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Page No 282:

Question 1:

In the given figure, AB || CD and O is the midpoint of AD.
Show that
(i) ΔAOB ≅ ΔDOC.
(ii) O is the midpoint of BC.

Answer:



Given: In the given figure, AB || CD and O is the midpoint of AD.

To prove:
(i) ΔAOB ≅ ΔDOC.
(ii) O is the midpoint of BC.

Proof:
(i) In ΔAOB and ΔDOC,
BAO = ∠CDO                    (Alternate interior angles, AB || CD)
AO = DO                              (Given, O is the midpoint of AD)
AOB = ∠DOC                   (Vertically opposite angles)

∴ By ASA congruence criteria,
ΔAOB ≅ ΔDOC

(ii) ∵ ΔAOB ≅ ΔDOC           [From (i)]
BO = CO                           (CPCT)
Hence, O is the midpoint of BC.

Page No 282:

Question 2:

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Answer:



Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.

To prove: CD bisects AB

Proof:
In ΔAOD and ΔBOC,
DAO = ∠CBO = 90°           (Given)
AD = BC                              (Given)
DOA = ∠COB                    (Vertically opposite angles)

∴ By AAS congruence criteria,
ΔAOD ≅ ΔBOC
∴ AO = BO                           (CPCT)
Hence, 
CD bisects AB.

Page No 282:

Question 3:

In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
Show that ΔABC ≅ ΔCDA.

Answer:



Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.

To prove: ΔABC ≅ ΔCDA

Proof:
In ΔABC and ΔCDA,

BACDCA           (Alternate interior angles, pq)
BCADAC           (Alternate interior angles, lm)
AC = CA                        (Common side)

 By ASA congruence criteria,
ΔABC ≅ ΔCDA



Page No 283:

Question 4:

AD is an altitude of an isosceles ΔABC in which AB = AC.
Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Answer:



Given: AD is an altitude of an isosceles ΔABC in which AB AC.

To prove: (i) AD bisects BC, (ii) AD bisects ∠A

Proof:
(i) In ΔABD and ΔACD,

ADB = ADC = 90°          (Given, ADBC)
AB = AC                                   (Given)
AD = AD                                  (Common side)

 By RHS congruence criteria,
ΔABD ≅ ΔACD

BD = CD       (CPCT)

Hence, AD bisects BC.

(ii)
 ΔABD ≅ ΔACD       [From (i)]
 BAD = CAD      (CPCT)
Hence, AD bisects ∠A.

Page No 283:

Question 5:

In the given figure, BE and CF are two equal altitudes of ΔABC.
Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Answer:

In ABE and ACF, we have:BE =CF     (Given)BEA=CFA=90°A=A  (Common)ABEACF   (AAS criterion)
AB = AC (CPCT)

Page No 283:

Question 6:

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.

Answer:



Given: ΔABC and ΔDBC are two isosceles triangles on the same base BC.

To prove:
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC

Proof:
(i) In ΔABD and ΔACD,

BD CD                                  (Given, ΔDBC is an isosceles triangles)
AB = AC                                   (Given, ΔABC is an isosceles triangles)
AD = AD                                  (Common side)

 By SSS congruence criteria,
ΔABD ≅ ΔACD

Also, BAD = CAD         (CPCT)
or, BAE = CAE     .....(1)

(ii)
In ΔABE and ΔACE,

AB = AC                                   (Given, ΔABC is an isosceles triangles)
BAE = CAE                       [From (i)]
AE = AE                                  (Common side)

 By SAS congruence criteria,
ΔABE ≅ ΔACE

Also, BE = CE                    (CPCT)    .....(2)
And, AEB = AEC         (CPCT)   .....(3)

(iii)
In ΔBED and ΔCED,

BD CD                                  (Given, ΔDBC is an isosceles triangles)
BECE                                   [From (2)]
DE = DE                                  (Common side)

 By SSS congruence criteria,
ΔBED ≅ ΔCED

Also, BDE = CDE         (CPCT)   .....(4)

(iv)
 BAE = CAE        [From (1)]
And, BDE = CDE   [From (4)]
 AE bisects ∠A as well as ∠D.

