Effective ICSE Class 9 Maths Solutions S Chand Chapter 11 Rectilinear Figures Ex 11(B) can help bridge the gap between theory and application.

## S Chand Class 9 ICSE Maths Solutions Chapter 11 Rectilinear Figures Ex 11(B)

Question 1.

(i) If in a parallelogram, the diagonals are equal in length and intersect at right angles, prove that the parallelogram will be a square.

(ii) Prove that in a rectangle, the diagonals are of equal lengths.

Solution:

(i) Given : ABCD is a parallelogram in which diagonal AC = diagonal BD and AC and BD intersect each other at O at right angles

To prove : ABCD is a square

Proof: In ∆ABC and ∆ADB

AB = AB (common)

AC = BD (given)

BC = AD (opposite sides of || gm)

∴ ∆ABC ≅ ∆ADC (SSS axiom)

∴ ∠ABC = ∠BAD (c.p.c.t.)

But ∠ABC + ∠B AD = 180° (co-interior angles)

∴ ∠ABC = ∠BAD = 90°

Again in ∆AOB and ∆BOC

∠AOB = ∠BOC (each 90°)

BO = BO (common)

AO = OC (∵ diagonals bisect each other)

∴ ∆AOB ≅ ∆BOC (SAS axiom)

∴ AB = BC

But AB = CD and BC = AD (opposite sides of ||gm)

∴ AB = BC = CD = DA

and each angle is 90°

Hence ABCD is a square

Hence proved.

(ii) Given : ABCD is a rectangle and AC and BD are its diagonals

To Prove: AC = BD

Proof: In ∆ABC and ∆ADB,

AB = AB (common)

BC = AD (opposite sides of rectangle)

∠ABC = ∠BAD (each 90°

∴ ∆ABC ≅ ∆ADB (SAS axiom)

AC = BD (c.p.c.t.)

Hence proved.

Question 2.

In the figure, ABCD is a parallelogram. M is the mid-point of AC; X, Y are points on AB and DC respectively such that AX = CY.

Prove that:

(i) ∆AXM ≅ ∆CYM

(ii) XMY is a straight line.

Solution:

Given : In parallelogram ABCD, AC is its diagonal X and Y are the mid-points of AB and CD respectively

M is mid-point of AC and AX = CY

To prove :

(i) ∆AXM ≅ ∆CYM

(ii) XMY is a straight line

Construction : Join XM and YM

Proof: In ∆AMX and ∆CMY

AX = CY (given)

AM = CM (M is mid-point of AC)

and ∠YCM = ∠MAX (alternate angles)

∴ ∆AMX = ∆CMY (SSA axiom)

∴ ∠AMX = ∠CMY

But these are vertically opposite angles

∴ XMY is a straight line

Hence proved.

Question 3.

Prove that :

(i) A diagonal of a square makes an angle of 45° with the side of the square.

(ii) The diagonals of a rhombus are at right’ angles.

(iii) A diagonal of a rhombus bisects the angles at vertices.

Solution:

Given :

(i) In a square ABCD, BD is its diagonal

To prove : ∠ABD = 45°

Proof: In ∆ABD and ABCD

AB = DC (sides of square)

AD = BC (sides of square)

BD = BD (common)

∴ ∆ABD ≅ ∆BCD (SSS axiom)

∴ ∠ABD =∠CBD (c.p.c.t.)

But ∠ABD + ∠CBD = 90°

∴ ∠ABD = \(\frac { 90° }{ 2 }\) = 45°

Hence diagonal BD makes an angle of 45° with the side of the square

(ii) Given : In rhombus ABCD its diagonals AC and BD bisect each other at O

To prove : ∠AOB = ∠BOC = ∠COD = ∠DOA – 90°

Proof: In ∆AOB and ∆COB

AQ = OC (diagonals bisect each other)

BO = BO (common)

AB = BC (sides of rhombus)

∴ ∆AOB = ∆COB (SSS axiom)

∴ ∠AOB ≅ ∠COB (c.p.c.t.)

But ∠AOB + ∠COB = 180° (Linear pair)

∴ ∠AOB = ∠COB = 90°

Similarly we can prove that

∠COD = ∠DOA = 90°

Hence diagonals of a rhombus bisect each other at right angles

(iii) Given : In rhombus ABCD, AC is its diagonal

To prove : AC, bisect ∠A and ∠C

Proof: In ∆ABC and ∆ADC

AB = CD (opposite sides of rhombus)

BC=AD (opposite sides of rhombus)

AC = AC

(common)

∴ ∆ABC ≅ ∆ADC (SSS axiom)

∴ ∠CAB = ∠CAD (c.p.c.t.)

and ∠ACB = ∠ACD (c.p.c.t.)

