## 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 110Â°, 50Â° and 40Â°. Find its fourth angle.

Solution:

Sum of four angles of a quadrilateral = 360Â°

Three angles are = 110Â°, 50Â° and 40Â°

âˆ´ Fourth angle = 360Â° – (110Â° + 50Â° + 40Â°)

= 360Â° – 200Â° = 160Â°

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 = 360Â°

Ratio in angles = 1 : 2 : 4 : 5

Let first angle = x

Second angle = 2x

Third angle = 4x

and fourth angle = 5x

âˆ´ x + 2x + 4x + 5x = 360Â°

â‡’ 12x = 360Â° â‡’ x = \(\frac { { 360 }^{ \circ } }{ 12 }\)Â = 30Â°

âˆ´ First angle = 30Â°

Second angle = 30Â° x 2 = 60Â°

Third angle = 30Â° x 4 = 120Â°

Fourth angle = 30Â° x 5 = 150Â°

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 = 360Â°

Ratio in the angles = 3 : 5 : 9 : 13

Let first angle = 3x

Then second angle = 5x

Third angle = 9x

and fourth angle = 13x

âˆ´ 3x + 5x + 9x+ 13x = 360Â°

â‡’ 30x = 360Â°

â‡’ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)Â = 12Â°

âˆ´ First angle = 3x = 3 x 12Â° = 36Â°

Second angle = 5x = 5 x 12Â° = 60Â°

Third angle = 9x = 9 x 12Â° = 108Â°

Fourth angle = 13 x 12Â° = 156Â°

Question 4.

In a quadrilateral ABCD, CO and DO are the bisectors of âˆ C and âˆ D respectively.

Prove that âˆ COD = \(\frac { 1 }{ 2 }\) (âˆ A + âˆ B).

Solution:

### 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)Â° and (50 – x)Â°. Find the measure of each angle of the parallelogram.

Solution:

âˆµ Opposite angles of a parallelogram are equal

âˆ´ 3x – 2 = 50 – x

â‡’ 3x + x – 50 + 2

â‡’ 4x = 52

â‡’ x = \(\frac { 52 }{ 4 }\) = 13

âˆ´ âˆ A = 3x – 2 = 3 x 13 – 2 = 39Â° – 2 = 37Â°

âˆ C = 50Â° -x = 50Â° – 13 = 37Â°

Butâˆ A + B = 180Â°

âˆ´ 37Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â° – 37Â° = 143Â°

and âˆ D = âˆ B (Opposite angles of a ||gm)

âˆ´ âˆ D = 143Â°

Hence angles and 37Â°, 143Â°, 37Â°, 143Â°

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,

âˆ A =x

Then âˆ B = \(\frac { 2 }{ 3 }\) x

But, âˆ A + âˆ B = 180Â° (Sum of two adjacent angles of a ||gm)

â‡’ x + \(\frac { 2 }{ 3 }\)x = 180Â°

â‡’ \(\frac { 5 }{ 3 }\)x = 180Â°

â‡’ x = 180Â° x \(\frac { 3 }{ 5 }\) = 108Â°

âˆ´ âˆ A = 108Â°

and âˆ B = 108Â° x \(\frac { 2 }{ 3 }\) = 72Â°

But, âˆ A = âˆ C and âˆ B = âˆ D (Opposite angles of a ||gm)

âˆ´ âˆ C = 108Â°, âˆ D = 72Â°

Hence angles are 108Â°, 72Â°, 108Â°, 72Â°

Question 3.

Find the measure of all the angles of a parallelogram, if one angle is 24Â° less than twice the smallest angle.

Solution:

Let smallest angle of a ||gm = x

Then second angle = 2x – 24Â°

But these are consecutive angles

âˆ´ x + (2x- 24Â°) = 180Â°

â‡’ x + 2x – 24Â° = 180Â°

â‡’ 3x = 180Â° + 24Â° = 204Â°

â‡’ x =\(\frac { { 204 }^{ \circ } }{ 3 }\)Â = 68Â°

âˆ´ Smallest angle = 68Â°

and second angle = 2x 68Â° – 24Â°

= 136Â°-24Â° = 112Â°

âˆµ The opposite angles of a ||gm are equal Other two angles will be 68Â° and 112Â°

âˆ´ Hence angles are 68Â°, 112Â°, 68Â°, 112Â°

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,

Perimeter = 22cm

and longest side = 6.5 cm

Let shorter side = x

âˆ´ 2x (6.5 + x) = 22

â‡’ 13 + 2x = 22

â‡’ 2x = 22 – 13 = 9

â‡’ x = \(\frac { 9 }{ 2 }\) = 4.5

âˆ´ shorter side = 4.5cm

Question 5.

In a parallelogram ABCD, âˆ D = 135Â°, determine the measure of âˆ A and âˆ B.

Solution:

In ||gm ABCD,

âˆ D = 135Â°

But, âˆ A + âˆ D = 180Â° (Sum of consecutive angles)

â‡’âˆ A+ 135Â° = 180Â°

â‡’ âˆ A = 180Â° – 135Â° = 45Â°

âˆµ âˆ B = âˆ D (Opposite angles of a ||gm)

âˆ´ âˆ B = 135Â°

Question 6.

ABCD is a parallelogram in which âˆ A = 70Â°. Compute âˆ B, âˆ C and âˆ D.

