Accessing OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Ex 8(C) can be a valuable tool for students seeking extra practice.

## S Chand Class 9 ICSE Maths Solutions Chapter 8 Triangles Ex 8(C)

Question 1.

In a ∆PQR, side PQ is produced to S so that QS = RQ. If ∠PQR = 60° and ∠RPQ = 70°, prove that

(i) PS > RS

(ii) PS > PR

Solution:

Given : In the figure, ∆PQR, whose side PQ is produced to S such that QS = RQ

∠PQR = 60° and ∠RPQ = 70°

To prove :

(i) PS > RS

(ii) PS > PR

Proof: In ∆PQR,

∵ ∠PQR = 60° and ∠RPQ = 70°

∴ ∠QRP = 180° – (60° + 70°)

= 180° – 130° = 50°

and Ext. ∠RQS =180° – 60° = 120°

∵ In ∆RSQ,

QS = QR (given)

∴ ∠QSR = ∠QRS

But ∠QSR + ∠QRS + ∠RQS = 180°

∴ ∠QSR + ∠QRS + 120°= 180°

⇒ ∠QSR + ∠QRS =180°- 120° = 60°

∴ ∠QSR = ∠QRS = \(\frac { 60° }{ 2 }\) = 30°

(i) In ∆PSR,

∠PRS = ∠PRQ + ∠QRS = 50° + 30° = 80°

∵ ∠PRS > ∠RPS

∴ PS > RS

(∵ side opposite to greater angle is greater)

(ii) Similarly

∵ ∠PRS > ∠PSR (∵ ∠PSR = 30°)

∴ PS > PR

Hence proved.

Question 2.

In a ∆PQR, ∠Q = 35°, ∠R = 61°, the bisector of ∠QPR cuts QR at X. Arrange in descending order PX, QX, RX.

Solution:

Given : In ∆PQR, PX is the bisector of ∠P meeting QR at X

∠Q = 35°, ∠R = 61°

∴ ∠P= 180° – (35° + 61°)

= 180° – 96° = 84°

∵ PX is the bisector of ∠QPR

∴ ∠RPX = ∠QPX = \(\frac { 84° }{ 2 }\) = 42°

Now ∠PXQ = ∠RPX + ∠PRQ

= 42° + 61° = 103°

and ∠PXR = 180° – ∠PXQ = 180° – 103° = 77°

Now we have to arrange PX, QX and RS in descending order

In ∆PXQ, > PQS (42° > 35°)

In ∆PXQ, QPX > PQS

∴ XQ > PX … (i)

and in ∆PRX,

∠PRX > ∠RPX (∵ 610 > 42°)

∴ XP > XR … (ii)

From (i) and (ii),

XQ > PX > XR

⇒ QX > PX > RX

Which are in descending order

Question 3.

In the figure, AB < BC. If the base angles of ∆ABC are 70° and 50°, state which of the two is 70°. AM bisects the exterior angle BAX and AP is parallel to CB, find ∠MAP.

Solution:

Given : In ∆ABC,

AB < BC

and base angles are 70° and 50°

AM is the bisector of ∠BAX and AP || BC

Now we have to find ∠MAP

∵ AB < BC

∴ ∠C < ∠B

∴ ∠C will be = 50° and ∠B = 70°

(angles opposite to greater side is greater)

In ∆ABC

Ext. BAX = ∠B + ∠C

= 70°+ 50° = 120°

∵ AM is the bisector of ∠BAX (given)

∠MAX = ∠MAB = \(\frac { 120° }{ 2 }\) = 60°

∵ AD || BC (given)

∴ ∠PAB = ∠ABC = 70°

Now ∠MAP = ∠PAB – ∠MAB

= 70° – 60° = 10°

Question 4.

In the figure, ABCD is a straight line which is greater?

(i) RB or RC

(ii) PB or PR?

Solution:

In the figure, ∆BCD is a straight line

∠BPR = ∠CQR = 90°, ∠ABP =114° and ∠DCQ = 112°

To find : Which is greater

(i) RB or RC

(ii) PB or PR?

(i) In ∆BPR, ∠P = 90° and in ∠CQR = 90°

∠PRB = ∠QRC (vertically opposite angles)

∠PBR = ∠QCR

But ∠PBA + ∠PBC = 180° (linear pair)

⇒ 114° +∠PBC = 180°

⇒ ∠PBC = 180°- 114° = 66°

Similarly ∠QCB = 180° – 112° = 68°

∴ ∠QCB > ∠PBC ⇒ ∠QCR + ∠RCB> ∠PBR + ∠RBC

But ∠PBR = ∠QCR (proved)

∴ ∠RCB > ∠RBC

∴ RB > RC

(ii) ∵ ∠PBR > ∠PRB

∴ PR > PB

Question 5.