(v)
AEB+AEC=180°      Linear pairAEB+AEB=180°      From 32AEB=180°AEB=180°2AEB=90°          .....5

From (2) and (5), we get
AE is the perpendicular bisector of BC.

Page No 283:

Question 7:

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Answer:

Consider the triangles AEB and CDB.

EBA=DBC  (Common angle)  ...(i)

Further, we have:

BEA=180-yBDC=180-xSince x = y, we have:180-x =180-y BEA=BDC    ...(ii)

AB = CB     (Given) ...(iii)
From (i), (ii) and (iii), we have:
BDCBEA   (AAS criterion)
∴ AE = CD (CPCT)
Hence, proved.

Page No 283:

Question 8:

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Answer:



Given: In the given figure, ∠BAQ = ∠BAPBPAP and BQAQ.

To prove:
(i) ΔAPB ≅ ΔAQB
(ii) BP BQ, i.e., B is equidistant from the arms of ∠A.

Proof:
(i) In ΔAPB and ΔAQB,

BAQ = ∠BAP,                   (Given)
APB = ∠AQB = 90°          (Given, BPAP and BQAQ)
AB = AB                               (Common side)

 By AAS congruence criteria,
ΔAPB ≅ ΔAQB

(ii)
 ΔAPB ≅ ΔAQB            [From (i)]
BP = BQ                       (CPCT)
Hence, is equidistant from the arms of ∠A.

Page No 283:

Question 9:

ABCD is a quadrilateral such that diagonal AC bisects the angles ∠A and ∠C. Prove that AB = AD and CB = CD.

Answer:



Given: In quadrilateral ABCDAC bisects the angles ∠A and ∠C.

To prove: AB AD and CB CD

Proof:
In ABC and ADC,

BACDAC            (Given, AC bisects the angles ∠A)
AC = AC                        (Common side)
BCADCA            (Given, AC bisects the angles ∠C)

 By ASA congruence criteria,
ABC ADC

Hence, AB AD and CB CD.     (CPCT)

Page No 283:

Question 10:

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersects the side AB at D. Prove that AC + AD = BC.

Answer:


Given: In right triangle ΔABC, ∠BAC = 90°AB AC and ∠ACD = ∠BCD

To prove: AC AD BC

Proof:
Let AB = AC = x and AD = y.

In ABC,

BC2=AB2+AC2BC2=x2+x2BC2=2x2BC=2x2BC=x2

Now,

BDAD=BCAC                  (An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.)
x-yy=x2xxy-yy=2xy-1=2xy=2+1y=x2+1
y=x2+1×2-12-1y=x2-122-12y=x2-x2-1y=x2-xx+y=x2

Hence, AC + AD = BC.



Page No 284:

Question 11:

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.

Answer:

Proof:
In OQA and OPB, we have:OQ=OP        (Given)OA=OB        (Given)AOQ=BOP        (Common)OQAOPB       (SAS criterion)
OAQ=OBP        (Corresponding angles of congruent triangles)
Now, consider triangles BQX and APX.
Given:OA=OBOP=OQ OA-OP=OB-OQAP=BQ
Further, BXQ=AXP     (Vertically opposite angles)
Also, we have proven that QBX=PAX.
BQXAPX   (AAS criterion)

 PX=QX     (corresponding sides of congruent triangles)
Also, AX=BX    (corresponding sides of congruent triangles)    
Hence, proved.

Page No 284:

Question 12:

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Answer:

Since ABC is an equilateral , thenABC = BCA = CAB = 60°Since PQCA and PC is a transversal, thenBPQ=BCA=60°   Corresponding anglesSince PQCA and QA is a transversal, thenBQP=BAC=60°  Corresponding anglesFurther, B=60°In BPQ,B = BPQ = BQP = 60°BPQis an equilateral triangle.i.e., BP=PQ=BQNow,  BP=CRPQ=CR     ....1Considering MPQ and MCR, we get:PQM=MRC     Alternate interior anglesPMQ=CMR     Vertically opposite angles   PQ=CR        using 1MPQMCR AAS criterionMP=MC    Corresponding parts of congruent triangles are equalQR bisects PC.

