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Page No 372:

Question 1:

Answer:

(i)
In  ABC, it is given that DEBC.
Applying Thales' theorem, we get:
ADDB = AEEC

 AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
 DB = 10 - 3.6 = 6.4 cm
or, 3.66.4= 4.5ECor, EC = 6.4×4.53.6or, EC =8 cmThus, AC = AE +EC                    = 4.5 + 8 = 12.5 cm


(ii)

In ABC, it is given that DE BC.Applying Thales' Theorem, we get:ADDB = AEECAdding 1 to both sides, we get:ADDB+1 = AEEC+1ABDB= ACEC13.3DB = 11.95.1DB = 13.3×5.111.9 = 5.7 cm
Therefore, AD = AB - DB = 13.5 - 5.7 = 7.6 cm


(iii)
In ABC, it is given that DEBC.Applying Thales' theorem, we get:ADDB = AEEC47= AEECAdding 1 to both the sides, we get:117= ACECEC = 6.6×711 = 4.2 cmTherefore, AE = AC -EC= 6.6-4.2 = 2.4 cm


(iv)

In ABC, it is given that DEBC.Applying Thales' theorem, we get: ADAB=AEAC815= AEAE + EC815 = AEAE + 3.58AE + 28 = 15AE7AE = 28AE = 4 cm
 

Page No 372:

Question 2:

(i)
In  ABC, it is given that DEBC.
Applying Thales' theorem, we get:
ADDB = AEEC

 AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
 DB = 10 - 3.6 = 6.4 cm
or, 3.66.4= 4.5ECor, EC = 6.4×4.53.6or, EC =8 cmThus, AC = AE +EC                    = 4.5 + 8 = 12.5 cm


(ii)

In ABC, it is given that DE BC.Applying Thales' Theorem, we get:ADDB = AEECAdding 1 to both sides, we get:ADDB+1 = AEEC+1ABDB= ACEC13.3DB = 11.95.1DB = 13.3×5.111.9 = 5.7 cm
Therefore, AD = AB - DB = 13.5 - 5.7 = 7.6 cm


(iii)
In ABC, it is given that DEBC.Applying Thales' theorem, we get:ADDB = AEEC47= AEECAdding 1 to both the sides, we get:117= ACECEC = 6.6×711 = 4.2 cmTherefore, AE = AC -EC= 6.6-4.2 = 2.4 cm


(iv)

In ABC, it is given that DEBC.Applying Thales' theorem, we get: ADAB=AEAC815= AEAE + EC815 = AEAE + 3.58AE + 28 = 15AE7AE = 28AE = 4 cm
 

Answer:

(i)
In ABC, it is given that DEBC.Applying Thales' theorem, we have:ADDB = AEECxx-2=x+2x-1xx-1 = x-2x+2x2-x = x2-4x=4 cm 


(ii)
In ABC, it is given that DEBC.Applying Thales' theorem, we have: ADDB  = AEEC4x-4 = 83x-1943x-19 = 8x-412x -76 = 8x - 324x = 44x = 11 cm


(iii)
In ABC, it is given that DEBC.Applying Thales' theorem, we have:ADDB = AEEC7x-43x+4 = 5x-23x3x7x-4 =5x-23x+421x2 - 12x = 15x2 +14 x-86x2-26x+8 = 0(x-4)(6x-2) = 0x = 4, 13 x13     as if x=13 then AE will become negative x =4 cm

Page No 372:

Question 3:

(i)
In ABC, it is given that DEBC.Applying Thales' theorem, we have:ADDB = AEECxx-2=x+2x-1xx-1 = x-2x+2x2-x = x2-4x=4 cm 


(ii)
In ABC, it is given that DEBC.Applying Thales' theorem, we have: ADDB  = AEEC4x-4 = 83x-1943x-19 = 8x-412x -76 = 8x - 324x = 44x = 11 cm


(iii)
In ABC, it is given that DEBC.Applying Thales' theorem, we have:ADDB = AEEC7x-43x+4 = 5x-23x3x7x-4 =5x-23x+421x2 - 12x = 15x2 +14 x-86x2-26x+8 = 0(x-4)(6x-2) = 0x = 4, 13 x13     as if x=13 then AE will become negative x =4 cm

Answer:

(i) We have:

ADDB = 5.79.5 = 0.6 cmAEEC= 4.88 = 0.6 cmHence,ADDB=AEECApplying the converse of Thales' theorem, we conclude that DEBC.

(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 - 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 - 4.2 = 7 cm

Now,ADDB = 5.26.5=45AEEC = 4.27Thus, ADDBAEEC

Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 - 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 - 4 = 5.6 cm
Now,
ADDB=6.34.5=75AEEC=5.64=75ADDB=AEECApplying the converse of Thales' theorem, we conclude that DEBC.

(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 - 7.2 =  4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 - 6.4 = 3.6 cm
Now,
ADDB = 7.24.8=32AEEC = 6.43.6= 169Thus, ADDBAEECApplying the converse of Thales' theorem, we conclude that DE is not parallel to BC.

Page No 372:

Question 4:

(i) We have:

ADDB = 5.79.5 = 0.6 cmAEEC= 4.88 = 0.6 cmHence,ADDB=AEECApplying the converse of Thales' theorem, we conclude that DEBC.

(ii)
We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 - 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 - 4.2 = 7 cm

Now,ADDB = 5.26.5=45AEEC = 4.27Thus, ADDBAEEC

Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

(iii)
We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 - 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 - 4 = 5.6 cm
Now,
ADDB=6.34.5=75AEEC=5.64=75ADDB=AEECApplying the converse of Thales' theorem, we conclude that DEBC.

(iv)
We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 - 7.2 =  4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 - 6.4 = 3.6 cm
Now,
ADDB = 7.24.8=32AEEC = 6.43.6= 169Thus, ADDBAEECApplying the converse of Thales' theorem, we conclude that DE is not parallel to BC.

Answer:

(i)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:BDDC=ABAC5.6DC=6.48DC = 8×5.66.4 = 7 cm


(ii)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC =ABACLet BD be x cm.Therefore, DC = (6-x) cmx6-x = 101414x = 60-10x24x = 60x = 2.5 cmThus, BD = 2.5 cmDC = 6-2.5 = 3.5 cm 

(iii)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC=ABACBD = 3.2 cm, BC = 6 cmTherefore, DC = 6-3.2 = 2.8 cm3.22.8=5.6AC

AC = 5.6×2.83.2=4.9 cm


(iv)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC = ABACBD3 = 5.64BD = 5.6×34BD = 4.2 cm

Hence,  BC = 3 + 4.2 = 7.2 cm



Page No 373:

Question 5:

(i)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:BDDC=ABAC5.6DC=6.48DC = 8×5.66.4 = 7 cm


(ii)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC =ABACLet BD be x cm.Therefore, DC = (6-x) cmx6-x = 101414x = 60-10x24x = 60x = 2.5 cmThus, BD = 2.5 cmDC = 6-2.5 = 3.5 cm 

(iii)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC=ABACBD = 3.2 cm, BC = 6 cmTherefore, DC = 6-3.2 = 2.8 cm3.22.8=5.6AC

AC = 5.6×2.83.2=4.9 cm


(iv)
It is given that AD bisects A.Applying angle-bisector theorem in ABC, we get:

BDDC = ABACBD3 = 5.64BD = 5.6×34BD = 4.2 cm

Hence,  BC = 3 + 4.2 = 7.2 cm

Answer:

Given: ABCD is a parallelogram
To prove: 
iDMMN=DCBNiiDNDM=ANDC
Proof: In â–³DMC and â–³NMB
DMC =NMB      (Vertically opposite angle)
DCM =NBM       (Alternate angles)  
By AAA- similarity
â–³DMC ~ â–³NMB
DMMN=DCBN
Now,  MNDM=BNDC
Adding 1 to both sides, we get
MNDM+1=BNDC+1MN+DMDM=BN+DCDCMN+DMDM=BN+ABDC       ABCD is a parallelogramDNDM=ANDC        

Page No 373:

Question 6:

Given: ABCD is a parallelogram
To prove: 
iDMMN=DCBNiiDNDM=ANDC
Proof: In â–³DMC and â–³NMB
DMC =NMB      (Vertically opposite angle)
DCM =NBM       (Alternate angles)  
By AAA- similarity
â–³DMC ~ â–³NMB
DMMN=DCBN
Now,  MNDM=BNDC
Adding 1 to both sides, we get
MNDM+1=BNDC+1MN+DMDM=BN+DCDCMN+DMDM=BN+ABDC       ABCD is a parallelogramDNDM=ANDC        

Answer:

Let the trapezium â€‹be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.

In PAB, DC  AB.
Applying Thales' theorem, we get:
PDDA = PCCBNow, E and F are the midpoints of AD and BC, respectively. PD2DE = PC2CF PDDE = PCCFApplying the converse of Thales' theorem in PEF, we get that DC  EF.Hence, EF  AB.

Thus. EF is parallel to both AB and DC.
This completes the proof.

Page No 373:

Question 7:

Let the trapezium â€‹be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.

In PAB, DC  AB.
Applying Thales' theorem, we get:
PDDA = PCCBNow, E and F are the midpoints of AD and BC, respectively. PD2DE = PC2CF PDDE = PCCFApplying the converse of Thales' theorem in PEF, we get that DC  EF.Hence, EF  AB.

Thus. EF is parallel to both AB and DC.
This completes the proof.

Answer:

In trapezium ABCD, ABCD and the diagonals AC and BD intersect at O.
Therefore,

 AOOC=BOOD2x+15x-7=7x+17x-5(5x - 7)(7x + 1) = (7x - 5)(2x + 1)35x2 + 5x - 49x - 7 = 14x2 - 10x + 7x - 521x2 - 41x - 2 = 021x2 - 42x +x- 2 = 021x(x - 2) + 1(x - 2) = 0(x - 2)(21x+1) = 0x = 2,-121 x  -121 x = 2 

Page No 373:

Question 8:

In trapezium ABCD, ABCD and the diagonals AC and BD intersect at O.
Therefore,

 AOOC=BOOD2x+15x-7=7x+17x-5(5x - 7)(7x + 1) = (7x - 5)(2x + 1)35x2 + 5x - 49x - 7 = 14x2 - 10x + 7x - 521x2 - 41x - 2 = 021x2 - 42x +x- 2 = 021x(x - 2) + 1(x - 2) = 0(x - 2)(21x+1) = 0x = 2,-121 x  -121 x = 2 

Answer:


In â–³ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN           (∵BM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In â–³ABC,
∠A + ∠B + ∠C = 180o                     .....(1)
                                                               (Angle Sum Property of triangle)
Again In In â–³AMN,
∠A +∠AMN + ∠ANM = 180o          ......(2)
                                                               (Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B  = 2∠AMN  
⇒∠B  = ∠AMN
Since, ∠B  and ∠AMN are corresponding angles.
∴ MN∥BC

Page No 373:

Question 9:


In â–³ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN           (∵BM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In â–³ABC,
∠A + ∠B + ∠C = 180o                     .....(1)
                                                               (Angle Sum Property of triangle)
Again In In â–³AMN,
∠A +∠AMN + ∠ANM = 180o          ......(2)
                                                               (Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B  = 2∠AMN  
⇒∠B  = ∠AMN
Since, ∠B  and ∠AMN are corresponding angles.
∴ MN∥BC

Answer:

In CAB , PQ  AB.Applying Thales' theorem, we get:CPPB = CQQA       ...(1)
Similarly, applying Thales' theorem in BDC, where PR  BD, we get:
CPPB = CRRD               ...(2)Hence, from (1) and (2), we have:CQQA = CRRD

Applying the converse of Thales' theorem, we conclude that QRAD in ADC.
This completes the proof.

Page No 373:

Question 10:

In CAB , PQ  AB.Applying Thales' theorem, we get:CPPB = CQQA       ...(1)
Similarly, applying Thales' theorem in BDC, where PR  BD, we get:
CPPB = CRRD               ...(2)Hence, from (1) and (2), we have:CQQA = CRRD

Applying the converse of Thales' theorem, we conclude that QRAD in ADC.
This completes the proof.

Answer:


Join BX and CX.
It is given that BC is bisected at D.
BD = DC â€‹
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX  bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO  CXand BX  CO
BX  CF and CX  BE
BX  OF and CX  OE
Applying Thales' theorem in ABX, we get:
AOAX = AFAB            ...(1)Also, in ACX, CX  OE.Therefore by Thales' theorem, we get:AOAX = AEAC           ...(2)
From (1) and (2), we have:
AFAB = AEAC
Applying the converse of Thales' theorem in ABC, EFCB.
This completes the proof.

Page No 373:

Question 11:


Join BX and CX.
It is given that BC is bisected at D.
BD = DC â€‹
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX  bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO  CXand BX  CO
BX  CF and CX  BE
BX  OF and CX  OE
Applying Thales' theorem in ABX, we get:
AOAX = AFAB            ...(1)Also, in ACX, CX  OE.Therefore by Thales' theorem, we get:AOAX = AEAC           ...(2)
From (1) and (2), we have:
AFAB = AEAC
Applying the converse of Thales' theorem in ABC, EFCB.
This completes the proof.

Answer:


Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = 12AC      ...(i)
Also, it is given that CQ = 14AC        ...(ii)
Dividing equation (ii) by (i), we get:
CQCS = 14AC12AC
or, CQ = 12CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in CSD
PQDS
 if PQ DS ,  we can say that QRSB 

In CSB, Q is midpoint of CS and QRSB.
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.

Page No 373:

Question 12:


Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = 12AC      ...(i)
Also, it is given that CQ = 14AC        ...(ii)
Dividing equation (ii) by (i), we get:
CQCS = 14AC12AC
or, CQ = 12CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in CSD
PQDS
 if PQ DS ,  we can say that QRSB 

In CSB, Q is midpoint of CS and QRSB.
Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB.
This completes the proof.

Answer:

Given:
AD = AE     ...(i)
AB = AC     ...(ii)
Subtracting AD from both sides, we get:
AB - AD = AC - AD
AB - AD = AC - AE  (Since, AD = AE)
BD = EC    ...(iii)
Dividing  equation  (i) by equation (iii), we get:

ADDB=AEEC

Applying the converse of Thales' theorem, DEBC

DEC + ECB  = 180°   (Sum of interior angles on the same side of a transversal line is 180°.)    DEC + CBD =180°  (Since, AB = ACB =C)

Hence, quadrilateral BCED is cyclic.

Therefore, B,C,E and D are concyclic points.