In the figure, arrange the angles in descending order of magnitude.

Solution:

In AABC, AB = 3 cm, AC = 5 cm and BC = 6 cm

∴ ∠A > ∠B and ∠A > ∠C

But AC > AB

∴ ∠B > ∠C

∴ ∠A > ∠B > ∠C or

∴ ∠A, ∠B and ∠C are in descending order

Question 6.

In the figure, which is longer

(i) LM or MN

(ii) PQ or PR

(iii) AB || DC; AB or BD

(iv) PB or PC, given PB and PC bisect ∠ABC and ∠ACB respectively.

(v) QM or QR if LM > LR and ∠LMQ = ∠LRQ.

Solution:

(i) In ∆LMN,

∠M = 58°, ∠N = 63°

∴ ∠L = 180° – (58° + 63°)

= 180° – 121° = 59°

∵ ∠N > ∠L

∴ LM > MN

or LM is longer

(ii) In ∆PQR,

∵ RT || QP

∴ ∠PQR = ∠TQS (corresponding angles)

= 42°

and ∠QPR = ∠PRT (alternate angles)

= 80°

∴ ∠PRQ = 180° – (80° + 42°) = 180° – 122°

= 58°

∵ ∠PRQ > ∠PQR

∴ PQ > PR

∴ PQ is longer

(iii) ABCD is a quadrilateral and DC is a produced to E

∠BCE = 127°

∠A = 73° and ∠CBD = 95°

∵ ∠BCD + ∠BCE = 180° (linear pair)

∴ ∠BCD = 180°- 127° = 53°

∴ ∠BDC = 180° – (95° + 53°) = 180° – 148°

= 32°

∴ ∠ADB + ∠BDC =127° (∵ ∠ADC = ∠BCE)

⇒ ∠ADB + 32°= 127°(corresponding angles)

⇒ ∠ADB = 127° – 32° = 95°

∵ ∠ADB > ∠BAD

∴ BA > BD

BA is longer

(iv) In the ∆ABC,

PB and PC are the bisectors of

∠ABC and ∠ACB respectively

Ext. ∠LAB = 126° and ∠MBA = 118°

Now ∠ABC + ∠ABM = 180° (linear pair)

⇒ 118° + ∠ABC = 180° ⇒ ∠ABC = 180° – 118° = 62°

∴ ∠PBC = \(\frac { 1 }{ 2 }\) ∠ABC = \(\frac { 1 }{ 2 }\) x 62° = 31°

and ∠BAL = ∠ABC + ∠ACB = 126°

⇒ 126° = 62° +∠ACB

⇒ ∠ACB = 126° – 62° = 64°

∴ ∠PCB = \(\frac { 1 }{ 2 }\) ∠ACB = \(\frac { 1 }{ 2 }\) x 64° = 32°

∴ In ∆PBC

∠PCB > ∠PBC

∴ PB > PC

or PB is longer

(v) In the figure,

LM > LR and

∠LMQ = ∠LRQ

∵ LM > LR

∴ ∠LRM > ∠LMR

(angle opposite to greater side is greater)

But ∠M = ∠R (given)

∴ ∠QRM > ∠QMR

∴ QM > QR

Hence QM is longer

Question 7.

Answer true or false

(a) If three sides of a triangle are 3, 4 and 5, then the greatest angle is opposite to the side 5 units.

(b) If two sides of a triangle are unequal, the greater side has the greater angle’ opposite to it.

(c) The sides of a certain triangle are 36,46 and 84 cm.

Solution:

(a) Three sides of a ∆ are 3, 4, 5 units in which side of 5 units is greater

∴ angle opposite to it greater Yes

(b) True : Greater side of a triangle has greater angle opposite to it.

(c) Sides of a triangle are 36, 46 and 84 cm It is not possible as

36 + 46 = 82

and 82 > 84

as sum of any two sides of a triangle is greater than the third side.

Question 8.

The side BC of a triangle ABC is produced to D, so that CD = AC. If the angle BAD = 109° and the angle ACD = 72°, prove that BC is greater than AC.