Page No 284:

Question 13:

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Answer:

In ABP and QCP, we have:BPA=CPQ   (Vertically opposite angle)PAB=PQC    (Alternate angles)PB = PC     (P is the midpoint)ABPQCP   (AAS criterion)

∴ AB = CQ            (CPCT)

DQ=DC+CQ DQ=DC+AB    (Proved, AB = CQ)
Hence, proved.

Page No 284:

Question 14:

In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Answer:

In PAD and PAB, we have:AD=AB   (Side of a square)AP=AP    (Common)PD =PB    (Given)PADPAB   (SSS criterion)APD=APB

In CPD and CPB, we have:CD = CB (Sides of square)CP=CP   (Common)PD=PB   (Given)CPDCPB   (SSS test)CPD=CPB

APD+CPD=APB+CPBButAPD+CPD+APB+CPB=360°APD+CPD=180°

So, CPA is a straight line.
Hence, proved.

Page No 284:

Question 15:

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Answer:



Given: In square ABCD, ΔOAB is an equilateral triangle.

To prove: ΔOCD is an isosceles triangle.

Proof:

DAB=CBA=90°        Angles of square ABCDAnd, OAB=OBA=60°       Angles of equilateral OABDAB-OAB=CBA-OBA=90°-60°OAD=OBC=30°         .....i

Now, in ΔDAO and ΔCBO,

AD = BC                   (Sides of square ABCD)
DAOCBO      [From (i)]
AO = BO                  (Sides of equilateral ΔOAB)

 By SAS congruence criteria,
ΔDAO  ΔCBO

So, OD = OC         (CPCT)
Hence, ΔOCD is an isosceles triangle.

Page No 284:

Question 16:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ABC such that AX = AY. Prove that CX = BY.

Answer:

In AXC and AYB, we have:AC=AB  (Given)AX=AY  (Given)BAC=CAB   (Angle common to AXC and AYB)AXCAYB    (SAS criterion)So, CX=BY   (CPCT)
Hence, proved.



Page No 285:

Question 17:

In ABC, D is the midpoint of BC. If DLAB and DMAC such that DL = DM, prove that AB = AC.

Answer:

In BDL and CDM, we have:BD=CD    (D is midpoint)DL=DM    (Given)BLD=CMD    (90° each)Thus,BDLCDM  (RHS criterion)
BL=MC (CPCT)
B=C
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

Page No 285:

Question 18:

In ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠A.

Answer:

In triangle ABC, we have:
AB = AC   (Given)
B=C12B=12COBC=OCBBO=CO

Now, in AOB and AOC, we have:OB=OC   (Proved)AB=AC    (Given)AO=AO    (Common)AOBAOC   (SSS criterion)
i.e., BAO=CAO (Corresponding angles of congruent triangles)

So, it shows that ray AO is the bisector of A.
Hence, proved.

Page No 285:

Question 19:

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Answer:


Given: In trapezium ABCDand are mid-points of AB and DC, MNAB and MNDC.

To prove: AD = BC

Construction: Join CM and DM.

Proof:

In ΔCMN and ΔDMN,

MN = MN                           (Common sides)
CNM = DNM = 90°     (Given, MNDC)
CN = DN                             (Given, N is the mid-point DC)

 By SAS congruence criteria,
ΔCMN  ΔDMN

So, CM = DM                  (CPCT)    .....(i)
And, CMNDMN    (CPCT)
But, AMNBMN = 90°      (Given, MNAB)
AMN - CMNBMN - DMN
AMD = BMC           .....(ii)

Now, in ΔAMD and ΔBMC,

DM = CM                            [From (i)]
AMD = BMC                [From (ii)]
AM = BM                            (Given, M is the mid-point AB)

 By SAS congruence criteria,
ΔAMD  ΔBMC

Hence, AD = BC    (CPCT)

Page No 285:

Question 20:

The bisectors of B and C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that MOC = ABC.

Answer:



Given: In isosceles ABC, AB = AC; OB and OC are bisectors of B and C, respectively.