Solution:

In ||gm ABCD,

âˆ A = 70Â°

But âˆ A + âˆ B = 180Â° (Sum of consecutive angles)

â‡’ 70Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â° – 70Â° = 110Â°

But âˆ C = âˆ A and âˆ D = âˆ B (Opposite angles of a ||gm)

âˆ C = 70Â° and âˆ D = 110Â°

Hence âˆ B = 110Â°, âˆ C = 70Â° and âˆ D = 110Â°

Question 7.

In the figure, ABCD is a parallelogram in which âˆ DAB = 75Â° and âˆ DBC = 60Â°. Compute âˆ CDB and âˆ ADB.

Solution:

In ||gm ABCD,

âˆ A + âˆ B = 180Â°

(Sum of consecutive angles) But, âˆ A = 75Â°

âˆ´ âˆ B = 180Â° – âˆ A = 180Â° – 75Â° = 105Â°

âˆ´ DBA = 105Â° -60Â° = 45Â°

But âˆ CDB = âˆ DBA (alternate angles)

= 45Â°

and âˆ ADB = âˆ DBC = 60Â°

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 âˆ A = 60Â°. If the bisectors of âˆ A and âˆ B meet at P, prove that AD = DP, PC= BC and DC = 2AD.

Solution:

Given : In ||gm ABCD,

âˆ A = 60Â°

Bisector of âˆ A and âˆ B meet at P.

To prove :

(i) AD = DP

(ii) PC = BC

(iii) DC = 2AD

Construction : Join PD and PC

Proof : In ||gm ABCD,

âˆ A = 60Â°

But âˆ A + âˆ B = 180Â° (Sum of excutive angles)

â‡’ 60Â° + âˆ B = 180Â°

âˆ´ âˆ B = 1809 – 60Â° = 120Â°

âˆµ DC || AB

âˆ PAB = âˆ DPA (alternate angles)

â‡’ âˆ PAD = âˆ DPA (âˆµ âˆ PAB = âˆ PAD)

âˆ´ AB = DP

(PA is its angle bisector, sides opposite to equal angles)

(ii) Similarly, we can prove that âˆ PBC = âˆ PCB (âˆµ âˆ PAB = âˆ BCA alternate angles)

âˆ´ PC = BC

(iii) DC = DP + PC

= AD + BC [From (i) and (ii)]

= AD + AB = 2AB (âˆµBC = 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

To prove : AF = 2AB

Proof : In âˆ†CDE and âˆ†EBF

âˆ DEC = âˆ BEF (vertically opposite angles)

CE = EB (E is mid point of BC)

âˆ DCE = âˆ EBF (alternate angles)

âˆ´ âˆ†CDE â‰… âˆ†EBF (SAS Axiom)

âˆ´ DC = BF (c.p.c.t.)

But AB = DC (opposite sides of a ||gm)

âˆ´ 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,

âˆ C + âˆ D = 180Â°

(Sum of consecutive angles)

Question 2.

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

Solution:

In a ||gm ABCD, âˆ B = 135Â°

âˆ´ âˆ D = âˆ B = 135Â° (Opposite angles of a ||gm)

But âˆ A + âˆ B = 180Â° (Sum of consecutive angles)

â‡’ âˆ B + 135Â° = 180Â°

âˆ´ âˆ A = 180Â° – 135Â° = 45Â°

Butâˆ C = âˆ B = 45Â° (Opposite angles of a ||gm)

âˆ´ Angles are 45Â°, 135Â°, 45Â°, 135Â°.

Question 3.

ABCD is a square, AC and BD intersect at O. State the measure of âˆ AOB.

Solution:

In a square ABCD,

Diagonal AC and BD intersect each other at O

âˆµ Diagonals of a square bisect each other at right angle

âˆµâˆ AOB = 90Â°

Question 4.

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

Solution:

In rectangle ABCD,

âˆ B = 90Â°, BD is its diagonal

But âˆ ABD = 40Â°

and âˆ ABD + âˆ DBC = 90Â°

â‡’ 40Â° + âˆ DBC = 90Â°

â‡’ âˆ DBC = 90Â° – 40Â° = 50Â°

Hence âˆ DBC = 50Â°

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 : âˆµ ABCD is a ||gm

âˆ´ AB = CD and AB || CD

(Opposite sides of a ||gm are equal and parallel)

âˆ´ EB || DF and EB = DF (âˆµ E and F are mid points of AB and CD)

âˆ´ 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: âˆµ Diagonals of a parallelogram bisect each other

âˆ´ AO = OC and BO = OD

âˆ´ P and Q are point of trisection of BD

âˆ´ BP = PQ = QD …(i)

âˆµ BO = OD and BP = QD …(ii)

Subtracting, (ii) from (i) we get

OB – BP = OD – QD

â‡’ OP = OQ

In quadrilateral APCQ,

OA = OC and OP = OQ (proved)

Diagonals AC and PQ bisect each other at O

âˆ´ 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 âˆ†AEH and âˆ†BFE

AE = BF (given)

âˆ A = âˆ B (each 90Â°)

AH = BE (proved)

âˆ´ âˆ†AEH â‰… âˆ†BFE (SAS criterion)

âˆ´ âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4 (c.p.c.t.)