Page No 374:

Question 13:

Given:
AD = AE     ...(i)
AB = AC     ...(ii)
Subtracting AD from both sides, we get:
AB - AD = AC - AD
AB - AD = AC - AE  (Since, AD = AE)
BD = EC    ...(iii)
Dividing  equation  (i) by equation (iii), we get:

ADDB=AEEC

Applying the converse of Thales' theorem, DEBC

DEC + ECB  = 180°   (Sum of interior angles on the same side of a transversal line is 180°.)    DEC + CBD =180°  (Since, AB = ACB =C)

Hence, quadrilateral BCED is cyclic.

Therefore, B,C,E and D are concyclic points.

Answer:

In triangle ​BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
QRPR = BQBP
 BP × QR = BQ × PR
This completes the proof.



Page No 399:

Question 1:

In triangle ​BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
QRPR = BQBP
 BP × QR = BQ × PR
This completes the proof.

Answer:

(i)
We have:
BAC = PQR = 50°ABC =QPR = 60°ACB =PRQ=70°

Therefore, by AAA similarity theorem, ABC ~QPR

(ii)
We have:
 
ABDF=36=12 and BCDE=4.59=12

But, ABCEDF (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

(iii)
We have:
CAQR=86=43 and CBPQ=64.5=43CAQR=CBPQ
Also, ACB =PQR=80°
Therefore, by SAS similarity theorem, ACB ~RQP.

(iv)
 We have
DEQR=2.55=12EFPQ=24=12DFPR=36=12DEQR=EFPQ=DFPR

Therefore, by SSS similarity theorem, FED~PQR

(v)
In ABCA+B+C=180°      Angle Sum Property80°+B+70°=180°B=30°  
A=M and B=N
Therefore, by AA similarity theorem, ABC~MNR



Page No 400:

Question 2:

(i)
We have:
BAC = PQR = 50°ABC =QPR = 60°ACB =PRQ=70°

Therefore, by AAA similarity theorem, ABC ~QPR

(ii)
We have:
 
ABDF=36=12 and BCDE=4.59=12

But, ABCEDF (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

(iii)
We have:
CAQR=86=43 and CBPQ=64.5=43CAQR=CBPQ
Also, ACB =PQR=80°
Therefore, by SAS similarity theorem, ACB ~RQP.

(iv)
 We have
DEQR=2.55=12EFPQ=24=12DFPR=36=12DEQR=EFPQ=DFPR

Therefore, by SSS similarity theorem, FED~PQR

(v)
In ABCA+B+C=180°      Angle Sum Property80°+B+70°=180°B=30°  
A=M and B=N
Therefore, by AA similarity theorem, ABC~MNR

Answer:

(i)
It is given that DB is a straight line.
Therefore,
DOC + COB = 180°DOC= 180° - 115° = 65°

(ii)
In DOC, we have:
ODC + DCO + DOC =180°Therefore,70°+ DCO + 65° = 180°DCO = 180 - 70 - 65 = 45°

(iii)
It is given that ODC ~OBA
Therefore,
OAB =OCD = 45°

(iv)
Again, ODC ~OBA
Therefore,
OBA =ODC= 70°

Page No 400:

Question 3:

(i)
It is given that DB is a straight line.
Therefore,
DOC + COB = 180°DOC= 180° - 115° = 65°

(ii)
In DOC, we have:
ODC + DCO + DOC =180°Therefore,70°+ DCO + 65° = 180°DCO = 180 - 70 - 65 = 45°

(iii)
It is given that ODC ~OBA
Therefore,
OAB =OCD = 45°

(iv)
Again, ODC ~OBA
Therefore,
OBA =ODC= 70°

Answer:

(i) Let OA be x cm.
 OAB ~OCD
 OAOC=ABCDx3.5=85x = 8 × 3.55 = 5.6 
Hence, OA = 5.6 cm

(ii)  Let OD be y cm

 OAB ~OCD
 ABCD=OBOD85 = 6.4yy = 6.4 × 58 = 4
Hence, DO = 4 cm



Page No 401:

Question 4:

(i) Let OA be x cm.
 OAB ~OCD
 OAOC=ABCDx3.5=85x = 8 × 3.55 = 5.6 
Hence, OA = 5.6 cm

(ii)  Let OD be y cm

 OAB ~OCD
 ABCD=OBOD85 = 6.4yy = 6.4 × 58 = 4
Hence, DO = 4 cm

Answer:

Given:
ADE = ABC and A = A  
Let DE be x cm
Therefore, by AA similarity theorem, ADE~ABC

ADAB = DEBC3.83.6 + 2.1 = x4.2x = 3.8 × 4.25.7 = 2.8

Hence, DE = 2.8 cm

Page No 401:

Question 5:

Given:
ADE = ABC and A = A  
Let DE be x cm
Therefore, by AA similarity theorem, ADE~ABC

ADAB = DEBC3.83.6 + 2.1 = x4.2x = 3.8 × 4.25.7 = 2.8

Hence, DE = 2.8 cm

Answer:

It is given that triangles ABC and PQR are similar.
Therefore,

Perimeter (ABC)Perimeter (PQR)=ABPQ3224=AB12AB=32×1224=16 cm

Page No 401:

Question 6:

It is given that triangles ABC and PQR are similar.
Therefore,

Perimeter (ABC)Perimeter (PQR)=ABPQ3224=AB12AB=32×1224=16 cm

Answer:

It is given that ABC~DEF.
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.

Perimeter of ABCPerimeter of DEF=BCEF

Let the perimeter of ∆ABC be x cm.

Therefore,
x25=9.16.5x=9.1×256.5=35

Thus, the perimeter of ∆ABC is 35 cm.

Page No 401:

Question 7:

It is given that ABC~DEF.
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.

Perimeter of ABCPerimeter of DEF=BCEF

Let the perimeter of ∆ABC be x cm.

Therefore,
x25=9.16.5x=9.1×256.5=35

Thus, the perimeter of ∆ABC is 35 cm.

Answer:

In BDA and BAC, we have:BDA =BAC = 90°DBA = CBA          (Common)Therefore, by AA similarity theorem, BDA~BAC

ADAC=ABBC

AD0.75=11.25AD=0.751.25=0.6 m or 60 cm

Page No 401:

Question 8:

In BDA and BAC, we have:BDA =BAC = 90°DBA = CBA          (Common)Therefore, by AA similarity theorem, BDA~BAC

ADAC=ABBC

AD0.75=11.25AD=0.751.25=0.6 m or 60 cm

Answer:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

InBDC and ABC, we have:ABC = BDC = 90° (given)C = C  (common)By AA similarity theorem, we get:BDC~ABC

ABBD = BCDC5.73.8 = BC5.4BC = 5.73.8 × 5.4= 8.1

Hence, BC = 8.1 cm

Page No 401:

Question 9:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

InBDC and ABC, we have:ABC = BDC = 90° (given)C = C  (common)By AA similarity theorem, we get:BDC~ABC

ABBD = BCDC5.73.8 = BC5.4BC = 5.73.8 × 5.4= 8.1

Hence, BC = 8.1 cm

Answer:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

In DBA and DCB, we have:BDA = CDBDBA= DCB = 90°Therefore, by AA similarity theorem, we get:DBA~DCBBDCD = ADBDCD = BD2AD

CD  = 8×84=16 cm

Page No 401:

Question 10:

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

In DBA and DCB, we have:BDA = CDBDBA= DCB = 90°Therefore, by AA similarity theorem, we get:DBA~DCBBDCD = ADBDCD = BD2AD

CD  = 8×84=16 cm

Answer:

We have:

APAB = 26 = 13 and AQAC = 39 = 13APAB = AQACIn APQ and ABC, we have:APAB = AQACA = ATherefore, by AA similarity theorem, we get:APQ~ABCHence, PQBC = AQAC = 13PQBC = 13BC = 3PQ
This completes the proof.



Page No 402:

Question 11:

We have:

APAB = 26 = 13 and AQAC = 39 = 13APAB = AQACIn APQ and ABC, we have:APAB = AQACA = ATherefore, by AA similarity theorem, we get:APQ~ABCHence, PQBC = AQAC = 13PQBC = 13BC = 3PQ
This completes the proof.

Answer:

We have: 
AFD = EFB    (Vertically Opposite angles)
 DA  BC
 DAF =BEF      (Alternate angles)
DAF~BEF          (AA similarity theorem)
 AFEF = FDFB
or, AF × FB = FD × EF
This completes the proof.

Page No 402:

Question 12:

We have: 
AFD = EFB    (Vertically Opposite angles)
 DA  BC
 DAF =BEF      (Alternate angles)
DAF~BEF          (AA similarity theorem)
 AFEF = FDFB
or, AF × FB = FD × EF
This completes the proof.

Answer:

In BED and ACB, we have:

BED = ACB = 90° B + C = 180° BD  ACEBD = CAB (Alternate angles)Therefore, by AA similarity theorem, we get:BED~ACB BEAC = DEBC BEDE = ACBC
This completes the proof.

Page No 402:

Question 13:

In BED and ACB, we have:

BED = ACB = 90° B + C = 180° BD  ACEBD = CAB (Alternate angles)Therefore, by AA similarity theorem, we get:BED~ACB BEAC = DEBC BEDE = ACBC
This completes the proof.

Answer:

Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m. 

In ABC and PQR, we have:
ABC = PQR = 90°ACB = PRQ   (Angular elevation of the Sun at the same time)
Therefore, by AA similarity theorem, we get:
ABC~PQR
ABBC=PQQR

7.55=x24x=7.55×24=36 m
Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.

Page No 402:

Question 14:

Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m. 

In ABC and PQR, we have:
ABC = PQR = 90°ACB = PRQ   (Angular elevation of the Sun at the same time)
Therefore, by AA similarity theorem, we get:
ABC~PQR
ABBC=PQQR

7.55=x24x=7.55×24=36 m
Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.

Answer:

Disclaimer: It should be APC~BCQ instead of ∆ACP ∼ ∆BCQ
It is given that ABC is an isosceles triangle.
Therefore,
CA = CB
 CAB = CBA 180° - CAB = 180° - CBA CAP = CBQ

Also,
AP × BQ = AC2 APAC = ACBQ APAC= BCBQ ( AC = BC)

Thus, by SAS similarity theorem, we get:
APC~BCQ
This completes the proof.

Page No 402:

Question 15:

Disclaimer: It should be APC~BCQ instead of ∆ACP ∼ ∆BCQ
It is given that ABC is an isosceles triangle.
Therefore,
CA = CB
 CAB = CBA 180° - CAB = 180° - CBA CAP = CBQ

Also,
AP × BQ = AC2 APAC = ACBQ APAC= BCBQ ( AC = BC)

Thus, by SAS similarity theorem, we get:
APC~BCQ
This completes the proof.

Answer:

We have:
ACBD = CBCEACCB = BDCEACCB = CDCE  (Since, BD= DC as 1 = 2)Also, 1 = 2i.e, DBC = ACBTherefore, by SAS similarity theorem, we get:ACB~DCE

Page No 402:

Question 16:

We have:
ACBD = CBCEACCB = BDCEACCB = CDCE  (Since, BD= DC as 1 = 2)Also, 1 = 2i.e, DBC = ACBTherefore, by SAS similarity theorem, we get:ACB~DCE

Answer:

In â–³ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and PQ=12BC                                              .....(1)

Similarly, In â–³ADC, QR=12AD=12BC                          .....(2)
Now, In â–³BCD, SR=12BC                                               .....(3)

Similarly, In â–³ABD, PS=12AD=12BC                           .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.

Page No 402:

Question 17:

In â–³ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and PQ=12BC                                              .....(1)

Similarly, In â–³ADC, QR=12AD=12BC                          .....(2)
Now, In â–³BCD, SR=12BC                                               .....(3)

Similarly, In â–³ABD, PS=12AD=12BC                           .....(4)
Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.

Answer:

Given: AB and CD are two chords
To Prove:
aPAC~PDBbPA.PB=PC.PD

Proof: In PAC and PDB
APC=DPB  (Vertically Opposite angles)
CAP=BDP  (Angles in the same segment are equal)
By AA similarity-criterion PAC~PDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
PAPD=PCPBPA.PB=PC.PD



Page No 403:

Question 18:

Given: AB and CD are two chords
To Prove:
aPAC~PDBbPA.PB=PC.PD

Proof: In PAC and PDB
APC=DPB  (Vertically Opposite angles)
CAP=BDP  (Angles in the same segment are equal)
By AA similarity-criterion PAC~PDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
PAPD=PCPBPA.PB=PC.PD

Answer:

Given: AB and CD are two chords
To Prove:
aPAC~PDBbPA.PB=PC.PD

Proof:
ABD+ACD=180           .....(1)
                                                                         (Opposite angles of a cyclic quadrilateral are supplementary)
PCA+ACD=180           ....(2)
                                                                        (Linear Pair Angles)
Using (1) and (2), we get
ABD=PCA
A=A          (Common)
By AA similarity-criterion PAC~PDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
PAPD=PCPBPA.PB=PC.PD

Page No 403:

Question 19:

Given: AB and CD are two chords
To Prove:
aPAC~PDBbPA.PB=PC.PD

Proof:
ABD+ACD=180           .....(1)
                                                                         (Opposite angles of a cyclic quadrilateral are supplementary)
PCA+ACD=180           ....(2)
                                                                        (Linear Pair Angles)
Using (1) and (2), we get
ABD=PCA
A=A          (Common)
By AA similarity-criterion PAC~PDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
PAPD=PCPBPA.PB=PC.PD

Answer:

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse  then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In â–³BDC, we get

â–³CQD ∼ â–³DQB
CQDQ=DQQBDQ2=QB.CQ

Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
DQ2=DP.CQ

(b)

Similarly, â–³APD ∼ â–³DPB
APDP=PDPBDP2=AP.PBDP2=AP.DQ      DQ=PB

Page No 403:

Question 20:

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse  then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.
(a) Now using the same property in In â–³BDC, we get

â–³CQD ∼ â–³DQB
CQDQ=DQQBDQ2=QB.CQ

Now. since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
DQ2=DP.CQ

(b)

Similarly, â–³APD ∼ â–³DPB
APDP=PDPBDP2=AP.PBDP2=AP.DQ      DQ=PB

Answer:

Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DCBC2 and QM = MR = QR2       ...(1)

Now,
ΔABC ~ ΔPQR

As we know, corresponding sides of similar triangles are proportional.
Thus, ABPQ=BCQR=ACPR                                        ...(2)

Also, A=P, B=Q and C=R            ...(3)

From (1) and (2), we get
ABPQ=BCQRABPQ=2BD2QMABPQ=BDQM      ...4

Now, in ΔABD and ΔPQM

ABPQ=BDQM               From 4B=Q                  From 3


By SAS similarity,
ΔABD ~ ΔPQM

Therefore, ABPQ=BDQM=ADPM.

Hence, ABPQ=ADPM.