To prove: MOC = ABC

Proof:

In ∆ABC,

AB = AC             (Given)
ABC = ∠ACB   (Angles opposite to equal sides are equal)
12ABC = 12ACB
OBC = ∠OCB         (Given, OB and OC are the bisectors of ∠B and ∠C, respectively)    .....(i)

Now, in ∆OBC, ∠MOC is an exterior angle
MOC = ∠OBC + ∠OCB    (An exterior angle is equal to the sum of two opposite interior angles)
MOC = ∠OBC + ∠OBC    [From (i)]
MOC = 2∠OBC
Hence, ∠MOC = ∠ABC    (Given, OB is the bisector of ∠B)

Page No 285:

Question 21:

The bisectors of B and C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ABC is equal to BOC.

Answer:



Given: In an isosceles ΔABC, AB = AC, BO and CO are the bisectors of 
ABC and ∠ACB, respectively.

To prove: ∠ABD = ∠BOC


Construction: Produce CB to point D.

Proof:

In ΔABC,

 AB = AC                    (Given)
 ACB = ABC            (Angle opposite to equal sides are equal)

12ACB=12ABCOCB=OBC           .....iGiven, BO and CO are angle bisector of ABC and ACB, respectively

In ΔBOC,

OBC+OCB+BOC=180°        By angle sum property of triangleOBC+OBC+BOC=180°    From i2OBC+BOC=180°ABC+BOC=180°           BO is the angle bisector of ABC      .....ii

Also, DBC is a straight line.
So, ABC+DBA=180°               Linear pair        .....iii

From (ii) and (iii), we get
ABC+BOC=ABC+DBA BOC=DBA

Page No 285:

Question 22:

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.

Answer:



Given: BP is the bisector of ∠ABC, and BAQP

To prove: ΔBPQ is an isosceles triangle

Proof:

1=2           Given, BP is the bisector of ABCAnd, 1=3      Alternate interior angles2=3So, PQ=BQ         In a triangle, sides opposite to equal sides are equal.

But these are sides of BPQ.

Hence, BPQ is an isosceles triangle.

Page No 285:

Question 23:

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Answer:


Given: An object is placed at a point A, the image of the object is seen at the point B, an observer is at point D, and LM is a plane mirror.

To Prove: The image is as far behind the mirror as the object is in front of the mirror, i.e. BT = AT.

Proof:

LM is a plane mirror

 i = r              (Angle of incidence is always equal to angle of reflection)    .....(1)

Also, ABCN                            (Both AB and CN are perpendicular to LM)

TAC=ACN=i                (Alternate interior angles)      .....(2)

And, CBT=NCD=r            (Corresponding angles)         .....(3)

From (1), (2) and (3), we get

TAC = CBT        .....(4)

Now,

In TAC and CBT,TAC=CBT               From 4ATC=BTC=90°CT=CT                       Common side By AAS congruence criteria,TACCBT

Hence, AT = BT    (CPCT)

Page No 285:

Question 24:

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Answer:

Let AB be the breadth of the river.
M is any point situated on the bank of the river.
Let O be the mid point of BM.
​Moving along perpendicular to point such that A,O and N are in a straight line.
Then MN is the required breadth of the river.

In OBA and OMN, we have:OB=OM     (O is midpoint)OBA=OMN    (Each 90°)AOB=NOM    (Vertically opposite angle)OBAOMN    (ASA criterion)

Thus, MN = AB (CPCT)
If MN is known, one can measure the width of the river without actually crossing it.

Page No 285:

Question 25:

In a ΔABC, D is the midpoint of side AC such that BD = 12AC. Show that ∠ABC is a right angle.

Answer:



Given: In ABCD is the midpoint of side AC such that BD 12AC.

To prove: ∠ABC is a right angle.

Proof:

In ADB,AD=BD              Given, BD = 12ACDAB=DBA=x Let            Angles opposite to equal sides are equal

Similarly, in DCB,BD=CD               GivenDBC=DCB=y  Let

In ABC,

ABC+BCA+CAB=180°   Angle sum propertyx+x+y+y=180°2(x+y)=180°x+y=90°ABC=90°

Hence, ∠ABC is a right angle.



