CBSE Solutions for Class 10 English

જવાબ : In △ ABC, it is given that DE∥BC.
Applying Thales' theorem, we get:
AD/DB = AE/EC
∵AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
∴ DB = 10 −- 3.6 = 6.4 cm
or, 3.6/6.4= 4.5EC or, EC = 6.4×4.5/3.6 or, EC =8 cm

જવાબ :

In △ABC, it is given that DE ∥BC.

Applying Thales' Theorem, we get:

AD/DB = AE/EC

Adding 1 to both sides, we get:

AD/DB +1 = AE/EC+1

⇒AB/DB= AC/EC

⇒13.3/DB = 11.9/5.1

⇒DB = 13.3×5.1/11.9 = 5.7 cm Therefore, AD = AB − DB = 13.5 − 5.7 = 7.6 cm

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we get:

AD/DB = AE/EC

⇒4/7= AE/EC

Adding 1 to both the sides, we get:

11/7= AC/EC

⇒EC = 6.6×7/11 = 4.2 cm Therefore, AE = AC −EC= 6.6−4.2 = 2.4 cm

જવાબ : In △ABC, it is given that DE∥BC. Applying Thales' theorem, we get:  AD/AB=AE/AC ⇒8/15= AE/AE + EC⇒ 8/15 = AE/(AE+3.5) ⇒8AE + 28 = 15AE ⇒7AE = 28 ⇒AE = 4 cm

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

AD/DB = AE/EC

⇒x/x−2=x+2/x−1

⇒x(x−1) = (x−2)(x+2)

⇒x²−x = x²−4

⇒x=4 cm

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have: 

AD/DB  = AE/EC

⇒4x−4 = 83x−19

⇒4(3x−19) = 8(x−4)

⇒12x −76 = 8x – 32

⇒4x = 44

⇒x = 11 cm

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

AD/DB = AE/EC

⇒7x−43x+4 = 5x−23x

⇒3x(7x−4) =(5x−2)(3x+4)

⇒21x² − 12x = 15x² +14 x−8

⇒6x²−26x+8 = 0

⇒(x−4)(6x−2) = 0

⇒x = 4, 13∵ x≠13     (as if x=13 then AE will become negative)

∴ x =4 cm

જવાબ : We have:
AD/DB = 5.7/9.5 = 0.6 cm AE/EC= 4.8/8 = 0.6 cm Hence,AD/DB=AE/EC Applying the converse of Thales' theorem, we conclude that DE∥BC

જવાબ : 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,AD/DB = 5.2/6.5=4/5

AE/EC = 4.2/7

Thus, AD/DB≠AE/EC Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.
 

જવાબ : 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,
AD/DB=6.3/4.5=7/5
AE/EC=5.6/4=7/5
⇒AD/DB=AE/EC
Applying the converse of Thales' theorem, 
we conclude that DE∥BC. 
 

જવાબ : 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,
AD/DB = 7.2/4.8=3/2
AE/EC = 6.4/3.6= 16/9
Thus, AD/DB≠AE/EC
Applying the converse of Thales' theorem,
 we conclude that DE is not parallel to BC.

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
⇒5.6/DC=6.4/8
⇒DC = 8×5.6/6.4 = 7 cm

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC =AB/AC
Let BD be x cm.
Therefore, DC = (6−x) cm
⇒x6−x = 10/14
⇒14x = 60−10x
⇒24x = 60
⇒x = 2.5 cm
Thus, BD = 2.5 cm
DC = 6−2.5 = 3.5 cm  

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
BD = 3.2 cm, BC = 6 cm
Therefore, DC = 6−3.2 = 2.8 cm
 ⇒3.2/2.8=5.6/AC ⇒AC = 5.6×2.8/3.2=4.9 cm

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get: 
BD/DC = AB/AC
⇒BD/3 = 5.6/4
⇒BD = 5.6×3/4
⇒BD = 4.2 cm
Hence,  BC = 3 + 4.2 = 7.2 cm
 

