# Question-1

**In ΔPQR, PQ = QR = RP = 7 cm, then find each angle of ΔPQR.**

**Solution:**

In ΔPQR, PQ = PR ...... (Given)

.

^{.}. ∠Q = ∠R ...... (angles opposite to equal sides) ... (i)

Again, QR = RP

... ∠Q = ∠P ...... (ii)

From eqns (i) and (ii), we get

... ∠P = ∠Q = ∠R

But in ΔPQR, we have

∠P + ∠Q + ∠R = 180°

... 3 ∠P = 180° ⇒ ∠P = 60°

Hence ∠P = ∠Q = ∠R = 60°

# Question-2

**In the figure AB = AC and**

**∠ACD = 110° find A.**

**Solution:**

∠ACB + ∠ACD = 180° ...... (Linear pair)

... ÐACB + 110° = 180°

⇒ ∠ACB = 180° - 110° = 70°

Since in ΔABC, AB = AC

ÐB = ∠C = 70°

Now, in ΔABC, we have

∠A + ∠B + ∠C = 180° ...... (Sum angles)

... ÐA + 70° + 70° = 180°

...ÐA = 180° - 140° = 40°

# Question-3

**In the figure, AP and BQ are perpendiculars to PQ and AP = BQ, prove that R is the mid-point of PQ and AB.**

**Solution:**

In ΔAPR and ΔBQR,

AP = BQ ...... (Given)

∠ARP = ∠BRQ ...... (Vertically opposite angles)

∠APR = ∠BQR ...... (each 90°)

ΔAPR ≅ ΔBQR ...... (RHS Criterion)

^{ }∴PR = RQ ..... (c.p.c.t.)

and AR = RB ...... (c.p.c.t.)

Hence R is the mid-point of AB and PQ.

# Question-4

**In the figure,***l***ïï***m*and M is the mid-point of AB. Prove that M is also the mid-point of CD.**Solution:**

Since

*l*ïï

*m*, and AB is the transversal

∴ ÐCAM = ∠DBM ...... (Alternate interior angles)

Now in ΔACM and ΔBDM, we have

∴ ∠CAM = ∠DBM ..... (proved above)

AM = BM ...... (M is the mid-point)

ÐAMC = ∠BMD ............(Vertically opposite angles)

∴ ΔACM ≅ ΔBDM

∴ DM = MC .............(CPCT)

Hence, M is the mid-point of DC.

# Question-5

**ΔABC is an isosceles triangle with AB = AC. Side BA is produced to D such that AB = AD. Prove that ∠BCD = 90°.****Solution:**

In ΔABC, AB = AC

^{ }∴ ÐB = ∠C= ∠4 ......(i)

Since, AB = AC ...... (Given)

and AB = AD ...... (Produced)

∴ AD = AC

Now, in ΔACD, AD = AC

∴ ∠D = ∠C = ∠3 ...... (ii)

Adding eqn (i) and eqn (ii), we get

∠B + ∠D = ∠4 + ∠3

⇒ ∠B + ∠D = ∠BCD

Now in ΔBCD, we have

∠B + ∠BCD + ∠D = 180°

⇒ (∠B + ∠D) + ∠BCD = 180°

⇒ ∠BCD + ∠BCD = 180°

⇒ 2 ∠BCD = 180°

... ÐBCD =

^{ }= 90°

# Question-6

**In****ΔABC, if AB = AC and BE, CF are the bisectors of ∠B and ∠C****respectively****. Prove that ΔEBC ≅ ΔFCB and BE = CF.****Solution:**

Since in ΔABC, AB = AC ∠ABC = ∠ACB ......(i)

Since CF and BE are angle bisectors of ∠C and ∠B,

we get ∠ABE = ∠EBC ...... (ii)

and ∠ACF = ∠FCB ...... (iii)

Now from eqns (i), (ii) and (iii), we get

__1__∠ABC =

__1__∠ACB

2 2

ÐEBC = ∠FCB ...... (iv)

Now in ΔFBC and ΔECB,

we have ∠FBC = ∠ECB ...... (∠B = ∠C)

BC = BC ...... (Common)

ÐFCB = ∠EBC ...... [From (iv)]

ΔEBC ≅ ΔFCB

BE = CF ...... (c.p.c.t.)