Page No 286:

Question 26:

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.”
Is the statement true? Why?

Answer:

No, the statement is not true because the two triangles are congruent only by SAS congruence condition but the statement contains ASS or SSA condition as well which are not any condition for congruence of triangles.

Page No 286:

Question 27:

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?

Answer:

Yes, the statement is true because the two triangles can be congruent by either AAS or ASA congruence criteria.

Disclaimer: If corresponding angles of two triangles are equal, then the by angle sum property, the third corresponding angle will be equal. So, if we have two corresponding angles and a corresponding side are equal, then the triangles can be proved congruent by SAS congruence criteria. Therefore, ASA and SAS criteria are treated as same.



Page No 296:

Question 1:

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
(i) 5 cm, 4 cm, 9 cm
(ii) 8 cm, 7 cm, 4 cm
(iii) 10 cm, 5 cm, 6 cm
(iv) 2.5 cm, 5 cm, 7 cm
(v) 3 cm, 4 cm, 8 cm

Answer:

(i) No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9

(ii) Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7

(iii) Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6

(iv) Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5

(v) No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8

Page No 296:

Question 2:

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

Answer:

Given: In ΔABC, ∠A = 50° and ∠B = 60°

In ΔABC,
A + ∠B + ∠C = 180°           (Angle sum property of a triangle)
50° + 60° + ∠C = 180°
110° + ∠C = 180°
C = 180° - 110°
C = 70°

Hence, the longest side will be opposite to the largest angle (∠C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle (∠A = 50° ) i.e. BC.

Page No 296:

Question 3:

(i) In ABC, A = 90°. Which is its longest side?
(ii)
In ABC, A = ∠B = 45°. Which is its longest side?
(iii) In ABC, A = 100° and ∠C = 50°. Which is its shortest side?

Answer:

(i) Given: In ABCA = 90°

So, sum of the other two angles in triangle ∠B + ∠C = 90°

i.e. ∠B, ∠C < 90°

Since, ∠A is the greatest angle.

So, the longest side is BC.

(ii) Given: ∠A = ∠B = 45°

Using angle sum property of triangle,

C = 90°

Since, ∠C is the greatest angle.

So, the longest side is AB.

(iii) Given: ∠A = 100° and ∠C = 50°

Using angle sum property of triangle,

B = 30°

Since, ∠A is the greatest angle.

So, the shortest side is BC.



Page No 297:

Question 4:

In ABC, side AB is produced to D such that BD = BC. If B = 60° and ∠A = 70°, prove that (i) AD > CD and (ii) AD > AC.

Answer:

In triangle CBA, CBD is an exterior angle.

i.e., CBA+CBD=180°60°+CBD=180°CBD=120°

Triangle BCD is isosceles and BC = BD.
Let BCD=BDC = x°.
In CBD, we have:BCD+CBD+CDB=180°x+120°+x=1802x=60°x=30°BCD=BDC=30°

In triangle ADC, C=ACB + BCD = 50°+30°=80°
A=70°and D=30°

C>AAD>CD    ...(1)

Also, C>DAD>AC   ...(2)

Page No 297:

Question 5:

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.

Answer:


Given: ∠B < ∠and ∠C < ∠D

To prove: AD > BC

Proof: 

In AOB,B<AAO<BO       Side opposite to the greater angle is longer  .....1

In COD,C<DOD<OC       Side opposite to the greater angle is longer  .....2

Adding (1) and (2), we getAO+OD<BO+OCAD<BC

Page No 297:

Question 6:

AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ∠A > ∠C and
B > ∠D.

Answer:

Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.

To prove:
​(i) ∠A > ∠C
(ii) ∠> ∠D


Construction: Join AC.

Proof:

In ABC, BC>AB           Given, AB is the smallest side 1>2            ...1

In ADC, CD>AD           Given, CD is the largest side 3>4            ...2

Adding 1 and 2, we get1+3>2+4 A>C


(ii)

Construction: Join BD.