 



Page No 417:

Question 1:

Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DCBC2 and QM = MR = QR2       ...(1)

Now,
ΔABC ~ ΔPQR

As we know, corresponding sides of similar triangles are proportional.
Thus, ABPQ=BCQR=ACPR                                        ...(2)

Also, A=P, B=Q and C=R            ...(3)

From (1) and (2), we get
ABPQ=BCQRABPQ=2BD2QMABPQ=BDQM      ...4

Now, in ΔABD and ΔPQM

ABPQ=BDQM               From 4B=Q                  From 3


By SAS similarity,
ΔABD ~ ΔPQM

Therefore, ABPQ=BDQM=ADPM.

Hence, ABPQ=ADPM.


 

Answer:

It is given that ABC~DEF.
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

ar(ABC)ar(DEF) = BC2EF2Let BCbe x cm. 64121 = x2(15.4)2 x2 = 64 × 15.4 × 15.4121 x = 64 × 15.4 × 15.4121= 8 × 15.411= 11.2

Hence, BC = 11.2 cm

Page No 417:

Question 2:

It is given that ABC~DEF.
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.

ar(ABC)ar(DEF) = BC2EF2Let BCbe x cm. 64121 = x2(15.4)2 x2 = 64 × 15.4 × 15.4121 x = 64 × 15.4 × 15.4121= 8 × 15.411= 11.2

Hence, BC = 11.2 cm

Answer:

It is given that ABC~PQR.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
ar(ABC)ar(PQR) = BC2QR2 916 = 4.52QR2 QR2 = 4.5 × 4.5 × 169 QR = 4.5 × 4.5 × 169 = 4.5 × 43= 6 cm

Hence, QR = 6 cm

Page No 417:

Question 3:

It is given that ABC~PQR.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
ar(ABC)ar(PQR) = BC2QR2 916 = 4.52QR2 QR2 = 4.5 × 4.5 × 169 QR = 4.5 × 4.5 × 169 = 4.5 × 43= 6 cm

Hence, QR = 6 cm

Answer:

Given:ar(ABC)= 4ar(PQR)ar(ABC)ar(PQR) = 41 ABC~PQR ar(ABC)ar(PQR) = BC2QR2 BC2QR2 = 41
 QR2 = 1224 QR2 = 36 QR = 6 cm
Hence, QR = 6 cm

Page No 417:

Question 4:

Given:ar(ABC)= 4ar(PQR)ar(ABC)ar(PQR) = 41 ABC~PQR ar(ABC)ar(PQR) = BC2QR2 BC2QR2 = 41
 QR2 = 1224 QR2 = 36 QR = 6 cm
Hence, QR = 6 cm

Answer:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.

ar(Larger triangle)ar(Smaller triangle)=(Longest side of larger traingle)2(Longest side of smaller traingle)2 169121 = 262x2 x = 26 × 26 × 121169= 22 

Hence, the longest side of the smaller triangle is 22 cm.

Page No 417:

Question 5:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.

ar(Larger triangle)ar(Smaller triangle)=(Longest side of larger traingle)2(Longest side of smaller traingle)2 169121 = 262x2 x = 26 × 26 × 121169= 22 

Hence, the longest side of the smaller triangle is 22 cm.

Answer:



It is given that âˆ†ABC ∼ ∆DEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the altitude of
∆ABC be AP, drawn from A to BC to meet BC at P and the altitude of ∆DEF be DQ, drawn from D to meet EF at Q.

Then,
ar(ABC)ar(DEF) = AP2DQ2 10049 = 52DQ2 10049 = 25DQ2 DQ2 = 49 × 25100 DQ = 49 × 25100 DQ = 3.5 cm


Hence, the altitude of ∆DEF is 3.5 cm

Page No 417:

Question 6:



It is given that âˆ†ABC ∼ ∆DEF.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the altitude of
∆ABC be AP, drawn from A to BC to meet BC at P and the altitude of ∆DEF be DQ, drawn from D to meet EF at Q.

Then,
ar(ABC)ar(DEF) = AP2DQ2 10049 = 52DQ2 10049 = 25DQ2 DQ2 = 49 × 25100 DQ = 49 × 25100 DQ = 3.5 cm


Hence, the altitude of ∆DEF is 3.5 cm

Answer:

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively​.

It is given that ABC~DEF.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes. 

 ar(ABC)ar(DEF) = (AP)2(DQ)2 ar(ABC)ar(DEF) = 6292= 3681 = 49

Hence, the ratio of their areas is 4 : 9



Page No 418:

Question 7:

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively​.

It is given that ABC~DEF.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes. 

 ar(ABC)ar(DEF) = (AP)2(DQ)2 ar(ABC)ar(DEF) = 6292= 3681 = 49

Hence, the ratio of their areas is 4 : 9

Answer:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively. 



ar(ABC)ar(PQR) = AP2DQ2 8149 = 6.32DQ2DQ2 = 4981 × 6.32 DQ2 = 4981 × 6.3 × 6.3= 4.9 cm

Hence, the altitude of the other triangle is 4.9 cm.

Page No 418:

Question 8:

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively. 



ar(ABC)ar(PQR) = AP2DQ2 8149 = 6.32DQ2DQ2 = 4981 × 6.32 DQ2 = 4981 × 6.3 × 6.3= 4.9 cm

Hence, the altitude of the other triangle is 4.9 cm.

Answer:

Let the two triangles be ABC and PQR with medians AM and PN, respectively.

Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.
 ar(ABC)ar(PQR) = AM2PN2 64100 = 5.62PN2 PN2 = 64100 × 5.62 PN2 = 10064 × 5.6 × 5.6= 7 cm
Hence, the median of the larger triangle is 7 cm.

Page No 418:

Question 9:

Let the two triangles be ABC and PQR with medians AM and PN, respectively.

Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding medians.
 ar(ABC)ar(PQR) = AM2PN2 64100 = 5.62PN2 PN2 = 64100 × 5.62 PN2 = 10064 × 5.6 × 5.6= 7 cm
Hence, the median of the larger triangle is 7 cm.

Answer:

We have:
APAB = 11 + 3 = 14 and AQAC = 1.51.5 + 4.5 = 1.56 = 14 APAB = AQAC

Also, A = A
By SAS similarity , we can conclude that âˆ†APQ~∆ABC.

ar(APQ)ar(ABC) = AP2AB2 = 1242 = 116 ar(APQ)ar(ABC) = 116 ar(APQ) = 116 × ar(ABC)

Hence proved.

Page No 418:

Question 10:

We have:
APAB = 11 + 3 = 14 and AQAC = 1.51.5 + 4.5 = 1.56 = 14 APAB = AQAC

Also, A = A
By SAS similarity , we can conclude that âˆ†APQ~∆ABC.

ar(APQ)ar(ABC) = AP2AB2 = 1242 = 116 ar(APQ)ar(ABC) = 116 ar(APQ) = 116 × ar(ABC)

Hence proved.

Answer:

It is given that DE âˆ¥ BC

 ADE = ABC   (Corresponding angles)    AED = ACB  (Corresponding angles)

By AA similarity , we can conclude that ADE~ABC.

 ar(ADE)ar(ABC) = DE2BC2 15ar(ABC) = 3262 ar(ABC)  = 15 × 369= 60 cm2

Hence, area of triangle ABC is 60 cm2.​

Page No 418:

Question 11:

It is given that DE âˆ¥ BC

 ADE = ABC   (Corresponding angles)    AED = ACB  (Corresponding angles)

By AA similarity , we can conclude that ADE~ABC.

 ar(ADE)ar(ABC) = DE2BC2 15ar(ABC) = 3262 ar(ABC)  = 15 × 369= 60 cm2

Hence, area of triangle ABC is 60 cm2.​

Answer:

In ABC and ADC, we have:
BAC=ADC=90°ACB=ACD    common
By AA similarity, we can conclude that BAC~ADC.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

 ar(BAC)ar(ADC) = BC2AC2 ar(BAC)ar(ADC) = 13252= 16925

 Hence, the ratio of areas of both the triangles is 169 : 25

Page No 418:

Question 12:

In ABC and ADC, we have:
BAC=ADC=90°ACB=ACD    common
By AA similarity, we can conclude that BAC~ADC.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their corresponding sides.

 ar(BAC)ar(ADC) = BC2AC2 ar(BAC)ar(ADC) = 13252= 16925

 Hence, the ratio of areas of both the triangles is 169 : 25

Answer:

It is given that DE  BC.
 ADE=ABC  (Corresponding angles)   AED=ACB   (Corresponding angles)
Applying AA similarity theorem, we can conclude that ADE~ABC.

 ar(ABC)ar(ADE) = BC2DE2Subtracting 1 from both sides, we get:ar(ABC)ar(ADE) - 1 = 5232 - 1 ar(ABC) - ar(ADE)ar(ADE)=25 - 99 ar(BCED)ar(ADE) = 169or, ar(ADE)ar(BCED)=916

Page No 418:

Question 13:

It is given that DE  BC.
 ADE=ABC  (Corresponding angles)   AED=ACB   (Corresponding angles)
Applying AA similarity theorem, we can conclude that ADE~ABC.

 ar(ABC)ar(ADE) = BC2DE2Subtracting 1 from both sides, we get:ar(ABC)ar(ADE) - 1 = 5232 - 1 ar(ABC) - ar(ADE)ar(ADE)=25 - 99 ar(BCED)ar(ADE) = 169or, ar(ADE)ar(BCED)=916

Answer:

It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE  BC.
Hence, by B.P.T., we get:
 ADAB = AEAC
Also, A = A.

Applying SAS similarity theorem, we can conclude that ADE~ABC.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
 ar(ADE)ar(ABC) = DE2BC2= 12BC2BC2= 14



Page No 441:

Question 1:

It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE  BC.
Hence, by B.P.T., we get:
 ADAB = AEAC
Also, A = A.

Applying SAS similarity theorem, we can conclude that ADE~ABC.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
 ar(ADE)ar(ABC) = DE2BC2= 12BC2BC2= 14

Answer:

For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
a2 + b2 = 92 + 162 = 81 + 256 = 337c2 = 192 = 361a2 + b2  c2

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
a2 + b2 = 72 + 242= 49 + 576 = 625c2 = 252= 625a2 + b2 = c2
Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
a2 + b2 = (1.4)2 + (4.8)2= 1.96 + 23.04 = 25c2 = 52= 25a2 + b2 = c2
Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
a2 + b2 = (1.6)2 + (3.8)2= 2.56 + 14.44 = 17c2 = 42= 16a2 + b2  c2
Thus, the given triangle is not right-angled.

(v)
p = (a - 1) cm,  q = 2a cm and r = (a + 1) cm
Then,
p2 + q2 = (a - 1)2 + (2a)2             = a2 + 1 - 2a + 4a            = a2 + 1 + 2a             = (a + 1)2r2 = (a + 1)2 p2 + q2 = r2
Thus, the given triangle is right-angled.

Page No 441:

Question 2:

For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
a2 + b2 = 92 + 162 = 81 + 256 = 337c2 = 192 = 361a2 + b2  c2

Thus, the given triangle is not right-angled.

(ii)
a = 7 cm, b = 24 cm and c = 25 cm
Then,
a2 + b2 = 72 + 242= 49 + 576 = 625c2 = 252= 625a2 + b2 = c2
Thus, the given triangle is a right-angled.

(iii)
a = 1.4 cm, b = 4.8 cm and c = 5 cm
Then,
a2 + b2 = (1.4)2 + (4.8)2= 1.96 + 23.04 = 25c2 = 52= 25a2 + b2 = c2
Thus, the given triangle is right-angled.

(iv)
a = 1.6 cm, b = 3.8cm and c = 4 cm
Then,
a2 + b2 = (1.6)2 + (3.8)2= 2.56 + 14.44 = 17c2 = 42= 16a2 + b2  c2
Thus, the given triangle is not right-angled.

(v)
p = (a - 1) cm,  q = 2a cm and r = (a + 1) cm
Then,
p2 + q2 = (a - 1)2 + (2a)2             = a2 + 1 - 2a + 4a            = a2 + 1 + 2a             = (a + 1)2r2 = (a + 1)2 p2 + q2 = r2
Thus, the given triangle is right-angled.

Answer:

Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.



We need to find AC.
In right-angled triangle ABC, we have:

AC2 = AB2 + BC2AC = 802 + 1502  = 6400 + 22500 = 28900 = 170 m

Hence, the man is 170 m away from the starting point.

Page No 441:

Question 3:

Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.



We need to find AC.
In right-angled triangle ABC, we have:

AC2 = AB2 + BC2AC = 802 + 1502  = 6400 + 22500 = 28900 = 170 m

Hence, the man is 170 m away from the starting point.

Answer:

Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right DEF, we have:
DE = 10 m, EF = 24 m



DF2 = EF2 + DE2DF = 102 + 242 = 100 + 576 = 676= 26 m

Hence, the man is 26 m away from the starting point.

Page No 441:

Question 4:

Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F.
In right DEF, we have:
DE = 10 m, EF = 24 m



DF2 = EF2 + DE2DF = 102 + 242 = 100 + 576 = 676= 26 m

Hence, the man is 26 m away from the starting point.

Answer:

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:



AB2 = AC2 + BC2    BC = 132 - 122              = 169 - 144             = 25             = 5 m

Hence, the distance of the foot of the ladder from the building is 5 m

Page No 441:

Question 5:

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:



AB2 = AC2 + BC2    BC = 132 - 122              = 169 - 144             = 25             = 5 m

Hence, the distance of the foot of the ladder from the building is 5 m

Answer:

Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:



AC2 = AB2 + BC2 AC =202 + 152              = 400 + 225               = 625              = 25 m
Hence, the length of the ladder is 25 m.

Page No 441:

Question 6:

Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.
We have:
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled triangle ABC, we get:



AC2 = AB2 + BC2 AC =202 + 152              = 400 + 225               = 625              = 25 m
Hence, the length of the ladder is 25 m.

Answer:

Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m  and AC = 5 m
We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled triangle ACD, we have: ​

AD2 = AC2 + DC2AD2 = 52 + 122 = 25 + 144 = 169 AD = 169 = 13 m

Hence, the distance between the tops of the two poles is 13 m.



Page No 442:

Question 7:

Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m  and AC = 5 m
We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled triangle ACD, we have: ​

AD2 = AC2 + DC2AD2 = 52 + 122 = 25 + 144 = 169 AD = 169 = 13 m

Hence, the distance between the tops of the two poles is 13 m.

Answer:


Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2242=182+CA2CA2=576-324CA2=252CA=67 m
Hence, the stake should be driven 67m far from the base of the pole.

Page No 442:

Question 8:


Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2242=182+CA2CA2=576-324CA2=252CA=67 m
Hence, the stake should be driven 67m far from the base of the pole.