But âˆ 1 + âˆ 3 = 90Â° and âˆ 2 + âˆ 4 = 90Â° (âˆ A = âˆ B = 90Â°)

â‡’ âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 90Â° + 90Â° = 180Â°

â‡’ âˆ 1 + âˆ 4 + âˆ 1 + âˆ 4 = 180Â°

â‡’ 2(âˆ 1 + âˆ 4) = 180Â°

â‡’ âˆ 1 + âˆ 4 = \(\frac { { 180 }^{ \circ } }{ 2 }\)Â = 90Â°

âˆ´ âˆ HEF = 180Â° – 90Â° = 90Â°

Similarly, we can prove that

âˆ F = âˆ G = âˆ H = 90Â°

Since sides of quad. EFGH is are equal and each angle is of 90Â°

âˆ´ 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: âˆ EGF = 90Â°

Proof : âˆµ Diagonals of a rhombus bisect

each other at right angles

AO = OC, BO = OD

âˆ AOD = âˆ COD = 90Â°

and âˆ AOB = âˆ BOC = 90Â°

In âˆ†BDE,

A and O are the mid-points of BE and BD respectively.

âˆ´ AO || ED

Similarly, OC || DG

In âˆ† CFA, B and O are the mid-points of AF and AC respectively

âˆ´ OB || CF and OD || GC

Now in quad. DOCG

OC || DG and OD || CG

âˆ´ DOCG is a parallelogram.

âˆ´ âˆ DGC = âˆ DOC (opposite angles of ||gm)

âˆ´ âˆ DGC = 90Â° (âˆµ âˆ DOC = 90Â°)

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 âˆ†ACE,

O and D are the mid points of sides AC and AE

âˆ´ DO || EC and DB || FC

â‡’ BD || EF

âˆ´ AB = BF

But AB = DC (Opposite sides of ||gm)

âˆ´ DC = BF

Now in âˆ†EDC and âˆ†CBF,

DC = BF (proved)

âˆ EDC = âˆ CBF

(âˆµâˆ EDC = âˆ DAB corresponding angles)

âˆ DAB = âˆ CBF (corresponding angles)

âˆ ECD = âˆ CFB (corresponding angles)

âˆ´ AEDC â‰… ACBF (ASA criterion)

âˆ´ DE = BC (c.p.c.t.)

â‡’ DC = BC

â‡’ AB = BC

â‡’ BF = BC (âˆµAB = 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 âˆ†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 âˆ†DEF.

Solution:

In âˆ†ABC, D, E and F are the mid-points of sides,

BC, CA, AB respectively

AB = 7cm, BC = 8cm and CA = 9cm

âˆµ D and E are the mid points of BC and CA

âˆ´ DE || AB and DE =\(\frac { 1 }{ 2 }\) AB =\(\frac { 1 }{ 2 }\) x 7 = 3.5cm

Similarly,

Question 2.

In a triangle âˆ ABC, âˆ A = 50Â°, âˆ B = 60Â° and âˆ C = 70Â°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Solution:

In âˆ†ABC,

âˆ A = 50Â°, âˆ B = 60Â° and âˆ C = 70Â°

D, E and F are the mid points of sides BC, CA and AB respectively

DE, EF and ED are joined

âˆµ D, E and F are the mid points of sides BC, CA and AB respectively

âˆ´ EF || BC

DE || AB and FD || AC

âˆ´ BDEF and CDEF are parallelogram

âˆ´ âˆ B = âˆ E = 60Â° and âˆ C = âˆ F = 70Â°

Then âˆ A = âˆ D = 50Â°

Hence âˆ D = 50Â°, âˆ E = 60Â° and âˆ F = 70Â°

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 = 30Â°

âˆµ P, Q, R and the mid points of sides BC, CA and AB respectively.

âˆ´ PQ || AB and PQ = \(\frac { 1 }{ 2 }\) AB

Question 4.

In a âˆ†ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.

Solution:

Given : In âˆ†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 âˆ†ABD and âˆ†CDX

AD = DX (Given)

BD = DC (D is mid points)

âˆ ADB = âˆ CDX (Vertically opposite angles)

âˆ´ âˆ†ABD â‰… âˆ†CDX (SAS criterian)

âˆ´ AB = CX (c.p.c.t.)

and âˆ ABD = âˆ DCX

But these are alternate angles

âˆ´ AB || CX and AB = CX

âˆ´ ABXC is a parallelogram.

Question 5.

In a âˆ†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 âˆ†ABC, E and F are the mid-points of AC and AB respectively.

EF are joined.

AP âŠ¥ BC is drawn which intersects EF at Q and meets BC at P.

To prove: AQ = QP

proof : In âˆ†ABC

E and F are the mid points of AC and AB

âˆ´ EF || BC and EF = \(\frac { 1 }{ 2 }\)BC

âˆ´ âˆ F = âˆ B

In âˆ†ABP,

F is mid point of AB and Q is the mid point of FE or FQ || BC

âˆ´ Q is mid point of AP,

âˆ´ AQ = QP

Question 6.

In a âˆ†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 âˆ†ABC,

BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.

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 âˆ†ADC

Solution:

In âˆ†ABC, âˆ B = 90Â°

AC =15 cm, AB = 9cm

D and E are the mid points of sides AB and AC respectively and D, E are joined.

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.

Solution:

In âˆ†ABC,

M, N and P are the mid points of side, AB, AC and BC respectively.

Question 9.

In the figure, AB = AC and CP || BA and AP is the bisector of exterior âˆ CAD of âˆ†ABC. Prove that (i) âˆ PAC = âˆ BCA (ii) ABCP is a parallelogram.

