Class 9

RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Miscellaneous Exercise is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Miscellaneous Exercise.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 7
Chapter Name	Congruence and Inequalities of Triangles
Exercise	Miscellaneous Exercise
Number of Questions Solved	40
Category	RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Miscellaneous Exercise

Multiple Choice Questions (Q1 to Q16)

Question 1.
Which of the following is not a criterion (RBSESolutions.com) for congruence of triangle?
(A) SAS
(B) ASA
(C) SSA
(D) SSS
Solution.
(C) SSA

Question 2.
If AB = QR; BC = PR and CA = PQ, then:
(A) ΔABC = ΔPQR
(B) ΔCBA = ΔPRQ
(C) ΔBAC = ΔRPQ
(D) ΔPQR = ΔBCA
Solution.
(B) ΔCBA = ΔPRQ

Question 3.
In ΔABC, AB = AC and ∠B = 50°, then ∠C is equal to:
(A) 40°
(B) 50°
(C) 80°
(D) 130°
Solution.
(B) 50°

Question 4.
In ΔABC, BC = AB and ∠B = 80°, then ∠A is equal to:
(A) 80°
(B) 40°
(C) 50°
(D) 100°
Solution.
(C) 50°

Question 5.
In ΔPQR. ∠R = ∠P and QR = 4 cm and PR = 5 cm, then length of PQ is:
(A) 4 cm
(B) 5 cm
(C) 2 cm
(D) 2.5 cm
Solution.
(A) 4 cm

Question 6.
In ΔABC, D is a point on the side BC is situated in (RBSESolutions.com) such a way that AD is the bisector of ∠BAC then
(A) BD = CD
(B) BA > BD
(C) BD > BA
(D) CD > CA
Solution.
(A) BD = CD

Question 7.
Given that ΔABC = ΔFDE and AB = 5 cm, ∠B = 40° and ∠A = 80° then which of the following relation is true?
(A) DF = 5 cm, ∠F = 60°
(B) DF = 5 cm, ∠E = 60°
(C) DE = 5 cm, ∠E = 60°
(D) DE = 5 cm, ∠D = 40°
Solution.
(B) DF = 5 cm, ∠E = 60°

Question 8.
The length of the two sides of (RBSESolutions.com) a triangle are 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(A) 3.6 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm
Solution.
(D) 3.4 cm

Question 9.
In ΔPQR, if ∠R > ∠Q, then
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR Solution. (B) PQ > PR

Question 10.
In Δ’s ABC and PQR, AB = AC, ∠C = ∠P, and ∠B = ∠Q then these (RBSESolutions.com) two triangles are:
(A) Isosceles but not congruent
(B) Isosceles and congruence
(C) Congruent but not isosceles
(D) Neither isosceles nor congruent
Solution.
(A) Isosceles but not congruent

Question 11.
In Δ’s ABC and DEF, AB = FD and ∠A = ∠D, the two triangles will be congruent by SAS congruence rule if:
(A) BC = EF
(B) AC = DE
(C) AC = EF
(D) BC = DF
Solution.
(B) AC = DE

Question 12.
If ∠C is right angle in ΔABC, then the (RBSESolutions.com) largest side is:
(A) AB
(B) BC
(C) CA
(D) None
Solution.
(A) AB

Question 13.
The difference of any two sides of a triangle is ………. than the third side:
(A) greater
(B) equal
(C) less
(D) half
Solution.
(C) less

Question 14.
If two sides of a triangle are unequal (RBSESolutions.com) then opposite angle of larger side is:
(A) greater
(B) less
(C) equal
(D) half
Solution.
(A) greater

Question 15.
The perimeter of the triangle is ………. than the sum of its three medians.
(A) greater
(B) less
(C) equal
(D) half
Solution.
(A) greater

Question 16.
The sum of altitudes of a triangle is ………. than the perimeter of the triangle.
(A) greater
(B) equal
(C) half
(D) less
Solution.
(D) less

Question 17.
In an ∆ABC, if AB = AC and ∠A < 60° then write (RBSESolutions.com) the relation between BC and AC.
Solution.
BC < AC.

Question 18.
In figure, what is the relation between AB and AC?

Solution.
∠ABC = 180° – 135° = 45°,
∠ACB = 65°
i.e. ∠ACB > ∠ABC,
⇒ AB > AC.

Question 19.
In ∆ABC, ∠A > ∠B and ∠B > ∠C, then write (RBSESolutions.com) the smallest side.
Solution.
According to given relation ∠A > ∠C.
Side AB will be least because ∠C is least.

Question 20.
Find each angle of an equilateral triangle.
Solution.
All sides are equal to each angle will be equal to 60°.

Question 21.
P is a point on the bisectors of ∠ABC. If the line (RBSESolutions.com) through P, parallel to BA, meet BC at Q. Prove that BPQ is an isosceles triangle.
Solution.
P lie on the bisector of ∠ABC.
QT || BA
(∠1 + ∠2) = ∠3 (corresponding angles)

But BL is the bisector of ∠ABC
∠1 = ∠2
Also ∠3 = ∠2 + ∠BPQ
∠1 + ∠2 = ∠2 + ∠BPQ
⇒ ∠1 = ∠BPQ
⇒ ∠2 = ∠BPQ [∵ ∠1 = ∠2]
⇒ QP = BQ
∆BPQ is an isosceles triangle.

Question 22.
ABC is a right (RBSESolutions.com) triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.
Solution.
In ∆ABD and ∆ACD
AB = AC (given)
∠BAD = ∠CAD

As AD is bisector of ∠A and AD = AD
⇒ ∆DAB = ∆DAC (by SAS congruency rule)
⇒ ∠ADB = ∠ADC (by c.p.c.t)
⇒ ∠ADB = ∠ADC = 90°
and BD = DC
In ∆ABD,
AD² + BD² = AB² …(i)
⇒ AD² + DC² = AC² …(ii)
Adding (i) and (ii), we get
2 AD² + BD² + DC² = AB² + AC²
⇒ 2AD² + BD² + DC² = BC²
⇒ 2 AD² + 2BD² = BC²
⇒ 2 (AD² + BD²) = BC²

⇒ 4 AD² = BC²
⇒ 2 AD = BC
i.e. BC = 2AD
Hence proved.