Answer:

Applying Pythagoras theorem in right-angled triangle POR, we have:
       PR2 = PO2 + OR2 PR2 = 62 + 82 = 36 + 64 = 100 PR = 100 = 10 cm

In ∆ PQR,

PQ2 + PR2 = 242 + 102 = 576 + 100 = 676and QR2 = 262 = 676 PQ2 + PR2 = QR2

Therefore, by applying Pythagoras theorem, we can say that âˆ†PQR is right-angled at P.

Page No 442:

Question 9:

Applying Pythagoras theorem in right-angled triangle POR, we have:
       PR2 = PO2 + OR2 PR2 = 62 + 82 = 36 + 64 = 100 PR = 100 = 10 cm

In ∆ PQR,

PQ2 + PR2 = 242 + 102 = 576 + 100 = 676and QR2 = 262 = 676 PQ2 + PR2 = QR2

Therefore, by applying Pythagoras theorem, we can say that âˆ†PQR is right-angled at P.

Answer:

It is given that ABC is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
ADB and ADC are right-angled triangles.
Applying Pythagoras theorem, we have:

AB2 = AD2 + BD2BD2 = AB2 - AD2 = 132 - 52BD2 = 169 - 25 = 144BD= 144 = 12

Hence,
BC = 2(BD) = 2 × 12 = 24 cm

Page No 442:

Question 10:

It is given that ABC is an isosceles triangle.
Also, AB = AC = 13 cm
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD = 5 cm
ADB and ADC are right-angled triangles.
Applying Pythagoras theorem, we have:

AB2 = AD2 + BD2BD2 = AB2 - AD2 = 132 - 52BD2 = 169 - 25 = 144BD= 144 = 12

Hence,
BC = 2(BD) = 2 × 12 = 24 cm

Answer:

In isosceles ∆ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC = a2 units
Applying Pythagoras theorem in right-angled ∆ABD, we have:

AB2 = AD2 + BD2AD2 = AB2 - BD2 = (2a)2 - a22AD2 = 4a2 - a24 = 15a24AD = 15a24  = a152 units

Page No 442:

Question 11:

In isosceles ∆ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC = a2 units
Applying Pythagoras theorem in right-angled ∆ABD, we have:

AB2 = AD2 + BD2AD2 = AB2 - BD2 = (2a)2 - a22AD2 = 4a2 - a24 = 15a24AD = 15a24  = a152 units

Answer:



Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.

In right-angled âˆ†ABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
AB2 = AD2 + BD2AD2 = AB2 - BD2 = (2a)2 - a2AD2 = 4a2 - a2 = 3a2AD = 3a units

Similarly,
BE = a3 units and CF = a3 units

Page No 442:

Question 12:



Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.
Then, D, E and F are the midpoints of BC, AC and AB, respectively.

In right-angled âˆ†ABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
AB2 = AD2 + BD2AD2 = AB2 - BD2 = (2a)2 - a2AD2 = 4a2 - a2 = 3a2AD = 3a units

Similarly,
BE = a3 units and CF = a3 units

Answer:

Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:



AB2 = AD2 + BD2 AD2 = 122 - 62    ( BD= 12BC = 6) AD2 = 144 - 36 = 108 AD = 108 = 63 cm

Hence, the height of the given triangle is 63 cm.

Page No 442:

Question 13:

Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:



AB2 = AD2 + BD2 AD2 = 122 - 62    ( BD= 12BC = 6) AD2 = 144 - 36 = 108 AD = 108 = 63 cm

Hence, the height of the given triangle is 63 cm.

Answer:

Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 302 + 162 = 900 + 256 = 1156AC = 1156 = 34 cm

Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm

Page No 442:

Question 14:

Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 302 + 162 = 900 + 256 = 1156AC = 1156 = 34 cm

Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm

Answer:

Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52AB2 = 144 + 25 = 169AB = 169 = 13 cm


Hence, the length of each side of the rhombus is 13 cm.

Page No 442:

Question 15:

Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at right angles.
Applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52AB2 = 144 + 25 = 169AB = 169 = 13 cm


Hence, the length of each side of the rhombus is 13 cm.

Answer:

In right-angled triangle AEB, applying Pythagoras theorem, we have:
AB2 = AE2 + EB2    ...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:

AD2 = AE2 + ED2
 AE2 = AD2 - ED2     ...(ii)

Therefore, AB2 = AD2 - ED2  + EB2   (from (i) and (ii))AB2 = AD2 - ED2 + (BD - DE)2         = AD2 - ED2 + (12BC - DE)2         = AD2  - DE2 + 14BC2 + DE2 - BC.DE          = AD2 + 14BC2 - BC.DE

This completes the proof.

Page No 442:

Question 16:

In right-angled triangle AEB, applying Pythagoras theorem, we have:
AB2 = AE2 + EB2    ...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:

AD2 = AE2 + ED2
 AE2 = AD2 - ED2     ...(ii)

Therefore, AB2 = AD2 - ED2  + EB2   (from (i) and (ii))AB2 = AD2 - ED2 + (BD - DE)2         = AD2 - ED2 + (12BC - DE)2         = AD2  - DE2 + 14BC2 + DE2 - BC.DE          = AD2 + 14BC2 - BC.DE

This completes the proof.

Answer:

Given: ∠ACB = 900 and CD ⊥ AB
To Prove: BC2AC2=BDAD
Proof:
In ACB and CDB
ACB=CDB=90  (Given)
ABC=CBD  (Common)
By AA similarity-criterion ACB~CDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
BCBD=ABBCBC2=BD.AB            .....1
In ACB and ADC
ACB=ADC=90  (Given)
CAB=DAC  (Common)
By AA similarity-criterion ACB~ADC
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
ACAD=ABACAC2=AD.AB         .....2
Dividing (2) by (1), we get
BC2AC2=BDAD

Page No 442:

Question 17:

Given: ∠ACB = 900 and CD ⊥ AB
To Prove: BC2AC2=BDAD
Proof:
In ACB and CDB
ACB=CDB=90  (Given)
ABC=CBD  (Common)
By AA similarity-criterion ACB~CDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
BCBD=ABBCBC2=BD.AB            .....1
In ACB and ADC
ACB=ADC=90  (Given)
CAB=DAC  (Common)
By AA similarity-criterion ACB~ADC
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
ACAD=ABACAC2=AD.AB         .....2
Dividing (2) by (1), we get
BC2AC2=BDAD

Answer:

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

AC2 = AE2 + EC2  b2 = h2  + x + a22 = h2 + x2 + a24 + ax  ...(i)                In right-angled triangle AED, we have:AD2 = AE2 + ED2 p2 = h2 + x2   ...(ii)Therefore,from (i) and (ii), b2 = p2 + ax + a24     

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
AB2 = AE2 + EB2 c2 = h2 + a2 - x2   (BD = a2 and BE = BD - x) c2 = h2 + x2 - ax + a24    (h2 + x2 = p2) c2 = p2 - ax + a24

(iii)

Adding (i) and (ii), we get:
 b2 + c2 = p2 + ax + a24 + p2 - ax + a24               = 2p2 + ax - ax + a2 + a24                = 2p2 + a22

(iv)
Subtracting (ii) from (i), we get:
b2 - c2 = p2 + ax + a24 - (p2 - ax + a24)              = p2 - p2 + ax + ax + a24 - a24               = 2ax



Page No 443:

Question 18:

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

AC2 = AE2 + EC2  b2 = h2  + x + a22 = h2 + x2 + a24 + ax  ...(i)                In right-angled triangle AED, we have:AD2 = AE2 + ED2 p2 = h2 + x2   ...(ii)Therefore,from (i) and (ii), b2 = p2 + ax + a24     

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:
AB2 = AE2 + EB2 c2 = h2 + a2 - x2   (BD = a2 and BE = BD - x) c2 = h2 + x2 - ax + a24    (h2 + x2 = p2) c2 = p2 - ax + a24

(iii)

Adding (i) and (ii), we get:
 b2 + c2 = p2 + ax + a24 + p2 - ax + a24               = 2p2 + ax - ax + a2 + a24                = 2p2 + a22

(iv)
Subtracting (ii) from (i), we get:
b2 - c2 = p2 + ax + a24 - (p2 - ax + a24)              = p2 - p2 + ax + ax + a24 - a24               = 2ax

Answer:

Draw AEBC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD

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Question 19:

Draw AEBC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD

Answer:

We have, ABC as an isosceles triangle, right angled at B.Now, AB = BC

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 2AB2      (AB = AC)   ...(i)

 ACD~ABE

We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides.

 ar(ABE)ar(ACD) = AB2AC2 = AB22AB2 from i = 12 = 1 : 2 

Page No 443:

Question 20:

We have, ABC as an isosceles triangle, right angled at B.Now, AB = BC

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 2AB2      (AB = AC)   ...(i)

 ACD~ABE

We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides.

 ar(ABE)ar(ACD) = AB2AC2 = AB22AB2 from i = 12 = 1 : 2 

Answer:


Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in 112 hours = 1000×32=1500 km
Distance covered by plane B in 112 hours = 1200×32=1800 km
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2=18002+15002=3240000+2250000=5490000AB2=5490000AB=30061 m
Hence, the distance between two planes after 112 hours is 30061 m.

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Question 21:


Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in 112 hours = 1000×32=1500 km
Distance covered by plane B in 112 hours = 1200×32=1800 km
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2=18002+15002=3240000+2250000=5490000AB2=5490000AB=30061 m
Hence, the distance between two planes after 112 hours is 30061 m.

Answer:

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
AC2=AL2+LC2=AD2-DL2+DL+DC2       Using1=AD2-DL2+DL+BC22                               AD is a median=AD2-DL2+DL2+BC22+BC.DLAC2=AD2+BC.DL+BC22            .....(2)
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
AB2=AL2+LB2=AD2-DL2+LB2        Using3=AD2-DL2+BD-DL2=AD2-DL2+12BC-DL2=AD2-DL2+BC22-BC.DL+DL2AB2=AD2-BC.DL+BC22              .....(4)
(c) Adding (2) and (4), we get

AC2+AB2=AD2+BC.DL+BC22 +AD2-BC.DL+BC22=2AD2+BC24+BC24=2AD2+12BC2

Page No 443:

Question 22:

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
AC2=AL2+LC2=AD2-DL2+DL+DC2       Using1=AD2-DL2+DL+BC22                               AD is a median=AD2-DL2+DL2+BC22+BC.DLAC2=AD2+BC.DL+BC22            .....(2)
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL2 = AD2 − DL2                 .....(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
AB2=AL2+LB2=AD2-DL2+LB2        Using3=AD2-DL2+BD-DL2=AD2-DL2+12BC-DL2=AD2-DL2+BC22-BC.DL+DL2AB2=AD2-BC.DL+BC22              .....(4)
(c) Adding (2) and (4), we get

AC2+AB2=AD2+BC.DL+BC22 +AD2-BC.DL+BC22=2AD2+BC24+BC24=2AD2+12BC2

Answer:


Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In â–³BMC
By using Pythagoras theorem, we have
BC2 = CM2 + MB2
= (2.4)2 + (1.8)2
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In â–³BC'M
By using Pythagoras theorem, we have
C'M2 = BC'2 − MB2
= (2.4)2 − (1.8)2
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m



Page No 446:

Question 1:


Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In â–³BMC
By using Pythagoras theorem, we have
BC2 = CM2 + MB2
= (2.4)2 + (1.8)2
= 9
∴ BC = 3 m
Now, BC' = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In â–³BC'M
By using Pythagoras theorem, we have
C'M2 = BC'2 − MB2
= (2.4)2 − (1.8)2
= 2.52
∴ C'M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C'A = C'M + MA
= 1.6 + 1.2
= 2.8 m

Answer:

The two triangles are similar if and only if

1. The corresponding sides are in proportion.
2. The corresponding angles are equal.

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Question 2:

The two triangles are similar if and only if

1. The corresponding sides are in proportion.
2. The corresponding angles are equal.

Answer:

If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

Page No 446:

Question 3:

If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

Answer:

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

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Question 4:

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Answer:

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.

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Question 5:

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.

Answer:

If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.

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Question 6:

If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar.

Answer:

If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

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Question 7:

If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

Answer:

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

Page No 446:

Question 8:

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

Answer:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

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Question 9:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.

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Question 10:

The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and it's always opposite the right angle.

Answer:

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle

Page No 446:

Question 11:

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle

Answer:


By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
DFBC andDF=12BCDF=BE
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In â–³ABC and â–³EFD
∠ABC = ∠EFD         (Opposite angles of a parallelogram)
∠BCA = ∠EDF       (Opposite angles of a parallelogram)
By AA similarity criterion, â–³ABC ∼ â–³EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaDEFareaABC=DFBC2=DF2DF2=14
Hence, the ratio of the areas of â–³DEF and â–³ABC is 1 : 4.



Page No 447:

Question 12:


By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
DFBC andDF=12BCDF=BE
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In â–³ABC and â–³EFD
∠ABC = ∠EFD         (Opposite angles of a parallelogram)
∠BCA = ∠EDF       (Opposite angles of a parallelogram)
By AA similarity criterion, â–³ABC ∼ â–³EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaDEFareaABC=DFBC2=DF2DF2=14
Hence, the ratio of the areas of â–³DEF and â–³ABC is 1 : 4.

Answer:

Now, In â–³ABC and â–³PQR
∠A = ∠P = 700          (Given)
ABPQ=ACPR   34.5=6911.5=11.5
By SAS similarity criterion, â–³ABC ∼ â–³PQR

Page No 447:

Question 13:

Now, In â–³ABC and â–³PQR
∠A = ∠P = 700          (Given)
ABPQ=ACPR   34.5=6911.5=11.5
By SAS similarity criterion, â–³ABC ∼ â–³PQR

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, â–³ABC ∼ â–³DEF
ABDE=BCEFAB2AB=6EFEF=12 cm

Page No 447:

Question 14:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, â–³ABC ∼ â–³DEF
ABDE=BCEFAB2AB=6EFEF=12 cm

Answer:

In â–³ADE and â–³ABC
∠ADE = ∠ABC         (Corresponding angles in DE∥BC)
∠AED = ∠ACB         (Corresponding angles in DE∥BC)
By AA similarity criterion, â–³ADE ∼ â–³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
ADAB=AEAC
ADAD+DB=AEAE+ECxx+3x+4=x+3x+3+3x+19x4x+4=x+34x+22
x2x+2=x+32x+112x2+11x=2x2+2x+6x+63x=6x=2
Hence, the value of x is 2.

Page No 447:

Question 15:

In â–³ADE and â–³ABC
∠ADE = ∠ABC         (Corresponding angles in DE∥BC)
∠AED = ∠ACB         (Corresponding angles in DE∥BC)
By AA similarity criterion, â–³ADE ∼ â–³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
ADAB=AEAC
ADAD+DB=AEAE+ECxx+3x+4=x+3x+3+3x+19x4x+4=x+34x+22
x2x+2=x+32x+112x2+11x=2x2+2x+6x+63x=6x=2
Hence, the value of x is 2.

