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Page No 368:

Question 1:

D and E are points on the sides AB and AC, respectively, of a ABC, such that DEBC.


(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.
(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.
(iii) If ADDB=47 and AC=6.6 cm, find AE.
(iv) If ADAB=815and EC=3.5 cm, find AE.

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 368:

Question 2:

D and E are points on the sides AB and AC respectively of a ABC such that DEBC. Find the value of x, when


(i) AD=x cm, DB=(x-2)cm,AE=(x+2) cm and EC=(x-1) cm.
(ii) AD=4 cm, DB=(x-4) cm, AE=8 cmand EC=(3x-19) cm.
(iii) AD=(7x-4) cm, AE=(5x-2) cm,DB=(3x+4) cm and EC=3x 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 368:

Question 3:

D and E are points on the sides AB and AC respectively of a ABC. In each of the following cases, determine whether DEBC or not.


(i) AD=5.7 cm, DB=9.5 cm, BD=4.8 cm and EC=8 cm.
(ii) AB=11.7 cm, AC=11.2 cm, BD=6.5 cm and AE=4.2 cm.
(iii) AB=10.8 cm, AD=6.3 cm, AC=9.6 cm and EC=4 cm.
(iv) AD=7.2 cm, AE=6.4 cm, AB=12 cm and AC=10 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 368:

Question 4:

In a ABC, AD is the bisector or A.


(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.
(ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
(iii) If AB = 5.6 cm, BD = 3.8 cm and BC = 6 cm, find AC.
(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find 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 369:

Question 5:

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that

iDMMN=DCBNiiDNDM=ANDC

Answer:

Given: ABCD is a parallelogram
To prove: 
iDMMN=DCBNiiDNDM=ANDC
Proof: In △DMC and △NMB
DMC =NMB      (Vertically opposite angle)
DCM =NBM       (Alternate angles)  
By AAA- similarity
△DMC ~ △NMB
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 369:

Question 6:

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

Answer:

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:
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 369:

Question 7:

In the adjoining figure, ABCD is a trapezium in which CDAB and its diagonals intersect at O. If AO = (2x + 1) cm, OC = (5x – 7) cm, DO = (7x − 5) cm and OB = (7x + 1) cm, find the value of x.

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 369:

Question 8:

In △ABC, M and N are points on AB and AC respectively such that BM = CN. If ∠B = ∠C then show that MN∥BC.

Answer:


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 = 180o                     .....(1)
                                                               (Angle Sum Property of triangle)
Again In In △AMN,
∠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 369:

Question 9:

ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQAB and PR BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QRAD.

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 369:

Question 10:

In the given figure, side BC of ∆ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F, respectively, and AD is produced to X, so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EFBC.

Answer:


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  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 369:

Question 11:

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ=14AC. If PQ produced meets BC at R, prove that R is the midpoint BC.

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 369:

Question 12:

In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.

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 370:

Question 13:

In ∆ABC, the bisector of ∠B meets AC at D. A line PQAC meets AB, BC and BD at P, Q and R respectively.
Show that PR × BQ = QR × BP.

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 395:

Question 1:

Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form.
 

(i)
(ii)
(iii)
(iv)                         
(v)                        

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 396:

Question 2:

In the given figure ODC~OBA,BOC=115° and CDO=70°. Find


(i) ∠DOC
(ii) ∠DCO
(iii) ∠OAB
(iv) ∠OBA.

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 396:

Question 3:

In the given figure OAB~OCD. If AB=8 cm, BO=6.4 cm, OC=3.5 and CD=5 cm, Find
(i) OA
(ii) DO.

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 397:

Question 4:

In the given figure, if ∠ADE = ∠B, show that ∆ADE ∼ ∆ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

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 397:

Question 5:

The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm. find AB.

Answer:

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

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

Page No 397:

Question 6:

The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. If the perimeter of ∆ DEF is 25 cm, find the perimeter of ∆ABC.