Question 23.
∆ABC and ∆DBC are two triangles (RBSESolutions.com) on the same base BC such that A and D lie on the opposite side of BC. AB = AC and BD = DC. Show that AD is the ⊥ bisector of BC.
Solution.
In ∆BAD and ∆CAD
AB = AC (given)
BD = DC (given)
and AD = AD (common)
∆BAD = ∆CAD (by SSS congruence rule)

∠1 = ∠2 (by c.p.c.t)
Also AB = AC (given)
∠3 = ∠4 (angle opposite to equal sides are equal)
∆BAO = ∆CAO
⇒ BO = OC (by c.p.c.t)
or AO bisects BC (by c.p.c.t)
Also ∠AOB = ∠AOC (by c.p.c.t)
But ∠AOB + ∠AOC = 180°
⇒ ∠AOB = ∠AOC = 90°
⇒ AD is perpendicular bisector of BC.

Question 24.
ABC is an isosceles (RBSESolutions.com) triangle in which AC = BC and AD and BE are altitude on the sides BC and AC respectively. Prove that AE = BD.
Solution.
In ∆ABC,
AC = BC (given)
∠EAB = ∠DBA …(i) (angles opposite to equal sides are equal)

Now in ∆EAB and ∆DAB
∠AEB = ∠ADB = 90° (given)
∠EAB = ∠DBA [using (i)]
and AB = AB (common)
∆DAB = ∆EAB (by AAS congruencey rule)
⇒ AE = BD (by c.p.c.t)

Question 25.
Prove that the sum of any two sides (RBSESolutions.com) of a triangle is greater than twice the median drawn to the third side?
Solution.
Given: In ∆ABC, AD is the median
To prove: AB + AC > 2AD
Construction: Produce AD up to E, such that AD + DE and Join EC.

Proof: In ∆ADB and ∆EDC
AD = DE (by construction)
BD = DC (given)
and ∠ADB = ∠EDC (vertically opposite angles)
∆ADB = ∆EDC (by SAS congruency property)
⇒ AB = CE (by c.p.c.t)
Now in ∆ACE
AC + CE > AE
⇒ AC + AB > AE (∵ AB = CE) (Proved above)
⇒ AC + AB > 2AD (∵ AE = 2AD)
Hence proved.

Question 26.
In ∆ABC, D is the mid-point (RBSESolutions.com) of side AC such that BD = \(\frac { 1 }{ 2 }\) AC. Show that ∠ABC = 90°.
Solution.
D is mid-point of AC
⇒ AD = DC = \(\frac { 1 }{ 2 }\) AC

But BD = \(\frac { 1 }{ 2 }\) AC (given)
⇒ AD = DC = BD
In ∆ABD
AD = BD
⇒ ∠2 = ∠1 = x (angles opposite to equal sides are equal)
Also BD = DC
∠3 = ∠4 = x
In ∆ABC
∠1 + (∠2 + ∠3) + ∠4 = 180°
⇒ x + (x + x) + x = 180°
⇒ 4x = 1 80°
⇒ x = 45°
⇒ ∠ABC = x + x = 45° + 45° = 90°
Hence proved.

Question 27.
Prove that the line segment joining (RBSESolutions.com) the mid-point of the hypotenuse of a right triangle to the vertex of the right angle is equal to half the hypotenuse.
Solution.
Let ABC is a right triangle right angled at B.
Let P be the mid-point of the hypotenuse AC.
Through P, draw PQ || BC

PQ || BC
⇒ ∠1 = ∠B (corresponding angle)
⇒ ∠2 = ∠1 = 90°
In ∆ABC, P is the (RBSESolutions.com) mid-point of AC and PQ || BC
Q must be mid-point of AB (converse of mid-point theorem)
⇒ AQ = QB …(i)
Now in ∆AQP and ∆BPQ
AQ = QB
∠2 = ∠1 = 90°
PQ = PQ
∆AQP = ∆BPQ (by SAS congruency rule)
⇒ AP = PB (by c.p.c.t)
Thus, AP = PB = PC
Hence in other words, we can say that the point P is equi-distance from (RBSESolutions.com) the vertices of a right angled triangle ABC.

Question 28.
In figure, if AB = AC, what is the relation between AB and AD.

Solution.
AD > AB

Question 29.
AD is the median of any ∆ABC. Is it true to say that AB + BC + CA > 2AD. Give reason for your answer.
Solution.
Yes, if AD is median of ∆ABC

⇒ AB + BD > AD …(i) (sum of any two sides of a A is greater than third side)
and AC + DC > AD …(iii) (reason as above)
⇒ AB + (BD + DC) + AC > 2 AD
⇒ AB + BC + CA > 2AD

Question 30.
M is any point on the side BC of ∆ABC in such (RBSESolutions.com) a way that AM is the bisector of ∠BAC. Is it true to say that perimeter of ∆ABC is greater than 2 AM? Give reason for your answer.
Solution.
Yes, AB + BM > AM …(i)
also AC + CM > AM …(ii)

Adding (i) and (ii), we get
AB + (BM + CM) + AC > 2AM
⇒ AB + BC + CA > 2AM
⇒ Perimeter of ∆ABC > 2AM

Question 31.
In figure, Q is a point on side SR of ∆PSR such that PQ = PR. Prove that PS > PQ.

Solution.
In ∆PQR
PQ = PR
∠PQR = ∠PRQ …(i) (angles opposite to (RBSESolutions.com) equal sides of a triangle are equal)
In ∆PSQ
Ext. ∠PQR > ∠PSQ …(ii)
From (i) and (ii), we get
∠PRQ > ∠PSQ
⇒ PS > PR (side opposite to greater angle is longer)
⇒ PS > PQ (∵ PQ = PR)

Question 32.
In ∆PQR, S is any point on side QR. Show that PQ + QR + RP > 2PS.
Solution.
In ∆PQR,
PQ + QS > PS …(i) (∵ the sum of any two sides of a triangle is greater than the third side)

In ∆PRS,
PR + RS > PS (∵ The sum of (RBSESolutions.com) any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we get
(PQ + QS) + (PR + RS) > 2PS
⇒ PQ + (QS + RS) + PR > 2PS
⇒ PQ + QR + PR > 2PS
⇒ PQ + QR + RP > 2PS
Hence proved.

Question 33.
In an ∆ABC with AB = AC, D is any point on the side AC. Show that CD < BD.
Solution.
In ∆ABC

AB = AC (given)
⇒∠B = ∠C
⇒ ∠ADB = ∠DBC + ∠C (exterior angle (RBSESolutions.com) is equal to sum of opposite interior angles)
⇒ ∠ADB > ∠C
⇒ AB > BD
But AB = AC
⇒ AD + DC > BD
⇒ BD > CD
Hence proved.