Hence diagonal of a rhombus bisects the angles at the vertices.

Question 4.

In the figure, PQRS is a parallelogram. QU and ST are perpendicular on the diagonal PR. Prove that

(i) ∆STR ≅ ∆QUP

(ii) ST = QU

Solution:

Given : PQRS is a parallelogram in which PR is its diagonal ST ⊥ PR and QU ⊥ PR

To prove :

(i) ∆STR ≅ ∆QUP

(ii) ST = QU

Proof: In ∆STR and ∆QUP

PS = SR (Opposite sides of parallelogram)

∠T = ∠U (each 90°)

∠SPT = ∠URQ (alternate angles)

(i) ∴ ∆STR ≅ ∆QUP (AAS axiom)

(ii) ∴ ST = QU (c.p.c.t.)

Question 5.

In the figure, PQRS is a parallelogram PO and QO are respectively the angle bisectors of ∠P and ∠Q. Line LOM is drawn parallel to PQ. Prove that

(i) PL = QM

(ii) LO = OM

Solution:

Given : In ||gm ABCD, PO and QO are the angle bisectors of ∠P and ∠Q respectively meeting each other at O. LOM is drawn parallel to PQ through O

To prove :

(i) PL = QM

(ii) LO = OM

Proof: ∵ PO is the bisector of ∠P

∴ ∠1 = ∠2

Similarly QO is the bisector of ∠Q

∴ ∠5 = ∠6

∵ LM || PQ

LPQM is a parallelogram

and ∠3 = ∠2 = ∠1 (alternate angles)

and ∠4 = ∠5 = ∠6

(i) LP = QM (∵ LPQM is a parallelogram)

(ii) PL = LO (∵ ∠1 = ∠3)

and OM = OM (∵ ∠4 = ∠6)

But PL = OM (proved)

∴ LO = OM

Hence proved.

Question 6.

(i) ABCD is a parallelogram. L and M are points on AB and DC respectively and AL = CM. Prove that LM and BD bisect each other.

(ii) ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC.

Solution:

(i) Given : In ||gm ABCD, BD is its diagonal L and M are points on AB and CD respectively such that AL = CM

To prove : LM and BD bisect each other i.e. BO = OD and LO = OM

Proof : In ||gm ABCD,

AB = CD

But AL = CM

∴ AB – AL = CD – CM

⇒ LB = MD

Now in ∆LOB and ∆DOM

LB = MD (proved)

∠LBO = ∠MDO (alternate angles)

∠LOB = ∠DOM

(vertically opposite angles)

∴ ∆LOB ≅ ∆DOM (AAS axiom)

∴ OB = OD and OL = OM(c.p.c.t.)

Hence BD and LM bisect each other

(ii) Given : In a parallelogram ABCD, AB is produced to E such that BE = AB

To prove : ED bisects BC

Proof:

∵ AB = DC (opposite sides of ||gm)

and AB = BE (given)

∴ DC = BE

Now in ∆BOE and ∆COD

BE = DC (Proved)

∠BOE = ∠COD (vertically opposite angles)

∠BEO = ∠CDO (alternate angles)

∴ ∆BOE ≅ ∆COD (AAS axiom)

∴ BO = OC (c.p.c.t.)

Hence DE bisects BC

Question 7.

(i) The diagonals of a parallelogram intersect at O. A line through O intersects AB at X and DC at Y. Prove that OX = OY.

(ii) In atriangle ABC, median AD is produced to X, such that AD = DX. Prove that ABXC is a parallelogram.

Solution:

(i) Given : In parallelogram ABCD diagonals AC and BD intersect each other at O

A line XOY is drawn which meets AB in X and CD in Y

To prove : OX = OY

Proof: In ∆XOB and ∆YOD

OB = OD (diagonals bisect each other)

∠BOX = ∠DOY

(vertically opposite angles)

∠OBX = ∠ODY (alternate angles)

∴ ∆XOB ≅ ∆YOD (AAS axiom)

∴ OX = OY (c.p.c.t.)

Hence proved.

(ii) Given : In AABC, AD is the median which is produced to X such that DX = AD BX and XC are joined

To prove : ABXC is a parallelogram

Proof: In ∆ADB and ∆XDC

AD = DX (given)

BD = DC (D is mid-point)

∠ADB = ∠XDC

(vertically opposite angles)

∴ ∆ADB ≅ ∆XDC (SAS axiom)

∴ AB = CX (c.p.c.t.)

