RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables

In this chapter, we provide RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables pdf, free RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

Question 1.
Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
Sum of four angles of a quadrilateral = 360°
Three angles are = 110°, 50° and 40°
∴ Fourth angle = 360° – (110° + 50° + 40°)
= 360° – 200° = 160°

Question 2.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measures of each angle of the quadrilateral.
Solution:
Sum of angles of a quadrilateral ABCD = 360°
Ratio in angles = 1 : 2 : 4 : 5
Let first angle = x
Second angle = 2x
Third angle = 4x
and fourth angle = 5x
∴ x + 2x + 4x + 5x = 360°
⇒ 12x = 360° ⇒ x = (frac { { 360 }^{ circ } }{ 12 })  = 30°
∴ First angle = 30°
Second angle = 30° x 2 = 60°
Third angle = 30° x 4 = 120°
Fourth angle = 30° x 5 = 150°

Question 3.
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. [NCERT]
Solution:
Sum of four angles of a quadrilateral = 360°
Ratio in the angles = 3 : 5 : 9 : 13
Let first angle = 3x
Then second angle = 5x
Third angle = 9x
and fourth angle = 13x
∴ 3x + 5x + 9x+ 13x = 360°
⇒ 30x = 360°
⇒ x = (frac { { 360 }^{ circ } }{ 30 }) = 12°
∴ First angle = 3x = 3 x 12° = 36°
Second angle = 5x = 5 x 12° = 60°
Third angle = 9x = 9 x 12° = 108°
Fourth angle = 13 x 12° = 156°

Question 4.
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
Prove that ∠COD = (frac { 1 }{ 2 }) (∠A + ∠B).
Solution:
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables

RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.2

Question 1.
Two opposite angles of a parallelogram are (3x- 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
∵ Opposite angles of a parallelogram are equal
∴ 3x – 2 = 50 – x
⇒ 3x + x – 50 + 2
⇒ 4x = 52
⇒ x = (frac { 52 }{ 4 }) = 13
∴ ∠A = 3x – 2 = 3 x 13 – 2 = 39° – 2 = 37°
∠C = 50° -x = 50° – 13 = 37°
But∠A + B = 180°
∴ 37° + ∠B = 180°
⇒ ∠B = 180° – 37° = 143°
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables
and ∠D = ∠B (Opposite angles of a ||gm)
∴ ∠D = 143°
Hence angles and 37°, 143°, 37°, 143°

Question 2.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let in ||gm ABCD,
∠A =x
Then ∠B = (frac { 2 }{ 3 }) x
Linear Equations in Two Variables Class 9 RD Sharma Solutions
But, ∠A + ∠B = 180° (Sum of two adjacent angles of a ||gm)
⇒ x + (frac { 2 }{ 3 })x = 180°
⇒ (frac { 5 }{ 3 })x = 180°
⇒ x = 180° x (frac { 3 }{ 5 }) = 108°
∴ ∠A = 108°
and ∠B = 108° x (frac { 2 }{ 3 }) = 72°
But, ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 108°, ∠D = 72°
Hence angles are 108°, 72°, 108°, 72°

Question 3.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.
Solution:
Let smallest angle of a ||gm = x
Then second angle = 2x – 24°
But these are consecutive angles
∴ x + (2x- 24°) = 180°
⇒ x + 2x – 24° = 180°
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables
⇒ 3x = 180° + 24° = 204°
⇒ x =(frac { { 204 }^{ circ } }{ 3 })  = 68°
∴ Smallest angle = 68°
and second angle = 2x 68° – 24°
= 136°-24° = 112°
∵ The opposite angles of a ||gm are equal Other two angles will be 68° and 112°
∴ Hence angles are 68°, 112°, 68°, 112°

Question 4.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?
Solution:
In a ||gm ABC,
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
Perimeter = 22cm
and longest side = 6.5 cm
Let shorter side = x
∴ 2x (6.5 + x) = 22
⇒ 13 + 2x = 22
⇒ 2x = 22 – 13 = 9
⇒ x = (frac { 9 }{ 2 }) = 4.5
∴ shorter side = 4.5cm

Question 5.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In ||gm ABCD,
∠D = 135°
But, ∠A + ∠D = 180° (Sum of consecutive angles)
⇒∠A+ 135° = 180°
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions
⇒ ∠A = 180° – 135° = 45°
∵ ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠B = 135°

Question 6.
ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Solution:
In ||gm ABCD,
∠A = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables

Question 7.
In the figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables
Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles) But, ∠A = 75°
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions
∴ ∠B = 180° – ∠A = 180° – 75° = 105°
∴ DBA = 105° -60° = 45°
But ∠CDB = ∠DBA (alternate angles)
= 45°
and ∠ADB = ∠DBC = 60°

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) In a parallelogram, the diagonals are equal.
(ii) In a parallelogram, the diagonals bisect each other.
(iii) In a parallelogram, the diagonals intersect each other at right angles.
(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.
(v) If all the angles of a quadrilateral are equal, it is a parallelogram.
(vi) If three sides of a quadrilateral are equal, it is a parallelogram.
(vii) If three angles of a quadrilateral are equal, it is a parallelogram.
(viii)If all the sides of a quadrilateral are equal it is a parallelogram.
Solution:
(i) False, Diagonals of a parallelogram are not equal.
(ii) True.
(iii) False, Diagonals bisect each other at right angles is a rhombus or a square only.
(iv) False, In a quadrilateral, if opposite sides are equal and parallel, then it is a ||gm.
(v) False, If all angles are equal, then it is a square or a rectangle.
(vi) False, If opposite sides are equal and parallel then it is a ||gm
(vii) False, If opposite angles are equal, then it is a parallelogram.
(viii)False, If all the sides are equal then it is a square or a rhombus but not parallelogram.