Question 34.
In figure, ∠B > ∠A and ∠D > ∠E then show that AE > BD.

Solution.
∠B > ∠A
⇒ AC > BC …(i)
Also ∠D > ∠E
CE > CD …(ii) (greater angle has larger side opposite to it)
Adding (i) and (ii), we get AC + CE > BC + CD
⇒ AE > BD
Hence proved.

Question 35.
In any triangle ABC if AB > AC and D is (RBSESolutions.com) any point on BC then prove that AB > AD.
Solution.
In ∆ABC
AB > AC
⇒ ∠ACB > ∠ABC …(i) (∵ angle opposite to larger side is greater)

Now in ∆ACD, CD is produced to B forming an ext ∠ADB.
∠ADB > ∠ACD (exterior angle of a A is greater than each interior opposite angle)
⇒ ∠ADB > ∠ACB …(ii)
∠ACD = ∠ACB
From (i) and (ii)
∠ADB > ∠ABC
⇒ ∠ADB > ∠ABD [∵ ∠ABC = ∠ABD]
⇒ AB > AD [∵ side opposite to greater angle is larger]
Hence proved.

Question 36.
Prove that the (RBSESolutions.com) sum of three sides of a triangle is greater than the sum of its three medians.
Solution.
Given: In ∆ABC, AD, BE and CF are its medians.
To prove:
AB + BC + CA > AD + BE + CF

Proof: As we know that the sum of any two sides of (RBSESolutions.com) a triangle is greater than twice the median bisecting the third side. Therefore AD is the median bisecting BC
⇒ AB + AC > 2 AD
BE is the median bisecting AC
⇒ AB + BC > 2BE
Similarly for median CF
BC + AC > 2CF
Adding (i), (ii) and (iii), we get
2 (AB + BC + CA) > 2 (AD + BE + CF)
⇒ AB + BC + CA > AD + BE + CF
Hence proved.

Question 37.
In figure, O is an interior point of ∆ABC. Show that: AB + BC + CA < 2(OA + OB + OC).
Solution.
In triangle ABC, O is a point interior of ∆ABC.

As we know that “The sum of any (RBSESolutions.com) two sides of a triangle is greater than the third side”.
OA + OB > AB …(i)
OA + OC > AC …(ii)
and OB + OC > BC …(iii)
Now, adding (i), (ii) and (iii), we get
2(OA + OB + OC) > AB + BC + CA
or AB + BC + CA < 2(OA + OB + OC)
Hence proved.

Question 38.
Prove that the sum of three altitudes of a triangle is less than the perimeter of the triangle.
Solution.
Given: A triangle ABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.
To prove:
AD + BE + CF < AB + BC + CA
or AD + BE + CF < Perimeter of ∆ABC

Proof: As we know that from all the segments that can (RBSESolutions.com) be drawn to a given line, from a point not lying on it, the perpendi-cular line segment is the shortest one.
AD ⊥ BC
⇒ AB > AD and AC > AD
⇒ AB + AC > 2AD …(i)
BE ⊥ AC
⇒ BC > BE and BA > BE
⇒ BC + BA > 2AB …(ii)
Also CF ⊥ AB
⇒ AC > CF and BC > CF
⇒ AC + BC > 2CF …(iii)
Adding (i), (ii) and (iii), we get
(AB + AC) + (AB + BC) + (AC + BC) > 2AD + 2BE + 2CF
⇒ 2(AB + BC + CA) > 2(AD + BE + CF)
⇒ AB + BC + CA > AD + BE + CF
⇒ Perimeter of ∆ABC > AD + BE + CF
Hence proved.

Question 39.
Prove that the’difference of any two sides of (RBSESolutions.com) a triangle is less than the third side.
Solution.
Given: A ∆ABC

To prove:
(i) AC – AB < BC
(ii) BC – AC < AB
(iii) BC – AB < AC
Construction: Take a point D on AC such that AB = AD. Join BD.
Proof: In ∆ABD AB = AD ⇒ ∠1 = ∠2 …(i)
In ∆ABD, side AD is produced to C ∠3 > ∠1 …(ii)
(ext. angle is greater than one of the opposite interior angle)
Also in ∆BCD
∠2 > ∠4 …(iii) (reason as above)
From (i) and (ii), we get
∠3 > ∠2 …(iv)
From (iii) and (iv), we get
∠3 > ∠4
⇒ BC > CD (side opposite to greater angle is larger)
⇒ CD < BC
⇒ AC – AD < BC
⇒ AC – AB < BC [∵ AD = AB]
Similarly we can prove
BC – AC < AB and BC – AB < AC
Hence proved.

Question 40.
In figure, AD is bisector of ∠BAC then prove that AB > BD.

Solution.
In ∆ABC
AD is the bisector of ∠BAC
⇒ ∠1 = ∠2 …(i)
Also in ∆ADC
∠3 = ∠2 + ∠C (ext. angle is equal to (RBSESolutions.com) sum of opposite interior angles)
⇒ ∠3 > ∠2 (ext. angle is greater than one of the interior angles)
⇒ But ∠1 = ∠2
⇒ ∠3 > ∠1
⇒ AB > BD
Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Miscellaneous Exercise will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Additional Questions.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 7
Chapter Name	Congruence and Inequalities of Triangles
Exercise	Additional Questions
Number of Questions Solved	32
Category	RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Additional Questions

Multiple Choice Questions

Question 1.
In figure, if AB = AC and ∠B = 70°, the (RBSESolutions.com) value of ∠A will be:

(A) 70°
(B) 40°
(C) 55°
(D) 90°
Solution:
(B) 40°

Question 2.
In figure, if AB = AC, then the value of ∠C is equal to:

(A) 60°
(B) 36°
(C) 72°
(D) 108°
Solution:
(C) 72°

Question 3.
If the perpendicular drawn from the (RBSESolutions.com) mid-point of one side of a triangle to its other two sides are equal, then triangle is:
(A) Equilateral
(B) Isosceles
(C) Equiangular
(D) Scalene
Solution:
(B) Isosceles

Question 4.
In figure, AB = AC and AD ⊥ BC, then AD bisects the:

(A) ∠A
(B) Side BC
(C) ∠A and side BC
(D) None of these
Solution:
(C) ∠A and side BC

Question 5.
∆ABC, shown (RBSESolutions.com) in figure, in which AD = BD, and AC = DC and ∠C = 44°, then ∠A is:

(A) 68°
(B) 112°
(C) 34°
(D) 102°
Solution:
(D) 102°

Question 6.
In an isosceles triangle, AB = AC and side BA is (RBSESolutions.com) produced up to D such that AB = AD, then ∠BCD is:
(A) 70°
(B) 90°
(C) 60°
(D) 45°
Solution:
(B) 90°

Question 7.
In figure, AB = AC and ∠ABD = ∠ACD then ∆BDC is:

(A) Equilateral
(B) Isosceles
(C) Equiangular
(D) Scalene
Solution:
(B) Isosceles

Question 8.
In figure, AB = AC and AD is the (RBSESolutions.com) bisector of ∠BAC meet BC at D. If ∠BAC = 60° then ∠ADC is equal to:

(A) 30°
(B) 60°
(C) 90°.
(D) 120°
Solution:
(C) 90°

Question 9.
In two triangles ABC and DEF, if AC = DF, BC = EF and ∠ABC = ∠DEF = 90° then two triangles are said to be congruent by:
(A) RHS property
(B) SAS property
(C) ASA property
(D) SSS property
Solution:
(A) RHS property

Question 10.
In the adjoining figure, if PQ = PR and QS = RT then ∆PST is equal to:

(A) Isosceles triangle
(B) Equilateral triangle
(C) Scalene triangle
(D) Isosceles (RBSESolutions.com) right angled triangle
Solution:
(A) Isosceles triangle

Very Short Answer Type Questions

Question 1.
In figure, AB = AC, CD = CA and ∠ADC = 20°, find ∠ABC.

Solution.
CD = CA
⇒ ∠CAD = ∠ADC = 20°
⇒ ∠ACD = 180° – (20° + 20°) = 140°
∠ACB = 180° – 140° = 40°
But AB also equal to AC
⇒ ∠ABC = ∠ACB = 40°

Question 2.
In Δ’s ABC and DEF, if AB = DF, BC = DE, AC = EF and ∠D = 55°. Find ∠B.
Solution.
According to above (RBSESolutions.com) information ΔABC = ΔDEF
(by c.p.c.t) ∠D = ∠B = 55°

Question 3.
In figure, ∠B = ∠D = 90° and BC = CD. Is AB = DE? Why?

Solution.
In right angled triangles ABC and CDE
∠B = ∠D = 90° (given)
∠ACB = ∠DCE (vertically opposite angles)
and BC = CD (given)
ΔABC = ΔCDE (by ASA congruency property)
AB = DE (by c.p.c.t)
⇒ Yes, AB and DE are equal.

Question 4.
In a ΔABC, if ∠A = ∠B = 45°, which is the (RBSESolutions.com) longest side?
Solution.
∠A = ∠B = 45° (given)
∠C = 90°
Hence AB will be the longest side.

Question 5.
In figure, AB = 7 cm, BC = 8 cm and AC = 7.6 cm then write the
(i) greatest angle of the triangle
(ii) smallest angle (RBSESolutions.com) of the triangle

Solution.
(i) ∠CAB
(ii) ∠ACB

Question 6.
In ΔPQR, ∠Q = 35°, ∠R = 61° and the (RBSESolutions.com) bisector of ∠QPR meet QR at x. Then arrange the sides PX, QX and RX in descending order of their length.
Solution.
QX > PX > XR

Short Answer Type Questions

Question 1.
In the given figure, AB = AC and BD = EC then prove that ΔADE is an isosceles triangle.

Solution.
In ΔABD and ΔAEC
AB = AC (given) …(i)
∠ABC = ∠ACB …(ii) (angle opposite to equal sides are equal)
Also BD = EC (given) …(iii)
From (i), (ii) and (iii), we have
ΔABD = ΔAEC (by SAS congruency property)
⇒ AD = AE (by c.p.c.t)
Hence, ΔADE is (RBSESolutions.com) an isosceles triangle.
Hence proved.

Question 2.
In figure, BA ⊥ AC, DE ⊥ EF, such that BA = DE and BF = CD, prove that AC = EF.

Solution.
To prove: AC = EF, we have to prove
ΔABC = ΔDEF
Here BA = DE (given) …(i)
BA ⊥ AC and DE ⊥ FE
∠BAC = ∠DEF = 90° (each) …(ii)
Also BF = CD (given) …(iii)
Adding FC in equation (iii), we get
BF + FC = CD + FC
⇒ BC = FD …(iv)
From (i), (ii) and (iv), we get
ΔABC = ΔDEF (by RHS congruency property)
⇒ AC = EF (by c.p.c.t)
Hence proved.

Question 3.
In figure, CB = AD and AB = CD. Can we say ∠ABC and ∠ADC are equal? (RBSESolutions.com) Why?

Solution.
In ΔABC and ΔADC
AB = DC
AD = CB
and AC = AC
ΔABC = ΔADC (by SSS congruency property)
So, ∠ABC = ∠ADC (by c.p.c.t)
Hence proved.

Question 4.
In figure, CN ⊥ AB, DM ⊥ AB and CN = DM. Are OC and OD equal? Why?

Solution.
In ΔOMD and ΔONC
∠OMD = ∠ONC (each 90°)
∠MOD = ∠NOC (vertically opposite angles)
and CN = DM (given)
ΔOMD = ΔONC (by AAS property)
⇒ OC = OD (by c.p.c.t)
Hence proved.

Question 5.
In figure, ∠A = 35°, ∠ABC = 100° and BD ⊥ AC, prove that ΔBDC is an isosceles triangle.

Solution.
In ΔABC,
∠A + ∠B + ∠C = 180° (the sum of the (RBSESolutions.com) interior angles of a Δ is equal to 180°)
35° + 100° + ∠C= 180°
⇒ ∠C = 180° – 135°
⇒ ∠C = 45°
But in ΔABD, BD ⊥ AC
i.e., ∠ADB = 90°
and ∠A = 35° (given)
∠ABD = 180° – 125° = 55°
∠DBC = 100° – 55° = 45°
Now, ∠DCB = ∠DBC = 45° (Proved above)
Hence BDC is an isosceles triangle.

Question 6.
In the given figure, O is the (RBSESolutions.com) middle point of both AB and CD. Prove that AC = BD and AC || BD.

Solution.
In Δ’s AOC and BOD
OA = OB
OC = OD
∠AOC = ∠BOD (vertically opposite angles)
ΔAOC = ΔBOD (by SAS congruence property)
Hence AC = BD (by c.p.c.t)
Also ∠OAC = ∠OBD (alt. angle)
AC || BD
Hence proved.

