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Page No 179:

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 179:

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 179:

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 179:

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 180:

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 180:

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 180:

Question 7:

In the adjoining figure, ABCD is a trapezium in which CDAB and its diagonals intersect at O. If AO = (5x − 7) cm, OC = (2x + 1) 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=BOOD5x-72x+1=7x-57x+1(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 180:

Question 8:

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 180:

Question 9:

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 180:

Question 10:

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 180:

Question 11:

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:

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 180:

Question 12:

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:

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 181:

Question 13:

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

Answer:

In ABC, P and Q are midpoints of AB and AC, respectively.
According to the midpoint theorem,
 PQBC and PQ = 12BC = 12DA  (It is given that AD = BC)

In CDA, R and Q are midpoints of DC and AC, respectively
According to the midpoint theorem,
RQDA and RQ = 12DA

In BDAP and S are midpoints of AB and BD, respectively.
According to the midpoint theorem,
SPDA and SP = 12DA

Similarly, in CDB, R and S are midpoints of DC and BD, respectively.
According to the midpoint theorem,
SRBC and SR = 12BC = 12DA

Therefore, SPRQ, PQSR  and PQ = RQ = SP = SR

Hence, PQRS is a rhombus with all sides equal and opposite sides parallel to each other.

Page No 181:

Question 14:

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 200:

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 201:

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 201:

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 202:

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 202:

Question 5:

ABC ∼ ∆ PQR and the perimeters of ∆ ABC and ∆ PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, find AB.

Answer:

We know that the ratio of perimeters of two similar triangles is same as the ratio of their corresponding sides.
Therefore,
Perimeter ofABCPerimeter of PQR=ABPQLet AB be x cm.3624=x10

x =36024=15

Hence, AB = 15 cm

Page No 202:

Question 6:

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 202:

Question 7:

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 202:

Question 8:

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 202:

Question 9:

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 202:

Question 10:

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 203:

Question 11:

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 203:

Question 12:

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 203:

Question 13:

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 203:

Question 14:

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 216:

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 216:

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 216:

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 216:

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 216:

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 216:

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 216:

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 216:

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 216:

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 217:

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 217:

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 217:

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 217:

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 236:

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 236:

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 236:

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 236:

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 236:

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 236:

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 236:

Question 7:

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 236:

Question 8:

ABC is an isosceles triangle, right-angled at C.Prove that AB2 = 2AC2.

Answer:

∆ABC is an isosceles triangle with AC = BC.
Also, ∆ABC is a right-angled triangle.
Therefore, applying Pythagoras theorem, we get:

AB2 = AC2 + BC2 AB2 = AC2 + AC2  ( AC = BC) AB2 = 2AC2

This completes the proof.

Page No 236:

Question 9:

ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ∆ABC is a right triangle

Answer:

Given:
AC = BC
Also,AB2 = 2AC2 = AC2 + AC2 AB2 = AC2 + BC2           ( AC = BC)

It is evident that the sum of squares of the two sides of triangle ABC is equal to square of the third side.

Therefore, by applying Pythagoras theorem, we conclude that  ∆ABC is right-angled at C.

Page No 236:

Question 10:

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 236:

Question 11:

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 236:

Question 12:

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 236:

Question 13:

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 236:

Question 14:

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 237:

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 = 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 237:

Question 16:

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 237:

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 237:

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 237:

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 239:

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 239:

Question 2:

A man goes 12 due south and then 35 m due west. How far is he from the starting point?
(a) 47 m
(b) 23.5 m
(c) 23 m
(d) 37 m

Answer:

(d) 37 m



Suppose, the man starts from point A and goes 12 m due south to point B. From here, he goes 35 m due west and stops at C.
In right triangle ABC, we have:
AB = 12 m, BC = 35 m
Applying Pythagoras theorem, we get:
AC2 = BC2 + AB2           = 352 + 122 = 1225 + 144 = 1369AC = 1369 = 37 m

Page No 239:

Question 3:

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 239:

Question 4:

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 239:

Question 5:

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 240:

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 240:

Question 7:

A ladder 15 m long reaches a window that is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. The width of the street is
(a) 27 m
(b) 21 m
(c) 24 m
(d) 18 m

Answer:

(b) 21 m

Let the ladder be at point C on the ground.
At first, the ladder is placed towards the left side of the street and it reaches the window AB, which is 9 m high from the ground.
Secondly, the ladder is placed towards the right side of the street and it reaches the window DE, which is 12 m high from the ground.