Question 9.
In the figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC= BC and DC = 2AD.
RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables
Solution:
Given : In ||gm ABCD,
∠A = 60°
Bisector of ∠A and ∠B meet at P.
To prove :
(i) AD = DP
(ii) PC = BC
(iii) DC = 2AD
Construction : Join PD and PC
Proof : In ||gm ABCD,
∠A = 60°
But ∠A + ∠B = 180° (Sum of excutive angles)
⇒ 60° + ∠B = 180°
∴ ∠B = 1809 – 60° = 120°
∵ DC || AB
∠PAB = ∠DPA (alternate angles)
⇒ ∠PAD = ∠DPA (∵ ∠PAB = ∠PAD)
∴ AB = DP
(PA is its angle bisector, sides opposite to equal angles)
(ii) Similarly, we can prove that ∠PBC = ∠PCB (∵ ∠PAB = ∠BCA alternate angles)
∴ PC = BC
(iii) DC = DP + PC
= AD + BC [From (i) and (ii)]
= AD + AB = 2AB (∵BC = AD opposite sides of the ||gm)
Hence DC = 2AD

Question 10.
In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove thatAF = 2AB.
Solution:
Given : In ||gm ABCD,
E a mid point of BC
DE is joined and produced to meet AB produced at F
RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables
To prove : AF = 2AB
Proof : In ∆CDE and ∆EBF
∠DEC = ∠BEF (vertically opposite angles)
CE = EB (E is mid point of BC)
∠DCE = ∠EBF (alternate angles)
∴ ∆CDE ≅ ∆EBF (SAS Axiom)
∴ DC = BF (c.p.c.t.)
But AB = DC (opposite sides of a ||gm)
∴ AB = BF
Now, AF = AB + BF = AB + AB = 2AB
Hence AF = 2AB

Linear Equations in Two Variables Class 9 RD Sharma Solutions Ex 13.3

Question 1.
In a parallelogram ABCD, determine the sum of angles ZC and ZD.
Solution:
In a ||gm ABCD,
∠C + ∠D = 180°
(Sum of consecutive angles)
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables

Question 2.
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
Solution:
In a ||gm ABCD, ∠B = 135°
Linear Equations in Two Variables Class 9 RD Sharma Solutions
∴ ∠D = ∠B = 135° (Opposite angles of a ||gm)
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ ∠B + 135° = 180°
∴ ∠A = 180° – 135° = 45°
But∠C = ∠B = 45° (Opposite angles of a ||gm)
∴ Angles are 45°, 135°, 45°, 135°.

Question 3.
ABCD is a square, AC and BD intersect at O. State the measure of ∠AOB.
Solution:
In a square ABCD,
Diagonal AC and BD intersect each other at O
∵ Diagonals of a square bisect each other at right angle
∵∠AOB = 90°
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables

Question 4.
ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.
Solution:
In rectangle ABCD,
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
∠B = 90°, BD is its diagonal
But ∠ABD = 40°
and ∠ABD + ∠DBC = 90°
⇒ 40° + ∠DBC = 90°
⇒ ∠DBC = 90° – 40° = 50°
Hence ∠DBC = 50°

Question 5.
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Solution:
Given : In ||gm ABCD, E and F are the mid points of the side AB and CD respectively
DE and BF are joined
To prove : EBFD is a ||gm
Construction : Join EF
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions
Proof : ∵ ABCD is a ||gm
∴ AB = CD and AB || CD
(Opposite sides of a ||gm are equal and parallel)
∴ EB || DF and EB = DF (∵ E and F are mid points of AB and CD)
∴ EBFD is a ||gm.

Question 6.
P and Q are the points of trisection of the diagonal BD of the parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Solution:
Given : In ||gm, ABCD. P and Q are the points of trisection of the diagonal BD
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables
To prove : (i) CQ || AP
AC bisects PQ
Proof: ∵ Diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
∴ P and Q are point of trisection of BD
∴ BP = PQ = QD …(i)
∵ BO = OD and BP = QD …(ii)
Subtracting, (ii) from (i) we get
OB – BP = OD – QD
⇒ OP = OQ
In quadrilateral APCQ,
OA = OC and OP = OQ (proved)
Diagonals AC and PQ bisect each other at O
∴ APCQ is a parallelogram
Hence AP || CQ.

Question 7.
ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:
Given : In square ABCD
E, F, G and H are the points on AB, BC, CD and DA respectively such that AE = BF = CG = DH
To prove : EFGH is a square
Proof : E, F, G and H are points on the sides AB, BC, CA and DA respectively such that
AE = BF = CG = DH = x (suppose)
Then BE = CF = DG = AH = y (suppose)
Now in ∆AEH and ∆BFE
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables
AE = BF (given)
∠A = ∠B (each 90°)
AH = BE (proved)
∴ ∆AEH ≅ ∆BFE (SAS criterion)
∴ ∠1 = ∠2 and ∠3 = ∠4 (c.p.c.t.)
But ∠1 + ∠3 = 90° and ∠2 + ∠4 = 90° (∠A = ∠B = 90°)
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 90° + 90° = 180°
⇒ ∠1 + ∠4 + ∠1 + ∠4 = 180°
⇒ 2(∠1 + ∠4) = 180°
⇒ ∠1 + ∠4 = (frac { { 180 }^{ circ } }{ 2 })  = 90°
∴ ∠HEF = 180° – 90° = 90°
Similarly, we can prove that
∠F = ∠G = ∠H = 90°
Since sides of quad. EFGH is are equal and each angle is of 90°
∴ EFGH is a square.