જવાબ : Given: ABCD is a parallelogram
To prove: DM/MN=DC/BN 
Proof: In △DMC and △NMB
∠DMC =∠NMB      (Vertically opposite angle)
∠DCM =∠NBM       (Alternate angles)  
By AAA- similarity
△DMC ~ △NMB
∴DM/MN=DC/BN 

જવાબ : Given: ABCD is a parallelogram
To prove: DN/DM=AN/DC  
Proof: In △DMC and △NMB
∠DMC =∠NMB      (Vertically opposite angle)
∠DCM =∠NBM       (Alternate angles)  
By AAA- similarity
△DMC ~ △NMB
∴DM/MN=DC/BN 
Now,  MN/DM=BN/DCNow,  MNDM=BNDC
Adding 1 to both sides, we get
MNDM+1=BNDC+1
⇒(MN+DM)/DM=(BN+DC)/DC
    [∵ABCD is a parallelogram]
⇒DN/DM=AN/DC        
 

જવાબ : Let the trapezium ​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:
PD/DA = PC/CB
Now, E and F are the midpoints of AD and BC, respectively.
⇒ PD2/DE = PC2/CF
⇒ PD/DE = PC/CF
Applying 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.
 

જવાબ : In trapezium ABCD, AB∥CDAB∥CD and the diagonals AC and BD intersect at O.
Therefore,  AO/OC=BO/OD
⇒2x+1/5x−7=7x+1/7x−5
⇒(5x − 7)(7x + 1) = (7x − 5)(2x + 1)
⇒35x2 + 5x − 49x − 7 = 14x2 − 10x + 7x – 5
⇒21x2 − 41x − 2 = 0
⇒21x2 − 42x +x− 2 = 0
⇒21x(x − 2) + 1(x − 2) = 0
⇒(x − 2)(21x+1) = 0
⇒x = 2,−121∵ x ≠ −121∴ x = 2 

જવાબ :

In △ABC, ∠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 △ABC,
∠A + ∠B + ∠C = 180^o                     .....(A)
                                                               (Angle Sum Property of triangle)
Again In △AMN,
∠A +∠AMN + ∠ANM = 180^o          ......(B)
                                                               (Angle Sum Property of triangle)
From (A) and (B), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B  = 2∠AMN  
⇒∠B  = ∠AMN
Since, ∠B  and ∠AMN are corresponding angles.
∴ MN∥BC

જવાબ : In △CAB , PQ ∥ AB.
Applying Thales' theorem, we get:
CPPB = CQQA       ...(A)
 Similarly, applying Thales' theorem in △BDC, 
where PR ∥ BD we get:
CPPB = CRRD               ...(B)
Hence, from (A) and (B), we have:
CQ/QA = CR/RD
Applying the converse of Thales' theorem, we conclude that QR∥AD in △ADC.
This completes the proof.
 

જવાબ :

Join BX and CX.
It is given that BC is bisected at D.
∴ BD = DC ​
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX  bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO ∥ CX and BX ∥ CO
BX ∥ CF and CX ∥ BE
BX ∥ OF and CX ∥ OE
Applying Thales' theorem in Δ∆ABX, we get:
AOAX = AFAB            ...(A)

Also, in △ACX, CX ∥ OE.

Therefore by Thales' theorem, we get: AO/AX = AE/AC           ...(B)
From (A) and (B), we have:
AF/AB = AE/AC
Applying the converse of Thales' theorem in △ABC,

 EF∥CB.
This completes the proof.

જવાબ :

Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = ½ AC      ...(A)
Also, it is given that CQ = ¼ AC        ...(B)
Dividing equation (A) by (A), we get:
CQ/CS = ¼AC/ ½ AC
or, CQ = ½CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in △CSD
PQ∥DS
 if PQ ∥DS ,  we can say that QR∥SB  In △CSB, Q is midpoint of CS and QR∥SB
Applying converse of midpoint theorem ,

we conclude that R is the midpoint of CB.
This completes the proof.