∠BAD = ∠CXD (c.p.c.t.)

But these are alternate angles

∴ AB || XC

∵ AB = XC and AB || XC

∴ ABXC is a parallelogram

Hence proved.

Question 8.

In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that (i) BE bisects ∠B, (ii) ∠AEB is a right angle.

Solution:

Given : (i) In parallelogram ABCD, AE is the bisector of ∠A which meets CD in E, AB = 2AD

EB is joined

To prove :

(i) BE bisects ∠E

(ii) ∠AEB is a right angle

Construction : Take F, a mid-point of AB. Join EF

Proof :

∵ AB = 2AD and F is mid-point of AB

∴ AF = AD

∵ ∠1 = ∠2 (AE is the bisector of ∠A)

∴ AFED is a rhombus

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

∴ FBCE is also a rhombus

whose BE is diagonal

But diagonal of a rhombus bisects the angles at the vertices

∴ BE bisects ∠E

(ii) ∵ ∠1 = ∠2 and ∠3 = ∠4

But ∠A + ∠B = 180° (co-interior angles)

2∠2 + 2∠3 = 180°

⇒ ∠2 + ∠3 = 90° (dividing by 2)

In ∆AEB,

∵ ∠2 + ∠3 + ∠AEB = 180° (angles of a triangle)

⇒ 90° + ∠AEB = 180°

⇒ ∠AEB = 180° – 90°

⇒ ∠AEB = 90°

Hence ∠AEB is a right angle

Question 9.

(i) In the figure, ABCD is a ||gm. X is the mid-point of AD and Y is the mid-point of BC. Prove that AYCX is a ||gm and that XY and BD bisect each other.

(ii) In the figure, ABCD is a ||gm. BM bisects ∠ABC, and DN bisects ∠ADC. Prove that:

(i) BNDM is a ||gm,

(ii) BM = DN.

(iii) In the figure, BM bisects ∠B and AN bisects ∠A of ||gm ABCD. Prove that :

(i) MN = CD

(ii) ABNM is a rhombus

Solution:

(i) Given : In ||gm ABCD,

X and Y are mid-points of the side AD and BC respectively. BD is its diagonal. AY and CX are joined

To prove :

(i) ATCX is a ||gm

(ii) XY and BD bisect each other

Proof:

∵ AD || BC (opposite sides of ||gm)

⇒ AX || CY

But X and Y are mid-points of AD and BC

respectively and AD = BC

AX = CY (half of equal sides)

AXCY is a ||gm

(ii) In ∆XOD and ∆BOY

XD = BY (half of equal sides)

∠XOD = ∠BOY

(vertically opposite angles) ∠ADO = OBY (alternate angles)

∴ ∆XOD ≅ ∆BOY (AAS axiom)

∴ XO = OY (c.p.c.t.)

and DO = OB (c.p.c.t.)

Hence XY and BD bisect each other

Hence proved.

(ii) Given : In ||gm ABCD,

BM is the bisector of ∠B and DN is the bisector of ∠D

To prove :

(i) BNDM is a ||gm

(ii) BM = DN

Proof: ∵ Bisector of ∠ABC meets AD at M

∴ M is mid-point of AD

Similarly DN is the bisector of ∠ADC

∴ N is mid-point of BC

∴ MD || BN (∵ AD || BC)

and DM = BN (half of equal sides)

∴ BNDM is a ||gm

∴ BM = DM (opposite sides of || gm BNDM)

Hence proved.

(iii) Given : In || gm ABCD, BM is the bisector of ∠B and AN is the bisector of ∠A

To prove :

(i) MN = CD

(ii) ABNM is a rhombus

Proof :∠A + ∠B = 180° (co-interior angles)

∵ AN and BM are the bisectors of ∠A and ∠B respectively

∴ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B = 90°

⇒ ∠OAB + ∠OBA = 90°

∴ In ∆AOB,

∠AOB = 180° – 90° = 90°

∴ AN and BM are perpendicular to each other

Now in ∆AOB and ∆MON

AB = MN (opposite sides of ||gm)

∠AOB = ∠MON

(vertically opposite angles)

∠BAN = ∠ANM (alternate angle)

∴ ∆AOB ≅ ∆MON (AAS axiom)

∴ AO = ON and BO = OM

∵ Diagonals BM and AN bisect each other at right angles

∴ ABNM is a rhombus

Question 10.

A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that:

(i) ACBD is a rectangle.

(ii) CD is parallel to the original parallel lines.