Question 8.
ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.
Solution:
Given : ABCD is a rhombus, EABF is a straight line such that
EA = AB = BF
ED and FC are joined
Which meet at G on producing
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions
To prove: ∠EGF = 90°
Proof : ∵ Diagonals of a rhombus bisect
each other at right angles
AO = OC, BO = OD
∠AOD = ∠COD = 90°
and ∠AOB = ∠BOC = 90°
In ∆BDE,
A and O are the mid-points of BE and BD respectively.
∴ AO || ED
Similarly, OC || DG
In ∆ CFA, B and O are the mid-points of AF and AC respectively
∴ OB || CF and OD || GC
Now in quad. DOCG
OC || DG and OD || CG
∴ DOCG is a parallelogram.
∴ ∠DGC = ∠DOC (opposite angles of ||gm)
∴ ∠DGC = 90° (∵ ∠DOC = 90°)
Hence proved.

Question 9.
ABCD is a parallelogram, AD is produced to E so that DE = DC = AD and EC produced meets AB produced in F. Prove that BF = BC.
Solution:
Given : In ||gm ABCD,
AB is produced to E so that
DE = DA and EC produced meets AB produced in F.
To prove : BF = BC
Proof: In ∆ACE,
RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables
O and D are the mid points of sides AC and AE
∴ DO || EC and DB || FC
⇒ BD || EF
∴ AB = BF
But AB = DC (Opposite sides of ||gm)
∴ DC = BF
Now in ∆EDC and ∆CBF,
DC = BF (proved)
∠EDC = ∠CBF
(∵∠EDC = ∠DAB corresponding angles)
∠DAB = ∠CBF (corresponding angles)
∠ECD = ∠CFB (corresponding angles)
∴ AEDC ≅ ACBF (ASA criterion)
∴ DE = BC (c.p.c.t.)
⇒ DC = BC
⇒ AB = BC
⇒ BF = BC (∵AB = BF proved)
Hence proved.

RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables Ex 13.4

Question 1.
In a ∆ABC, D, E and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of ∆DEF.
Solution:
In ∆ABC, D, E and F are the mid-points of sides,
BC, CA, AB respectively
AB = 7cm, BC = 8cm and CA = 9cm
∵ D and E are the mid points of BC and CA
∴ DE || AB and DE =(frac { 1 }{ 2 }) AB =(frac { 1 }{ 2 }) x 7 = 3.5cm
Similarly,
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables

Question 2.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Solution:
In ∆ABC,
∠A = 50°, ∠B = 60° and ∠C = 70°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and ED are joined
∵ D, E and F are the mid points of sides BC, CA and AB respectively
∴ EF || BC
DE || AB and FD || AC
∴ BDEF and CDEF are parallelogram
∴ ∠B = ∠E = 60° and ∠C = ∠F = 70°
Then ∠A = ∠D = 50°
Hence ∠D = 50°, ∠E = 60° and ∠F = 70°

Question 3.
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.
Solution:
P, Q, R are the mid points of sides BC, CA and AB respectively
AC = 21 cm, BC = 29 cm and AB = 30°
∵ P, Q, R and the mid points of sides BC, CA and AB respectively.
∴ PQ || AB and PQ = (frac { 1 }{ 2 }) AB
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables

Question 4.
In a ∆ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
Solution:
Given : In ∆ABC, AD is median and AD is produced to X such that DX = AD
To prove : ABXC is a parallelogram
Construction : Join BX and CX
Proof : In ∆ABD and ∆CDX
AD = DX (Given)
BD = DC (D is mid points)
∠ADB = ∠CDX (Vertically opposite angles)
∴ ∆ABD ≅ ∆CDX (SAS criterian)
∴ AB = CX (c.p.c.t.)
and ∠ABD = ∠DCX
But these are alternate angles
Linear Equations in Two Variables Class 9 RD Sharma Solutions
∴ AB || CX and AB = CX
∴ ABXC is a parallelogram.

Question 5.
In a ∆ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Solution:
Given : In ∆ABC, E and F are the mid-points of AC and AB respectively.
EF are joined.
AP ⊥ BC is drawn which intersects EF at Q and meets BC at P.
To prove: AQ = QP
proof : In ∆ABC
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables
E and F are the mid points of AC and AB
∴ EF || BC and EF = (frac { 1 }{ 2 })BC
∴ ∠F = ∠B
In ∆ABP,
F is mid point of AB and Q is the mid point of FE or FQ || BC
∴ Q is mid point of AP,
∴ AQ = QP

Question 6.
In a ∆ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 - 6a

Question 7.
In the figure triangle ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the mid points of the sides AB and AC respectively, calculate.
(i) The length of BC
(ii) The area of ∆ADC
Solution:
In ∆ABC, ∠B = 90°
AC =15 cm, AB = 9cm
D and E are the mid points of sides AB and AC respectively and D, E are joined.
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions

Question 8.
In the figure, M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables
Solution:
In ∆ABC,
M, N and P are the mid points of side, AB, AC and BC respectively.
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables

Question 9.
In the figure, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.
Solution:
Given : In ABC, AB = AC
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions
nd CP || BA, AP is the bisector of exterior ∠CAD of ∆ABC
To prove :
(i) ∠PAC = ∠BCA
(ii) ABCP is a ||gm
Proof : (i) In ∆ABC,
∵ AB =AC
∴ ∠C = ∠B (Angles opposite to equal sides) and ext.
∠CAD = ∠B + ∠C
= ∠C + ∠C = 2∠C ….(i)
∵ AP is the bisector of ∠CAD
∴ 2∠PAC = ∠CAD …(ii)
From (i) and (ii)
∠C = 2∠PAC
∠C = ∠CAD or ∠BCA = ∠PAC
Hence ∠PAC = ∠BCA
(ii) But there are alternate angles,
∴ AD || BC
But BA || CP
∴ ABCP is a ||gm.