Solution:

Given : In ABC, AB = AC

nd CP || BA, AP is the bisector of exterior âˆ CAD of âˆ†ABC

To prove :

(i) âˆ PAC = âˆ BCA

(ii) ABCP is a ||gm

Proof : (i) In âˆ†ABC,

âˆµ AB =AC

âˆ´ âˆ C = âˆ B (Angles opposite to equal sides) and ext.

âˆ CAD = âˆ B + âˆ C

= âˆ C + âˆ C = 2âˆ C ….(i)

âˆµ AP is the bisector of âˆ CAD

âˆ´ 2âˆ PAC = âˆ CAD …(ii)

From (i) and (ii)

âˆ C = 2âˆ PAC

âˆ C = âˆ CAD or âˆ BCA = âˆ PAC

Hence âˆ PAC = âˆ BCA

(ii) But there are alternate angles,

âˆ´ AD || BC

But BA || CP

âˆ´ 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.

Proof: In âˆ†ABD,

P and S are mid points of AB and AD

âˆ´ PS || BD and PS = \(\frac { 1 }{ 2 }\) BD …(i)

Similarly in âˆ†BCD,

Q and R the mid points of BC and CD

âˆ´ QR || BD and

QR = \(\frac { 1 }{ 2 }\) BD …(ii)

âˆ´ Similarly, we can prove that PQ || SR and PQ = SR …(iii)

From (i) and (ii) and (iii)

PQRS is a parallelogram,

âˆµ AC and BD intersect each other at right angles.

âˆ´ 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 âˆ†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.

Proof : âˆµ D, E and F are the mid-points of

the sides BC, CA and AB respectively

âˆ´ AFDE is a ||gm

âˆ´ AF = DE and AE = DF

But AF = AE

(âˆµ E and F are mid-points of equal sides AB and AC)

âˆ´ AF = DF = DE = AE

âˆ´AFDE is a rhombus

âˆµ The diagonals of a rhombus bisect each other at right angle.

âˆ´ 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 âˆ†ABC

P and Q are mid-points of AB and BC

âˆ´ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC …(i)

Similarly is âˆ†ADC,

S and R are the mid-points of AD and CD

âˆ´ SR || AC and SR = \(\frac { 1 }{ 2 }\) AC ..(ii)

from (i) and (ii)

PQ = SQ and PQ || SR

PQRS is a ||gm (âˆµ opposite sides are equal area parallel)

But the diagonals of a ||gm bisect each other.

âˆ´ 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.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.

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 âˆ†PQR is double the perimeter of âˆ†ABC.

Solution:

Given : In âˆ†ABC,

Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.

To prove : Perimeter of âˆ†PQR = 2 x perimeter of âˆ†ABC

Proof : âˆµ PQ || BC and QR || AB

âˆ´ ABCQ is a ||gm

âˆ´ BC = AQ

Similarly, BCAP is a ||gm

âˆ´ BC = AP …(i)

âˆ´ AQ = AP = BL

â‡’ PQ = 2BC

Similarly, we can prove that

QR = 2AB and PR = 2AC

Now perimeter of âˆ†PQR.

= PQ + QR + PR = 2AB + 2BC + 2AC

= 2(AB + BC + AC)

= 2 perimeter of âˆ†ABC.

Hence proved

Question 15.

In the figure, BE âŠ¥ 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 = 90Â°.

Solution:

Given: In âˆ†ABC, BE âŠ¥ 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.

To prove : âˆ PQR = 90Â°

Proof: In âˆ†ABC,

Q and R the mid points of AB and BC 1

âˆ´ QR || AC and QR = \(\frac { 1 }{ 2 }\) AC

Similarly, in âˆ†ABH,

Q and P are the mid points of AB and AH

âˆ´ QP || BH or QP || BE

But AC âŠ¥ BE

âˆ´ QP âŠ¥ QR

âˆ´ âˆ PQR = 90Â°

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 âˆ†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.

To prove : DE = \(\frac { 1 }{ 4 }\) BC

Construction : Take P and Q the mid points of AB and AC and join them

Proof: In âˆ†ABC,

âˆµ P and Q are the mid-points of AB and AC

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.

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.

To prove : R is mid point of BC

Construction : Join BD

Proof : âˆµ In ||gm ABCD,

âˆµ Diagonal AC and BD bisect each other at O

âˆ´ AO = OC = \(\frac { 1 }{ 2 }\) AC …(i)

In âˆ†OCD,

P and Q the mid-points of CD and CO

âˆ´ PQ || OD and PQ = \(\frac { 1 }{ 2 }\) OD

In âˆ†BCD,

P is mid-poiht of DC and PQ || OD (Proved above)

Or PR || BD

âˆ´ 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.

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.

Proof : (i) In âˆ†ACD,

Q is mid-point of AC and QP || AD (Sides of rectangles)

âˆ´ P is mid-point of CD

âˆ´ DP = PC

(ii) âˆµPR and QC are the diagonals of rectangle PQRC

âˆ´ PR = QC

But Q is the mid-point of AC

âˆ´ 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

To prove : GP = PH

Proof: âˆµ E and F are the mid-points of AB and CD

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 âˆ†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 Class 9 RD Sharma Solutions VSAQS

Question 1.

In a parallelogram ABCD, write the sum of angles A and B.