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 397:

Question 7:

In the given figure, ∠CAB = 90° and AD BC. Show that ∆ BDA ∼ ∆ BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

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 397:

Question 8:

In the given figure, ∠ABC = 90° and BD AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

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 397:

Question 9:

In the given figure, ∠ABC = 90° and BD AC. If BD = 8 cm, AD = 4 cm, find CD.

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 397:

Question 10:

P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ.

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 398:

Question 11:

ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.

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 398:

Question 12:

In the given figure, DBBC, DEAB and ACBC.
Prove that BEDE=ACBC.

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 398:

Question 13:

A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

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 398:

Question 14:

In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC2.
Prove that ∆ ACP ∼ ∆ BCQ.

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 398:

Question 15:

In the given figure, 1=2 and ACBD=CBCE.
Prove that ∆ ACB ∼ ∆ DCE.

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 398:

Question 16:

ABCD is a quadrilateral in which AD = BC. If P, Q, R , S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Answer:

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

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

Similarly, In △ABD, 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 398:

Question 17:

In a circle, two chords AB and CD intersect at a point inside the circle. Prove that
aPAC~PDBbPA.PB=PC.PD

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 399:

Question 18:

Two chords AB and CD of a circle intersect at a point outside the circle. Prove that
aPAC~PDBbPA.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 399:

Question 19:

In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC. If  DP⊥AB and DQ⊥BC then
prove that
(a) DQ2 = DP.QC
(b) DP2 = DQ.AP

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 △BDC, we get

△CQD ∼ △DQB
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, △APD ∼ △DPB
APDP=PDPBDP2=AP.PBDP2=AP.DQ      DQ=PB



Page No 413:

Question 1:

ABC ∼ ∆ DEF and their areas are respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

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 413:

Question 2:

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

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 413:

Question 3:

ABC ∼ ∆PQR and ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, find QR.

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 413:

Question 4:

The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm. find the longest side of the smaller triangle.

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 413:

Question 5:

ABC ∼ ∆DEF and their areas are respectively 100 cm2 and 49 cm2. If the altitude of ∆ABC is 5 cm, find the corresponding altitude of ∆DEF.

Answer:



It is given that ∆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 413:

Question 6:

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

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 414:

Question 7:

The areas of two similar triangles are 81 cm2 and 49 cm2, respectively. If the altitude of one triangle is 6.3 cm, find the corresponding altitude of the other triangle.

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 414:

Question 8:

The areas of two similar triangles are 100 cm2 and 64 cm2 respectively. If a median of the similar triangle is 5.6 cm, find the corresponding median of the other.

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 414:

Question 9:

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that are of ∆APQ is 116 of the area of ∆ABC.

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 ∆APQ~∆ABC.

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

Hence proved.

Page No 414:

Question 10:

In the given figure, DEBC. If DE = 3 cm, BC = 6 cm and ar(∆ADE) = 15 cm2, find the area of ∆ABC.

Answer:

It is given that DE ∥ 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 414:

Question 11:

ABC is right-angled at A and AD ⊥ BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ∆ABC and ∆ADC.

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 414:

Question 12:

In the given figure, DEBC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

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 414:

Question 13:

In ∆ABC, D and E are the midpoints of AB and AC, respectively. Find the ratio of the areas of ∆ADE and ∆ABC.

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 437:

Question 1:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) 9 cm, 16 cm, 18 cm
(ii) 7 cm, 24 cm, 25 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
(iv) 1.6 cm, 3.8 cm, 4 cm
(v) (a-1) cm, 2a cm, (a+1) cm

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 437:

Question 2:

A man goes 80 m due east and then 150 m due north. How far is he from the starting point?

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 437:

Question 3:

A man goes 10 m due south and then 24 due west. How far is he 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 437:

Question 4:

A 13 m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

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 437:

Question 5:

A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground. Find the length of the ladder.

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 437:

Question 6:

Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their bases is 12 m, find the distance between their tops.