# Question-7

**In the figure AC = BC, ∠DCA = ∠ECB and ∠DBC = ∠EAC. Prove that triangles DBC and EAC are congruent and hence DC = EC.**

**Solution:**

In ΔACE and ΔBCD, we have

AC = BC ...... (given)

ÐEAC = ∠DBC ...... (given)

∠DCA = ∠ECB ...... (given)

∠DCA + ∠DCE = ∠ECB + ∠DCE ........(Adding ∠DCE on on both sides)

⇒ ∠ACE = ∠BCD

ΔACE ≅ ΔBCD ...... (ASA criterion)

DC = EC ...... (c.p.c.t.)

# Question-8

**Prove that the medians of an equilateral triangle are equal.**

**Solution:**

Let ABC be an equilateral triangle and let AD, BE and CF be the medians.

Since ΔABC is an equilateral triangle ∠A = ∠B = ∠C = 60°

Now in ΔADC and ΔABE,

we have BC = AC

DC = AE

∠C = ∠A ...... (each 60°)

AC = AB

ΔADC ≅ ΔABE

AD = BE ...... (c.p.c.t.)

Similarly, we can prove that BE = CF

AD = BE = CF

Hence medians of equilateral triangle are equal.

# Question-9

**In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that**

(i) PT = QT

(ii)(i) PT = QT

(ii)

**∠TQR = 15°**

**Solution:**

Since PQRS is a square

∠PSR = ∠QRS ...... (each 90°)

Now again in equilateral ΔSRT,

we have ∠TSR = ∠TRS ...... (each 60°)

∠PSR + ∠TSR = ∠QRS + ∠TRS

⇒ ∠TSP = ∠TRQ

Now in ΔTSP and ΔTRQ, we have

TS = TR ...... (sides of equilateral triangle)

∠TSP = ∠TRQ ...... (Proved above)

PS = QR ...... (sides of a square)

ΔTSP ≅ ΔTRQ

∴ PT = QT

Now in ΔTQR, we have

TR = QR ...... (QR = RS = TR)

∠TQR = ∠QTR and ∠TQR + ∠QTR + ∠TRQ = 180°

∠TQR + ∠QTR + ∠TRS + ∠SRQ = 180°

2 ∠TQR + 60° + 90° = 180°...........(∵∠TQR = ∠QTR)

2 ∠TQR = 180° - 150° = 30°

∠TQR =

__30__= 15°

2

# Question-10

**In the figure x is a point in the interior of square ABCD, AXYZ is also a square. Prove that BX = DZ.****Solution:**

Since ABCD and AXYZ both are squares

∠AZY = ∠AXY

= ∠AXB ... (each 90°) and AX = AZ and AB = AD

Now in ΔABX and ΔADZ,

∠AZD = ∠AXB ...... (each 90°)

AZ = AX ...... (sides of a square)

AB = AD ...... (sides of a square)

ΔABX ≅ ΔADZ

BX = DZ ....... (c.p.c.t.)

# Question-11

**In the figure, the sides AB and BC of square ABCD are produced to P and Q respectively so that BP = CQ. Prove that DP and AQ are perpendicular to each other.**

**Solution:**

Since ABCD is a square, AB = BC

Also BP = CQ ...... (given)

.

^{.}. AB + BP = BC + CQ

.

^{.}. AP = BQ

Now in ΔAPD and ΔBQA,

AP = BQ ...... (proved above)

∠ABQ = ∠DAP ... (each 90°)

and AB = AD

.

^{.}. ΔAPD ≅ ΔBQA

.

^{.}. ∠APD = ∠BQA ...... (c.p.c.t.)

and ∠ADP = ∠QAP ...... (c.p.c.t.)

Also ∠DAQ = ∠AQB

.

^{.}. ∠DAO = ∠APO

Now in ΔAOD and ΔAOP,

∠ADO = ∠OAP,

and ∠DAO = ∠APO

.

^{.}. 3rd ∠DOA = 3rd ∠AOP [Since, two angles of DAOD and DAOP are equal, the third angle is also equal]

but ∠DOA + ∠AOP = 180°

.

^{.}. 2 ∠DOA = 180°

.

^{.}. ∠DOA = 90°

.

^{.}. DO is perpendicular to AO or DP is perpendicular to AQ.

# Question-12

**In the figure,**

**∠B = ∠C and AB = AC. Prove that BD = CE.**

**Solution:**

In ΔABD and ΔACE,

AB = AC ...... (given)

∠B = ∠C ...... (given)

∠A = ∠A ...... (common)

.

^{.}. ΔABD ≅ ΔACE ...... (By ASA criterion)

.

^{.}. BD = CE ...... (c.p.c.t.)