Solution:

In ||gm ABCD,

âˆ A + âˆ B = 180Â°

(Sum of consecutive angles of a ||gm)

Question 2.

In a parallelogram ABCD, if âˆ D = 115Â°, then write the measure of âˆ A.

Solution:

In ||gm ABCD,

âˆ D = 115Â°

But âˆ A + âˆ D = 180Â°

(Sum of consecutive angles of a ||gm)

â‡’ âˆ A + 115Â°= 180Â° âˆ A = 180Â°- 115Â°

âˆ´ âˆ A = 65Â°

Question 3.

PQRS is a square such that PR and SQ intersect at O. State the measure of âˆ POQ.

Solution:

In a square PQRS,

Diagonals PR and QS intersects each other at O.

âˆµ The diagonals of a square bisect each other at right angles.

âˆ´ âˆ POQ = 90Â°

Question 4.

If PQRS is a square then write the measure of âˆ SRP.

Solution:

In square PQRS,

Join PR,

âˆµDiagonals of a square bisect are opposite angles

âˆ´âˆ SRP = \(\frac { 1 }{ 2 }\)x âˆ SRQ

= \(\frac { 1 }{ 2 }\) x 90Â° = 45Â°

Question 5.

If ABCD is a rhombus with âˆ ABC = 56Â°, find the measure of âˆ ACD.

Solution:

In rhombus ABCD,

Diagonals bisect each other at 0 at right angles.

âˆ ABC = 56Â°

But âˆ ABC + âˆ BCD = 180Â° (Sum of consecutive angles)

â‡’ 56Â° + âˆ BCD = 180Â°

â‡’ âˆ BCD = 180Â° – 56Â° = 124Â°

âˆµ Diagonals of a rhombus bisect the opposite angle

âˆ´ âˆ ACD = \(\frac { 1 }{ 2 }\) âˆ BCD = \(\frac { 1 }{ 2 }\) x 124Â°

= 62Â°

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

âˆ´ Sum of two consecutive sides = \(\frac { 22 }{ 2 }\)

= 11cm

i.e. AB + BC = 11 cm

AB = 6.5 cm and let BC = x cm

âˆ´ 6.5 + x = 11 cm

x = 11 – 6.5 = 4.5

âˆ´ Shorter side = 4.5 cm

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

âˆµ The sum of angles of a quadrilateral = 360Â°

âˆ´ 3x + 5x + 9x + 13x = 360Â°

â‡’ 30x = 360Â° â‡’ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)Â = 12

âˆ´ Smallest angle = 3x = 3 x 12Â° = 36Â°

Question 8.

In parallelogram ABCD if âˆ A = (3x – 20Â°), âˆ B = (y + 15)Â°, âˆ C = (x + 40Â°), then find the value of x and y.

Solution:

In a ||gm ABCD,

âˆ A = (3x – 20Â°), âˆ B = y + 15Â°,

âˆ C = x + 40Â°

Now, âˆ A = âˆ C (Opposite angles of a ||gm)

â‡’ 3x – 20 = x + 40Â°

â‡’ 3x – x = 40Â° + 20Â° â‡’ 2x = 60Â°

â‡’ x = \(\frac { { 60 }^{ \circ } }{ 2 }\)Â = 30Â°

and âˆ A + âˆ B = 180Â° (Sum of the consecutive angles)

â‡’ 3x-20Â° + y + 15Â° = 180Â°

â‡’ 3x + y – 5Â° = 180Â°

â‡’ 3 x 30Â° +y- 5Â° = 180Â°

â‡’ 90Â° – 5Â° + y = 180

y = 180Â° – 90Â° + 5 = 95Â°

âˆ´ x = 30Â°, y = 95Â°

Question 9.

If measures opposite angles of a parallelogram are (60 – x)Â° and (3x – 4)Â°, then find the measures of angles of the parallelogram.

Solution:

Opposite angles of a ||gm ABCD are (60 – x)Â° and (3x – 4Â°)

But opposite angles of a ||gm are equal, the

60Â° – xÂ° = 3x – 4Â° â‡’ 60Â° + 4Â° = 3x + x

â‡’ 4x = 64Â° â‡’ x = \(\frac { { 64 }^{ \circ } }{{ 4 }^{ \circ } }\)Â = 16Â°

âˆ´ âˆ A = 60Â° – x = 60Â° – 16Â° = 44Â°

But âˆ A + âˆ B = 180Â° (sum of consecutive angle)

â‡’ 44Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â° – 44Â°

â‡’ âˆ B = 136Â°

But âˆ A = âˆ C and âˆ B = âˆ D (Opposite angles)

âˆ´ Angles are 44Â°, 136Â°, 44Â°, 136Â°

Question 10.

In a parallelogram ABCD, the bisectors of âˆ A also bisect BC at x, find AB : AD.

Solution:

In ||gm ABCD,

Bisectors of âˆ A meets BC at X and BX = XC

Draw XY ||gm AB meeting AD at Y

Question 11.

In the figure, PQRS in an isosceles trapezium find x and y.

Solution:

âˆµ PQRS is an isosceles trapezium in which

SP = RQ and SR || PQ

âˆ´ âˆ P + âˆ S = 180Â° (Sum of co-interior angles)

3x + 2x = 180Â° â‡’ 5x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)Â = 36Â°

But âˆ P = âˆ Qm (Base angles of isosceles trapezium)

y = 2x = 2 x 36Â° = 12Â°

âˆ´ y = 12Â°

Hence x = 36Â°, y = 12Â°

Question 12.