Proof:

In ABD, AD>AB                  Given, AB is the smallest side. 5>6                 ...3

In CBD, CD>BC                  Given, CD is the greatest side. 7>8                 ...4

Adding 3 and 4, we get5+7>6+8B>D

Page No 297:

Question 7:

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Answer:

Given: Quadrilateral ABCD

To prove: (AB + BC + CD + DA) > (AC + BD)

Proof:

In ABC,AB+BC>AC         ...iIn CAD,CD+AD>AC        ...iiIn BAD,AB+AD>BD        ...iiiIn BCD,BC+CD>BD        ...iv

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

2(AB + BC + CD + DA) < 2( AC + BD)

Hence, (AB + BC + CD + DA) < (AC + BD).

Page No 297:

Question 8:

In a quadrilateral ABCD, show that
AB+BC+CD+DA<2BD+AC

Answer:

Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof:

In ∆AOB, 
OA+OB>AB      .....i
In ∆BOC, 
OB+OC>BC      .....ii
In ∆COD, 
OC+OD>CD     .....iii
In ∆AOD, 
OD+OA>AD     .....iv
Adding i,ii,iii and iv, we get
     2OA+OB+OC+OD>AB+BC+CD+DA 2OB+OD+OA+OC>AB+BC+CD+DA2BD+AC>AB+BC+CD+DA

Page No 297:

Question 9:

In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

Answer:



Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

In ABX, BAX>ABX BX>AX                 ...i

Similarly, in ACX, ACX>XAC AX>CX                 ...ii

From i and ii, we getBX>AX>CX

Page No 297:

Question 10:

In the given figure, PQ > PR and QS and RS are the bisectors of Q and ∠R respectively. Show that SQ > SR.

Answer:

Since the angle opposite to the longer side is greater, we have:

PQ>PRR>Q12R>12QSRQ>RQSQS>SR

SQ>SR

Page No 297:

Question 11:

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.

Answer:


Given: In ABC, AB = AC

To prove: CD BD

Proof:

In ABC,

Since, AB = AC       (Given)

So, ABC=ACB      ...(i)

In ABC and DBC, ABC>DBC ACB> DBC         From i BD>CD               Side opposite to greater angle is longer. CD<BD

Page No 297:

Question 12:

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 23 of a right angle.

Answer:


Given: In ABC, BC is the longest side.

To prove: BAC > 23 of a right angle, i.e., BAC > 60°

Construct: Mark a point D on side AC such that AD = AB = BD.

Proof:

In ABD,

 AD = AB = BD    (By construction)

1=3=4=60°

Now,BAC=1+2=60°+2but 60° =23 of a right angleSo, BAC=23 of a right angle + 2

Hence, BAC > 23 of a right angle.



Page No 298:

Question 13:

In the given figure, prove that
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC.

Answer:



Given: Quadrilateral ABCD

To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC

Proof:
(i)
In ACD,CD+DA>CA        ...1In ABC,AB+CA>BC         ...2
Adding (1) and (2), we getCD+DA+AB+CA>CA+BC AB+CD+DA>BC

(ii)
 In CDA,CD+DA>CA          ...3In BCA,BC + AB > CA       ...4
Adding (3) and (4), we getCD+AD+BC+AB>CA+CA CD+AD+BC+AB>2CA

Page No 298:

Question 14:

If O is a point within ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA+OB+OC>12(AB+BC+CA)

Answer:


Given:
In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.​
AB>OB      ...(1)AC>OC       ...(2)Thus, AB+AC>OB+OC      ...[Adding (1) and(2)]

(ii) AB>OA......(3)BC>OB.....(4)CA>OC.....(5)
Adding the above three equations, we get:
Thus, AB+BC+CA>OA+OB+OC              ...(6)

OA cannot be longer than both AB and CA.​
AB>OB.....(5)AC>OC.....(6)AB+AC>OB+OC..........[On adding (5) and (6)]
Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:
 OA+OC>ACOA+OB>ABOB+OC>BCAdding the above three equations, we get:OA+OC+OA+OB+OB+OC>AB+AC+BC2OA+OB+OC>AB+AC+BCThus, OA+OB+OC>12AB+BC+CA

Page No 298:

Question 15:

In the given figure, AD BC and CD > BD. Show that AC > AB.

