 60o = 180o
xo + 105o = 180o
xo = 180o - 105o = 75o
x = 75
(v)
In AEF,
Exterior BED = EAF + EFA
100o = 40o + EFA
EFA = 100o - 40o = 60o
Also, CFD = EFA [Vertically Opposite angles]
CFD = 60o
Now in FCD,
Exterior BCF = CFD + CDF
90o = 60o + xo
xo = 90o - 60o = 30o
x = 30
(vi)
In ABE, we have,
A + B + E = 180o
75o + 65o + E = 180o
140o + E = 180o
E = 180o - 140o = 40o
Now, CED = AEB [Vertically opposite angles]
CED = 40o
Now, in CED, we have,
C + E + D = 180o
110o + 40o + xo = 180o
150o + xo = 180o
xo = 180o - 150o = 30o
x = 30
Solution 17
AB ∥ CD and AC is the transversal.
⇒ ∠BAC = ∠ACD = 60° (alternate angles)
i.e. ∠BAC = ∠GCH = 60°
Now, ∠DHF = ∠CHG = 50° (vertically opposite angles)
In ΔGCH, by angle sum property,
∠GCH + ∠CHG + ∠CGH = 180°
⇒ 60° + 50° + ∠CGH = 180°
⇒ ∠CGH = 70°
Now, ∠CGH + ∠AGH = 180° (linear pair)
⇒ 70° + ∠AGH = 180°
⇒ ∠AGH = 110°
Solution 18
Produce CD to cut AB at E.
Now, in BDE, we have,
Exterior CDB = CEB + DBE
xo = CEB + 45o .....(i)
In AEC, we have,
Exterior CEB = CAB + ACE
= 55o + 30o = 85o
Putting CEB = 85o in (i), we get,
xo = 85o + 45o = 130o
x = 130
Solution 19
The angle BAC is divided by AD in the ratio 1 : 3.
Let BAD and DAC be y and 3y, respectively.
As BAE is a straight line,
BAC + CAE = 180o [linear pair]
BAD + DAC + CAE = 180o
y + 3y + 108o = 180o
4y = 180o - 108o = 72o
Now, in ABC,
ABC + BCA + BAC = 180o
y + x + 4y = 180o
[Since, ABC = BAD (given AD = DB) and BAC = y + 3y = 4y]
5y + x = 180
5 18 + x = 180
90 + x = 180
x = 180 - 90 = 90
Solution 20
Given : A ABC in which BC, CA and AB are produced to D, E and F respectively.
To prove : Exterior DCA + Exterior BAE + Exterior FBD = 360o
Proof : Exterior DCA = A + B(i)
Exterior FAE = B + C(ii)
Exterior FBD = A + C(iii)
Adding (i), (ii) and (iii), we get,
Ext. DCA + Ext. FAE + Ext. FBD
= A + B + B + C + A + C
= 2A +2B + 2C
= 2 (A + B + C)
= 2 180o
[Since, in triangle the sum of all three angle is 180o]
= 360o
Hence, proved.
Solution 21
In ACE, we have,
A + C + E = 180o (i)
In BDF, we have,
B + D + F = 180o (ii)
Adding both sides of (i) and (ii), we get,
A + C+E + B + D + F = 180o + 180o
A + B + C + D + E + F = 360o.
Solution 22
In ΔABC, by angle sum property,
∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 20° = 180°
⇒ ∠A = 90°
In ΔABM, by angle sum property,
∠BAM + ∠ABM + ∠AMB = 180°
⇒ ∠BAM + 70° + 90° = 180°
⇒ ∠BAM = 20°
Since AN is the bisector of ∠A,
Now, ∠MAN + ∠BAM = ∠BAN
⇒ ∠MAN + 20° = 45°
⇒ ∠MAN = 25°
Solution 23
BAD ∥ EF and EC is the transversal.
⇒ ∠AEF = ∠CAD (corresponding angles)
⇒ ∠CAD = 55°
Now, ∠CAD + ∠CAB = 180° (linear pair)
⇒ 55° + ∠CAB = 180°
⇒ ∠CAB = 125°
In ΔABC, by angle sum property,
∠ABC + ∠CAB + ∠ACB = 180°
⇒ ∠ABC + 125° + 25° = 180°
⇒ ∠ABC = 30°
Solution 24
In the given ABC, we have,
A : B : C = 3 : 2 : 1
Let A = 3x, B = 2x, C = x. Then,
A + B + C = 180o
3x + 2x + x = 180o
6x = 180o
x = 30o
A = 3x = 3 30o = 90o
B = 2x = 2 30o = 60o
and, C = x = 30o
Now, in ABC, we have,
Ext ACE = A + B = 90o + 60o = 150o
ACD + ECD = 150o
ECD = 150o - ACD
ECD = 150o - 90o [since , ]
ECD= 60o
Solution 25
∠FGH + ∠FGE = 180° (linear pair)
⇒ 120° + y = 180°
⇒ y = 60°
AB ∥ DF and BD is the transversal.
⇒ ∠ABC = ∠CDE (alternate angles)
⇒ ∠CDE = 50°
BD ∥ FG and DF is the transversal.
⇒ ∠EFG = ∠CDE (alternate angles)
⇒ ∠EFG = 50°
In ΔEFG, by angle sum property,
∠FEG + ∠FGE + ∠EFG = 180°
⇒ x + y + 50° = 180°
⇒ x + 60° + 50° = 180°
⇒ x = 70°
Solution 26
AB ∥ CD and EF is the transversal.
⇒ ∠AEF = ∠EFD (alternate angles)
⇒ ∠AEF = ∠EFG + ∠DFG
⇒ 65° = ∠EFG + 30°
⇒ ∠EFG = 35°
In ΔGEF, by angle sum property,
∠GEF + ∠EGF + ∠EFG = 180°
⇒ x + 90° + 35° = 180°
⇒ x = 55°
Solution 27
AB ∥ CD and AE is the transversal.
⇒ ∠BAE = ∠DOE (corresponding angles)
⇒ ∠DOE = 65°
Now, ∠DOE + ∠COE = 180° (linear pair)
⇒ 65° + ∠COE = 180°
⇒ ∠COE = 115°
In ΔOCE, by angle sum property,
∠OEC + ∠ECO + ∠COE = 180°
⇒ 20° + ∠ECO + 115° = 180°
⇒ ∠ECO = 45°
Solution 28
AB ∥ CD and EF is the transversal.
⇒ ∠EGB = ∠GHD (corresponding angles)
⇒ ∠GHD = 35°
Now, ∠GHD = ∠QHP (vertically opposite angles)
⇒ ∠QHP = 35°
In DQHP, by angle sum property,
∠PQH + ∠QHP + ∠QPH = 180°
⇒ ∠PQH + 35° + 90° = 180°
⇒ ∠PQH = 55°
Solution 29
AB ∥ CD and GE is the transversal.
⇒ ∠EGF + ∠GED = 180° (interior angles are supplementary)
⇒ ∠EGF + 130° = 180°
⇒ ∠EGF = 50°