Solution:

Given : In ∆ABC, BC is produced to D such that CD = AC

∠BAD = 109° and ∠ACD = 72°

To prove : BC > AC

Proof: ∵ AC = CD

∴ ∠CAD = ∠CDA

But ∠CAD + ∠CDA + ∠ACD = 180°

(angles of a triangle)

⇒ ∠CAD + ∠CDA + 72° = 180°

⇒ ∠CAD + ∠CDA =180° – 72° = 108°

∴ ∠CAD = ∠CDA =\(\frac { 108° }{ 2 }\) = 54°

∴ ∠BAC = ∠BAD – ∠CAD = 109° – 54° = 55°

and ∠B = 180° – (∠A + ∠D) = 180° – (109° + 54°)

∠B = 180° – 163° = 17°

∵ ∠BAO > ∠B

∴ BC > AC

Hence proved.

Question 9.

Can you draw triangle with sides (i) 5 cm, 7 cm, 8 cm (ii) 2.5 cm, 1 cm, 3.5 cm (iii) 2 cm, 3 cm, 5.6 cm (iv) 3.5 cm, 3.5 cm, 4.1 cm Give reasons in each case.

Solution:

We know that in a triangle,

Sum of any two sides is greater than the third side

(i) In triangle sides are 5 cm, 7 cm, 8 cm

∵ 5 + 7= 12

and 12 > 8 (third side)

∴ This triangle can be drawn

(ii) Here 2.5 cm + 1 cm = 3.5 cm

and 3.5 cm = 3.5 cm (third side)

∴ It can not be drawn

(iii) 2 + 3 = 5 cm

and 5 cm < 5.6 cm

∴ It cannot be drawn

(iv) 3.5 + 3.5 = 8.0 cm

and 7.0 > 4.1 cm (third side)

∴ This triangle can be drawn

Question 10.

In the figure, XY is a diameter. Prove that XY > XZ.

Solution:

Given : In the figure, XOY is the diameter of the circle XZ is any chord

To prove : XY > XZ

Construction : Join ZX

Proof:

∵ XOY is the diameter of the circle

∴ ∠XZY = 90° (angle in a semicircle)

∵ ∠ZXY + ∠ZYX = 90°

or ∠XZY > ∠XYZ and also > ∠ZXY

∴ XY > XZ

Hence proved.

Question 11.

In the figure, prove that AH < HC and DC > DH.

Solution:

Given: In ∆ABC,

AD ⊥ BC, CE ⊥ AB

∠BAC = 58°, ∠ABC = 67°

AD and CE intersect each other at H To prove :

(i) AH < HC

(ii) DC > DH

Proof:

In ∆AEC,

∠AEC = 90°, ∠CAE = 58°

∴ ∠ACE = 180° – (90° + 58°) = 180° – 148° = 32°

Similarly in ∆ADB,

∠ADB = 90°, ∠ABD = 69°

∴ ∠BAD =180°- (90° + 67°) = 180° – 157° = 23°

∠CAD = 58° – 23° = 35°

Now in ∆AHC,

∠ACE or ∠ACH < ∠CAD or ∠CAH

∴ AH < HC

(ii) In ∆HDC

∠D = 90°

∴ ∠DHC + ∠DCH = 90°

In ∆BCE, ∠E = 90°

∴ ∠BCE = 90° – 67° = 23°

∴ In ∆DHC,

∠DHC = 90° – ∠BCE

= 90° – ∠DCH = 90° – 23° = 67°

∵ ∠DHC > ∠HCD

∴ DC > DH

Hence proved.

Question 12.

PQRS is a convex quadrilateral. Prove that PQ + QR + RS > PS.

Solution:

Given : PQRS is a convex quadrilateral

To prove : PQ + QR + RS > PS

Construction : Join PR

Proof: We know that sum of any two sides of a triangle is greater than its third side

∴ In ∆PQR

PQ + QR > PR … (i)

and in APRS

PR + RS > PS … (ii)

From (i) and (ii)

PQ + QR + RS > PS

Hence proved.

Question 13.

In the figure, M is any point inside the triangle PQR. PM is produced to meet QR in N. Prove that

(i) ∠QMN > ∠QPN

(ii) ∠QMR > ∠QPR

Solution:

Given : In ∆PQR, M is any point inside the triangle. PM is.joined and produced to meet QR in N

To prove :

(i) ∠QMN > ∠QPN

(ii) ∠QMR > ∠QPR

Construction : Join MQ and MR

Proof:

(i) In ∆PQM,

Ext. ∠QMN = ∠MQP + ∠QPM

∴ ∠QMN > ∠QPM or ∠QPN … (i)

(ii) Similarly in ∆PMR,

Ext. ∠RMN > ∠RPM or ∠RPN … (ii)

Adding (i) and (ii)

∠QMN + ∠RMN > ∠QPN + ∠RPN

⇒ ∠QMR > ∠QPR

Hence proved.