Solution:

Given : Two parallel line PQ and RS and a transversal LM intersects them at A and B respectively

The bisectors of interipr angles at A and B, meet each other at C and D as shown in the figure fonning a quadrilateral ABCD

To prove :

(i) ACBD is a rectangle

(ii) CD is parallel to the original parallel lines PQ and RS

Construction : Join CD

Proof:

∵ (i) AC and BD are the bisectors of ∠PAB and ∠ABS and AC || BD

Similarly AD || BC

ACBD is a parallelogram

Now ∠PAB + ∠ABR = 180°

∴ \(\frac { 1 }{ 2 }\) ∠PAB +\(\frac { 1 }{ 2 }\) ∠ABR = 90°

⇒ ∠CAB + ∠CBA = 90°

In ∆ABC,

∠ACB = 90°

∴ A parallelogram with each angle a right angle is a rectangle

Hence ACBD is a rectangle

(ii) ∵ Diagonals of a rectangle are equal and bisect each other

Diagonal AB and CD of rectangle ACBD bisect each other at O

∴ OA = OC

∴ ∠OAC = ∠ACO

⇒ ∠CAP = ∠ACO

But these are alternate angles

∴ CD || PQ or RS

Question 11.

In the figures,

(i) BY bisects ∠B of AABC. Prove that B∠YX is a rhombus.

(ii) HJKL is a square, HO = HX. Prove that ∠HOX = 3∠XOJ.

(iii) ABCD is a || gm. ABML and ADXY are squares. Prove that ACXM is isosceles.

(iv) ABCD and ALMN are squares. Prove that BL = DN.

Solution:

(i) Given : In ∆ABC, BY is the bisector of ∠B meeting AC at Y

Through Y. YX || AB and YZ || BC are drawn

To prove : B∠YX is a rhombus

Proof: ∵ XY || BA or BZ

and YZ || BC or BX

∴ B∠YX is a parallelogram

But BY is the diagonal which bisects ∠B (given)

∴ B∠YX is a rhombus

(ii) Given : HJKL is a square in which diagonals

HK and JL intersect each other at O

X is a point on HJ such that HX = HO

XO is joined

To prove : ∠HOX = 3∠XOJ

Proof: ∵ Diagonals of square HJKL bisect each other at right angle at O

∴ ∠HOJ = 90° and ZKHJ or ∠OLIX = 45°

∵ In AOHX

OH = HX (given)

∴ ∠HOX = ∠HXO

(angles opposite to equal side)

But ∠OHX + ∠HOX + ∠HXO = 180°

(angles of a triangle)

⇒ 45° + ∠HOX + ∠HOX = 180°

⇒ 2∠HOX = 180° – 45°= 135°

⇒∠HOX= \(\frac { 135 }{ 2 }\) = 67\(\frac { 1° }{ 2 }\) … (i)

But ∠XOT = ∠HOJ = ∠HOX

90° – 67\(\frac { 1° }{ 2 }\) = 22\(\frac { 1° }{ 2 }\) … (ii)

From (i) and (ii)

∠HOX = 67 \(\frac { 1° }{ 2 }\) = 3 x 22\(\frac { 1° }{ 2 }\)

= 3∠XOJ

Hence proved.

(iii) Given : In the figure,

ABCD is a ||gm

ABL and ∆DXY are squares

To prove : ∆CXM is an isosceles triangle

Construction : Join MC, MX and CX

Proof: In parallelogram ABCD

AB = CD and BC = AD (opposite sides)

and ∠ABC = ∠CDA and ∠BCD = ∠BAD (opposite angles)

∠MBC = ∠MBA + ∠ABC = 90° + ∠ABC … (i)

∠CDX = ∠CDA + ∠ADX = ∠ABC+ 90° … (ii)

(∵ ∠CDA = ∠ABC)

∴ From (i) and (ii),

∠MBC = ∠CDX

Now in ∆BMC and ∆DCX

BM = BA = CD (given)

BC = AD = DX (given)

and ∠MBC = ∠CDX (proved)

∆BAC ≅ ∆DCX (SAS axiom)

∴ CM = CX

Hence ACMX is an isosceles triangle

(iv) Given : ABCD and ALMN are two squares as shown in the figure

To prove : BL = DN

Construction : Join DM and BL

Proof: In ∆ADN and ∆ALB

AD = AB (sides of square ABCD)

AN = AL (sides of square ALMN)

∠DAN = ∠LAB (each = 90° – ∠LAD)

∴ ∆ADN ≅ ∆ALB (SAS axiom)

∴ DN = BL (c.p.c.t.)