Question 10.
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.
Solution:
Given : In fne figure, ABCD is a kite in which AB = AD and BC = CD.
P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
To prove : PQRS is a rectangle.
Construction : Join AC and BD.
RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables
Proof: In ∆ABD,
P and S are mid points of AB and AD
∴ PS || BD and PS = (frac { 1 }{ 2 }) BD …(i)
Similarly in ∆BCD,
Q and R the mid points of BC and CD
∴ QR || BD and
QR = (frac { 1 }{ 2 }) BD …(ii)
∴ Similarly, we can prove that PQ || SR and PQ = SR …(iii)
From (i) and (ii) and (iii)
PQRS is a parallelogram,
∵ AC and BD intersect each other at right angles.
∴ PQRS is a rectangle.

Question 11.
Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Solution:
In ∆ABC, AB = AC
D, E and F are the mid points of the sides BC, CA and AB respectively,
AD and EF are joined intersecting at O
To prove : AD and EF bisect each other at right angles.
Construction : Join DE and DF.
RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables
Proof : ∵ D, E and F are the mid-points of
the sides BC, CA and AB respectively
∴ AFDE is a ||gm
∴ AF = DE and AE = DF
But AF = AE
(∵ E and F are mid-points of equal sides AB and AC)
∴ AF = DF = DE = AE
∴AFDE is a rhombus
∵ The diagonals of a rhombus bisect each other at right angle.
∴ AO = OD and EO = OF
Hence, AD and EF bisect each other at right angles.

Question 12.
Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.
Solution:
Given : In quad. ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.
PR and QS to intersect each other at O
To prove : PO = OR and QO = OS
Construction: Join PQ, QR, RS and SP and also join AC.
Proof: In ∆ABC
P and Q are mid-points of AB and BC
∴ PQ || AC and PQ = (frac { 1 }{ 2 }) AC …(i)
Similarly is ∆ADC,
S and R are the mid-points of AD and CD
∴ SR || AC and SR = (frac { 1 }{ 2 }) AC ..(ii)
Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables
from (i) and (ii)
PQ = SQ and PQ || SR
PQRS is a ||gm (∵ opposite sides are equal area parallel)
But the diagonals of a ||gm bisect each other.
∴ PR and QS bisect each other.

Question 13.
Fill in the blanks to make the following statements correct :
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is …
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is …
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is …
Solution:
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is an isosceles triangle.
RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.
RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.
Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables

Question 14.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.
Solution:
Given : In ∆ABC,
Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables
To prove : Perimeter of ∆PQR = 2 x perimeter of ∆ABC
Proof : ∵ PQ || BC and QR || AB
∴ ABCQ is a ||gm
∴ BC = AQ
Similarly, BCAP is a ||gm
∴ BC = AP …(i)
∴ AQ = AP = BL
⇒ PQ = 2BC
Similarly, we can prove that
QR = 2AB and PR = 2AC
Now perimeter of ∆PQR.
= PQ + QR + PR = 2AB + 2BC + 2AC
= 2(AB + BC + AC)
= 2 perimeter of ∆ABC.
Hence proved

Question 15.
In the figure, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90°.
Solution:
Given: In ∆ABC, BE ⊥ AC
AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q and R are respectively mid points of AH, AB and BC. PQ and QR are joined B.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables
To prove : ∠PQR = 90°
Proof: In ∆ABC,
Q and R the mid points of AB and BC 1
∴ QR || AC and QR = (frac { 1 }{ 2 }) AC
Similarly, in ∆ABH,
Q and P are the mid points of AB and AH
∴ QP || BH or QP || BE
But AC ⊥ BE
∴ QP ⊥ QR
∴ ∠PQR = 90°

Question 16.
ABC is a triangle. D is a point on AB such that AD = (frac { 1 }{ 4 }) AB and E is a point on AC such that AE = (frac { 1 }{ 4 }) AC. Prove that DE = (frac { 1 }{ 4 }) BC.
Solution:
Given : In ∆ABC,
D is a point on AB such that
AD = (frac { 1 }{ 4 }) AB and E is a point on AC such 1
that AE = (frac { 1 }{ 4 }) AC
DE is joined.
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
To prove : DE = (frac { 1 }{ 4 }) BC
Construction : Take P and Q the mid points of AB and AC and join them
Proof: In ∆ABC,
∵ P and Q are the mid-points of AB and AC
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables

Question 17.
In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (frac { 1 }{ 4 }) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
Linear Equations in Two Variables Class 9 RD Sharma Solutions
Solution:
Given : In ||gm ABCD,
P is the mid-point of DC and Q is a point on AC such that CQ = (frac { 1 }{ 4 }) AC. PQ is produced meets BC at R.
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables
To prove : R is mid point of BC
Construction : Join BD
Proof : ∵ In ||gm ABCD,
∵ Diagonal AC and BD bisect each other at O
∴ AO = OC = (frac { 1 }{ 2 }) AC …(i)
In ∆OCD,
P and Q the mid-points of CD and CO
∴ PQ || OD and PQ = (frac { 1 }{ 2 }) OD
In ∆BCD,
P is mid-poiht of DC and PQ || OD (Proved above)
Or PR || BD
∴ R is mid-point BC.