Answer:


Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2102=82+CA2CA2=100-64CA2=36CA=6 m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

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Question 16:


Let AB be a ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
AB2=BC2+CA2102=82+CA2CA2=100-64CA2=36CA=6 m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

Answer:


We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
AC2=CD2+DA22a2=a2+DA2DA2=4a2-a2DA2=3a2
DA=3 a
Hence, the length of the altitude of an equilateral triangle of side 2a cm is 3a cm.

Page No 447:

Question 17:


We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
∴DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
AC2=CD2+DA22a2=a2+DA2DA2=4a2-a2DA2=3a2
DA=3 a
Hence, the length of the altitude of an equilateral triangle of side 2a cm is 3a cm.

Answer:

We have â–³ABC ∼ â–³DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaABCareaDEF=BCEF264169=BCEF28132=4EF2
813=4EFEF=6.5 cm

Page No 447:

Question 18:

We have â–³ABC ∼ â–³DEF
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaABCareaDEF=BCEF264169=BCEF28132=4EF2
813=4EFEF=6.5 cm

Answer:



In â–³AOB and COD
∠ABO = ∠CDO         (Alternte angles in AB∥CD)
∠AOB = ∠COD         (Vertically opposite angles)
By AA similarity criterion, â–³AOB ∼ â–³COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaAOBareaCOD=ABCD284areaCOD=2CDCD2areaCOD=12 cm2

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Question 19:



In â–³AOB and COD
∠ABO = ∠CDO         (Alternte angles in AB∥CD)
∠AOB = ∠COD         (Vertically opposite angles)
By AA similarity criterion, â–³AOB ∼ â–³COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaAOBareaCOD=ABCD284areaCOD=2CDCD2areaCOD=12 cm2

Answer:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
area of smaller trianglearea of larger triangle=Side of smaller triangleSide of larger triangle248area of larger triangle=232area of larger triangle=108 cm2

Page No 447:

Question 20:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
area of smaller trianglearea of larger triangle=Side of smaller triangleSide of larger triangle248area of larger triangle=232area of larger triangle=108 cm2

Answer:


We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
DC=12a
Now, In right triangle ADC
By using Pythagoras theorem, we have
AC2=CD2+DA2a2=12a2+DA2DA2=a2-14a2DA2=34a2
DA=32 aNow, areaABC=12×BC×AD=12×a×32 a=34 a2

Page No 447:

Question 21:


We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides..
Suppose ABC is an equilateral triangle having AB = BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
DC=12a
Now, In right triangle ADC
By using Pythagoras theorem, we have
AC2=CD2+DA2a2=12a2+DA2DA2=a2-14a2DA2=34a2
DA=32 aNow, areaABC=12×BC×AD=12×a×32 a=34 a2

Answer:


Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
∴AB2 = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm

Page No 447:

Question 22:


Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 12 cm and BO = 5 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + BO2
= 122 + 52
= 144 + 25
= 169
∴AB2 = 169
⇒ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm

Answer:

If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, â–³DEF ∼ â–³GHK
∴∠E = ∠H = 57∘ 
Now, In â–³DEF
∠D + ∠E + ∠F = 180∘      (Angle sum property of triangle)
⇒ ∠F = 180∘  −  48∘  −  57∘  =  75∘ 

Page No 447:

Question 23:

If two traingle are similar then the corresponding angles of the two tringles are equal.
Here, â–³DEF ∼ â–³GHK
∴∠E = ∠H = 57∘ 
Now, In â–³DEF
∠D + ∠E + ∠F = 180∘      (Angle sum property of triangle)
⇒ ∠F = 180∘  −  48∘  −  57∘  =  75∘ 

Answer:

We have
AM : MB = 1 : 2
MBAM=21
Adding 1 to both sides, we get
MBAM+1=21+1MB+AMAM=2+11ABAM=31
Now, In â–³AMN and â–³ABC
∠AMN = ∠ABC         (Corresponding angles in MN∥BC)
∠ANM = ∠ACB         (Corresponding angles in MN∥BC)
By AA similarity criterion, â–³AMN ∼ â–³ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaAMNareaABC=AMAB2=132=19

Page No 447:

Question 24:

We have
AM : MB = 1 : 2
MBAM=21
Adding 1 to both sides, we get
MBAM+1=21+1MB+AMAM=2+11ABAM=31
Now, In â–³AMN and â–³ABC
∠AMN = ∠ABC         (Corresponding angles in MN∥BC)
∠ANM = ∠ACB         (Corresponding angles in MN∥BC)
By AA similarity criterion, â–³AMN ∼ â–³ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaAMNareaABC=AMAB2=132=19

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, â–³BMP ∼ â–³CNR
BMCN=BPCR=MPNR          ....1Now, BMCN=MPNR          Using 1CN=BM×NRMP=9×96=13.5 cmAgain, BMCN=BPCR          Using 1CR=BP×CNBM=5×13.59=7.5 cm
Perimeter of â–³CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm

Page No 447:

Question 25:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, â–³BMP ∼ â–³CNR
BMCN=BPCR=MPNR          ....1Now, BMCN=MPNR          Using 1CN=BM×NRMP=9×96=13.5 cmAgain, BMCN=BPCR          Using 1CR=BP×CNBM=5×13.59=7.5 cm
Perimeter of â–³CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm

Answer:


We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
AB2=BD2+DA2252=72+DA2DA2=625-49DA2=576DA=24 cm



Page No 448:

Question 26:


We know that the altitude drawn from the vertex opposite to the non equal side bisects the non equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
AB2=BD2+DA2252=72+DA2DA2=625-49DA2=576DA=24 cm

Answer:


In right triangle SOW
By using Pythagoras theorem, we have
OW2=WS2+SO2=352+122=1225+144=1369OW2=1369OW=37 m
Hence, the man is 37 m away from the starting point.

Page No 448:

Question 27:


In right triangle SOW
By using Pythagoras theorem, we have
OW2=WS2+SO2=352+122=1225+144=1369OW2=1369OW=37 m
Hence, the man is 37 m away from the starting point.

Answer:

Let DC = x
∴ BD = ax
By using angle bisector theore in â–³ABC, we have
ABAC=BDDCcb=a-xxcx=ab-bxxb+c=abx=abb+c
Now,
a-x=a-abb+c=ab+ac-abb+c=aca+b

Page No 448:

Question 28:

Let DC = x
∴ BD = ax
By using angle bisector theore in â–³ABC, we have
ABAC=BDDCcb=a-xxcx=ab-bxxb+c=abx=abb+c
Now,
a-x=a-abb+c=ab+ac-abb+c=aca+b

Answer:

In â–³AMN and â–³ABC
∠AMN = ∠ABC = 760     (Given)
∠A = ∠A         (Common)
By AA similarity criterion, â–³AMN ∼ â–³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
AMAB=MNBC
AMAM+MB=MNBCaa+b=MNcMN=aca+b

Page No 448:

Question 29:

In â–³AMN and â–³ABC
∠AMN = ∠ABC = 760     (Given)
∠A = ∠A         (Common)
By AA similarity criterion, â–³AMN ∼ â–³ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
AMAB=MNBC
AMAM+MB=MNBCaa+b=MNcMN=aca+b

Answer:


Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + OB2
= 202 + 212
= 400 + 441
= 841
∴AB2 = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm

Page No 448:

Question 30:


Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 900, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB2 = AO2 + OB2
= 202 + 212
= 400 + 441
= 841
∴AB2 = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm

Answer:

(ii) True
Two circles of any radii are similar to each other.

(i) False
Two rectangles are similar if their corresponding sides are proportional.

(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.


(iv) True
Suppose ABC is a triangle and M, N are



Construction: DE is expanded to F sich that EF = DE
To Prove = DE=12BC
Proof: In â–³ADE and â–³CEF
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, â–³ADE ≅ â–³CEF
CF = AD   (CPCT)
⇒ BD = CF
∠ADE = ∠EFC    (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ∥ CF and BD ∥ CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
∵DF = BC and BD ∥ CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
DE=12BC

(v) False
In â–³ABC, AB = 6 cm, ∠A = 45∘ and AC = 8 cm and in â–³DEF, DF = 9 cm, ∠D = 45∘ and DE = 12 cm, then â–³ABC ∼ â–³DEF.

In â–³ABC and â–³DEF
∠A = ∠D = 45∘ 
ABACDEDF
So â–³ABC is not similar to â–³DEF

(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.

(vii) True



Given: â–³ABC ∼ â–³DEF
To Prove = ArABCArDEF=APDQ2
Proof: In â–³ABP and â–³DEQ
∠BAP = ∠EDQ        (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E                   (â–³ABC ∼ â–³DEF)
By AA criterion, â–³ABP ∼ â–³DEQ
ABDE=APDQ           .....(1)
Since, â–³ABC ∼ â–³DEF
ArABCArDEF=ABDE2ArABCArDEF=APDQ2          Using 1

(viii) True



Given: â–³ABC ∼ â–³DEF
To Prove = PerimeterABCPerimeterDEF=APDQ
Proof: In â–³ABP and â–³DEQ
∠B = ∠E                  (∵△ABC ∼ â–³DEF)
∵△ABC ∼ â–³DEF
ABDE=BCEFABDE=2BP2EQABDE=BPEQ

By SAS criterion, â–³ABP ∼ â–³DEQ
ABDE=APDQ           .....(1)
Since, â–³ABC ∼ â–³DEF
PerimeterABCPerimeterDEF=ABDEPerimeterABCPerimeterDEF=APDQ         Using 1




(ix) True

Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ∥ AB ∥ DC and PR ∥ BC ∥ AD
To prove: OA2 + OC2 = OB2 + OD2 
Proof:
OA2 + OC2 = (AS2 + OS2) + (OQ2 + QC2)             [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ2 + OS2) + (OQ2 + DS2
= (BQ2 + OQ2) + (OS2 + DS2)                                 [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB2 + OD2
Hence, LHS = RHS



(x) True

Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
AB2=AO2+OB2AB2=AC22+BD22   Diagonals of a rhombus perpendicularly bisect each otherAB2=AC24+BD244AB2=AC2+BD2AB2+AB2+AB2+AB2=AC2+BD2AB2+BC2+CD2+DA2=AC2+BD2    All sides of a rhombus are equal



Page No 451:

Question 1:

(ii) True
Two circles of any radii are similar to each other.

(i) False
Two rectangles are similar if their corresponding sides are proportional.

(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.


(iv) True
Suppose ABC is a triangle and M, N are



Construction: DE is expanded to F sich that EF = DE
To Prove = DE=12BC
Proof: In â–³ADE and â–³CEF
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, â–³ADE ≅ â–³CEF
CF = AD   (CPCT)
⇒ BD = CF
∠ADE = ∠EFC    (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ∥ CF and BD ∥ CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
∵DF = BC and BD ∥ CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
DE=12BC

(v) False
In â–³ABC, AB = 6 cm, ∠A = 45∘ and AC = 8 cm and in â–³DEF, DF = 9 cm, ∠D = 45∘ and DE = 12 cm, then â–³ABC ∼ â–³DEF.

In â–³ABC and â–³DEF
∠A = ∠D = 45∘ 
ABACDEDF
So â–³ABC is not similar to â–³DEF

(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.

(vii) True



Given: â–³ABC ∼ â–³DEF
To Prove = ArABCArDEF=APDQ2
Proof: In â–³ABP and â–³DEQ
∠BAP = ∠EDQ        (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E                   (â–³ABC ∼ â–³DEF)
By AA criterion, â–³ABP ∼ â–³DEQ
ABDE=APDQ           .....(1)
Since, â–³ABC ∼ â–³DEF
ArABCArDEF=ABDE2ArABCArDEF=APDQ2          Using 1

(viii) True



Given: â–³ABC ∼ â–³DEF
To Prove = PerimeterABCPerimeterDEF=APDQ
Proof: In â–³ABP and â–³DEQ
∠B = ∠E                  (∵△ABC ∼ â–³DEF)
∵△ABC ∼ â–³DEF
ABDE=BCEFABDE=2BP2EQABDE=BPEQ

By SAS criterion, â–³ABP ∼ â–³DEQ
ABDE=APDQ           .....(1)
Since, â–³ABC ∼ â–³DEF
PerimeterABCPerimeterDEF=ABDEPerimeterABCPerimeterDEF=APDQ         Using 1




(ix) True

Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ∥ AB ∥ DC and PR ∥ BC ∥ AD
To prove: OA2 + OC2 = OB2 + OD2 
Proof:
OA2 + OC2 = (AS2 + OS2) + (OQ2 + QC2)             [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ2 + OS2) + (OQ2 + DS2
= (BQ2 + OQ2) + (OS2 + DS2)                                 [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB2 + OD2
Hence, LHS = RHS



(x) True

Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
AB2=AO2+OB2AB2=AC22+BD22   Diagonals of a rhombus perpendicularly bisect each otherAB2=AC24+BD244AB2=AC2+BD2AB2+AB2+AB2+AB2=AC2+BD2AB2+BC2+CD2+DA2=AC2+BD2    All sides of a rhombus are equal

Answer:

(c) 26 m

Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 = 242 + 102AC2 = 576 + 100 = 676AC = 676 = 26

Page No 451:

Question 2:

(c) 26 m

Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 = 242 + 102AC2 = 576 + 100 = 676AC = 676 = 26

Answer:

(b) 10 m

Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:
AD2= DC2 + AC2          = 82 + 62 = 64 + 36 = 100AD = 100 = 10 m

Page No 451:

Question 3:

(b) 10 m

Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Applying Pythagoras theorem in right-angled triangle ACD, we have:
AD2= DC2 + AC2          = 82 + 62 = 64 + 36 = 100AD = 100 = 10 m

Answer:

(c) 1.5 m

Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m 
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
BAC = EDF = 90°ACB = DFE              (Angular elevation of the Sun at the same time)Therefore, by AA similarity theorem, we get:ABC~DEF

ABAC=DEDF1.80.45=6DFDF =6×0.451.8=1.5 m

Page No 451:

Question 4:

(c) 1.5 m

Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m 
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
BAC = EDF = 90°ACB = DFE              (Angular elevation of the Sun at the same time)Therefore, by AA similarity theorem, we get:ABC~DEF

ABAC=DEDF1.80.45=6DFDF =6×0.451.8=1.5 m

Answer:

(d) 30 m

Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
​DE = ?
Now, in right-angled triangles ABC and DEF, we have:
BAC = EDF = 90°ACB = DFE               (Angular elevation of the Sun at the same time)Therefore, by AA similarity theorem, we get:ABC~DEF

ABAC = DEDF63.6 = DE18DE = 6 × 183.6 = 30 m

Page No 451:

Question 5:

(d) 30 m

Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
​DE = ?
Now, in right-angled triangles ABC and DEF, we have:
BAC = EDF = 90°ACB = DFE               (Angular elevation of the Sun at the same time)Therefore, by AA similarity theorem, we get:ABC~DEF

ABAC = DEDF63.6 = DE18DE = 6 × 183.6 = 30 m

Answer:


Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
Now, In â–³ABC and â–³ADE
∠ABC= ∠ADE = 900      
∠A = ∠A             (Common)
By AA-similarity criterion
â–³ABC ∼ â–³ADE

If two triangles are similar, then the the ratio of their corresponding sides are equal.
ABAD=BCDEAB2=12.55AB=5 cm
Hence, the correct answer is option (d).