Answer:



Given: AD ⊥ BC and CD BD

To prove: AC AB

Proof:

ADB=ADC=90°       ADBC      ...1BAD<DAC                  CD>BD      ...2

In ABD,

Using angle sum property of a triangle,

B=180°-ADB-BADB=90°-BAD         ...3

In ADC,

Using angle sum property of a triangle,

ACD=90°-DAC      ...4

From (2), (3) and (4), we get

B>C

Therefore, AC>AB.

Page No 298:

Question 16:

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Answer:



Given: CD DE

To prove: AB AC BE

Proof:

In ABC,

AB+AC>BC       ...1

In BED,

BD+CD>BEBC>BE        ...2

From (1) and (2), we get

AB AC BE



Page No 300:

Question 1:

Which of the following is not a criterion for congruence of triangles?
(a) SSA
(b) SAS
(c) ASA
(d) SSS

Answer:

(a) SSA
SSA is not a criterion for congruence of triangles.

Page No 300:

Question 2:

If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ABC ≅ ∆PQR
(b) CBA ≅ ∆PQR
(c) CAB ≅ ∆PQR
(d) BCA ≅ ∆PQR

Answer:

(c) CABPQR
As,
AB = QR,   (given)
BC = RP,     (given)
CA = PQ     (given)


CABPQR



Page No 301:

Question 3:

If ABC ≅ ∆PQR  then which of the following is not true?
(a) BC = PQ
(b) AC = PR
(c) BC = QR
(d) AB = PQ

Answer:

(a) BC = PQ

If ABCPQR, then
BC = QR
Hence, the correct answer is option (a).

Page No 301:

Question 4:

In ABC, AB = AC and ∠B = 50°. Then, ∠A = ?
(a) 40°
(b) 50°
(c) 80°
(d) 130°

Answer:

In ABC, we have:
AB = AC 
B = 50°
Since ABC is an isosceles triangle, we have:
C=B
C=50°
In triangle ABC, we have:
A+B+C=180°A+50+50=180°A=180°-100°A=80°
Hence, the correct answer is option (c).

Page No 301:

Question 5:

In ABC, BC = AB and ∠B = 80°. Then, ∠A = ?
(a) 50°
(b) 40°
(c) 100°
(d) 80°

 

Answer:

Given: In ABCBC AB and ∠B = 80°. 

In ABC,

As, AB=BC

A=C

Let A=C=x

Using angle sum property of a triangle,

A+B+C=180°x+80°+ x=180°2x=180°-80°2x=100°

x=100°2x=50°A=50°

Hence, the correct option is (a).

Page No 301:

Question 6:

In ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?
(a) 4 cm
(b) 5 cm
(c) 8 cm
(d) 2.5 cm

Answer:


Given: In ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm.

In ABC,

As, ∠C = ∠A                   (Given)

Therefore, BC=AB       (Sides opposite to equal angles.)

BC=AB=4 cm

Hence, the correct option is (a).

Page No 301:

Question 7:

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be
(a) 6 cm
(b) 6.5 cm
(c) 5.5 cm
(d) 6.3 cn

Answer:

Since, 4 + 2.5 = 6.5

So, 6.5 cm cannot be the third side of the triangle, as the sum of two sides of a triangle is always greater than the third side.

Hence, the correct option is (b).

Page No 301:

Question 8:

In ABC, if ∠C > ∠B, then
(a) BC > AC
(b) AB > AC

(c) AB > AC
(d) BC > AC

Answer:

(b) AB > AC

In ABC, we have:
C>B
The side opposite to the greater angle is larger.
AB>AC

Page No 301:

Question 9:

It is given that ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, A = 80° and FD = 5 cm. Then, which of the following is true?
(a) ∠D = 60°
(b) ∠E = 60°

(c) ∠F = 60°
(d) D = 80°

Answer:

(b)​ E=60°

ABCFDE
AB = 5cm, B=40°,A=80° and FD = 5cm

Then A+B+C=180°80°+40°+C=180°C=60°Also, C=E
E=60°

Page No 301:

Question 10:

In ABC, ∠A = 40° and ∠B = 60°. Then the longest side of ∆ABC is
(a) BC
(b) AC
(c) AB
(d) cannot be determined

Answer:

(c) AB

In triangle ABC, we have:
A=40°,B=60°        ...(Given)
Here, A+B+C=180°60°+40°+C=180°C=80°
∴ The side opposite toC, i.e., AB, is the longest side of triangle ABC.