# Question-13

**In the figure, AD = BE, BC = DF and****∠ABC = ∠EDF. Prove that AC ïï EF and AC = EF.****Solution:**

Since AD = BE

.

^{.}. AD + DB = BE + DB

⇒ AB = DE

Now in ΔABC and ΔEDF

AB = DE ...... (proved above)

BC = DF ...... (given)

and ∠ABC = ∠EDF ...... (given)

.

^{.}. ΔABC ≅ ΔEDF

.

^{.}. AC = EF ...... (c.p.c.t.)

and ∠BAC = ∠DEF

but these are alternate interior angles of AC and EF with transversal AE

.

^{.}. AC ïï EF

# Question-14

Given ABCD is a parallelogram, BC is produced to F and BD is produced to E and AE=CF Prove that:

(i) BE || DF

(ii) BD and EF bisect each other.

**Solution:**

To Prove: BE ïï DF

and EO = OF

and DO = OB

Proof: In ΔABE and ΔCDF

AB = DC ...... (opposite sides of parallelogram)

AE = CF ...... (given)

and BE = DF ...... (given)

... ΔABE ≅ ΔCDF

... ∠ABE = ∠CDF ...... (i) (c.p.c.t.)

Now since AB ïï DC and DB is a transversal

... ∠ABD = ∠BDC ...... (ii) (Alternate interior angles)

Adding equations (i) and (ii), we get

∠ABE + ∠ABD = ∠CDF + ∠BDC

⇒ ∠EBD = ∠BDF ⇒ BE ïï DF

Now in ΔOBE and ΔODF,

BE = DF ...... (given)

∠EBO = ∠FDO ...... (proved above)

and ∠BOE = ∠DOF ...... (Vertically opposite angles)

.^{.}.ΔOBE ≅ ΔODF

.^{.}. OE = OF

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

Therefore O is the mid-point of EF and DB.

# Question-15

O is a point in the interior of a rhombus ABCD. If OA = OC then prove that DOB is a straight line.

**Solution:**

**Given**: A rhombus ABCD and a point O in it such that OA = OC

**To Prove**: DOB is a straight line.

**Construction**: Join OB and OD

**Proof**: In ΔAOD and ΔCOD

AO = CO ...... (given)

OD = OD ...... (common)

AD = CD ...... (sides of a rhombus)

... ΔAOD ≅ ΔCOD

.

^{.}.∠AOD = ∠COD ...... (i)

Similarly ΔAOB ≅ ΔCOB

... ∠AOB = ∠COB ...... (ii)

Adding eqns (i) and (ii), we get

∠AOD + ∠AOB = ∠COD + ∠COB

But ∠AOD + ∠AOB + ∠COD + ∠COB = 360° ...... (Angles at a point)

... ∠AOD + ∠AOB + ∠AOD + ∠AOB = 360°

⇒ 2 (∠AOD + ∠AOB) = 360°

⇒ ∠AOD + ∠AOB = 180° but it is a linear pair

.

^{.}. OD and OB are in a line

Hence DOB is a straight line.

# Question-16

**In the figure PQ > PR, QM and RM are the bisectors of****∠Q and ∠R respectively. Prove that QM > RM.**

**Solution:**

Since QM and RM are angle bisectors of ∠Q and ∠R.

**∴**∠1 = ∠2 and ∠3 = ∠4

Now in ΔPQR, PQ > PR

*.*∠PRQ > ∠PQR ...... (Angles opposite to larger side)

^{.}.*.*∠PRM + ∠MRQ > ∠PQM + ∠MQR

^{.}.*.*∠4 + ∠3 > ∠1 + ∠2

^{.}.Þ 2 ∠3 > 2 ∠2

⇒ ∠3 > ∠2

*.*QM > RM.

^{.}.# Question-17

**In the figure, PQ = PR, show that PS > PQ.****Solution:**

In ΔPQR, PQ = PR

*⇒*∠PQR = ∠PRQ ..(i)

Now in ΔPSQ, ∠PQR is the exterior angle

*⇒*

*∠PQR = ∠PSQ + ∠SPQ*

*⇒*

*∠PQR > ∠PSQ*

*⇒*

*∠PRQ > ∠PSQ ...... (∠PQR = ∠PRQ)*

*⇒*

*PS > PR ...... (side opposite to greater angle)*

*.*PS > PQ ...... (PR = PQ)

^{.}.# Question-18

**In****ΔABC, AC > AB and AD is the bisector of ∠A show that y > x.****Solution:**

In ΔABC, since AC > AB

∴

*∠B > ∠C ...... (i)*

Also ∠1 = ∠2 ...... [given]