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 438:

Question 7:

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

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 438:

Question 8:

In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ∆PQR is right-angled.
    

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 ∆PQR is right-angled at P.

Page No 438:

Question 9:

ABC is an isosceles triangle with AB = AC = 13 cm. The length of the altitude from A on BC is 5 cm. Find BC.

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 438:

Question 10:

Find the length of altitude AD of an isosceles ∆ABC in which AB = AC = 2a units and BC = a units.

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 438:

Question 11:

ABC is an equilateral triangle of a side 2a units. Find each of its altitudes.

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 ∆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 438:

Question 12:

Find the height of an equilateral triangle of side 12 cm.

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 438:

Question 13:

Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 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 438:

Question 14:

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

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 438:

Question 15:

In ∆ABC, D is the midpoint of BC and AE ⊥ BC. If AC > AB, show that
AB2=AD2-BC·DE+14BC2.

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 438:

Question 16:


In the given figure, ∠ACB = 900 and CD ⊥ AB. Prove that

BC2AC2=BDAD

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 438:

Question 17:

In the given figure, D is the midpoint of side BC and AEBC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that


(i) b2=p2+ax+a24
(ii) c2=p2-ax+a24
(iii) (b2+c2)=2p2+12a2
(iv) (b2-c2)=2ax

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 439:

Question 18:

In ∆ABC, AB = AC. Side BC is produced to D. Prove that
(AD2AC2)=BDCD.

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

Page No 439:

Question 19:

ABC is an isosceles triangle, right-angled at B. similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

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 439:

Question 20:

An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another plane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will the two planes be after 112 hours?

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.

Page No 439:

Question 21:


In △ABC, AD is a median and AL ⊥ BC.
Prove that
i AC2=AD2+BC.DL+BC22ii AB2=AD2-BC.DL+BC22iii AB2+AC2=2AD2-14BC2

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 439:

Question 22:

Naman is doing fly-fishing in a stream. The tip of his fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from him and 2.4 m from the point directly under the tip of the road. Assuming that the string(from the top of his road to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from him after seconds.

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 △BMC
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 △BC'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 442:

Question 1:

State the two properties which are necessary for given two triangles to be similar.

Answer:

The two triangles are similar if and only if

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

Page No 442:

Question 2:

State the basic proportionality theorem.

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 442:

Question 3:

State the converse of Thale's theorem.

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.

Page No 442:

Question 4:

State the mid point theorem.

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.

Page No 442:

Question 5:

State the AAA-similarity criterion.

Answer:

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

Page No 442:

Question 6:

State the AA-similarity criterion.

Answer:

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

Page No 442:

Question 7:

State the SSS-similarity criterion.

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 442:

Question 8:

State the SAS-similarity criterion.

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.

Page No 442:

Question 9:

State the Pythagoras' theorem.

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.

Page No 442:

Question 10:

State the converse of Pythagoras' theorem.

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 442:

Question 11:

If D, E and F are respectively the midpoints of sides AB, BC and CA of △ABC then what is the ratio of the areas of △DEF and △ABC?

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 △ABC and △EFD
∠ABC = ∠EFD         (Opposite angles of a parallelogram)
∠BCA = ∠EDF       (Opposite angles of a parallelogram)
By AA similarity criterion, △ABC ∼ △EFD
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 △DEF and △ABC is 1 : 4.



Page No 443:

Question 12:

Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, ∠A = 700 PR = 9 cm, ∠P = 700 and PQ = 4.5 cm. Show that
△ABC ∼ △PQR and state similarity theorem.

Answer:

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

Page No 443:

Question 13:

If △ABC ∼ △DEF such that 2AB = DE and BC = 6 cm, find EF

Answer:

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

Page No 443:

Question 14:

In the given figure, DE∥BC such that AD = x  cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x

Answer:

In △ADE and △ABC
∠ADE = ∠ABC         (Corresponding angles in DE∥BC)
∠AED = ∠ACB         (Corresponding angles in DE∥BC)
By AA similarity criterion, △ADE ∼ △ABC
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 443:

Question 15:

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall is 6 m.

