## RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

Other Exercises

- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

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)

Question 2.

In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.

Solution:

In a ||gm ABCD, ∠B = 135°

∴ ∠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°

Question 4.

ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.

Solution:

In rectangle ABCD,

∠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

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

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

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

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,

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.

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3 are helpful to complete your math homework.

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