In the figure ABCD is a trapezium. Find the values of x and y.

Solution:

In trapezium ABCD,

AB || CD

âˆ´ âˆ A + âˆ D = 180Â° (Sum of cointerior angles)

x + 20Â° + 2x + 10Â° = 180Â°

3x + 30Â° = 180Â°

â‡’ 3x= 180Â° – 30Â°

3x = 150Â°

x = \(\frac { { 150 }^{ \circ } }{ 3 }\)Â = 50Â°

Similarly, âˆ B + âˆ C = 180Â°

â‡’ y + 92Â° = 180Â°

â‡’ y = 180Â° – 92Â° = 88Â°

âˆ´ x = 50Â°, y = 88Â°

Question 13.

In the figure, ABCD and AEFG are two parallelograms. If âˆ C = 58Â°, find âˆ F.

Solution:

In the figure, ABCD and AEFG are two parallelograms âˆ C = 58Â°

âˆµ DC || GF and CB || FE (Sides of ||gms)

âˆ´ âˆ C = âˆ F

But âˆ C = 58Â°

âˆ´ âˆ F = 58Â°

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 âˆ AOB = 75Â°, then write the value of âˆ C + âˆ D.

Solution:

In quadrilateral ABCD,

Bisectors of âˆ A and âˆ B meet at O and âˆ AOB = 75Â°

In AOB, âˆ AOB = 75Â°

âˆ´ âˆ OAB + âˆ OBA = 180Â° – 75Â° = 105Â°

But OA and OB are the bisectors of âˆ A and âˆ B.

âˆ´ âˆ A + âˆ B = 2 x 105Â° = 210Â°

But âˆ A + âˆ B + âˆ C + âˆ D = 360Â° (Sum of angles of a quad.)

âˆ´ 210Â° + âˆ C + âˆ D = 360Â°

â‡’ âˆ C + âˆ D = 360Â° – 210Â° = 150Â°

Hence âˆ C + âˆ D = 150Â°

Question 16.

The diagonals of a rectangle ABCD meet at O. If âˆ BOC = 44Â° find âˆ OAD.

Solution:

In rectangle ABCD,

Diagonals AC and BD intersect each other at O and âˆ BOC = 44Â°

But âˆ AOD = âˆ BOC (Vertically opposite angles)

âˆ´ âˆ AOD = 44Â°

In âˆ†AOD,

âˆ AOD + âˆ OAD + âˆ ODA = 180Â° (Sum of angles of a triangle)

â‡’ 44Â° + âˆ OAD + âˆ OAD = 180Â° [âˆµ OA = OD, âˆ OAD = âˆ ODA]

â‡’ 2âˆ OAD = 180Â° – 44Â° = 136Â°

âˆ´ âˆ OAD = \(\frac { { 136 }^{ \circ } }{ 2 }\)Â = 68Â°

Question 17.

If ABCD is a rectangle with âˆ BAC = 32Â°, find the measure if âˆ DBC.

Solution:

In rectangle ABCD,

Diagonals bisect each other at O

âˆ BAC = 32Â°

âˆµ OA = OB

âˆ´ âˆ OBA Or âˆ DBA = âˆ BAC = 32Â°

But âˆ ABC = 90Â° (Angle of a rectangles)

âˆ´ âˆ DBC = âˆ ABC – âˆ DBA

= 90Â° – 32Â° = 58Â°

Question 18.

If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that âˆ C + âˆ D = k(âˆ AOB), then find the value of k.

Solution:

In quadrilateral ABCD,

Bisectors of âˆ A and âˆ B meet at O

Such that âˆ C + âˆ D = k (âˆ AOB)

Question 19.

In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If âˆ SRT = 152Â°, find x, y and z.

Solution:

In rhombus PQRS,

Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that âˆ SRT = 152Â°

But âˆ SRT + âˆ SRP = 180Â° (Linear pair)

â‡’ 152Â° +âˆ SRP = 180Â°

â‡’ âˆ SRP =180Â°- 152Â° = 28Â°

But âˆ SPR = âˆ SRP (âˆµ PR bisects âˆ P and âˆ R)

â‡’ z = 28Â°

y = 90Â° (âˆµ Diagonals bisect each other at right angles)

âˆ RPQ = z = 28Â°

âˆ´ In âˆ†POQ,

z + x = 90Â° â‡’ 28Â° + x = 90Â°

â‡’ x = 90Â° – 28Â° = 62Â°

âˆ´ x = 62Â°, y = 90Â°, z = 28Â°

Question 20.

In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If âˆ ECD = 146Â°, find âˆ AOB.