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.

Page No 443:

Question 16:

Find the length of the altitude of an equilateral triangle of side 2a cm.

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 443:

Question 17:

△ABC ∼ △DEF such that ar(△ABC) = 64 cm2 and ar(△DEF) = 169 cm2 If BC = 4 cm, find EF

Answer:

We have △ABC ∼ △DEF
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 443:

Question 18:

In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm2
Find ar(△COD)

Answer:



In △AOB and COD
∠ABO = ∠CDO         (Alternte angles in AB∥CD)
∠AOB = ∠COD         (Vertically opposite angles)
By AA similarity criterion, △AOB ∼ △COD
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

Page No 443:

Question 19:

The corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2 , find the area of larger triangle.

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 443:

Question 20:

In an equilateral triangle with side a prove that area=34a2.

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 443:

Question 21:

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

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 443:

Question 22:

Two traingles DEF and GHK are such that ∠D = 48and ∠H = 57 . If △DEF ∼ △GHK, then find the measure of  ∠F.

Answer:

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

Page No 443:

Question 23:

In the given figure, MN∥BC and AM : MB = 1 : 2.
FindareaAMNareaABC

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 △AMN and △ABC
∠AMN = ∠ABC         (Corresponding angles in MN∥BC)
∠ANM = ∠ACB         (Corresponding angles in MN∥BC)
By AA similarity criterion, △AMN ∼ △ABC
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 443:

Question 24:

In a triangle BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR = 9 cm. If △BMP ∼ △CNR, then find the perimeter of the △CNR

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
Here, △BMP ∼ △CNR
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 △CNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm

Page No 443:

Question 25:

Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 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 444:

Question 26:

A man goes 12 m due south and then 35 m due west. How far is he from the starting point?

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 444:

Question 27:

If the length of the sides BC, CA and AB of a △ABC are a, b and c respectively and AD is the bisector of ∠A then find the length of BD and DC.

Answer:

Let DC = x
∴ BD = ax
By using angle bisector theore in △ABC, 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 444:

Question 28:

In the given figure, ∠AMN = ∠MBC = 760 If p, q and r are the lengths of AM, MB and BC respectively, then express the length of MN  in terms of p, q and r.

Answer:

In △AMN and △ABC
∠AMN = ∠ABC = 760     (Given)
∠A = ∠A         (Common)
By AA similarity criterion, △AMN ∼ △ABC
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 444:

Question 29:

The length of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of each sides of the rhombus

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 444:

Question 30:

For each of the following statements state whether true(T) or false(F).
(i) Two circles with different radii are similar.
(ii) Any two rectangles are similar.
(iii) If two traingles are similar, the their corresponding angles are equal and their corresponding sides are equal.
(iv) The length of the line segment joining the mid points of any two sides of a trinagle is equal to the half the length of the third side.
(v) In △ABC, AB = 6 cm, ∠A = 450 and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 450 and DE = 12 cm, then △ABC ∼ △DEF.
(vi) The polygon formed by joining the mid points of the sides of a quadrilateral is a rhombus.
(vii) The ratio of areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
(viii) The ratio of the perimeters of two similar triangles is same as the ratio of their corresponding medians.
(ix) If O is any point inside a rectangle ABCD then OA2 + OC2 = OB2 + OD2
(x) The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.

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 △ADE and △CEF
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, △ADE ≅ △CEF
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 △ABC, AB = 6 cm, ∠A = 45 and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 45 and DE = 12 cm, then △ABC ∼ △DEF.