Question 7.
In figure, X and Y are the points (RBSESolutions.com) on equal sides of AB and AC such that AX = AY. Show that XC = BY.

Solution.
In ΔABC
AB = AC (given)
and AX = AY (given)
⇒ AB – AX = AC – AY
⇒ BX = CY ….(i)
Now in ΔBXC and ΔBYC
BX = CY [from (i)]
BC = BC (common)
∠B = ∠C (angle opp. to equal sides are equal)
ΔBXC = ΔBYC (by SAS property)
⇒ XC = BY (by c.p.c.t)
Hence proved.

Question 8.
In the following figures, two sides (RBSESolutions.com) of ΔABC, AB and BC and median AD are respectively equal to PQ, QR and median PM of ΔPQR.
Prove that: ΔABC = ΔPQR.

Solution.
According to question
AB = PQ (given)
Median AD = Median PM
and BC = QR
\(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) QR
BD = QM
ΔABD = ΔPQM (by SSS congruency property)
∠B = ∠Q (by c.p.c.t)
Now in ΔABC and ΔPQR
AB = PQ (given)
∠B = ∠Q (proved above)
and BC = QR (given)
ΔABC = ΔPQR (by SAS congruency property)
Hence proved.

Question 9.
In figure, side QR of ΔPQR is (RBSESolutions.com) produced on both sides such that ∠PQS = ∠PRT. Prove that PQ = PR.

Solution.
In ΔPQR
∠PQS + ∠PQR = 180° (linear pair angles)
⇒ ∠PQS + ∠q = 180°
⇒ ∠q = 180° – ∠PQS …(i)
Also ∠r = 180° – ∠PRT …(ii)
But ∠PQS = ∠PRT (given) …(iii)
From (iii), eqn (i) becomes
∠q = 180° – ∠PRT …(iv)
Now from (ii) and (iv), we have
∠q = ∠r
i.e., ∠PQR = ∠PRQ
⇒ PR = PQ (converse of isosceles A theorem)
Hence proved.

Question 10.
Prove that the medians bisecting (RBSESolutions.com) the equal sides of an isosceles triangle are equal.
Solution.
Given: In ΔABC, D and E are mid-points of AB and AC respectively.
To prove: BE = CD

Proof: Since triangle ABC is (RBSESolutions.com) an isosceles triangle, then
AB = AC …(i)
and ∠ABC = ∠ACB …(ii)
D and E are mid-points of AB and AC respectively,
then DB = DA and EC = AE …(iii)
Now, in ΔBCD and ΔBCE
BC = BC (common)
∠DBC = ∠ECB by (ii)
BD = CE by (iii)
ΔBCD = ΔBCE (by SAS congruency rule)
⇒ BE = CD
Hence proved.

Long Answer Type Questions

Question 1.
“If two sides of a triangle are unequal (RBSESolutions.com) then the longer side has greater angle opposite to it.” Prove it.
Solution.
Given: A ΔABC in which AC > AB (say)
To prove: ∠ABC > ∠ACB
Construction: Mark a point D on AC such that AB = AD.
Join BD.

Proof: In ΔABD
AB = AD (by construction)
⇒ ∠1 = ∠2 …(i) (angles opposite to equal sides are equal)
Now in ΔBCD
∠2 > ∠DCB (ext. angle is greater than (RBSESolutions.com) one of the opposite interior angles)
⇒ ∠2 > ∠ACB …(ii) [∵ ∠ACB = ∠DCB]
From (i) and (ii), we get
∠1 > ∠ACB …(iii)
But ∠1 is a part of ∠ABC
∠ABC > ∠1 …..(iv)
Now from (iii) and (iv), we get
∠ABC > ∠ACB
Hence proved.

Question 2.
In figure, PQR is a triangle and S is (RBSESolutions.com) any point in its interior, show that SQ + SR < PQ + PR.

Solution.
Given: S is any point in the interior of ΔPQR
To prove: SQ + SR < PQ + PR
Construction: Produce QS to meet PR in T.
Proof: In ΔPQT PQ + PT > QT (sum of two sides of Δ is greater than third side)
or, PQ + PT > QS + ST …(i)
Also in ΔRST
ST + TR > RS …(ii)
Adding (i) and (ii), we get
PQ + PT + ST + TR > SQ + ST + SR
⇒ PQ + (PT + TR) > SQ + SR
⇒ PQ + PR > SQ + SR
Hence proved.

Question 3.
In figure, T is a point on side QR of ΔPQR and S is (RBSESolutions.com) a point such that RT = ST. Prove that PQ + PR > QS.

Solution.
In ΔPQR, we have
PQ + PR > QR (the sum of two sides of a triangle is greater than third side).
⇒ PQ + PR > QT + TR [∵ QT + TR = QR]
⇒ PQ + PR > QT + ST …(i) [∵ TR = ST]
In ΔQST, we have
QT + ST > QS …(ii)
From (i) and (ii), we get
PQ + PR > QS.
Hence proved.

Question 4.
In figure, side AB and AC are produced (RBSESolutions.com) to point D and E respectively. Bisectors BO and CO of ∠DBC and ∠ECB respectively meet at O. If AB > AC. Prove that OC > OB.

Solution.
Given: ABC is a triangle in which AB and AC is produced up to D and E respectively and bisectors of ∠DBC and ∠ECB meet at O. Also AB > AC.
To prove: OC > OB
Proof: In ΔABC
AB > AC
⇒ ∠ACB > ∠ABC …(i)
Also BO and CO are the bisectors of ∠DBC and ∠BCE respectively.
∠OBD = ∠OBC
and ∠OCB = ∠OCE
∠ACB = 180° – ∠BCE
⇒ ∠ACB = 180° – 2∠OCB …(ii)
Similarly ∠ABC = 180° – 2 ∠OBC…(iii)
From (i), (ii) and (iii), we have
180° – 2∠OCB > 180° – 2∠OBC
⇒ – 2∠OCB > – 2∠OBC
⇒ ∠OBC > ∠OCB
⇒ OC > OB
Hence proved.

Question 5.
If figure, ABCD is a quadrilateral. Prove that AB + BC + CD + DA > AC + BD.

Solution.
In ΔABC
AB + BC > AC …(i) (the sum of (RBSESolutions.com) any two sides of a Δ is greater than third side)
Also in ΔADC
AD + DC > AC …(ii) (reason as above)
Similarly in ΔABD and ΔBCD
AB + AD > BD …(iii)
BC + CD >BD …(iv)
Adding (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2AD + 2CD > 2AC + 2BD
⇒ AB + BC + CD + DA > AC + BD
Hence proved.