Page No 301:

Question 11:

In the given figure, AB > AC. Then which of the following is true?
(a) AB < AD
(b) AB = AD
(c) AB > AD
(d) Cannot be determined

Answer:

(c) AB > AD

AB>AC is given.

ACB>ABC

 Now, ADB>ACD    (exterior angle)ADB>ACB>ABCADB>ABDAB>AD

Page No 301:

Question 12:

In the given figure, AB > AC. If BO and CO are the bisectors of B and ∠C respectively, then
(a) OB = OC
(b) OB > OC
(c) OB < OC

Answer:

(b) OB > OC

AB >AC    (Given)
C>B
12C>12B
OCB>OBC  (Given)
OB>OC

Page No 301:

Question 13:

In the given figure, AB = AC and OB = OC. Then, ABO : ∠ACO = ?
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) None of these

Answer:

(a) 1:1

​In OAB and OAC, we have:
AB=AC   (Given)OB=OC    (Given) OA =OA    (Common side)
Thus, OABOAC     (SSS criterion)
i.e., ABO=ACO
∴ ABO : ACO=1 : 1



Page No 302:

Question 14:

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer:

(b) isosceles


In ABC, BLAC.
CMAB such that BL=CM.

To prove: AB = AC

In ABL and ACM,BL=CM     (Given)BAL=CAM     (Common angle)ALB=AMC      (Each 90°)ABLACM   (AAS criterion) AB=AC   (CPCT)

Page No 302:

Question 15:

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have
(a) A = ∠D
(b) ∠B = ∠E
(c) ∠C = ∠F
(d) none of these

Answer:


(b) B=E

In ABC and DEF, we have:
AB = DE      (Given)
BC = EF       (Given)
In order that ABCDEF, we must have B=E.

Page No 302:

Question 16:

In ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have
(a) AB = DF
(b) AC = DE
(c) BC = EF
(d) ∠A = ∠D

Answer:


In order that ABCDEF, we must have BC = EF.
Hence, the correct answer is option (c).

Page No 302:

Question 17:

In ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

Answer:

(a) isosceles but not congruent

AB =AC C=BP=Q        [C=P and B=Q]
Thus, both the triangles are isosceles but not congruent.

Page No 302:

Question 18:

Which is true?
(a) A triangle can have two right angles.
(b) A triangle can have two obtuse angles.
(c) A triangle can have two acute angles.
(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Answer:

(c) A triangle can have two acute angles.
The sum of two acute angles is always less than 180o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

Page No 302:

Question 19:

Fill in the blanks with < or >.
(a) (Sum of any two sides of a triangle) ...... (the third side)
(b) (Difference of any two sides of a triangle) ...... (the third side)
(c) (Sum of three altitudes of a triangle) ...... (sum of its three side)
(d) (Sum of any two sides of a triangle) ...... (twice the median to the 3rd side)
(e) (Perimenter of a triangle) ...... (sum of its three medians)

Answer:


a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

Page No 302:

Question 20:

Fill in the blanks.
(a) Each angle of an equilateral triangle measures ...... .
(b) Medians of an equilateral triangle are ...... .
(c) In a right triangle the hypotenuse is the ...... side.
(d) Drawing a ABC with AB = 3 cm, BC = 4 cm and CA = 7 cm is ...... .

Answer:

a) Each angle of an equilateral triangle measures 60°.

b) Medians of an equilateral triangle are equal.

c) In a right triangle, the hypotenuse is the longest side.

d) Drawing a ABC with AB = 3cm, BC = 4cm and CA = 7cm is not possible.



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