In △ABC and △DEF
∠A = ∠D = 45 
ABACDEDF
So △ABC is not similar to △DEF

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

(vii) True



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

(viii) True



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

By SAS criterion, △ABP ∼ △DEQ
ABDE=APDQ           .....(1)
Since, △ABC ∼ △DEF
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 447:

Question 1:

A man goes 24 m due west and them 10 m due north. How far is he from the starting point?
(a) 34 m
(b) 17 m
(c) 26 m
(d) 28 m

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 447:

Question 2:

Two poles of height 13 m and 7 m respectively stand vertically on a plane ground at a distance of 8 m from each other. The distance between their tops is
(a) 9 m
(b) 10 m
(c) 11 m
(d) 12 m

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 447:

Question 3:

A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?
(a) 2.4 m
(b) 1.35 m
(c) 1.5 m
(d) 13.5 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 447:

Question 4:

A vertical pole 6 m long casts a shadow of length 3.6 m on the ground. What is the height of a tower which casts a shadow of length 18 m at the same time?
(a) 10.8 m
(b) 28.8 m
(c) 32.4 m
(d) 30 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 447:

Question 5:

The shadow of a 5-m-long stick is 2 m long. At the same time the length of the shadow of a 12.5 m high tree (in m) is
(a) 3.0
(b) 3.5
(c) 4.5
(d) 5.0

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 △ABC and △ADE
∠ABC= ∠ADE = 900      
∠A = ∠A             (Common)
By AA-similarity criterion
△ABC ∼ △ADE

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 447:

Question 6:

A ladder 25 m long just reaches the top of a building 24 m high from the ground. What is the distance of the foot of the ladder from the building?
(a) 7 m
(b) 14 m
(c) 21 m
(d) 24.5 m

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 447:

Question 7:

In the given figure, O is the point inside a △MNP such that ∠MOP = 900 , OM = 16 cm and OP = 12 cm. If MN = 21 cm and ∠NMP = 900 then
NP = ?
(a) 25 cm
(b) 29 cm
(c) 33 cm
(d) 35 cm

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 447:

Question 8:

The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer. The lengths of these sides are
(a) 10 cm, 15 cm
(b) 15 cm, 20 cm
(c) 12 cm, 17 cm
(d)13 cm, 18 cm

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 447:

Question 9:

The height of an equilateral triangle having each side 12 cm, is
(a) 62 cm
(b) 63 cm
(c) 36 cm
(d) 66 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



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Question 10:

ABC is an isosceles triangle with AB = AC = 13 cm and the length of altitude from A on BC is 5 cm. Then, BC = ?
(a) 12 cm
(b) 16 cm
(c) 18 cm
(d) 24 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

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Question 11:

In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of ∠A. Then, BD : DC =?


(a) 3 : 4
(b) 9 : 16
(c) 4 : 3
(d) 3:2

Answer:

(a) 3 : 4

In ∆ ABD and ∆ACD, we have:
BAD =CAD

Now,

BDDC = ABAC = 68 = 34BD: DC = 3 : 4

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Question 12:

In a ∆ABC, it is given that AD is the internal bisector of ∠A. If BD = 4 cm, DC = 5 cm and AB = 6 cm, then AC = ?


(a) 4.5 cm
(b) 8 cm
(c) 9 cm
(d) 7.5 cm

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

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Question 13:

In a △ABC, it is given that AD is the internal bisector of ∠A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, the CD = ?
(a) 4.8 cm
(b) 3.5 cm
(c) 7 cm
(d) 10.5 cm

Answer:

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

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Question 14:

In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is
Figure

(a) right-angle
(b) isosceles
(c) scalene
(d) obtuse-angled

Answer:

(b) isosceles

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

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Question 15:

In an equilateral triangle ABC, if AD ⊥ BC, then which of the following is true?


(a) 2AB2 = 3AD2
(b) 4AB2 = 3AD2
(c) 3AB2 = 4AD2
(d) 3AB2 = 2AD2

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

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Question 16:

In a rhombus of side 10 cm, one of the diagonals is 12 cm long. The length of the second diagonal is
(a) 20 cm
(b) 18 cm
(c) 16 cm
(d) 22 cm

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.