Question 6.
In the given figure, PQRS is a quadrilateral. PQ is its (RBSESolutions.com) longest side and RS is its shortest side. Prove that: ∠R > ∠P and ∠S > ∠Q.

Solution.
Given: PQRS is a quadrilateral. PQ is its (RBSESolutions.com) longest side and RS is its shortest side.
To prove: ∠R > ∠P and ∠S > ∠Q
Construction: Join PR and QS.
Proof: In ΔPQR
PQ is longest side (given)
PQ > QR
⇒ ∠5 > ∠2 …(i) (∠ opposite to longer side is greater)
In ΔPSR
RS is the shortest side (given)
PS > RS
⇒ ∠6 > ∠1 …(ii)
Adding (i) and (ii), we get
∠5 + ∠6 > ∠1 + ∠2
⇒ ∠R > ∠P
Now in ΔPQS, PQ is the (RBSESolutions.com) longest side
PQ > PS
⇒ ∠8 > ∠3 …(iii)
In ΔSRQ, RS is the shortest side
RQ > RS …(iv)
⇒ ∠7 > ∠4
Adding (iii) and (iv), we get
∠7 + ∠8 > ∠3 + ∠4
∠S > ∠Q
Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.2.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 7
Chapter Name	Congruence and Inequalities of Triangles
Exercise	Ex 7.2
Number of Questions Solved	11
Category	RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.2

Question 1.
In figure, AB = AC and ∠B = 58°, then find (RBSESolutions.com) the value of ∠A.

Solution.
∠A = 180° – (58° + 58°) = 180° – 116° = 64°

Question 2.
In figure, AD = BD and ∠C = ∠E. Prove that BC = AE.

Solution.
In ∆AED and ∆BCD
AD = BD (given)
∠ADE = ∠BDC
(vertically opposite angles)
∠E = ∠C
∴ ∆AED ≅ ∆BCD
(by AAS congruency property)
=> AE = BC (by c.p.c.t)

Question 3.
If AD be a median of (RBSESolutions.com) an isosceles ABC and ∠A = 120° and AB = AC then find ∠ADB.

Solution.
In ∆ABC
AB = AC (given)
∠B = ∠C = x (say)
x + x + 120° = 180°
=> 2x + 120 = 180°
=> 2x = 60° => x = 30°
∠ADB = 180° – (60° + 30°)
=> ∠ADB = 90°

Question 4.
If the bisector of an angle of a triangle (RBSESolutions.com) also bisects the opposite side, show that the triangle is isosceles.
Solution.
Given: In ∆ABC, the bisector of ∠BAC meets BC at D such that AD ⊥ BC

To prove: AB = AC or ∆ABC is an isosceles triangle.
Proof: In ∆’s ABD and ACD
∠BAD = ∠D AC (given)
∠ADB = ∠ADC
(each is a right angle)
and AD = AD (common)
∆ABD ≅ ∆ACD (by AAS congruency property)
=> AB = AC (c.p.c.t)
=> ABC is an isosceles ∆.
Hence proved.

Question 5.
In figure, AB = AC and BE = CD. Prove that AD = AE.

Solution.
In ∆ABD and ∆AEC
BE = DC (given) …(i)
Subtracting DE from (RBSESolutions.com) both side of (i), we get
BE – DE = DC – DE
=> BD = EC
AB = AC (given)
=> ∠B = ∠C
(angle opposite to equal sides are equal)
=> ∆ABD ≅ ∆AEC
=> AD = AE (by c.p.c.t)
Hence proved

Question 6.
Two points E and F of the sides AD and BC respectively of (RBSESolutions.com) a square ABCD such that AF = BE then prove that.
(i) ∠BAF = ∠ABE
(ii) BF = AE

Solution.
∵ ABCD is a square
=> ∠A = ∠B = 90°
Now in ∆AEB and ∆AFB
AF = BE
∠BAE = ∠ABF = 90° (given)
AB = AB (common side)
∴ ∆EAB ≅ ∆FBA
=> ∠FAB = ∠EBA (by c.p.c.t)
and also BF = AE (by c.p.c.t)
Hence proved.

Question 7.
AD and BC are equal perpendiculars to (RBSESolutions.com) a line segment AB (see figure). Show that CD bisects AB.

Solution.
In ∆OBC and ∆OAD
∵ ∠B = ∠A = 90°
(as AD and BC are perpendiculars)
and OB = OA (given)
and ∠BOC = ∠AOD
(vertically opposite angles)
∴ ∆OBC ≅ ∠OAD
(by AAS congruence rule)
=> OA = OB (by c.p.c.t)
=> CD bisects AB
Hence proved.

Question 8.
In an isosceles triangle (RBSESolutions.com) with AB = AC, the bisectors of ∠B and ∠C meet at O. Produce BO upto M then prove that ∠MOC = ∠ABC.
Solution.

In ∆ABC with AB = AC
∠B = ∠C (given)
In ∆BOC
∠MOC = ∠OBC + ∠OCB

Question 9.
Line l is the bisector of an angle ∠A and ∠B is any point on l. BP and BQ are (RBSESolutions.com) perpendicular from B to the arms of ∠A (see figure). Show that
(i) ∆APB = ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Solution.
(i) In ∆APB and ∆AQB
∠BQA = ∠BPA = 90° (given)
AB = AB (common side)
and ∠QAB = ∠PAB
(∵ l is the bisector of ∠A)
∴ ∆APB ≅ ∆AQB
(by AAS congruence rule)
(ii) ∵ ∆APB ≅ ∆AQB
=> BP = BQ (by c.p.c.t)
i.e. B is equidistant from the arms of ∠A.

Question 10.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC, show that BC = DE.

Solution.
We are given that ∠BAD = ∠EAC
Adding ∠DAC to both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
=> ∠BAC = ∠DAE …(i)
Now in ∆BAC and ∆EAD, we have
∠BAC = ∠DAE [using (i)]
AB = AD (given)
AC = AE (given)
∆BAC ≅ ∆EAD
(by SAS congruence rule)
=> BC = DE (by c.p.c.t)
Hence proved.

Question 11.
In right triangle ABC, right angled at C, M is the mid-point (RBSESolutions.com) of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure).