Hence proved.

Question 12.

(i) Prove that the bisectors of any two adjacent angles of a parallelogram are at right angles.

(ii) In the figure, PQRS is a parallelogram with bisectors PA, QD, RC and SB respectively of angles P, Q, R and S. Show that ABCD is an rectangle.

Solution:

(i) Given : In ||gm ABCD, AE and BE are the bisectors of adjacent ∠A and ∠B which meet at E

To prove : ∠E is a right angle

Proof: ∵ AE is the bisector of ∠A

∴ ∠EAB = \(\frac { 1 }{ 2 }\) ∠A

Similarly BE is the bisector of ∠B

∴ ∠EBA = \(\frac { 1 }{ 2 }\) ∠B

But ∠A + ∠B = 180° (co-interior angles)

∴ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B = 90°

⇒ ∠EAB + ∠EBA = 90°

But in ∆AEB

∠EAB + ∠EBA + ∠AEB = 180°

(sum of angles of a triangle)

⇒ 90° + ∠AEB = 180°

⇒ ∠AEB = 180° -90° = 90°

∴ ∠E is a right angle

(ii) Given : In ||gm ABCD, PA, QD, RC and SB are the bisectors of ∠P, ∠Q, ∠R and ∠S respectively which meet each other at A, C, D and B respectively forming a quad. ABCD

To prove : ABCD is a rectanlge

Proof:

∵ PD and QD are the bisectors of two adjacent angles ∠P and ∠Q respectively

∴ ∠D = 90°

Similarly ∠B = 90°

and PA and SA are the bisectors of ∠P and ∠S respectively

∴ ∠A = 90°

Similarly ∠C = 90°

∴ The quad. ABCD has its each angle equal to 90°

∴ ABCD is a rectangle

Hence proved.

Question 13.

In the figure, ABCD is a parallelogram and X is the midpoint of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:

(i) ∆ABX ≅ ∆QCX

(ii) DC = CQ = QP

Solution:

Given : ABCD is a ||gm and X is mid-point of BC

AX is joined and produced to meet BC produced at Q. The ||gm ABPQ is completed as shown in the figure

To prove :

(i) ∆ABX ≅ ∆QCX

(ii) DC = CQ = QP

Proof: In ∆ABX and ∆QCX

BX = XC (∵ Q is mid-point of BC)

∠AXB = ∠CXQ

(vertically opposite angles)

∠ABX = ∠XCQ (alternate angles)

∴ ∆ABX ≅ ∆QCX (AA side axiom),

∴ AB = CQ (c.p.c.t.)

But AB = DC and AB = QP

(opposite sides of ||gms)

∴ DC = CQ = QP

Hence proved.

Question 14.

ABCD is a rhombus with P, Q, R as mid-points of AB, BC and CD. Prove that PQ ⊥ QR.

Solution:

Given : P, Q and R are the mid points of the sides AB, BC and CD of rhombus ABCD PQ and QR are joined

To prove : PQ ⊥ QR

Construction : Join diagonals AC and BD of the rhombus ABCD

Proof: In ∆ABC,

P and Q are mid-points of AB and BC respectively

∴ PQ || AC … (i)

Similarly in ∆BCD,

Q and R the mid-points of BC and CD respectively

∴ QR || BD … (ii)

But AC and BD bisect each other at right angles

∴ PQ ⊥ QR

Question 15.

ABCD is a rhombus. RABS is a straight line such that RA = AB = BS. Prove that RD and SC when produced meet at right angles.

Solution:

Given : In a rhombus ABCD

RABS is a straight line such that RA = AB = RS

RD and SC are joined and produced to meet atT

To prove : ∠T = 90°

Proof: In ∆RAD,

RA = AB =AD

∴ ∠DRA = ∠RDA

and Ex. ∠DAB = ∠DRA + ∠RDA = ∠DRA + ∠DRA = 2∠DRA … (i)

Similarly in ∆BSC,

BS = AB = BC

∴ ∠BCS = ∠BSC

and Ext. ∠ABC = ∠BCS + ∠BSC

= ∠BSC + ∠BSC = 2∠BSC … (ii)

But in rhombus ABCD,

∠DOB + ∠ABC = 180° (co-interior angles)

⇒ 2∠DRA + 2∠BSC = 180°

∠DRA + ∠BSC = 90° (dividing by 2)

Now in ∆RTS,

∠DRA + ∠BSC = 90°

or ∠TRS + ∠RST = 90°

∴ ∠RTS = 180°- 90° = 90°

Hence ∠T = 90°