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Question 17:

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. The length of each side of the rhombus is
(a) 12 cm
(b) 13 cm
(c) 14 cm
(d) 17 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.



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Question 18:

If the diagonals of a quadrilateral divide each other proportionally, then it is a
(a) parallelogram
(b) trapezium
(c) rectangle
(d) square

Answer:

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

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Question 19:

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA = (3x −1) cm, OB = (2x + 1) cm, OC = (5x − 3) cm and OD = (6x − 5) cm. Then, x = ?


(a) 2
(b) 3
(c) 2.5
(d) 4

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 449:

Question 20:

The line segments joining the midpoints of the adjacent sides of a quadrilateral form
(a) a parallelogram
(b) a rectangle
(c) a square
(d) a rhombus

Answer:

(a) a parallelogram

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

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Question 21:

If the bisector of an angle of a triangle bisects the opposite side, then the triangle is
(a) scalene
(b) equilateral
(c) isosceles
(d) right-angled

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.

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Question 22:

In ∆ABC, it is given that ABAC=BDDC. IfB=70° and C=50°, then BAD=?


(a) 30°
(b) 40°
(c) 45°
(d) 50°

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°

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Question 23:

In ∆ABC, DE ∥ BC so that AD = 2.4 cm, AE = 3.2 cm and EC = 4.8 cm. Then, AB = ?


(a) 3.6 cm
(b) 6 cm
(c) 6.4 cm
(d) 7.2 cm

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

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Question 24:

In ∆ABCDE is drawn parallel to BC, cutting AB and AC at D and E, respectively, such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Find AE.


(a) 5.4 cm
(b) 4 cm
(c) 3.6 cm
(d) 3.2 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



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Question 25:

In ∆ABC, DEBC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have:


(a) x = 3
(b) x = 5
(c) x = 4
(d) x = 2.5

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

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Question 26:

In ABC,DEBC such that ADDB=35.
If AC = 5.6 cm, then AE = ?


(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm

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

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Question 27:

ABC ∼ ∆DEF and the perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm respectively. If BC = 9 cm, then EF = ?
(a) 6.3 cm
(b) 5.4 cm
(c) 7.2 cm
(d) 4.5 cm

Answer:

(b) 5.4 cm

ABC ∼ ∆DEF

Therefore,

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

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Question 28:

ABC ∼ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, what is the perimeter of ∆ABC?
(a) 35 cm
(b) 28 cm
(c) 42 cm
(d) 40 cm

Answer:

(a) 35 cm

 ∆ABC ∼ ∆DEF

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

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Question 29:

In ∆ABC, it is given that AB = 9 cm, BC = 6 cm and CA = 7.5 cm. Also, in ∆DEF , EF = 8 cm and ∆DEF ∼ ∆ABC. Then, perimeter of ∆DEF is
(a) 22.5 cm
(b) 25 cm
(c) 27 cm
(d) 30 cm

Answer:

(d) 30 cm

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

 ∆DEF ∼ ∆ABC

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

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Question 30:

ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1

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 △ABC and △EBD
∠BED = ∠BAC             (Corresponding angles)
∠B = ∠B             (Common)
By AA-similarity criterion
△ABC ∼ △EBD

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).

 

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Question 31:

It is given that ∆ABC ∼ ∆DEF. If ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm, then which of the following is true?
(a) DE = 12 cm, ∠F = 50°
(b) DE = 12 cm, ∠F = 100°
(c) EF = 12 cm, ∠D = 100°
(d) EF = 12 cm, ∠F = 30°

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°

 ∆ABC ∼ ∆DFE

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

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Question 32:

In the given figure ∠BAC = 90° and AD ⊥ BC. Then,


(a) BC CD = BC2
(b) AB AC = BC2
(c) BD CD = AD2
(d) AB AC = AD2

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

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Question 33:

In △ABC, AB = 63 cm, AC = 12 cm and BC = 6 cm. Then ∠B is
(a) 45o
(b) 60o
(c) 90o
(d) 120o

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 △ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)



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Question 34:

In ∆ABC and ∆DEF, it is given that ABDE=BCFD, then
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠F = ∠F

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

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Question 35:

In ∆DEF and ∆PQR, it is given that ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
(a) EFPR=DFPQ
(b) DEPQ=EFRP
(c) DEQR=DFPQ
(d) EFRP=DEQR

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

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Question 36:

If ∆ABC ∼ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true?
(a) BCEF = ACFD
(b) ABEF = ACDE
(c) BCDE = ABEF
(d) BCDE = ABFD

Answer:

(c) BCDE = ABEF

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

Page No 451:

Question 37:

In ∆ABC and ∆DEF, it is given that ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) similar as well as congruent

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.

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Question 38:

If in ∆ABC and ∆PQR, we have: ABQR=BCPR=CAPQ, then
(a) ∆PQR ∼ ∆CAB
(b) ∆PQR ∼ ∆ABC
(c) ∆CBA ∼ ∆PQR
(d) ∆BCA ∼ ∆PQR

Answer:

(a) ∆PQR ∼ ∆CAB

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

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Question 39:

In the given figure, two lines segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°, then ∠PBA = ?


(a) 50°
(b) 30°
(c) 60°
(d) 100°

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°

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Question 40:

Corresponding sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(a) 2 : 3
(b) 4 : 9
(c) 9 : 4
(d) 16 : 81

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).

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Question 41:

It is given that ∆ ABC ∼ ∆PQR and BCQR=23, thenar(PQR)ar(ABC)=?
(a) 23
(b) 32
(c) 49
(d) 94

Answer:

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

Page No 451:

Question 42:

In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE) = ?


(a) 2 : 1
(b) 4 : 1
(c) 1 : 2
(d) 1 : 4
 

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



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Question 43:

In ∆ABC and ∆DEF, we have: ABDE=BCEF=ACDF=57, then
ar(∆ABC) : ar(∆DEF) = ?

(a) 5 : 7
(b) 25 : 49
(c) 49 : 25
(d) 125 : 343

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

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Question 44:

ABC ∼ ∆DEF such that ar(∆ABC) = 36 cm2 and ar(∆DEF) = 49 cm2.
Then, the ratio of their corresponding sides is
(a) 36 : 49
(b) 6 : 7
(c) 7 : 6
(d) 6:7

Answer:

(b) 6:7

 ∆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. 

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Question 45:

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25 : 36. The ratio of their corresponding heights is
(a) 25 : 36
(b) 36 : 25
(c) 5 : 6
(d) 6 : 5

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

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Question 46:

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is
(a) congruent to the original triangle
(b) similar to the original triangle
(c) an isosceles triangle
(d) an equilateral triangle

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.

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Question 47:

If ABC~QRP,ar(ABC)ar(PQR)=94, AB = 18 cm and BC = 15 cm, then PR = ?
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 203 cm

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 452:

Question 48:

In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 45°. Then, ∆OAC and ∆ODB are


(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar

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.

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Question 49:

In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(d) 90°

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

Page No 452:

Question 50:

In ∆ABC, if AB = 16 cm, BC = 12 cm and AC = 20 cm, then ∆ABC is
(a) acute-angled
(b) right-angled
(c) obtuse-angled

Answer:

(b) right-angled

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

Hence, ∆ABC is a right-angled triangle.

Page No 452:

Question 51:

Which of the following is a true statement?
(a) Two similar triangles are always congruent.
(b) Two figures are similar if they have the same shape and size.
(c) Two triangles are similar if their corresponding sides are proportional.
(d) Two polygons are similar if their corresponding sides are proportional.