Show that:
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac { 1 }{ 2 }\) AB
Solution.
(i) In ∆AMC and ∆BMD
DM = CM (given)
∠3 = ∠4
(vertically opposite angles)
AM = MB
(as M is the mid-point of AB)
=> ∆AMC ≅ ∆BMD
(by SAS congruency rule)

(ii) ∵ ∆AMC ≅ ∆BMD
=> ∠1 = ∠2 (by c.p.c.t)
But they are alternate angles,
so AC || BD
=> ∠ACB + ∠DBC = 180
(sum of the interior angles on (RBSESolutions.com) the same side of a transversal is equal to 180°)
But ∠ACB = 90° (given)
=> ∠DBC = 180° = 90°
(iii) In ∆DBC and ∆ACB
∵ ∠DBC = ∠ACB = 90°
BC = BC (common side)
BD = CA
(∵ ∆AMC ≅ ∆BMD)
∴ ∆DBC ≅ ∆ACB
(by SAS congruency rule)
(iv) ∵ ∆DBC ≅ ∆ACB
=> DC = AB
But M mid-point of DC
=> 2CM = CD
=> 2CM = AB [∵ DC = AB]
=> CM = \(\frac { 1 }{ 2 }\) AB.
Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Miscellaneous Exercise is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Miscellaneous Exercise.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 6
Chapter Name	Rectilinear Figures
Exercise	Miscellaneous Exercise
Number of Questions Solved	32
Category	RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Miscellaneous Exercise

Multiple Choice Questions (Q1 to Q12)

Question 1.
If two angles of a triangle (RBSESolutions.com) measures 90° and 30°, the third angle is equal to:
(A) 90°
(B) 30°
(C) 60°
(D) 120°
Answer.
(C) 60°

Question 2.
The three angles of a triangle are in the ratio of 2 : 3 : 4, its greatest angle measures:
(A) 80°
(B) 60°
(C) 40°
(D) 180°
Answer.
(A) 80°

Question 3.
Each angle of an equilateral (RBSESolutions.com) triangle measures:
(A) 90°
(B) 30°
(C) 45°
(D) 60°
Answer.
(D) 60°

Question 4.
The angles of quadrilateral are in the ratio 1 : 2 : 3: 4, its smallest angle measures.
(A) 120°
(B) 36°
(C) 18°
(D) 10°
Answer.
(C) 18°

Question 5.
In figure, the side BC of a ΔABC has (RBSESolutions.com) been extended to D. If ∠A = 55° and ∠B = 60° then ∠ACD is:

(A) 120°
(B) 110°
(C) 115°
(D) 125°
Answer.
(C) 115°

Question 6.
The sum of interior angles of a hexagon is:
(A) 720°
(B) 360°
(C) 540°
(D) 1080°
Answer.
(A) 720°

Question 7.
The sum of n exterior (RBSESolutions.com) angles of an n-sided polygon is:
(A) n right angle
(B) 2n rigjjt angle
(C) (n – 4) right angle
(D) 4 right angle
Answer.
(D) 4 right angle

Question 8.
The number of sides in a regular polygon is n. Then measure of each interior angle is:
(A) \(\frac { 360 }{ n }\) degree
(B) (\(\frac { 2n – 4 }{ n }\)) right angle
(C) n right angle
(D) 2n right angle
Answer.
(B) (\(\frac { 2n – 4 }{ n }\)) right angle

Question 9.
If one angle of a triangle is (RBSESolutions.com) equal to the sum of the other two angles then the triangle is:
(A) Isosceles triangle
(B) Obtuse angled triangle
(C) Equilateral triangle
(D) Right angled triangle
Answer.
(D) Right angled triangle

Question 10.
One of the exterior angle of a triangle is 105° and its two opposite interior angles are equal, then each of the equal angle is:
(A) 37\(\frac { 1 }{ 2 }\)°
(B) 52\(\frac { 1 }{ 2 }\)°
(C) 72\(\frac { 1 }{ 2 }\)°
(D) 75°
Answer.
(B) 52\(\frac { 1 }{ 2 }\)°

Question 11.
The angles of a triangle are (RBSESolutions.com) in the ratio 5 : 3 : 7 then the triangle is:
(A) Acute angled triangle
(B) Obtuse angled triangle
(C) Right angled triangle
(D) Isosceles triangle
Answer.
(A) Acute angled triangle

Question 12.
If one of the angle of a triangle is 130° then the angle between the bisectors of the remaining two angles will be
(A) 50°
(B) 65°
(C) 145°
(D) 155°
Answer.
(D) 155°

Question 13.
In figure, find ∠A.

Solution.
∠ACB = 180° – 120° = 60° (due to linear pair)
Exterior ∠B = ∠A + ∠ACB
112° = ∠A + 60°
⇒ ∠A = 52°

Question 14.
In figure, ∠B = 60° and ∠C = 40°. Find the (RBSESolutions.com) measure of ∠A.

Solution.
∠A + ∠B + ∠C = 180°
⇒ ∠A + 60° + 40° = 180°
⇒ ∠A = 80°

Question 15.
In figure, m || QR then find ∠QPR.

Solution
∠QPR = 180° – (50° + 45°) = 180° – 95° = 85°

Question 16.
In figure, find ∠A.

Solution.
At point B,
100° + ∠ABC = 180° (linear pair angles)
∠ABC = 80°
At point C,
∠DCB + 95° = 180°
⇒ ∠DCB = 85°
Similarly at D,
82° + ∠CDA = 180° (linear pair axiom)
∠CDA = 180° – 82° = 98°
Now in quadrilateral ABCD
∠A + 80° + 85° + 98° = 360° (by angle sum property of a quadrilateral)
⇒ ∠A = 360° – 263°
⇒ ∠A = 97°.

Question 17.
The four angles of (RBSESolutions.com) a pentagon are 40°, 75°, 125°, and 135° respectively. Find the fifth angle.
Solution.
Suppose fifth angle be x
x + 40° + 75° + 125° + 135° = 540° (sum of interior angles of a pentagon)
⇒ x + 375° = 540°
⇒ x = 540° – 375°
⇒ x = 165°

Question 18.
Each exterior angle of a regular (RBSESolutions.com) polygon is 45°. Find the number of sides in the polygon.
Solution.
Measure of each exterior angle = \(\frac { { 360 }^{ 0 } }{ n }\)
n = \(\frac { 360 }{ 45 }\) = 8
Hence number of sides in the polygon = 8.

Question 19.
A regular polygon has 12 sides, find the measure of each of its interior angles.
Solution.