Now, in ΔABD, we have

∠1 + ∠B + ∠x = 180° ...... (ii)

And in ΔADC, we have

∠2 + ∠C + ∠y = 180° ...... (iii)

Comparing eqns (i) and (ii), we get

∠1 + ∠B + ∠x = ∠2 + ∠C + ∠y

∠B + ∠x = ∠C + ∠y

Now since ∠B > ∠C ...... [proved above]

*.*∠x < ∠y

^{.}.or ∠y > ∠x

# Question-19

**In****ΔPQR, PS ⊥ QR and SR > SQ, show that PR > PQ.****Solution:**

PR > PQ

**Construction**: Draw PT such that SQ = ST

**Proof**: In ΔPQS and ΔPTS,

PS = PS ...... [common]

SQ = ST ......[By construction]

∠PSQ = ∠PST ...... [each 90°]

*.*ΔPQS ≅ ΔPTS ...... [SAS criterion]

^{.}.*.*∠1 = ∠2 ...... [c.p.c.t.]

^{.}.Now in ΔPRT, we have the exterior angle ∠2

*.*∠2 = ∠3 + ∠TPR

^{.}.or ∠2 > ∠3

*.*∠1 > ∠3 ...... (∠1 = ∠2)

^{.}.Now in ΔPQR, since

∠1 > ∠3

*.*PR > PQ ...... [side opposite to greater angle]

^{.}.# Question-20

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

**Solution:**

**To Prove**: AC - AB < BC,

BC - AB < AC

and AC - BC < AB

**Construction**: Mark a point D on AC such that

AB = AD and join BD.

**Proof**:

In ΔABD, AB = AD

*⇒*∠1 = ∠2 ...... (i)

Now in ΔABD, AD is produced to C

*⇒*∠3 = ∠1 + ∠A ...... (Exterior angle)

*⇒*

*∠3 > ∠1 ...... (ii)*

Now in ΔBCD, CD is produced to A

*⇒*∠2 = ∠4 + C

*⇒*

*∠2 > ∠4 ...... (iii)*

From (i), (ii), and (iii), we get

∠3 > ∠4

*.*BC > CD

^{.}.or BC > AC - AD

or BC > AC - AB

Similarly we can prove other two inequalities

BC - AB < AC and AC - BC < AB.

# Question-21

**Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle.****Solution:**

ΔABC, in which altitudes AQ, BR and CP meet at M.

**To Prove:**

AB + BC + CA > BR + CP + AQ

**I Proof:**

In ΔABQ, we have ∠AQB = 90°

.

^{.}. ∠ABQ will be a acute angle

.

^{.}.∠AQB > ∠ABQ

.

^{.}.AB > AQ ...... (i)

Similarly, we can prove that:

BC > BR ...... (ii)

and CA > CP ...... (iii)

Adding three inequalities, we get

AB + BC + CA > AQ + BR + CP

**II Proof:**

AB > AQ, AC > AQ ...... [

^{ .}.

^{. }Perpendicular is shortest]

.

^{.}. AB + AC > 2AQ ...... (i)

Similarly BC + CA > 2PC ...... (ii)

and AB + BC > 2BR

Adding (i), (ii) and (iii), we get

2AB + 2BC + 2CA > 2AQ + 2BR + 2CP

or AB + BC + CA > AQ + BR + CP

# Question-22

**In a quadrilateral PQRS, the diagonals PR and QS intersect each other at O. Show that**

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

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

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

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

**Solution:**

We know that in a triangle the sum of two sides of a triangle is greater than the third side. Therefore in ΔPSR, we have

PS + SR > PR ...... (i)

In ΔQSR, QR + RS > QS ...... (ii)

In ΔPQS, PQ + SP > QS ...... (iii)

In ΔPQR,PQ + QR > PR ..... (iv)

Adding (i), (ii), (iii) and (iv), we get

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

Similarly in ΔPQO, ΔQRO, ΔRSO and ΔSPO,

we have

QO + PO > PQ ...... (v)

QO + RO > QR ...... (vi)

RO + SO > RS ...... (vii)

SO + PO > SP ...... (viii)

Adding (v), (vi), (vii) and (viii),

we get

2 (QO + SO + PO + OR) > PQ + QR + RS +SP 2 (QS + PR) > PQ + QR + RS + SP

or PQ + QR + RS + SP < 2 (QS + PR).