Page No 451:

Question 6:


Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppsose DA and BA are the shadows of DE and BC respectively.
Now, In â–³ABC and â–³ADE
∠ABC= ∠ADE = 900      
∠A = ∠A             (Common)
By AA-similarity criterion
â–³ABC ∼ â–³ADE

If two triangles are similar, then the the ratio of their corresponding sides are equal.
ABAD=BCDEAB2=12.55AB=5 cm
Hence, the correct answer is option (d).

Answer:

(a) 7 m

Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
  BC2 = AC2 + AB2 AB2 = BC2 - AC2 = 252 - 242 AB2 = 625 - 576 = 49 AB = 49 = 7 m

Page No 451:

Question 7:

(a) 7 m

Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
  BC2 = AC2 + AB2 AB2 = BC2 - AC2 = 252 - 242 AB2 = 625 - 576 = 49 AB = 49 = 7 m

Answer:

Now, In right triangle MOP
By using Pythagoras theorem, we have
MP2=PO2+OM2=122+162=144+256=400MP2=400MP=20 cm
Now, In right triangle MPN
By using Pythagoras theorem, we have
PN2=NM2+MP2=212+202=441+400=841MP2=841MP=29 cm
Hence, the correct answer is option (b).

Page No 451:

Question 8:

Now, In right triangle MOP
By using Pythagoras theorem, we have
MP2=PO2+OM2=122+162=144+256=400MP2=400MP=20 cm
Now, In right triangle MPN
By using Pythagoras theorem, we have
PN2=NM2+MP2=212+202=441+400=841MP2=841MP=29 cm
Hence, the correct answer is option (b).

Answer:

(b) 15 cm, 20 cm

It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x-5) cm.
Applying Pythagoras theorem, we get:

  252 = x2 + (x - 5)2 625 = x2 + x2 + 25 - 10x2x2 - 10x - 600 = 0 x2 - 5x - 300 = 0 x2 - 20x + 15x - 300 = 0 xx - 20 + 15x - 20 = 0 (x - 20)(x + 15) = 0 x - 20 = 0  or  x + 15 = 0 x = 20 or x = -15Side of a triangle cannot be negative.Therefore, x = 20 cm

Now,
x - 5 = 20 - 5 = 15 cm

Page No 451:

Question 9:

(b) 15 cm, 20 cm

It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x-5) cm.
Applying Pythagoras theorem, we get:

  252 = x2 + (x - 5)2 625 = x2 + x2 + 25 - 10x2x2 - 10x - 600 = 0 x2 - 5x - 300 = 0 x2 - 20x + 15x - 300 = 0 xx - 20 + 15x - 20 = 0 (x - 20)(x + 15) = 0 x - 20 = 0  or  x + 15 = 0 x = 20 or x = -15Side of a triangle cannot be negative.Therefore, x = 20 cm

Now,
x - 5 = 20 - 5 = 15 cm

Answer:

(b) 63 cm

Let ABC be the equilateral triangle with AD as its altitude from A.

In right-angled triangle ABD, we have:

AB2 = AD2 + BD2AD2 = AB2 - BD2       = 122 - 62         = 144 - 36 = 108AD = 108 = 63 cm



Page No 452:

Question 10:

(b) 63 cm

Let ABC be the equilateral triangle with AD as its altitude from A.

In right-angled triangle ABD, we have:

AB2 = AD2 + BD2AD2 = AB2 - BD2       = 122 - 62         = 144 - 36 = 108AD = 108 = 63 cm

Answer:

(d) 24 cm

In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
  AB2 = AD2 + BD2 BD2 = AB2 - AD2 BD2 = 132 - 52 BD2 = 169 - 25  BD2 = 144 BD= 144 = 12 cm

Therefore, BC = 2BD = 24 cm

Page No 452:

Question 11:

(d) 24 cm

In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
  AB2 = AD2 + BD2 BD2 = AB2 - AD2 BD2 = 132 - 52 BD2 = 169 - 25  BD2 = 144 BD= 144 = 12 cm

Therefore, BC = 2BD = 24 cm

Answer:

(a) 3 : 4

In ∆ ABD and ∆ACD, we have:
BAD =CAD

Now,

BDDC = ABAC = 68 = 34BD: DC = 3 : 4

Page No 452:

Question 12:

(a) 3 : 4

In ∆ ABD and ∆ACD, we have:
BAD =CAD

Now,

BDDC = ABAC = 68 = 34BD: DC = 3 : 4

Answer:

(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

BDDC = ABAC 45 = 6x x = 5 × 64 = 7.5

Hence, AC = 7.5 cm

Page No 452:

Question 13:

(d) 7.5 cm
It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

BDDC = ABAC 45 = 6x x = 5 × 64 = 7.5

Hence, AC = 7.5 cm

Answer:

By using angle bisector theore in â–³ABC, we have
ABAC=BDDC1014=6-xx10x=84-14x24x=84x=3.5
Hence, the correct answer is option (b).

Page No 452:

Question 14:

By using angle bisector theore in â–³ABC, we have
ABAC=BDDC1014=6-xx10x=84-14x24x=84x=3.5
Hence, the correct answer is option (b).

Answer:

(b) isosceles

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

Page No 452:

Question 15:

(b) isosceles

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

Answer:

(c) 3AB2 = 4AD2

Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:

   AB2 = AD2 + BD2 AB2 = 12AB2 + AD2     ( ABCis equilateral and AD=12AB) AB2 = 14AB2 + AD2 AB2 - 14AB2 = AD234AB2 = AD2 3AB2 = 4AD2

Page No 452:

Question 16:

(c) 3AB2 = 4AD2

Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:

   AB2 = AD2 + BD2 AB2 = 12AB2 + AD2     ( ABCis equilateral and AD=12AB) AB2 = 14AB2 + AD2 AB2 - 14AB2 = AD234AB2 = AD2 3AB2 = 4AD2

Answer:

(c) 16 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:

    AB2 = AO2 + BO2 BO2 = AB2 - AO2 BO2 = 102 - 62 = 100 - 36 = 64 BO = 64 = 8 BD = 2 × BO = 2 × 8 = 16 cm

Hence, the length of the second diagonal BD is 16 cm.

Page No 452:

Question 17:

(c) 16 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
If AC =12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. we get:

    AB2 = AO2 + BO2 BO2 = AB2 - AO2 BO2 = 102 - 62 = 100 - 36 = 64 BO = 64 = 8 BD = 2 × BO = 2 × 8 = 16 cm

Hence, the length of the second diagonal BD is 16 cm.

Answer:

(b) 13 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52            = 144 + 25 = 169AB = 169 = 13

Hence, the length of each side of the rhombus is 13 cm.



Page No 453:

Question 18:

(b) 13 cm

Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52            = 144 + 25 = 169AB = 169 = 13

Hence, the length of each side of the rhombus is 13 cm.

Answer:

(b) trapezium
Diagonals of a trapezium divide each other proportionally.

Page No 453:

Question 19:

(b) trapezium
Diagonals of a trapezium divide each other proportionally.

Answer:

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

OAOC = OBOD3X - 15X - 3 = 2X + 16X - 5 3X - 1 6X - 5 = 2X + 1 5X - 3 18X2 - 15X - 6X + 5 = 10X2 - 6X + 5X - 3 18X2 - 21X + 5 = 10X2 - X - 3 18X2 - 21X + 5 - 10X2 + X + 3 = 0 8X2 - 20X + 8 = 0 4 2X2 - 5X + 2 = 0 2X2 - 5X + 2 = 0 2X2 - 4X - X + 2 = 0 2XX - 2 - 1X - 2 = 0 X - 2 2X - 1 = 0 Either x - 2 = 0 or 2x-1 = 0 Either x = 2 or x = 12When x = 12,  6x - 5 = -2 < 0 , which is not possible.Therefore, x = 2

Page No 453:

Question 20:

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

OAOC = OBOD3X - 15X - 3 = 2X + 16X - 5 3X - 1 6X - 5 = 2X + 1 5X - 3 18X2 - 15X - 6X + 5 = 10X2 - 6X + 5X - 3 18X2 - 21X + 5 = 10X2 - X - 3 18X2 - 21X + 5 - 10X2 + X + 3 = 0 8X2 - 20X + 8 = 0 4 2X2 - 5X + 2 = 0 2X2 - 5X + 2 = 0 2X2 - 4X - X + 2 = 0 2XX - 2 - 1X - 2 = 0 X - 2 2X - 1 = 0 Either x - 2 = 0 or 2x-1 = 0 Either x = 2 or x = 12When x = 12,  6x - 5 = -2 < 0 , which is not possible.Therefore, x = 2

Answer:

(a) a parallelogram

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

Page No 453:

Question 21:

(a) a parallelogram

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

Answer:

(c) isosceles

Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
ABAC=BDDC

It is given that AD bisects BC.
Therefore, BD = DC
 ABAC = 1 AB = AC

Therefore, the triangle is isosceles.

Page No 453:

Question 22:

(c) isosceles

Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
ABAC=BDDC

It is given that AD bisects BC.
Therefore, BD = DC
 ABAC = 1 AB = AC

Therefore, the triangle is isosceles.

Answer:

(a) 30°
We have:

ABAC=BDDC

Applying angle bisector theorem, we can conclude that AD bisects A.

In ABC,A + B + C = 180° A = 180 - B - C A = 180 - 70 - 50 = 60° BAD =CAD =12BAC BAD = 12 × 60 = 30°

Page No 453:

Question 23:

(a) 30°
We have:

ABAC=BDDC

Applying angle bisector theorem, we can conclude that AD bisects A.

In ABC,A + B + C = 180° A = 180 - B - C A = 180 - 70 - 50 = 60° BAD =CAD =12BAC BAD = 12 × 60 = 30°

Answer:

(b) 6 cm

It is given that DEBC.

Applying basic proportionality theorem, we have:

ADBD = AEEC 2.4BD = 3.24.8 BD = 2.4 × 4.83.2 = 3.6 cm

Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm

Page No 453:

Question 24:

(b) 6 cm

It is given that DEBC.

Applying basic proportionality theorem, we have:

ADBD = AEEC 2.4BD = 3.24.8 BD = 2.4 × 4.83.2 = 3.6 cm

Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm

Answer:

(b) 4 cm

It is given that DEBC.

Applying basic proportionality theorem, we get:

    ADAB = AEAC 4.57.2 = AE6.4 AE = 4.5 × 6.47.2 = 4 cm



Page No 454:

Question 25:

(b) 4 cm

It is given that DEBC.

Applying basic proportionality theorem, we get:

    ADAB = AEAC 4.57.2 = AE6.4 AE = 4.5 × 6.47.2 = 4 cm

Answer:

(c) x = 4

It is given that DEBC.
Applying Thales' theorem, we get:

  ADBD = AEEC 7x - 43x + 4 = 5x - 23x 3x7x - 4 = 5x - 2 3x + 4 21x2 - 12x = 15x2 + 20x - 6x - 8 21x2 - 12x =  15x2 + 14x  - 8 21x2 - 12x -15x2 - 14x  + 8 = 0 6x2 - 26x + 8 = 02 3x2 - 13x + 4 = 0 3x2 - 13x + 4 = 0 3x2 - 12x - x + 4 = 0 3xx - 4 - 1 x - 4 = 0 x - 4 3x - 1 = 0 x - 4 = 0  or  3x -1 = 0 x = 4  or  x =13If x = 13, 7x - 4 = -53 < 0;  it is not possible.Therefore, x = 4

Page No 454:

Question 26:

(c) x = 4

It is given that DEBC.
Applying Thales' theorem, we get:

  ADBD = AEEC 7x - 43x + 4 = 5x - 23x 3x7x - 4 = 5x - 2 3x + 4 21x2 - 12x = 15x2 + 20x - 6x - 8 21x2 - 12x =  15x2 + 14x  - 8 21x2 - 12x -15x2 - 14x  + 8 = 0 6x2 - 26x + 8 = 02 3x2 - 13x + 4 = 0 3x2 - 13x + 4 = 0 3x2 - 12x - x + 4 = 0 3xx - 4 - 1 x - 4 = 0 x - 4 3x - 1 = 0 x - 4 = 0  or  3x -1 = 0 x = 4  or  x =13If x = 13, 7x - 4 = -53 < 0;  it is not possible.Therefore, x = 4

Answer:

(d) 2.1 cm

It is given that DEBC.
Applying Thales' theorem, we get:
ADDB = AEECLet AE be x cm.Therefore, EC = 5.6 - x cm 35 = x5.6 - x 3(5.6 - x) = 5x 16.8 -  3x = 5x 8x = 16.8 x = 2.1 cm

Page No 454:

Question 27:

(d) 2.1 cm

It is given that DEBC.
Applying Thales' theorem, we get:
ADDB = AEECLet AE be x cm.Therefore, EC = 5.6 - x cm 35 = x5.6 - x 3(5.6 - x) = 5x 16.8 -  3x = 5x 8x = 16.8 x = 2.1 cm

Answer:

(b) 5.4 cm

∆ABC ∼ ∆DEF

Therefore,

  PerimeterABCPerimeterDEF = BCEF3018 = 9EF EF = 9 × 1830 = 5.4 cm

Page No 454:

Question 28:

(b) 5.4 cm

∆ABC ∼ ∆DEF

Therefore,

  PerimeterABCPerimeterDEF = BCEF3018 = 9EF EF = 9 × 1830 = 5.4 cm

Answer:

(a) 35 cm

 âˆ†ABC ∼ ∆DEF

 Perimeter(ABC)Perimeter(DEF) = ABDE Perimeter(ABC)25 = 9.16.5 Perimeter(ABC) = 9.1 × 256.5 =  35 cm

Page No 454:

Question 29:

(a) 35 cm

 âˆ†ABC ∼ ∆DEF

 Perimeter(ABC)Perimeter(DEF) = ABDE Perimeter(ABC)25 = 9.16.5 Perimeter(ABC) = 9.1 × 256.5 =  35 cm

Answer:

(d) 30 cm

Perimeter of ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm

 âˆ†DEF ∼ ∆ABC

 Perimeter(ABC)Perimeter(DEF)=BCEF 22.5Perimeter(DEF)= 68 Perimeter(DEF) = 22.5 × 86 = 30 cm

Page No 454:

Question 30:

(d) 30 cm

Perimeter of ABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm

 âˆ†DEF ∼ ∆ABC

 Perimeter(ABC)Perimeter(DEF)=BCEF 22.5Perimeter(DEF)= 68 Perimeter(DEF) = 22.5 × 86 = 30 cm

Answer:

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
DECA and DE=12CA
Now, In â–³ABC and â–³EBD
∠BED = ∠BAC             (Corresponding angles)
∠B = ∠B             (Common)
By AA-similarity criterion
â–³ABC ∼ â–³EBD

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaABCareaDBE=ACED2=2EDED2=41
Hence, the correct answer is option (d).