Applying Pythagoras theorem in right-angled triangles ABC and DEC, we get:

AC2 = AB2 + BC2 BC2 = AC2 - AB2 = 152 - 92 = 225 - 81 = 144BC = 144 = 12 mAlso, DC2 = DE2 + CE2CE2 = DC2 - DE2 = 152 - 122 = 225 - 144 = 81CE = 81 = 9 mBE = BC + CE = 12 + 9 = 21 m

Page No 240:

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 ∠QPR = 90°, then QR = ?


(a) 28 cm
(b) 25 cm
(c) 26 cm
(d) 32 cm

Answer:

(c) 26 cm

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

PR = 10 cm

Now, applying Pythagoras theorem in right-angled triangle QPR, we get:

QR2 = PQ2 + PR2       = 242 + 102 = 576 + 100 = 676QR = 676 = 26 cm

Page No 240:

Question 9:

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 240:

Question 10:

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

Page No 240:

Question 11:

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

Page No 240:

Question 12:

The measures of three angles of a triangle are in the ratio 1 : 2 : 3. Then, the triangle is
(a) right-angled
(b) equilateral
(c) isosceles
(d) obtuse-angled

Answer:

(a) right-angled

Let the angles be x°, 2x° and 3x°.
In a triangle, the sum of its angles is equal to 180°.
Therefore,
x + 2x + 3x = 180
or 6x = 180
or x = 30

Hence, the angles are 30°, 60° and 90°.

Page No 240:

Question 13:

For a ∆ABC, which is true?
(a) ABAC = BC
(b) (ABAC) > BC
(c) (AB AC) < BC
(d) None of these

Answer:

(c) (AB − AC) < BC

In a triangle, the difference of lengths of its two sides is always less than the third side.
Hence, (AB − AC) < BC is true.

Page No 240:

Question 14:

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



Page No 241:

Question 15:

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

Page No 241:

Question 16:

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, then CD = ?


(a) 4.8 cm
(b) 3.5 cm
(c) 7 cm
(d) 10.5 cm

Answer:

(b) 3.5 cm

It is given that AD bisects angle A.
Therefore, applying angle bisector theorem, we get:

  BDDC = 1014 6 - XX = 1014 14 ×6 - X = 10X 84 - 14X = 10X 84 = 10X + 14X 84 = 24X  24X  =  84 X  = 8424 = 3.5

Hence, CD = 3.5 cm

Page No 241:

Question 17:

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.

Page No 241:

Question 18:

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

Page No 241:

Question 19:

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.

Page No 241:

Question 20:

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.

Page No 241:

Question 21:

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.

Page No 241:

Question 22:

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 242:

Question 23:

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.

Page No 242:

Question 24:

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.

Page No 242:

Question 25:

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°

Page No 242:

Question 26:

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

Page No 242:

Question 27:

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

Page No 242:

Question 28:

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

Page No 242:

Question 29:

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

Page No 242:

Question 30:

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



Page No 243:

Question 31:

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

Page No 243:

Question 32:

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

Page No 243:

Question 33:

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

Page No 243:

Question 34:

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

Page No 243:

Question 35:

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

Page No 243:

Question 36:

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

Page No 243:

Question 37:

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 243:

Question 38:

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.

Page No 243:

Question 39:

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.