જવાબ : Given:
AD = AE     ...(A)
AB = AC     ...(B)
Subtracting AD from both sides, we get:
⇒ AB −AD = AC - AD
⇒ AB − AD = AC − AE  (Since, AD = AE)
⇒BD = EC    ...(C)
Dividing  equation  (A) by equation (C), we get: AD/DB=AE/EC  Applying the converse of Thales' theorem, DE∥∥BC ⇒∠DEC + ∠ECB  = 180°   (Sum of interior angles on the same side of a transversal line is 180°.)    ⇒∠DEC + ∠CBD =180°  (Since, AB = AC⇒∠B =∠C) Hence, quadrilateral BCED is cyclic. Therefore, B,C,E and D are concyclic points.
 

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

જવાબ : We have:
∠BAC = ∠PQR = 50°
∠ABC =∠QPR = 60°
∠ACB =∠PRQ=70°
Therefore, by AAA similarity theorem, △ABC ~QPR 

જવાબ : We have: 
AB/DF=3/6=1/2 and BC/DE=4.5/9=1/2 
But, ∠ABC ≠∠EDF (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.
 

જવાબ : We have:
CA/QR=8/6=4/3 and CB/PQ=6/4.5=4/3
⇒CA/QR=CB/PQ 
Also, ∠ACB =∠PQR=80°∠ACB =∠PQR=80°
Therefore, by SAS similarity theorem, △ACB ~△RQP.
 

જવાબ : We have
DE/QR=2.5/5=1/2 
EF/PQ=2/4=1/2
 DF/PR=3/6=1/2
⇒DE/QR=EF/PQ=DF/PR
Therefore, by SSS similarity theorem, △FED~△PQR 

જવાબ : In △ABC
∠A+∠B+∠C=180°      (Angle Sum Property)
⇒80°+∠B+70°=180°
⇒∠B=30° 
∠A=∠M and ∠B=∠N
Therefore, by AA similarity theorem,
 △ABC~△MNR

જવાબ : It is given that DB is a straight line.
Therefore,
∠DOC + ∠COB = 180°
∠DOC = 180° − 115° = 65°
 

જવાબ : In △DOC, we have:
∠ODC + ∠DCO + ∠DOC =180°
Therefore,70°+ ∠DCO + 65° = 180°
⇒∠DCO = 180 − 70 − 65 = 45°
 

જવાબ : It is given that △ODC ~△OBA
Therefore,
∠OAB =∠OCD = 45°
 

જવાબ : △ODC ~△OBA△ODC ~△OBA
Therefore,
∠OBA =∠ODC= 70°∠OBA =∠ODC= 70°

જવાબ :  Let OA be x cm.
∵ △OAB ~△OCD 
∴ OA/OC=AB/CD
⇒x/3.5=8/5
⇒x = 8 × 3.5/5 = 5.6  
Hence, OA = 5.6 cm

જવાબ : Let OD be y cm
∵∵ △OAB ~△OCD 
∴ AB/CD=OB/OD
⇒8/5 = 6.4y
⇒y = 6.4 × 5/8 = 4
Hence, DO = 4 cm

જવાબ : Given:
∠ADE = ∠ABC  and ∠A = ∠A  
Let DE be x cm
Therefore, by AA similarity theorem, △ADE~△ABC 
⇒AD/AB = DE/BC
⇒3.8/3.6 + 2.1 = x/4.2
⇒x = 3.8 × 4.2/5.7 = 2.8
Hence, DE = 2.8 cm
 

જવાબ : It is given that triangles ABC and PQR are similar.
Therefore,
Perimeter (△ABC)/Perimeter (△PQR) =AB/PQ
⇒32/24=AB/12
⇒AB=32×12/24=16 cm

જવાબ : Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.
⇒Perimeter of △ABC/Perimeter of △DEF=BC/EF
Let the perimeter of ∆ABC be x cm.
Therefore,
x/25=9.1/6.5
⇒x=9.1×25/6.5=35
Thus, the perimeter of ∆ABC is 35 cm.