 

Page No 454:

Question 31:

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
DECA and DE=12CA
Now, In â–³ABC and â–³EBD
∠BED = ∠BAC             (Corresponding angles)
∠B = ∠B             (Common)
By AA-similarity criterion
â–³ABC ∼ â–³EBD

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
areaABCareaDBE=ACED2=2EDED2=41
Hence, the correct answer is option (d).

 

Answer:

(b) DE = 12 cm, F = 100°

Disclaimer: In the question, it should be ∆ABC ∼ ∆DFE  instead of  ∆ABC ∼ ∆DEF.

In triangle ABC,
A + B + C = 180° B = 180 - 30 - 50 = 100°

 âˆ†ABC ∼ ∆DFE

 D = A = 30°,F = B = 100°and E = C = 50°Also,ABDF = ACDE     57.5 = 8DE     DE = 8 ×  7.55 = 12 cm

Page No 454:

Question 32:

(b) DE = 12 cm, F = 100°

Disclaimer: In the question, it should be ∆ABC ∼ ∆DFE  instead of  ∆ABC ∼ ∆DEF.

In triangle ABC,
A + B + C = 180° B = 180 - 30 - 50 = 100°

 âˆ†ABC ∼ ∆DFE

 D = A = 30°,F = B = 100°and E = C = 50°Also,ABDF = ACDE     57.5 = 8DE     DE = 8 ×  7.55 = 12 cm

Answer:

(c) BD CD = AD2

In BDA and ADC, we have:BDA = ADC =  90°ABD = 90° - DAB            = 90° - 90° - DAC           = 90° - 90° + DAC           = DACApplying AAsimilarity theorem, we conclude that BDA~ADC. BDAD = ADCD AD2 = BD.CD

Page No 454:

Question 33:

(c) BD CD = AD2

In BDA and ADC, we have:BDA = ADC =  90°ABD = 90° - DAB            = 90° - 90° - DAC           = 90° - 90° + DAC           = DACApplying AAsimilarity theorem, we conclude that BDA~ADC. BDAD = ADCD AD2 = BD.CD

Answer:

AB=63 cmAB2=108 cm2AC=12 cmAC2=144 cm2BC=6 cmBC2=36 cmAC2=AB2+BC2
Since, the square of the longest side is equal to the sum of the square of two sides, so â–³ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)



Page No 455:

Question 34:

AB=63 cmAB2=108 cm2AC=12 cmAC2=144 cm2BC=6 cmBC2=36 cmAC2=AB2+BC2
Since, the square of the longest side is equal to the sum of the square of two sides, so â–³ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)

Answer:

(c) B = D

Disclaimer: In the question, the ratio should be ABDE = BCFD = ACEF.We can write it as:ABED = BCDF = ACFETherefore, ABC ~ EDFHence, the corresponding angles, i.e., B and D, will be equal.i.e., B = D

Page No 455:

Question 35:

(c) B = D

Disclaimer: In the question, the ratio should be ABDE = BCFD = ACEF.We can write it as:ABED = BCDF = ACFETherefore, ABC ~ EDFHence, the corresponding angles, i.e., B and D, will be equal.i.e., B = D

Answer:


(b) DEPQ = EFRP

In ∆DEF and ∆PQR, we have:

D = Q and R = EApplying AA similarity theorem, we conclude that DEF~QRP.Hence, DEQR = DFQP = EFPR

Page No 455:

Question 36:


(b) DEPQ = EFRP

In ∆DEF and ∆PQR, we have:

D = Q and R = EApplying AA similarity theorem, we conclude that DEF~QRP.Hence, DEQR = DFQP = EFPR

Answer:

(c) BCDE = ABEF

∆ABC ∼ ∆EDF
Therefore,
ABDE = ACEF = BCDF BC.DE  AB.EF

Page No 455:

Question 37:

(c) BCDE = ABEF

∆ABC ∼ ∆EDF
Therefore,
ABDE = ACEF = BCDF BC.DE  AB.EF

Answer:

(b) similar but not congruent
 
In ∆ABC and ∆DEF, we have:
B = E and F = CApplying AA similarity theorem, we conclude that ABC~DEF.Also,AB = 3DE AB  DETherefore, ABC and DEF are not congruent.

Page No 455:

Question 38:

(b) similar but not congruent
 
In ∆ABC and ∆DEF, we have:
B = E and F = CApplying AA similarity theorem, we conclude that ABC~DEF.Also,AB = 3DE AB  DETherefore, ABC and DEF are not congruent.

Answer:

(a) ∆PQR ∼ ∆CAB

In ∆ABC and ∆PQR, we have:
ABQR = BCPR = CAPQ ABC~QRPWe can also write it as PQR~CAB.

Page No 455:

Question 39:

(a) ∆PQR ∼ ∆CAB

In ∆ABC and ∆PQR, we have:
ABQR = BCPR = CAPQ ABC~QRPWe can also write it as PQR~CAB.

Answer:

(d) 100°

In  APB and DPC, we have:APB = DPC = 50°APBP = 63 = 2DPCP = 52.5 = 2Hence, APBP = DPCP Applying SAS theorem, we conclude that  APB~DPC. PBA = PCDIn DPC, we have:CDP + CPD + PCD = 180° PCD = 180° - CDP - CPD PCD = 180° - 30° - 50° PCD = 100°Therefore, PBA = 100°

Page No 455:

Question 40:

(d) 100°

In  APB and DPC, we have:APB = DPC = 50°APBP = 63 = 2DPCP = 52.5 = 2Hence, APBP = DPCP Applying SAS theorem, we conclude that  APB~DPC. PBA = PCDIn DPC, we have:CDP + CPD + PCD = 180° PCD = 180° - CDP - CPD PCD = 180° - 30° - 50° PCD = 100°Therefore, PBA = 100°

Answer:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
area of first trianglearea of second triangle=Side of first triangleSide of second triangle2=492=1681
Hence, the correct answer is option (d).

Page No 455:

Question 41:

If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
area of first trianglearea of second triangle=Side of first triangleSide of second triangle2=492=1681
Hence, the correct answer is option (d).

Answer:

(d) 9 : 4
It is given that âˆ† ABC ∼ ∆PQR and BCQR  = 23.
Therefore,
ar(PQR)ar(ABC) = QR2BC2 = 322 = 94

Page No 455:

Question 42:

(d) 9 : 4
It is given that âˆ† ABC ∼ ∆PQR and BCQR  = 23.
Therefore,
ar(PQR)ar(ABC) = QR2BC2 = 322 = 94

Answer:

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DEBC.
Also, by Basic Proportionality Theorem,
ADDB=AEEC

Also, AB = AC = BC ABC is an equilateral triangleSo, ADDB = AEEC = 1In ABC and ADE, we have:A = A ADAB = AEAC = 12 ABC~ADE (SAS criterion) arABC : arADE = AB2 : AD2 arABC : arADE = 22 : 12 arABC : arADE = 4 : 1



Page No 456:

Question 43:

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DEBC.
Also, by Basic Proportionality Theorem,
ADDB=AEEC

Also, AB = AC = BC ABC is an equilateral triangleSo, ADDB = AEEC = 1In ABC and ADE, we have:A = A ADAB = AEAC = 12 ABC~ADE (SAS criterion) arABC : arADE = AB2 : AD2 arABC : arADE = 22 : 12 arABC : arADE = 4 : 1

Answer:

(b) 25 : 49

  • In ABC and DEF, we have:ABDE = BCEF = ACDF = 57Therefore, by SSS criterion, we conclude that ABC~DEF. ar(ABC)ar(DEF) = AB2DE2 = 572 = 2549 = 25 : 49

Page No 456:

Question 44:

(b) 25 : 49

  • In ABC and DEF, we have:ABDE = BCEF = ACDF = 57Therefore, by SSS criterion, we conclude that ABC~DEF. ar(ABC)ar(DEF) = AB2DE2 = 572 = 2549 = 25 : 49

Answer:

(b) 6:7

 âˆ†ABC ∼ ∆DEF

 ABDE = BCEF = ACDF    ...(i)Also,ar(ABC)ar(DEF) = AB2DE2 3649 = AB2DE2 67 = ABDE ABDE = BCEF = ACDF= 67    (from (i))Thus, the ratio of corresponding sides is 6 : 7. 

Page No 456:

Question 45:

(b) 6:7

 âˆ†ABC ∼ ∆DEF

 ABDE = BCEF = ACDF    ...(i)Also,ar(ABC)ar(DEF) = AB2DE2 3649 = AB2DE2 67 = ABDE ABDE = BCEF = ACDF= 67    (from (i))Thus, the ratio of corresponding sides is 6 : 7. 

Answer:

(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
ar(1)ar(2) = 2536 = x2y2 x2y2 = 2536 xy = 2536 = 56 = 5 : 6

Page No 456:

Question 46:

(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
ar(1)ar(2) = 2536 = x2y2 x2y2 = 2536 xy = 2536 = 56 = 5 : 6

Answer:

(b) similar to the original triangle

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Page No 456:

Question 47:

(b) similar to the original triangle

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Answer:

(b) 10 cm

 ABC~QRP ABQR = BCPRNow,ar(ABC)ar(QRP) = 94 (ABQR)2 = 94 ABQR = 32Therefore,ABQR = BCPR = 32Hence, 3PR = 2BC = 2 × 15 = 30PR = 10 cm

Page No 456:

Question 48:

(b) 10 cm

 ABC~QRP ABQR = BCPRNow,ar(ABC)ar(QRP) = 94 (ABQR)2 = 94 ABQR = 32Therefore,ABQR = BCPR = 32Hence, 3PR = 2BC = 2 × 15 = 30PR = 10 cm

Answer:

(c) isosceles and similar

In ∆AOC and ∆ODB, we have:

AOC = DOB   (Vertically opposite angles)and OAC = ODB     (Angles in the same segment)Therefore, by AA similarity theorem, we conclude that AOC~DOB. OCOB = OAOD = ACBDNow, OB = OD OCOA = OBOD = 1 OC = OAHence, OAC and ODB are isosceles and similar.

Page No 456:

Question 49:

(c) isosceles and similar

In ∆AOC and ∆ODB, we have:

AOC = DOB   (Vertically opposite angles)and OAC = ODB     (Angles in the same segment)Therefore, by AA similarity theorem, we conclude that AOC~DOB. OCOB = OAOD = ACBDNow, OB = OD OCOA = OBOD = 1 OC = OAHence, OAC and ODB are isosceles and similar.

Answer:

(d) 90°

Given:
AC = BC
AB2 = 2AC2 = AC2 + AC2 = AC2 + BC2
Applying Pythagoras theorem, we conclude that âˆ†ABC is right angled at C.
or, C = 90°

Page No 456:

Question 50:

(d) 90°

Given:
AC = BC
AB2 = 2AC2 = AC2 + AC2 = AC2 + BC2
Applying Pythagoras theorem, we conclude that âˆ†ABC is right angled at C.
or, C = 90°

Answer:

(b) right-angled

We have:
AB2 + BC2 = 162 + 122 = 256 + 144 = 400and,  AC2 = 202 = 400 AB2 + BC2 = AC2

Hence, âˆ†ABC is a right-angled triangle.

Page No 456:

Question 51:

(b) right-angled

We have:
AB2 + BC2 = 162 + 122 = 256 + 144 = 400and,  AC2 = 202 = 400 AB2 + BC2 = AC2

Hence, âˆ†ABC is a right-angled triangle.

Answer:

(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:
ABC~DEFif ABDE = ACDF = BCEF



Page No 457:

Question 52:

(c) Two triangles are similar if their corresponding sides are proportional.
According to the statement:
ABC~DEFif ABDE = ACDF = BCEF

Answer:

(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.

Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Page No 457:

Question 53:

(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.

Because the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Answer:

(a) - (s)
Let AE be x.
Therefore, EC = 5.6 -x
It is given that DE  BC.
Therefore, by B.P.T., we get:

ADDB = AEEC 35 = x5.6 - x 3(5.6 - x) = 5x 16.8 - 3x = 5x 8x = 16.8 x = 2.1 cm

(b) - (q)

 ABC~DEF ABDE = BCEF 32 = 6EFEF = 6 × 23 = 4 cm

(c) - (p)

 ABC~PQR ar(ABC)ar(PQR) = BC2QR2 916 = 4.52QR2 QR  = 4.5 × 4.5 × 169=4.5 × 43 = 6 cm

(d) - (r)

 AB  CD OAOB=OCOD  (Thales' theorem) 2x + 49x -21 = 2x - 13 3(2x + 4) = (2x - 1)(9x - 21) 6x +12 = 18x2 - 42x - 9x + 21 18x2 - 57x + 9 = 0 6x2 - 19x + 3 = 0 6x2 - 18x - x + 3 = 0 (6x - 1)(x - 3) = 0 x = 3 or x = -16But x =-16 makes (2x-1)<0, which is not possible.Therefore, x = 3



Page No 458:

Question 54:

(a) - (s)
Let AE be x.
Therefore, EC = 5.6 -x
It is given that DE  BC.
Therefore, by B.P.T., we get:

ADDB = AEEC 35 = x5.6 - x 3(5.6 - x) = 5x 16.8 - 3x = 5x 8x = 16.8 x = 2.1 cm

(b) - (q)

 ABC~DEF ABDE = BCEF 32 = 6EFEF = 6 × 23 = 4 cm

(c) - (p)

 ABC~PQR ar(ABC)ar(PQR) = BC2QR2 916 = 4.52QR2 QR  = 4.5 × 4.5 × 169=4.5 × 43 = 6 cm

(d) - (r)

 AB  CD OAOB=OCOD  (Thales' theorem) 2x + 49x -21 = 2x - 13 3(2x + 4) = (2x - 1)(9x - 21) 6x +12 = 18x2 - 42x - 9x + 21 18x2 - 57x + 9 = 0 6x2 - 19x + 3 = 0 6x2 - 18x - x + 3 = 0 (6x - 1)(x - 3) = 0 x = 3 or x = -16But x =-16 makes (2x-1)<0, which is not possible.Therefore, x = 3

Answer:


(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
AB2 + BC2 = AC2 AC = 102 + 202 = 100 + 200 = 103
Hence, the man is 103 m away from the starting point.