Solution:

In rectangle ABCD,

Diagonals AC and BD bisect each other at O

AC is produced to E and âˆ DCE = 146Â°

âˆ DCE + âˆ DCA = 180Â° (Linear pair)

â‡’ 146Â°+ âˆ DCA= 180Â°

â‡’ âˆ DCA = 180Â°- 146Â°

â‡’ âˆ DCA = 34Â°

âˆ´ âˆ CAB = âˆ DCA (Alternate angles)

= 34Â°

Now in âˆ†AOB,

âˆ AOB = 180Â° – (âˆ DAB + âˆ OBA)

= 180Â° – (34Â° + 34Â°)

= 1803 – 68Â° = 112Â°

### 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) âˆ P = 100Â°, âˆ Q = 80Â°, âˆ R = 100Â°

(b) âˆ P = 85Â°, âˆ Q = 85Â°, âˆ R = 95Â°

(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 âˆ P = 100Â°, âˆ Q = 80Â°, âˆ R = 100Â° (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 60Â°

Solution:

A quadrilateral is not a rhombus if one angle between the diagonals is 60Â°. (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) 30Â°

(b) 45Â°

(c) 60Â°

(d) 90Â°

Solution:

The bisectors of any two adjacent angles of a parallelogram intersect at 90Â°. (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 24Â° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is

(a) 176Â°

(b) 68Â°

(c) 112Â°

(d) 102Â°

Solution:

Let the smallest angle be x

The largest angle = 2x – 24Â°

But sum of two adjacent angles = 180Â°

Question 16.

In a parallelogram ABCD, If âˆ DAB = 75Â° and âˆ DBC = 60Â°, then âˆ BDC =

(a) 75Â°

(b) 60Â°

(c) 45Â°

(d) 55Â°

Solution:

In ||gm ABC,

âˆ A = 75Â°, âˆ DBC = 60Â°

But âˆ A + âˆ B = 180Â° (Sum of two consecutive angles)

â‡’ 75Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â°- 75â€œ= 105Â°

But âˆ DBC = 60Â°

âˆ´ âˆ DBA = 105Â°-60Â° = 45Â°

But âˆ BDC = âˆ DBA (Alternate angles)

âˆ´ âˆ BDC = 45Â° (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

E and F are the centroid of âˆ†ABD and âˆ†BCD

Now E and F trisect AC

i.e. AE = EF = FC

âˆ´ EF = AE (a)

Question 18.

ABCD is a parallelogram, M is the midÂ¬point of BD and BM bisects âˆ B. Then, âˆ AMB =

(a) 45Â°

(b) 60Â°

(c) 90Â°

(d) 75Â°

Solution:

In ||gm ABCD, M is mid-point of BD and

BM bisects âˆ B

AM is joined

âˆ´AM bisects âˆ A

But âˆ A + âˆ B = 180Â° (Sum of two consecutive angles)

âˆ´ âˆ AMB = 90Â° (c)

Question 19.

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

(a) 108Â°

(b) 54Â°

(c) 12Â°

(d) 81Â°

Solution:

Let adjacent angle of a ||gm = x

Then second angle = \(\frac { 2 }{ 3 }\) x

âˆ´ x+ \(\frac { 2 }{ 3 }\) x= 180Â°

(Sum of two adjacent angles of a ||gm is 180Â°)

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) 140Â°

(b) 150Â°

(c) 168Â°

(d) 180Â°

Solution:

Sum of the angles of a quadrilateral = 360Â°

âˆ´ 4x + 1x + 9x + 10x = 360Â°

â‡’ 30x = 360Â°

â‡’ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)Â = 12Â°

Now sum of smallest and largest angle = 4 x 12Â° + 10 x 12Â°

= 48Â° + 120Â° = 168Â° (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

Question 22.

ABCD is a parallelogram in which diagonal AC bisects âˆ BAD. If âˆ BAC = 35Â°, then âˆ ABC =

(a) 70Â°

(b) 110Â°

(c) 90Â°

(d) 120Â°

Solution:

In ||gm ABCD, AC is its diagonal which bisect âˆ BAD

âˆ BAD = 35Â°

âˆ´ âˆ BAD = 2 x 35Â° = 70Â°

But âˆ A + âˆ B = 180Â° (Sum of consecutive angles)

â‡’ 70Â° + âˆ B = 180Â°â‡’ âˆ B = 180Â° – 70Â°

âˆ´ âˆ B = 110Â°

â‡’ ABC = 110Â° (b)

Question 23.

In a rhombus ABCD, if âˆ ACB = 40Â°, then âˆ ADB =

(a) 70Â°

(b) 45Â°

(c) 50Â°

(d) 60Â°

Solution:

In rhombus ABCD, âˆ ACB = 40Â°

âˆ´ âˆ BCD = 2 x âˆ ACB

= 2 x 40Â° = 80Â°

But âˆ BCD + âˆ ADC = 180Â° (Sum of consecutive angles of ||gm)

â‡’ 80Â° + âˆ ADC = 180Â°

â‡’ âˆ ADC = 180Â° – 80Â° = 100Â°

âˆ´ âˆ ADB = \(\frac { 1 }{ 2 }\)âˆ ADC = \(\frac { 1 }{ 2 }\)x 100Â° = 50Â° (c)

Question 24.

In âˆ†ABC, âˆ A = 30Â°, âˆ B = 40Â° and âˆ C = 110Â°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

(a) 70Â°, 70Â°, 40Â°

(b) 60Â°, 40Â°, 80Â°

(c) 30Â°, 40Â°, 110Â°

(d) 60Â°, 70Â°, 50Â°

Solution:

In âˆ†ABC,

âˆ A = 30Â°, âˆ B = 40Â°, âˆ C = 110Â°

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

âˆ´ âˆ A = âˆ D = 30Â°

âˆ B = âˆ E = 40Â°

âˆ C = âˆ F= 110Â°

âˆ´ Angles are 30Â°, 40Â°, 110Â° (c)

Question 25.