જવાબ : In △BDA and △BAC, we have:
∠BDA =∠BAC = 90°
∠DBA = ∠CBA          (Common)
Therefore, by AA similarity theorem, 
△BDA~△BAC
⇒AD/AC=AB/BC  ⇒AD/0.75=1/1.25
⇒AD=0.75/1.25=0.6 m or 60 cm

જવાબ : It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In△BDC and △ABC, we have:
∠ABC = ∠BDC = 90° (given)
∠C = ∠C  (common)
By AA similarity theorem,
 we get:△BDC~△ABC
AB/BD = BC/DC
⇒5.7/3.8 = BC/5.4
⇒BC = 5.7/3.8 × 5.4 = 8.1
Hence, BC = 8.1 cm

જવાબ : 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 = ∠CDB
∠DBA = ∠DCB = 90°
Therefore, by AA similarity theorem, we get:
△DBA~△DCB
⇒BD/CD = AD/BD
⇒CD = BD2/AD
CD  = 8×8/4  = 16cm
 

જવાબ : We have:
AP/AB = 2/6 = 1/3 and AQ/AC = 3/9 = 1/3
⇒AP/AB = AQ/AC
In △APQ and △ABC, we have:
AP/AB = AQ/AC
∠A = ∠A
Therefore, by AA similarity theorem, we get:
△APQ~△ABC
Hence, PQ/BC = AQ/AC = 1/3
⇒PQ/BC = 1/3
⇒BC = 3PQ
This completes the proof.

જવાબ : In △BED and △ACB, we have: ∠BED = ∠ACB = 90°
∵ ∠B + ∠C = 180°
∴ BD ∥ AC
∠EBD = ∠CAB (Alternate angles)
Therefore, by AA similarity theorem, we get:
△BED~△ACB
⇒ BE/AC = DE/BC
⇒ BE/DE = AC/BC
This completes the proof.
 

જવાબ : 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
⇒AB/BC=PQ/QR 
⇒7.5/5=x/24=36 m 
Therefore, PQ = 36 m
Hence, the height of the tower is 36 m.
 

જવાબ : 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
⇒ AP/AC = AC/BQ (∵ AC = BC) Thus, by SAS similarity theorem, we get:
△APC~△BCQ 
This completes the proof.

જવાબ : We have:
AC/BD = CB/CE
⇒AC/CB = BD/CE
⇒AC/CB = CD/CE  (Since, BD = DC as ∠1 = ∠2)
Also, ∠1 = ∠2i.e, ∠DBC = ∠ACB
Therefore, by SAS similarity theorem, we get:△ACB~△DCE
 

જવાબ : In △ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and PQ=1/2 BC                                             .....(A) Similarly, In △ADC, QR=1/2AD=1/2BC                          .....(B)
Now, In △BCD, SR=1/2BC                                                .....(C) Similarly, In △ABD, PS=1/2AD =1/2 BC                           .....(D)
Using (A), (B), (C) and (D)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.
 

જવાબ : AB and CD are two chords
To Prove:
△PAC~△PDB  Proof: In △PAC and △PDB△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.
∴PA/PD=PC/PB
⇒PA.PB=PC.PD
 

જવાબ : We have: 
∠AFD = ∠EFB    (Vertically Opposite angles) 
∵ DA ∥ BC
∴ ∠DAF =∠BEF      (Alternate angles)
△DAF~△BEF          (AA similarity theorem)
⇒ AF/EF = FD/FB 
or, AF × FB = FD × EF 
This completes the proof.
 

જવાબ :

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

જવાબ :

1-C, 2-A, 3-B, 4-D
 

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-D, 2-C, 3-A, 4-B
 

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-C, 2-A, 3-B, 4-D