(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
AB2 = AD2 + BD2 AD2 = 102 - 52    ( BD= 12BC) AD = 100 - 25 = 75 = 53  cm


(c) - (p)
Area of an equilateral triangle with side a = 34a2 = 34 × 102 = 3 × 5 × 5= 253  cm2  

(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
AC2 = AB2 + BC2 = 82 + 62 = 64 + 36 AC = 100 = 10 m



Page No 462:

Question 1:


(a) - (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
AB2 + BC2 = AC2 AC = 102 + 202 = 100 + 200 = 103
Hence, the man is 103 m away from the starting point.


(b) - (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
AB2 = AD2 + BD2 AD2 = 102 - 52    ( BD= 12BC) AD = 100 - 25 = 75 = 53  cm


(c) - (p)
Area of an equilateral triangle with side a = 34a2 = 34 × 102 = 3 × 5 × 5= 253  cm2  

(d) - (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
AC2 = AB2 + BC2 = 82 + 62 = 64 + 36 AC = 100 = 10 m

Answer:

(b) 7.5 cm

 âˆ†ABC ∼ ∆DEF

 Perimeter(ABC)Perimeter(DEF) = ABDE 3224 = 10DE

 DE = 10 × 2432 = 7.5  cm

Page No 462:

Question 2:

(b) 7.5 cm

 âˆ†ABC ∼ ∆DEF

 Perimeter(ABC)Perimeter(DEF) = ABDE 3224 = 10DE

 DE = 10 × 2432 = 7.5  cm

Answer:

(a) 5.6 cm
 DE âˆ¥ BC

 ADAB = AEAC = DEBC        (Thales' theorem) 3.5AB = 58 AB = 3.5 × 85 = 5.6 cm

Page No 462:

Question 3:

(a) 5.6 cm
 DE âˆ¥ BC

 ADAB = AEAC = DEBC        (Thales' theorem) 3.5AB = 58 AB = 3.5 × 85 = 5.6 cm

Answer:

(b) 13 m


Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have: ​
BD2 = BE2 + ED2           = 122 + 52     ( ED = CD- CE = 11 - 6 )            = 144 + 25 = 169BD= 13 m

Page No 462:

Question 4:

(b) 13 m


Let the poles be AB and CD.
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m.
Draw a perpendicular from B on CD, meeting CD at E.
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have: ​
BD2 = BE2 + ED2           = 122 + 52     ( ED = CD- CE = 11 - 6 )            = 144 + 25 = 169BD= 13 m

Answer:

(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,

2536= 3.52h2
h2 = 3.52×3625h2 =17.64 h = 4.2 cm



Page No 463:

Question 5:

(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,

2536= 3.52h2
h2 = 3.52×3625h2 =17.64 h = 4.2 cm

Answer:

∆ABC ∼ ∆DEF

 ABDE = BCEF 12 = 6EF EF = 12 cm

Page No 463:

Question 6:

∆ABC ∼ ∆DEF

 ABDE = BCEF 12 = 6EF EF = 12 cm

Answer:

 DE âˆ¥ BC

 ADDB = AEEC                  (Basic proportionality theorem)x3x + 4=x + 33x + 19 x(3x + 19) =(x + 3)(3x + 4) 3x2 + 19x = 3x2 + 4x + 9x + 1219x - 13x = 12 6x = 12 x = 2

Page No 463:

Question 7:

 DE âˆ¥ BC

 ADDB = AEEC                  (Basic proportionality theorem)x3x + 4=x + 33x + 19 x(3x + 19) =(x + 3)(3x + 4) 3x2 + 19x = 3x2 + 4x + 9x + 1219x - 13x = 12 6x = 12 x = 2

Answer:

Let the ladder be AB and BC be the height of the window from the ground.


We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:

AB2 = AC2 + BC2 AC2 = AB2 - BC2 = 102 - 82 = 100 - 64 = 36 AC = 6 m

Hence, the foot of the ladder is 6 m away from the base of the wall.

Page No 463:

Question 8:

Let the ladder be AB and BC be the height of the window from the ground.


We have:
AB = 10 m and BC = 8 m
Applying Pythagoras theorem in right-angled triangle ACB, we have:

AB2 = AC2 + BC2 AC2 = AB2 - BC2 = 102 - 82 = 100 - 64 = 36 AC = 6 m

Hence, the foot of the ladder is 6 m away from the base of the wall.

Answer:

Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:



AB2 = AD2 + DB2 AD2 = AB2 - DB2 = 4a2 - a2                 ( BD= 12BC)            = 3a2AD = 3a

Hence, the length of the altitude of an equilateral triangle of side 2a cm is 3a cm.

Page No 463:

Question 9:

Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:



AB2 = AD2 + DB2 AD2 = AB2 - DB2 = 4a2 - a2                 ( BD= 12BC)            = 3a2AD = 3a

Hence, the length of the altitude of an equilateral triangle of side 2a cm is 3a cm.

Answer:

 âˆ†ABC ∼ ∆DEF

 ar(ABC)ar(DEF) = BC2EF2 64169 = 42EF2 EF2 = 16 × 16964 EF = 4 × 138 = 6.5 cm

Page No 463:

Question 10:

 âˆ†ABC ∼ ∆DEF

 ar(ABC)ar(DEF) = BC2EF2 64169 = 42EF2 EF2 = 16 × 16964 EF = 4 × 138 = 6.5 cm

Answer:

In ∆AOB and ∆COD, we have:



AOB = COD (Vertically opposite angles)OAB = OCD (Alternate angles as AB  CD)Applying AA similiarity criterion, we get:AOB~COD ar(AOB )ar(COD) = AB2CD2 84ar(COD) =  ABCD2 84ar(COD) = 2CDCD2 ar(COD) = 844 = 21 cm2

Page No 463:

Question 11:

In ∆AOB and ∆COD, we have:



AOB = COD (Vertically opposite angles)OAB = OCD (Alternate angles as AB  CD)Applying AA similiarity criterion, we get:AOB~COD ar(AOB )ar(COD) = AB2CD2 84ar(COD) =  ABCD2 84ar(COD) = 2CDCD2 ar(COD) = 844 = 21 cm2

Answer:

It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.

   48Area of larger triangle = 2232  48Area of larger triangle = 49 Area of larger triangle = 48 × 94 = 108 cm2

Page No 463:

Question 12:

It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.

   48Area of larger triangle = 2232  48Area of larger triangle = 49 Area of larger triangle = 48 × 94 = 108 cm2

Answer:

LM  CB and LN  CD

Therefore, applying Thales' theorem, we have:
 ABAM = ACAL and ADAN = ACAL ABAM = ADAN AMAB = ANAD
This completes the proof.

Page No 463:

Question 13:

LM  CB and LN  CD

Therefore, applying Thales' theorem, we have:
 ABAM = ACAL and ADAN = ACAL ABAM = ADAN AMAB = ANAD
This completes the proof.

Answer:

Let the triangle be ABC with AD as the bisector of A which meets BC at D.
We have to prove:
BDDC = ABAC  



Draw CE  DA, meeting BA produced at E.
CE DA
Therefore,
2 = 3     (Alternate angles)and 1 = 4             (Corresponding angles)But, 1 = 2Therefore,3 = 4 AE = ACIn BCE, DA CE.Applying Thales' theorem, we gave: BDDC = ABAE BDDC =  ABAC
This completes the proof.

Page No 463:

Question 14:

Let the triangle be ABC with AD as the bisector of A which meets BC at D.
We have to prove:
BDDC = ABAC  



Draw CE  DA, meeting BA produced at E.
CE DA
Therefore,
2 = 3     (Alternate angles)and 1 = 4             (Corresponding angles)But, 1 = 2Therefore,3 = 4 AE = ACIn BCE, DA CE.Applying Thales' theorem, we gave: BDDC = ABAE BDDC =  ABAC
This completes the proof.

Answer:



Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have: 
AB2 = AD2 + BD2 a2 = h2 + (a2)2 h2 = a2 - a24 = 34a2 h = 32a

Therefore,
Area of triangle ABC = 12 × base × height = 12 × a × 32a  = 34a2

This completes the proof.

Page No 463:

Question 15:



Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled triangle ABD, we have: 
AB2 = AD2 + BD2 a2 = h2 + (a2)2 h2 = a2 - a24 = 34a2 h = 32a

Therefore,
Area of triangle ABC = 12 × base × height = 12 × a × 32a  = 34a2

This completes the proof.

Answer:



Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
 If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52 = 144 + 25 = 169AB = 13 cm

Hence, the length of each side of the given rhombus is 13 cm.

Page No 463:

Question 16:



Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
 If AC = 24 cm and BD =10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
AB2 = AO2 + BO2 = 122 + 52 = 144 + 25 = 169AB = 13 cm

Hence, the length of each side of the given rhombus is 13 cm.

Answer:

Let the two triangles be ABC and PQR.
We have:
ABC~PQR,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:
ap = bq = cr = a + b + cp + q + r
  
ABC~PQR; therefore, their corresponding sides will be proportional.
 ap = bq = cr =  k   (say)         ...(i) a = kp, b = kq and  c = kr Perimeter of ABCPerimeter of PQR = a + b + cp + q + r = kp + kq + krp + q + r = k         ...(ii)From (i) and (ii), we get:ap = bq = cr = a + b + cp + q + r= Perimeter of ABCPerimeter of PQR

This completes the proof.



Page No 464:

Question 17:

Let the two triangles be ABC and PQR.
We have:
ABC~PQR,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p

We have to prove:
ap = bq = cr = a + b + cp + q + r
  
ABC~PQR; therefore, their corresponding sides will be proportional.
 ap = bq = cr =  k   (say)         ...(i) a = kp, b = kq and  c = kr Perimeter of ABCPerimeter of PQR = a + b + cp + q + r = kp + kq + krp + q + r = k         ...(ii)From (i) and (ii), we get:ap = bq = cr = a + b + cp + q + r= Perimeter of ABCPerimeter of PQR

This completes the proof.

Answer:



Construction: Draw AXCO and DYBO.As, ar (ABC)ar(DBC)= 12 × AX × BC12 × DY × BC ar (ABC)ar(DBC) = AXDY   ...(i)In ABC and DBC,AXY = DYO = 90° (By construction)AOX = DOY  (Vertically opposite angles)AXODYO (By AA criterion) AXDY = AODO    (Thales's theorem)  ...(ii)From (i) and (ii), we have: ar (ABC)ar(DBC)=AXDY = AODO  or, ar (ABC)ar(DBC) = AODO
This completes the proof.

Page No 464:

Question 18:



Construction: Draw AXCO and DYBO.As, ar (ABC)ar(DBC)= 12 × AX × BC12 × DY × BC ar (ABC)ar(DBC) = AXDY   ...(i)In ABC and DBC,AXY = DYO = 90° (By construction)AOX = DOY  (Vertically opposite angles)AXODYO (By AA criterion) AXDY = AODO    (Thales's theorem)  ...(ii)From (i) and (ii), we have: ar (ABC)ar(DBC)=AXDY = AODO  or, ar (ABC)ar(DBC) = AODO
This completes the proof.

Answer:

In ABC and BXY, we have:B = BBXY = BAC                 (Corresponding angles)Thus, ABC~BXY         (AA criterion) ar(ABC)ar(BXY) = AB2BX2 = AB2AB - AX2    ...(i)Also,  ar(ABC)ar(BXY) = 21   { ar(BXY) = ar(trapezium AXYC)}    ...(ii)From (i) and (ii), we have:AB2AB - AX2 = 21 ABAB - AX = 2 AB - AXAB = 12 1 - AXAB = 12 AXAB = 1 - 12 = 2 - 12 = 2 - 22

Page No 464:

Question 19:

In ABC and BXY, we have:B = BBXY = BAC                 (Corresponding angles)Thus, ABC~BXY         (AA criterion) ar(ABC)ar(BXY) = AB2BX2 = AB2AB - AX2    ...(i)Also,  ar(ABC)ar(BXY) = 21   { ar(BXY) = ar(trapezium AXYC)}    ...(ii)From (i) and (ii), we have:AB2AB - AX2 = 21 ABAB - AX = 2 AB - AXAB = 12 1 - AXAB = 12 AXAB = 1 - 12 = 2 - 12 = 2 - 22

Answer:


Applying Pythagoras theorem in right-angled triangle ADC, we get:

  AC2 = AD2 + DC2 AC2 - DC2 = AD2 AD2 = AC2 - DC2                        ...1

Applying Pythagoras theorem in right-angled triangle ADB, we get:

  AB2 = AD2 + DB2 AB2 - DB2 = AD2 AD2 = AB2 - DB2                  ...(2)

From equation (1) and (2), we have:    AC2 - DC2 = AB2 - DB2 AC2 = AB2 + DC2 - DB2 AC2 = AB2 + (DB + BC)2 - DB2             DB + BC = DC AC2 = AB2 + DB2 + BC2 + 2DB.BC - DB2 AC2 = AB2 + BC2 + 2BC.BD

This completes the proof.

Page No 464:

Question 20:


Applying Pythagoras theorem in right-angled triangle ADC, we get:

  AC2 = AD2 + DC2 AC2 - DC2 = AD2 AD2 = AC2 - DC2                        ...1

Applying Pythagoras theorem in right-angled triangle ADB, we get:

  AB2 = AD2 + DB2 AB2 - DB2 = AD2 AD2 = AB2 - DB2                  ...(2)

From equation (1) and (2), we have:    AC2 - DC2 = AB2 - DB2 AC2 = AB2 + DC2 - DB2 AC2 = AB2 + (DB + BC)2 - DB2             DB + BC = DC AC2 = AB2 + DB2 + BC2 + 2DB.BC - DB2 AC2 = AB2 + BC2 + 2BC.BD

This completes the proof.

Answer:

In PAC and QBC, we have:A = B           Both angles are 90°P = Q           Corresponding anglesandC = C           Common anglesTherefore, PAC ~QBC   APBQ = ACBC

 xz  = a + bb a + b = bxz               ...(1)

In RCA and QBA, we have:C = B           Both angles are 90°R = Q           Corresponding anglesandA = A           Common anglesTherefore, RCA ~QBA    RCBQ = ACAB yz = a + ba a +  b = ayz                        ...(2)
From equation (1) and (2), we have:

 bxz = ayz bx = ay ab = xy                               ...(3)Also, xz = a + bb xz = ab + 1Using the value of ab from equation (3), we have:  xz = xy + 1Dividing both sides by x, we get: 1z = 1y + 1x

 1x + 1y = 1zThis completes the proof.



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