Question 18.
In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.
Prove that (i) DP = PC (ii) PR = (frac { 1 }{ 2 }) AC.
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
Solution:
Given : ABCD are PQRC are rectangles and Q is the mid-point of AC.
To prove : (i) DP = PC (ii) PR = (frac { 1 }{ 2 }) AC
Construction : Join diagonal AC which passes through Q and join PR.
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions
Proof : (i) In ∆ACD,
Q is mid-point of AC and QP || AD (Sides of rectangles)
∴ P is mid-point of CD
∴ DP = PC
(ii) ∵PR and QC are the diagonals of rectangle PQRC
∴ PR = QC
But Q is the mid-point of AC
∴ QC = (frac { 1 }{ 2 }) AC
Hence PR = (frac { 1 }{ 2 }) AC

Question 19.
ABCD is a parallelogram, E and F are the mid points AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P and H respectively. Prove that GP = PH.
Solution:
Given : In ||gm ABCD,
E and F are mid-points of AB and CD
GH is any line intersecting AD, EF and BC at GP and H respectively
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables
To prove : GP = PH
Proof: ∵ E and F are the mid-points of AB and CD
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables

Question 20.
BM and CN are perpendiculars to a line passing, through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A.
L is the mid point of BC.
ML and NL are joined.
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions

Linear Equations in Two Variables Class 9 RD Sharma Solutions VSAQS

Question 1.
In a parallelogram ABCD, write the sum of angles A and B.
Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles of a ||gm)
Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables

Question 2.
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
Solution:
In ||gm ABCD,
∠D = 115°
But ∠A + ∠D = 180°
(Sum of consecutive angles of a ||gm)
RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables
⇒ ∠A + 115°= 180° ∠A = 180°- 115°
∴ ∠A = 65°

Question 3.
PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.
Solution:
In a square PQRS,
Diagonals PR and QS intersects each other at O.
RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables
∵ The diagonals of a square bisect each other at right angles.
∴ ∠POQ = 90°

Question 4.
If PQRS is a square then write the measure of ∠SRP.
Solution:
In square PQRS,
Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables
Join PR,
∵Diagonals of a square bisect are opposite angles
∴∠SRP = (frac { 1 }{ 2 })x ∠SRQ
= (frac { 1 }{ 2 }) x 90° = 45°

Question 5.
If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.
Solution:
In rhombus ABCD,
Diagonals bisect each other at 0 at right angles.
∠ABC = 56°
But ∠ABC + ∠BCD = 180° (Sum of consecutive angles)
⇒ 56° + ∠BCD = 180°
⇒ ∠BCD = 180° – 56° = 124°
∵ Diagonals of a rhombus bisect the opposite angle
∴ ∠ACD = (frac { 1 }{ 2 }) ∠BCD = (frac { 1 }{ 2 }) x 124°
= 62°
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables

Question 6.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side.
Solution:
Perimeter of a ||gm ABCD = 22cm
∴ Sum of two consecutive sides = (frac { 22 }{ 2 })
= 11cm
i.e. AB + BC = 11 cm
AB = 6.5 cm and let BC = x cm
∴ 6.5 + x = 11 cm
x = 11 – 6.5 = 4.5
∴ Shorter side = 4.5 cm
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables

Question 7.
If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Then find the measure of the smallest angle.
Solution:
Ratio in the angles of a quadrilateral = 3 : 5 : 9 : 13
Let first angle = 3x
Second angle = 5x
Third angle = 9x
and fourth angle = 13x
∵ The sum of angles of a quadrilateral = 360°
∴ 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360° ⇒ x = (frac { { 360 }^{ circ } }{ 30 })  = 12
∴ Smallest angle = 3x = 3 x 12° = 36°

Question 8.
In parallelogram ABCD if ∠A = (3x – 20°), ∠B = (y + 15)°, ∠C = (x + 40°), then find the value of x and y.
Solution:
In a ||gm ABCD,
∠A = (3x – 20°), ∠B = y + 15°,
∠C = x + 40°
Now, ∠A = ∠C (Opposite angles of a ||gm)
⇒ 3x – 20 = x + 40°
⇒ 3x – x = 40° + 20° ⇒ 2x = 60°
⇒ x = (frac { { 60 }^{ circ } }{ 2 })  = 30°
and ∠A + ∠B = 180° (Sum of the consecutive angles)
⇒ 3x-20° + y + 15° = 180°
⇒ 3x + y – 5° = 180°
⇒ 3 x 30° +y- 5° = 180°
⇒ 90° – 5° + y = 180
y = 180° – 90° + 5 = 95°
∴ x = 30°, y = 95°

Question 9.
If measures opposite angles of a parallelogram are (60 – x)° and (3x – 4)°, then find the measures of angles of the parallelogram.
Solution:
Opposite angles of a ||gm ABCD are (60 – x)° and (3x – 4°)
But opposite angles of a ||gm are equal, the
60° – x° = 3x – 4° ⇒ 60° + 4° = 3x + x
⇒ 4x = 64° ⇒ x = (frac { { 64 }^{ circ } }{{ 4 }^{ circ } })  = 16°
∴ ∠A = 60° – x = 60° – 16° = 44°
But ∠A + ∠B = 180° (sum of consecutive angle)
⇒ 44° + ∠B = 180°
⇒ ∠B = 180° – 44°
⇒ ∠B = 136°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
∴ Angles are 44°, 136°, 44°, 136°

Question 10.
In a parallelogram ABCD, the bisectors of ∠A also bisect BC at x, find AB : AD.
Solution:
In ||gm ABCD,
Bisectors of ∠A meets BC at X and BX = XC
Draw XY ||gm AB meeting AD at Y
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables

Question 11.
In the figure, PQRS in an isosceles trapezium find x and y.
Solution:
∵ PQRS is an isosceles trapezium in which
SP = RQ and SR || PQ
∴ ∠P + ∠S = 180° (Sum of co-interior angles)
3x + 2x = 180° ⇒ 5x = 180°
⇒ x = (frac { { 180 }^{ circ } }{ 5 })  = 36°
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables
But ∠P = ∠Qm (Base angles of isosceles trapezium)
y = 2x = 2 x 36° = 12°
∴ y = 12°
Hence x = 36°, y = 12°

Question 12.
In the figure ABCD is a trapezium. Find the values of x and y.
Linear Equations in Two Variables Class 9 RD Sharma Solutions
Solution:
In trapezium ABCD,
AB || CD
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables
∴ ∠A + ∠D = 180° (Sum of cointerior angles)
x + 20° + 2x + 10° = 180°
3x + 30° = 180°
⇒ 3x= 180° – 30°
3x = 150°
x = (frac { { 150 }^{ circ } }{ 3 })  = 50°
Similarly, ∠B + ∠C = 180°
⇒ y + 92° = 180°
⇒ y = 180° – 92° = 88°
∴ x = 50°, y = 88°

Question 13.
In the figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
Solution:
In the figure, ABCD and AEFG are two parallelograms ∠C = 58°
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
∵ DC || GF and CB || FE (Sides of ||gms)
∴ ∠C = ∠F
But ∠C = 58°
∴ ∠F = 58°

Question 14.
Complete each of the following statements by means of one of those given in brackets against each:
(i) If one pair of opposite sides are equal and parallel, then the figure is ……… (parallelogram, rectangle, trapezium)
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is …….. (square, rectangle, trapezium)
(iii) A line drawn from the mid-point of one side of a triangle ………. another side intersects the third side at its mid-point, (perpendicular to, parallel to, to meet)
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a …….. (rectangle, square, rhombus)
(v) Consecutive angle of a parallelogram are ……… (supplementary, complementary)
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ……… (rectangle, parallelogram, rhombus)
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ………. (parallelogram, rhombus, rectangle)
(viii)If consecutive sides of a parallelogram are equal, then it is necessarily a …….. (kite, rhombus, square)
Solution:
(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.
(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point,
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.
(v) Consecutive angle of a parallelogram are supplementary.
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.
(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

Question 15.
In a quadrilateral ABCD, bisectors of A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O and ∠AOB = 75°
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions
In AOB, ∠AOB = 75°
∴ ∠OAB + ∠OBA = 180° – 75° = 105°
But OA and OB are the bisectors of ∠A and ∠B.
∴ ∠A + ∠B = 2 x 105° = 210°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quad.)
∴ 210° + ∠C + ∠D = 360°
⇒ ∠C + ∠D = 360° – 210° = 150°
Hence ∠C + ∠D = 150°

Question 16.
The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44° find ∠OAD.
Solution:
In rectangle ABCD,
Diagonals AC and BD intersect each other at O and ∠BOC = 44°
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables
But ∠AOD = ∠BOC (Vertically opposite angles)
∴ ∠AOD = 44°
In ∆AOD,
∠AOD + ∠OAD + ∠ODA = 180° (Sum of angles of a triangle)
⇒ 44° + ∠OAD + ∠OAD = 180° [∵ OA = OD, ∠OAD = ∠ODA]
⇒ 2∠OAD = 180° – 44° = 136°
∴ ∠OAD = (frac { { 136 }^{ circ } }{ 2 })  = 68°

Question 17.
If ABCD is a rectangle with ∠BAC = 32°, find the measure if ∠DBC.
Solution:
In rectangle ABCD,
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables
Diagonals bisect each other at O
∠BAC = 32°
∵ OA = OB
∴ ∠OBA Or ∠DBA = ∠BAC = 32°
But ∠ABC = 90° (Angle of a rectangles)
∴ ∠DBC = ∠ABC – ∠DBA
= 90° – 32° = 58°

Question 18.
If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that ∠C + ∠D = k(∠AOB), then find the value of k.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O
Such that ∠C + ∠D = k (∠AOB)
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions
RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables

Question 19.
In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.
RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables
Solution:
In rhombus PQRS,
Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that ∠SRT = 152°
Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables
But ∠SRT + ∠SRP = 180° (Linear pair)
⇒ 152° +∠SRP = 180°
⇒ ∠SRP =180°- 152° = 28°
But ∠SPR = ∠SRP (∵ PR bisects ∠P and ∠R)
⇒ z = 28°
y = 90° (∵ Diagonals bisect each other at right angles)
∠RPQ = z = 28°
∴ In ∆POQ,
z + x = 90° ⇒ 28° + x = 90°
⇒ x = 90° – 28° = 62°
∴ x = 62°, y = 90°, z = 28°

Question 20.
In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.
RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables
Solution:
In rectangle ABCD,
Diagonals AC and BD bisect each other at O
AC is produced to E and ∠DCE = 146°
RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables
∠DCE + ∠DCA = 180° (Linear pair)
⇒ 146°+ ∠DCA= 180°
⇒ ∠DCA = 180°- 146°
⇒ ∠DCA = 34°
∴ ∠CAB = ∠DCA (Alternate angles)
= 34°
Now in ∆AOB,
∠AOB = 180° – (∠DAB + ∠OBA)
= 180° – (34° + 34°)
= 1803 – 68° = 112°

RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables MCQS

Mark the correct alternative in each of the following:
Question 1.
The opposite sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The opposite sides of a quadrilateral have no common point. (a)

Question 2.
The consecutive sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The consecutive sides of a quadrilateral have one common point. (b)

Question 3.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?
(a) ∠P = 100°, ∠Q = 80°, ∠R = 100°
(b) ∠P = 85°, ∠Q = 85°, ∠R = 95°
(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm
(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm
Solution:
PQRS is a quadrilateral, PR and QS intersect each other at O. PQRS is a parallelogram if ∠P = 100°, ∠Q = 80°, ∠R = 100° (a)