The diagonals of a parallelogram ABCD intersect at O. If âˆ BOC = 90Â° and âˆ BDC = 50Â°, then âˆ OAB =

(a) 40Â°

(b) 50Â°

(c) 10Â°

(d) 90Â°

Solution:

In ||gm ABCD, diagonals AC and BD intersect each other at O

BOC = 90Â°, âˆ BDC = 50Â°

âˆµ âˆ BOC = 90Â°

âˆ´ Diagonals of ||gm bisect each other at 90Â°

âˆ´âˆ COD = 90Â°

In âˆ†COD,

âˆ OCD = 90Â° – 50Â° = 40Â°

But âˆ OAB = âˆ OCD (Alternate angles)

âˆ´âˆ OAB = 40Â° (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

âˆµ MN = \(\frac { 1 }{ 2 }\)(AB + CD)

â‡’ 2MN = AB + CD

â‡’ 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 âˆ A = 45Â°, then âˆ B =

(a) 115Â°

(b) 120Â°

(c) 125Â°

(d) 135Â°

Solution:

Diagonals AC and BD of quadrilateral ABCD bisect each other at O

âˆ´ AO = OC, BO = OD

âˆ´ ABCD is a ||gm âˆ A = 45Â°

But âˆ A + âˆ B = 180Â° (Sum of consecutive angles)

âˆ´ âˆ B = 180Â° – âˆ A = 180Â° – 45Â°

= 135Â° (d)

Question 28.

P is the mid-point of side BC of a paralleogram ABCD such that âˆ BAP = âˆ 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

âˆ BAP = âˆ DAP

Through P, draw PQ || AB

âˆ´ ABPQ is rhombus

âˆ´ AB = BP = AQ

= \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 10 = 5 cm

But CD = AB (Opposite sides of ||gm)

âˆ´ CD = 5 cm (a)

Question 29.

In âˆ†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 âˆ†ABC, E is the mid-point of median AD

Such that BE produced meets AC at F

AC = 10.5 cm

Draw DG || AF

In âˆ†ADG

E is mid-point of AD and EF || DG

âˆ´ F is mid-point of AG

â‡’ AF = FG …(i)

In âˆ†BCF

D is mid-point of BC and DG || BF

âˆ´ G is mid-point of FC

âˆ´ FG = GC …(i)

From (i) and (ii)

AF = FG = GC = \(\frac { 1 }{ 3 }\) AC

But AC = 10.5 cm

âˆ´ 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 =

Solution:

In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F

âˆµ E is mid point of BC

âˆ´ BE = EC

In âˆ†BEF and âˆ†CDE

BE = EC

âˆ BEF = âˆ CED (Vertically opposite angle)

and âˆ EBF = âˆ ECD (Alternate angles)

âˆ´ âˆ†BEF â‰… âˆ†CDE (ASA criterian)

âˆ´ DC = BF

But DC = AB

âˆ´ AB = BF

AF = AB + BF = AB + AB

= 2AB (b)

Question 31.

In a quadrilateral ABCD, âˆ A + âˆ C is 2 times âˆ B + âˆ D. If âˆ A = 140Â° and âˆ D = 60Â°, then âˆ B =

(a) 60Â°

(b) 80Â°

(c) 120Â°

(d) None of these

Solution:

In quadrilateral ABCD

â‡’ âˆ A + âˆ C = 2(âˆ B + âˆ D)

â‡’ âˆ A + âˆ C = 2âˆ B + 2âˆ D

Adding 2âˆ A + 2âˆ C both sides

2âˆ A + 2âˆ C + âˆ A + âˆ C = 2âˆ A + 2âˆ B + 2âˆ C + 2âˆ D

â‡’ 3âˆ A + 3âˆ C = 2(âˆ A + âˆ B + âˆ C + âˆ D)

â‡’ 3(âˆ A + âˆ C) = 2 x 360Â° = 720Â°

âˆ´ âˆ A + âˆ C = \(\frac { { 720 }^{ \circ } }{ 3 }\)Â = 240Â°

â‡’ 40Â° + âˆ C = 240Â° (âˆµ âˆ A = 40Â°)

âˆ C = 240Â° – 40Â° = 200Â°

Now 2(âˆ B + âˆ D) = âˆ A + âˆ C = 240Â°

âˆ B + âˆ D = \(\frac { { 240 }^{ \circ } }{ 2 }\)Â = 120Â°

âˆ´ âˆ B = 60Â° = 120Â°

âˆ´ âˆ B = 60Â° (a)

Question 32.

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If âˆ ABD = 50Â°, then âˆ DPC =

(a) 70Â°

(b) 90Â°

(c) 80Â°

(d) 100Â°

Solution:

In rectangle ABCD, diagonals AC and BD intersect each other at P

âˆ ABD = 50Â°

âˆ´ âˆ CAB = âˆ ABD = 50Â° (âˆµ AP = BP)

Now in âˆ†APB

âˆ CAB + âˆ ABD + âˆ APB = 180Â° (Angles of a triangle)

â‡’ âˆ PAB + âˆ PBA + âˆ APB = 180Â°

â‡’ 50Â° + 50Â° + âˆ APB = 180Â°

â‡’ âˆ APB = 180Â° – 50Â° – 50Â° = 80Â°

But âˆ DPC = ADB (Vertically opposite angles)

âˆ´ âˆ DPC = 80Â° (c)

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

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

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

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