Answer:

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



Page No 453:

Question 52:

Which of the following is false statement?
(a) If the areas of two similar triangles are equal, then the triangles are congruent.
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.
(c) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding medians.
(d) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

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 453:

Question 53:

Match the following columns:

Column I Column II
(a) In a given ∆ABC, DEBC and
ADDB=35. If AC = 5.6 cm, then
AE = ...... cm.
(p) 6
(b) If ∆ABC ∼ ∆DEF such that
2AB = 3DE and BC = 6 cm, then
EF = ...... cm.
(q) 4
(c) If ∆ABC ∼ ∆PQR such that
ar(∆ABC) : ar(∆PQR) = 9 : 16 and
BC = 4.5 cm, then
QR = ...... cm.
(r) 3
(d) In the given figure, ABCD and
OA = (2x + 4) cm, OB = (9x − 21) cm,
OC = (2x − 1) cm and OD = 3 cm.
Then x = ?
(s) 2.1

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 454:

Question 54:

Match the following columns:

Column I Column II
(a) A man goes 10 m due east and
then 20 m due north. His
distance from the starting point
is ...... m.
(p) 253
(b) In an equilateral triangle with
each side 10 cm, the altitude is
...... cm.
(q) 53
(c) The area of an equilateral
triangle having each side 10 cm
is ...... cm2.
(r) 105
(d) The length of diagonal of a
rectangle having length 8 m
and breadth 6 m is ...... m .
(s) 10

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 458:

Question 1:

ABC ∼ ∆DEF and their perimeters are 32 cm and 24 cm respectively. If AB = 10 cm, then DE =?
(a) 8 cm
(b) 7.5 cm
(c) 15 cm
(d) 53cm

Answer:

(b) 7.5 cm

 ∆ABC ∼ ∆DEF

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

 DE = 10 × 2432 = 7.5  cm

Page No 458:

Question 2:

In the given figure, DE ∥ BC. If DE = 5 cm, BC = 8 cm and AD = 3.5 cm, then AB = ?


(a) 5.6 cm
(b) 4.8 cm
(c) 5.2 cm
(d) 6.4 cm

Answer:

(a) 5.6 cm
 DE ∥ BC

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

Page No 458:

Question 3:

Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, then the distance between their tops is

(a) 12 m
(b) 13 m
(c) 14 m
(d) 15 m

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 458:

Question 4:

The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3.5 cm, then the corresponding altitude of the other triangle is
(a) 5.6 cm
(b) 6.3 cm
(c) 4.2 cm
(d) 7 cm

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 459:

Question 5:

If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.

Answer:

ABC ∼ ∆DEF

 ABDE = BCEF 12 = 6EF EF = 12 cm

Page No 459:

Question 6:

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer:

 DE 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 459:

Question 7:

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

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 459:

Question 8:

Find the length of the altitude of an equilateral triangle of side 2a cm.

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 459:

Question 9:

ABC ∼ ∆DEF such that ar(∆ABC) = 64 cm2 and ar(∆DEF) = 169 cm2. If BC = 4 cm, find EF.

Answer:

 ∆ABC ∼ ∆DEF

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

Page No 459:

Question 10:

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm2. Find ar(∆COD).

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 459:

Question 11:

The corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.

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 459:

Question 12:

In the given figure, LM  CB and LN  CD.
Prove that AMAB=ANAD.

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 459:

Question 13:

Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

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 459:

Question 14:

In an equilateral triangle with side a, prove that area = 34a2.

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 459:

Question 15:

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

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 459:

Question 16:

Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

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 460:

Question 17:

In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that ar(ABC)ar(DBC)=AODO.

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 460:

Question 18:

In the given figure, XYAC and XY divides ∆ABC into two regions, equal in area. Show that AXAB=(2-2)2.

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 460:

Question 19:

In the given figure, ∆ABC is an obtuse triangle, obtuse-angles at B. If AD CB, Prove that AC2 = AB2 + BC2 + 2BC ⋅ BD.

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 460:

Question 20:

In the given figure, PA, QB and RC are perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that 1x+1y=1z.

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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