Question 4.
Which of the following quadrilateral is not a rhombus?
(a) All four sides are equal
(b) Diagonals bisect each other
(c) Diagonals bisect opposite angles
(d) One angle between the diagonals is 60°
Solution:
A quadrilateral is not a rhombus if one angle between the diagonals is 60°. (d)

Question 5.
Diagonals necessarily bisect opposite angles in a
(a) rectangle
(b) parallelogram
(c) isosceles trapezium
(d) square
Solution:
Diagonals necessarily bisect opposite angles in a square. (d)

Question 6.
The two diagonals are equal in a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) trapezium
Solution:
The two diagonals are equal in a rectangle. (c)

Question 7.
We get a rhombus by joining the mid-points of the sides of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) triangle
Solution:
We get a rhombus by joining the mid points of the sides of a rectangle. (c)

Question 8.
The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
The bisectors of any two adjacent angles of a parallelogram intersect at 90°. (d)

Question 9.
The bisectors of the angle of a parallelogram enclose a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Solution:
The bisectors of the angles of a parallelogram enclose a rectangle. (c)

Question 10.
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) parallelogram
(b) rectangle
(c) square
(d) rhombus
Solution:
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. (a)

Question 11.
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) square
(b) rhombus
(c) trapezium
(d) none of these
Solution:
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus. (b)

Question 12.
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) square
(b) rectangle
(c) trapezium
(d) none of these
Solution:
The figure formed by the joining the mid-points of the adjacent sides of a rhombus is a rectangle. (b)

Question 13.
The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram
Solution:
Tire figure formed by joining the mid-points of the adjacent sides of a square is a square. (b)

Question 14.
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rectangle
(b) parallelogram
(b) rhombus
(d) square
Solution:
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram. (b)

Question 15.
If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
(a) 176°
(b) 68°
(c) 112°
(d) 102°
Solution:
Let the smallest angle be x
The largest angle = 2x – 24°
But sum of two adjacent angles = 180°
RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables

Question 16.
In a parallelogram ABCD, If ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =
(a) 75°
(b) 60°
(c) 45°
(d) 55°
Solution:
In ||gm ABC,
Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables
∠A = 75°, ∠DBC = 60°
But ∠A + ∠B = 180° (Sum of two consecutive angles)
⇒ 75° + ∠B = 180°
⇒ ∠B = 180°- 75“= 105°
But ∠DBC = 60°
∴ ∠DBA = 105°-60° = 45°
But ∠BDC = ∠DBA (Alternate angles)
∴ ∠BDC = 45° (c)

Question 17.
ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =
(a) AE
(b) BE
(c) CE
(d) DE
Solution:
In ||gm ABCD, BD is joined forming two triangles ABD and BCD
Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions
E and F are the centroid of ∆ABD and ∆BCD
Now E and F trisect AC
i.e. AE = EF = FC
∴ EF = AE (a)

Question 18.
ABCD is a parallelogram, M is the mid¬point of BD and BM bisects ∠B. Then, ∠AMB =
(a) 45°
(b) 60°
(c) 90°
(d) 75°
Solution:
In ||gm ABCD, M is mid-point of BD and
BM bisects ∠B
AM is joined
RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables
∴AM bisects ∠A
But ∠A + ∠B = 180° (Sum of two consecutive angles)
∴ ∠AMB = 90° (c)

Question 19.
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 12°
(d) 81°
Solution:
Let adjacent angle of a ||gm = x
Then second angle = (frac { 2 }{ 3 }) x
∴ x+ (frac { 2 }{ 3 }) x= 180°
(Sum of two adjacent angles of a ||gm is 180°)
RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables

Question 20.
If the degree measures of the angles of quadrilateral are Ax, lx, 9x and 10JC, what is the sum of the measures of the smallest angle and largest angle?
(a) 140°
(b) 150°
(c) 168°
(d) 180°
Solution:
Sum of the angles of a quadrilateral = 360°
∴ 4x + 1x + 9x + 10x = 360°
⇒ 30x = 360°
⇒ x = (frac { { 360 }^{ circ } }{ 30 })  = 12°
Now sum of smallest and largest angle = 4 x 12° + 10 x 12°
= 48° + 120° = 168° (c)

Question 21.
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
(a) 16 cm
(b) 15 cm
(c) 20 cm
(d) 17 cm
Solution:
Diagonals of a rhombus are 18 cm and 24 cm But diagonals of a rhombus bisect each other at right angles
Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions

Question 22.
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
(a) 70°
(b) 110°
(c) 90°
(d) 120°
Solution:
In ||gm ABCD, AC is its diagonal which bisect ∠BAD
RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables
∠BAD = 35°
∴ ∠BAD = 2 x 35° = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°⇒ ∠B = 180° – 70°
∴ ∠B = 110°
⇒ ABC = 110° (b)

Question 23.
In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =
(a) 70°
(b) 45°
(c) 50°
(d) 60°
Solution:
In rhombus ABCD, ∠ACB = 40°
RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables
∴ ∠BCD = 2 x ∠ACB
= 2 x 40° = 80°
But ∠BCD + ∠ADC = 180° (Sum of consecutive angles of ||gm)
⇒ 80° + ∠ADC = 180°
⇒ ∠ADC = 180° – 80° = 100°
∴ ∠ADB = (frac { 1 }{ 2 })∠ADC = (frac { 1 }{ 2 })x 100° = 50° (c)