Question 20.
The sum of the interior angles (RBSESolutions.com) of a polygon is 10 right angles. Find the number of sides.
Solution.
Sum of the interior angles of a polygon = 10 right angles = 10 x 90° = 900°
But sum of the interior angles of the polygon = (2n – 4) right angles
i.e. 900 = (2n – 4) x 90°
⇒ 180n – 360 = 900
⇒ 180n = 900 + 360
⇒ 180n = 1260
⇒ n = 7
Hence the required number of sides = 7.

Question 21.
Examine, if it is possible to have a regular (RBSESolutions.com) polygon whose each interior angle is 110°.
Solution.
Here, it is given that the measure of each interior angle of a regular polygon =110° (if possible)
Hence, measure of each exterior angle = 180° – 110° = 70°.
Suppose number of sides of the polygon = n
But sum of all the n exterior angles = 360°
⇒ n x 70° = 360°
⇒ n = \(\frac { 360 }{ 70 }\)
= 5\(\frac { 1 }{ 7 }\) ≠ a whole number.
Hence, a regular polygon having measure of each interior angle 110° can not exist.

Question 22.
In a ∆ABC, ∠A + ∠B = ∠C , then find the (RBSESolutions.com) greatest angle of the triangle ABC.
Solution.
∠A + ∠B + ∠C = 180° (due to angle sum property)
But ∠A + ∠B = ∠C
⇒ 2∠C = 180°
⇒ ∠C = 90°
Hence greatest angle = 90° which is right angle.

Question 23.
Find the sum of the (RBSESolutions.com) interior angles of an octagon.
Solution.
The sum of the interior angles of a polygon = (2n – 4) x 90°
Here n = 8
The sum of the interior angles = (2 x 8 – 4) x 90° = (16 – 4) x 90° = 12 x 90° = 1080°

Question 24.
Find the measure of each interior angle of a 10-sided regular polygon.
Solution.

Question 25.
The exterior angles of a triangle (RBSESolutions.com) obtained by producing, the sides in order are 110°, 130° and x. Find x.
Solution.
The sum of the exterior angles in a polygon = 360°
110° + 130° + x = 360°
⇒ x = 360° – 240°
⇒ x = 120°

Question 26.
One interior angle of (RBSESolutions.com) a hexagon is 165° and the remaining interior angles are x° each.
Solution.
Sum of the interior angles of a hexagon = 720°
165° + x + x + x + x + x = 720°
⇒ 165° + 5x = 720°
⇒ 5x = 720° – 165° = 555°
⇒ x = \(\frac { 555 }{ 11 }\) = 111°

Question 27.
In the given figure, if AB || DC then find ∠x, ∠y and ∠z.

Solution.
In ∆BCE
88° = 22° + ∠z (by ext. angle property)
∠z = 88° – 22°
⇒ ∠z = 66°
Also AB || DC
88° + ∠y = 180° (interior angle property)
∠y = 180° – 88°
⇒ ∠y = 92°
Again, 102° + ∠DAB = 180° (linear pair axiom)
∠DAB = 180° – 102° = 78°
AB || DC, then x + 78° = 180° (The sum of the interior angles on the same of a transversal is 180°)
⇒ x = 102°
Hence ∠x = 102°, ∠y = 92°, ∠z = 66°

Question 28.
In figure, if ∠x – ∠y = 10°, then find (RBSESolutions.com) the measure of ∠x and ∠y.

Solution.
According to figure
∠x + ∠y = 120° …(i) (by exterior angle property)
and ∠x – ∠y = 10° (given) …(ii)
2∠x = 130°
⇒ ∠x = 65°
Now, 65° + ∠y = 120°
⇒ ∠y = 120° – 65°
⇒ ∠y = 55°
Hence, ∠x = 65° and ∠y = 55°.

Question 29.
In a polygon two of its interior angles (RBSESolutions.com) are each equal to 90° and remaining angles are 150° each. Find the number of sides of the polygon.
Solution.
Suppose number of sides of the polygon be n.
According to problem
2 x 90° + (n – 2) x 150°=(2n – 4) x 90°
⇒ 180 + 150n – 300 = 180n – 360°
⇒ 360 – 300 + 180 = 180n – 150n
⇒ 240 = 30n
⇒ n = 8

Question 30.
From the given (RBSESolutions.com) figure, prove that ∠x + ∠y = ∠A + ∠C.
Solution.
Construction: Join A to C
In ∆ABC,
∠x = ∠1 + ∠2 …(i) (exterior angle is equal to sum of opposite interior angles)

Also in ∆ADC,
∠y = ∠3 + ∠4 …(ii) (reason as above)
Adding (i) and (ii), we get
∠x + ∠y = ∠1 + ∠2 + ∠3 + ∠4
⇒ ∠x + ∠y = (∠1 + ∠3) + (∠2 + ∠4)
Hence, ∠x + ∠y = ∠A + ∠C
Hence proved.

Question 31.
From figure, find ∠x. Here BO and CO are (RBSESolutions.com) bisectors of ∠B and ∠C respectively.

Solution.
We know that
∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A
∵ ∠A = 80°
⇒ ∠BOC = 90° + \(\frac { 1 }{ 2 }\) x 80° = 90° + 40° = 130°

Question 32.
In figure, ∠Q > ∠R and PA is the (RBSESolutions.com) bisector of ∠QPR and PM ⊥ QR, then prove that ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R).

Solution.
Given: In ΔPQR, PA is the bisector of ∠QPR and PM ⊥ QR.
To prove: ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)
Proof: Suppose ∠APR = ∠1, ∠APM = ∠2 and ∠QPM = ∠3
∠1 = ∠2 + ∠3 …(i) [As PA is the bisector of ∠QPR]
In ΔPMR,
∠MPR + ∠PMR + ∠PRM = 180° (angle sum property of a triangle)
⇒ (∠1 + ∠2) + 90° + ∠PRM = 180°
⇒ (∠1 + ∠2) + ∠PRM = 90° …(ii)
Similarly in ΔPQM,
∠3 + ∠Q = 90° …(iii) [As ∠PMQ = 90°]
From (ii) and (iii), we get
∠1 + ∠2 + ∠PRM = ∠3 + ∠Q
⇒ ∠1 + ∠2 + ∠R = ∠3 + ∠Q
⇒ ∠1 + ∠2 – ∠3 = ∠Q – ∠R
⇒ (∠1 – ∠3) + ∠2 = ∠Q – ∠R (as ∠Q > ∠R)
Using relation (i), we get
∠2 + ∠2 = ∠Q – ∠R
⇒ 2∠2 = ∠Q – ∠R
⇒ ∠2 = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)
⇒ ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)
Hence proved.

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