Question 14.

In the figure, if ∠1 = ∠2, prove that BA > BD.

Solution:

Given : In the figure, in ∆ABC,

AD is the bisector of ∠A

i.e., ∠1 = ∠2

To prove : BA > BD

Proof: In ∆ADC,

Ext. ∠3 > its interior opposite angle∠2

But ∠2 = ∠1 (AD is bisector of ∠A)

∴ ∠3 > ∠1

Now in ∆ABD

∵ ∠3 > ∠L

∴ AB > BD

(side opposite to greater angle is greater)

or BA > BD

Hence proved.

Question 15.

In figure, ABC is a triangle in which AC > AB and the bisectors of angles B and C intersect each other at O. Prove that OC > OB.

Solution:

Given : In ∆ABC, AC > AB

Bisectors of ∠B and ∠C meet each other at O

To prove : OC > OB

Proof: In ∆ABC,

∵ AC > AB (given)

∴ ∠B > ∠C

(angle opposite to greater side is greater)

⇒ \(\frac { 1 }{ 2 }\) < B > \(\frac { 1 }{ 2 }\) < C ⇒ ∠OBC > ∠OCB

(∵ BO and CO are bisectors of ∠B and ∠C)

∴ OC > OB

(side opposite to greater angle is greater)

Hence proved.

Question 16.

In the figure, If AB = AC, then prove that CD > BD.

Solution:

Given : In figure AB = AC

To prove : CD > BD

Proof: In ∆ABC,

AB = AC (given)

∴ ∠ABC = ∠ACB

Now ∠CBD = ∠ABC + ∠DBA

In ∆DBC,

∴ ∠CBD > ∠ABC

⇒ ∠CBD > ∠ACB (∵ ∠ABC = ∠ACB)

∴ CD > BD

(side opposite to greater angle is greater)

Hence proved.

Question 17.

Diagonals PR and QS of a quadrilateral PQRS intersect each other at O.

Prove that:

(i) PQ + QR + RS + SP > PR + QS

(ii) PQ + QR + RS + SP < 2 (PR + QS)

Solution:

Given : In quadrilateral PQRS, diagonals PR and QS intersect each other at O

To prove:

(i) PQ + QR + RS + SP > PR + QS

(ii) PQ + QR + RS + SP < 2 (PR + QS)

Proof:

In ∆PQR,

PQ + QR > PR … (i)

(sum of any two sides of a triangle is greater than the third side)

Similarly in ∆PRS

RS + SP > PR … (ii)

Adding (i) and (ii)

PQ + QR + RS + SP > PR + PR

⇒ PQ + QR + RS + SP > 2PR … (iii)

Similarly we can proved that

PQ + QR + RS + SP > 2QS (iv)

Adding (iii) and (iv)

2 (PQ + QR + RS + SP) > 2 (PR + QS)

⇒ PQ + QR + RS + SP > PR + QS

(ii) In ∆OPQ,

OP + OQ > PQ … (a)

Similarly in ∆QOR

OQ + OR > QR … (b)

In ∆ROS,

OR + OS > RS …(c)

and in ∆SOP,

OS + OP > SP … (d)

OP + OQ + OQ + OR + OR + OS + OS + OP > PQ + QR + RS + SP

⇒ 2 (OP + OQ + OR + OS) > PQ + QR + RS + SP

⇒ 2 (PR + QS) > PQ + QR + RS + SP

⇒ PQ + QR + RS + SP < 2 (PR + QS)

Hence proved.

Question 18.

Prove that any two sides of a triangle are together greater than the third side.

Solution:

Given: ∆ABC

To prove : AB + AC > BC

AC + BC > AB

BC + AB > AC

Construction: Produce BA to D such that

AD = AC

Join CD

Proof: In ∆ACD,

AC = AD (construction)

∴ ∠ACD = ∠ADC

(opposite angle to equal sides)

∴ ∠BCD > ∠ADC

∴ BD > BC ⇒ BA + AD > BC

⇒ AB + AC > BC (∵ AD = AC)

Similarly we can prove that

AC + BC > AB and BC + AB > AC

Question 19.

Prove that in a triangle, the difference of any two sides is less than the third side.

Solution:

Given : ∆ABC, AC > AB

To prove : AC – AB < BC

Construction : From AC, cut off AD = AB

Join BD

Proof: In ∆ABC, CD = AC – AD = AC – AB

Now AB + BC > AC ⇒ AB + BC > AD + DC

BC > DC (∵ AB = AD)

⇒ BC > AC – AB

Hence proved.