Question 24.
In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
(a) 70°, 70°, 40°
(b) 60°, 40°, 80°
(c) 30°, 40°, 110°
(d) 60°, 70°, 50°
Solution:
In ∆ABC,
Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables
∠A = 30°, ∠B = 40°, ∠C = 110°
D, E and F are mid-points of the sides of the triangle. By joining them in order,
DEF is a triangle formed
Now BDEF, CDFE and AFDE are ||gms
∴ ∠A = ∠D = 30°
∠B = ∠E = 40°
∠C = ∠F= 110°
∴ Angles are 30°, 40°, 110° (c)

Question 25.
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
(a) 40°
(b) 50°
(c) 10°
(d) 90°
Solution:
In ||gm ABCD, diagonals AC and BD intersect each other at O
RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables
BOC = 90°, ∠BDC = 50°
∵ ∠BOC = 90°
∴ Diagonals of ||gm bisect each other at 90°
∴∠COD = 90°
In ∆COD,
∠OCD = 90° – 50° = 40°
But ∠OAB = ∠OCD (Alternate angles)
∴∠OAB = 40° (a)

Question 26.
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm, MN = 14 cm, then CD =
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
In trapezium AB || DC
M and N are mid-points of sides AD and BC and MN are joined
AB = 12 cm, MN = 14 cm
RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables
∵ MN = (frac { 1 }{ 2 })(AB + CD)
⇒ 2MN = AB + CD
⇒ 2 x 14 = 12 + CD
CD = 2 x 14 – 12 = 28 – 12 = 16 cm (d)

Question 27.
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 45°, then ∠B =
(a) 115°
(b) 120°
(c) 125°
(d) 135°
Solution:
Diagonals AC and BD of quadrilateral ABCD bisect each other at O
Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables
∴ AO = OC, BO = OD
∴ ABCD is a ||gm ∠A = 45°
But ∠A + ∠B = 180° (Sum of consecutive angles)
∴ ∠B = 180° – ∠A = 180° – 45°
= 135° (d)

Question 28.
P is the mid-point of side BC of a paralleogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
(a) 5 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
Solution:
In ||gm ABCD, P is mid-point of BC
AD = 10cm
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables
∠BAP = ∠DAP
Through P, draw PQ || AB
∴ ABPQ is rhombus
∴ AB = BP = AQ
= (frac { 1 }{ 2 }) AB = (frac { 1 }{ 2 }) x 10 = 5 cm
But CD = AB (Opposite sides of ||gm)
∴ CD = 5 cm (a)

Question 29.
In ∆ABC, E is the mid-point of median AD such that BE produced meets AC at E If AC = 10.5 cm, then AF =
(a) 3 cm
(b) 3.5 cm
(c) 2.5 cm
(d) 5 cm
Solution:
In ∆ABC, E is the mid-point of median AD
Such that BE produced meets AC at F
AC = 10.5 cm
Draw DG || AF
RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables MCQS
In ∆ADG
E is mid-point of AD and EF || DG
∴ F is mid-point of AG
⇒ AF = FG …(i)
In ∆BCF
D is mid-point of BC and DG || BF
∴ G is mid-point of FC
∴ FG = GC …(i)
From (i) and (ii)
AF = FG = GC = (frac { 1 }{ 3 }) AC
But AC = 10.5 cm
∴ AF = (frac { 1 }{ 3 }) AC = (frac { 1 }{ 3 }) x 10.5 = 3.5 cm (b)

Question 30.
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables
Solution:
In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
∵ E is mid point of BC
∴ BE = EC
In ∆BEF and ∆CDE
BE = EC
∠BEF = ∠CED (Vertically opposite angle)
and ∠EBF = ∠ECD (Alternate angles)
∴ ∆BEF ≅ ∆CDE (ASA criterian)
∴ DC = BF
But DC = AB
∴ AB = BF
AF = AB + BF = AB + AB
= 2AB (b)

Question 31.
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =
(a) 60°
(b) 80°
(c) 120°
(d) None of these
Solution:
In quadrilateral ABCD
⇒ ∠A + ∠C = 2(∠B + ∠D)
⇒ ∠A + ∠C = 2∠B + 2∠D
Adding 2∠A + 2∠C both sides
2∠A + 2∠C + ∠A + ∠C = 2∠A + 2∠B + 2∠C + 2∠D
⇒ 3∠A + 3∠C = 2(∠A + ∠B + ∠C + ∠D)
⇒ 3(∠A + ∠C) = 2 x 360° = 720°
∴ ∠A + ∠C = (frac { { 720 }^{ circ } }{ 3 })  = 240°
⇒ 40° + ∠C = 240° (∵ ∠A = 40°)
∠C = 240° – 40° = 200°
Now 2(∠B + ∠D) = ∠A + ∠C = 240°
∠B + ∠D = (frac { { 240 }^{ circ } }{ 2 })  = 120°
∴ ∠B = 60° = 120°
∴ ∠B = 60° (a)

Question 32.
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =
(a) 70°
(b) 90°
(c) 80°
(d) 100°
Solution:
In rectangle ABCD, diagonals AC and BD intersect each other at P
RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables
∠ABD = 50°
∴ ∠CAB = ∠ABD = 50° (∵ AP = BP)
Now in ∆APB
∠CAB + ∠ABD + ∠APB = 180° (Angles of a triangle)
⇒ ∠PAB + ∠PBA + ∠APB = 180°
⇒ 50° + 50° + ∠APB = 180°
⇒ ∠APB = 180° – 50° – 50° = 80°
But ∠DPC = ADB (Vertically opposite angles)
∴ ∠DPC = 80° (c)

RD Sharma Solutions for Class 9 Chapter 13 Linear Equations in Two Variables Ex 13.1 Q1

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All Chapter RD Sharma Solutions For Class 9 Maths

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