Page No 244:

Question 40:

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°

Page No 244:

Question 41:

It is given that ∆ABC ∼ ∆DEF and the corresponding sides of these triangles are in the ratio 3 : 5. Then, ar(∆ABC) : ar(∆DEF) = ?
(a) 3 : 5
(b) 5 : 3
(c) 9 : 25
(d) 25 : 9

Answer:

(c) 9 : 25
It is given that ∆ABC ∼ ∆DEF.
Therefore,
ar(ABC)ar(DEF) = AB2DE2 ar(ABC)ar(DEF) = 352 ar(ABC)ar(DEF) = 925

Page No 244:

Question 42:

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 244:

Question 43:

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

Page No 244:

Question 44:

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

Page No 244:

Question 45:

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. 

Page No 244:

Question 46:

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

Page No 244:

Question 47:

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.



Page No 245:

Question 48:

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 245:

Question 49:

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.

Page No 245:

Question 50:

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 245:

Question 51:

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 245:

Question 52:

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 245:

Question 53:

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 246:

Question 54:

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 = ?
Figure
(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 246:

Question 55:

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 247:

Question 56:

Assertion (A)
In ∆ABC, if DE ∥ BC intersects AB in D and AC in E, then ADAB=AEAC.
Reason (R)
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then these sides are divided in the same ratio.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

The Reason (R) is clearly true by Thales' theorem.
It is given that DEBC.
Applying Thales' theorem, we have:

ADDB = AEEC DBAD = ECAE 1 + DBAD = 1 + ECAE ABAD = ACAE ADAB = AEACTherefore, Assertion (A) is true and R gives A.

Page No 247:

Question 57:

Assertion (A)
ABC ∼ ∆DEF such that BC = 4.5 cm and EF = 3.5 cm and ar(∆ABC) = 81 cm2,  then ar(∆DEF) = 49 cm2.
Reason (R)
The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding medians.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

The Reason(R) is clearly true.
ABC ∼ ∆DEF

 ar(ABC)ar(DEF) = BC2EF2 81ar(DEF) = 4.523.52 ar(DEF) = 81 × 3.5 × 3.54.5 × 4.5 = 49 cm2

Hence, Assertion (A) is true but Reason (R) is not the correct explanation of (A).

Page No 247:

Question 58:

Assertion (A)
ABC ∼ ∆DEF, such that ar(∆ABC) = 100 cm2 and ar(∆DEF) = 144 cm2. If AB = 24 cm, then DE = 36 cm.
Reason (R)
If ∆ABC ∼ ∆DEF, then ar(ABC)ar(DEF)=AB2DE2=BC2EF2=AC2DF2.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false and Reason (R) is true.

Reason (R) is clearly true.
 ABC ~DEF

 ar(ABC)ar(DEF) = AB2DE2
 100144= AB2DE2 100144 = 242DE2 DE2 = 144 × 576100 DE = 12 × 2410 DE = 28.8  36


Hence, Assertion (A) is false.



Page No 248:

Question 59:

Assertion (A)
In an isosceles ∆ABC, if ∠C = 90°, then AB2 = 3AC2.
Reason (R)
In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = 90°.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(d) Assertion (A) is false and Reason (R) is true.

In triangle ABC, let BC be equal to AC.
We have:
C = 90°
Thus,
AB2 = AC2 + BC2 = AC2 + AC2    ( AC = BC) AB2 = 2AC2

Assertion (A) is false.
This proves that Reason (R) is true.

Hence, Assertion (A) is false and Reason (R) is true.

Page No 248:

Question 60:

Look at the statements given below:
I. ∆ABC ∼ ∆DEF and the altitudes of these triangles are in the ratio 1 : 2, then ar(∆ABC) : ar(∆DEF) = 1 : 4.
II. In ∆ABC, DE ∥ BC and AD : DB = 1 : 2, then ar(ADE)ar(ABC)=14.
III. In a ∆ABC, P and Q are points on AB and AC respectively such that AB = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC = 3PQ.

Which is true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Answer:

(d) I and III  are true.

I .
 ∆ABC ∼ ∆DEF

 ar(ABC)ar(DEF) = 122 = 14 = 1 : 4

II.
We have:
AD : DB = 1 : 2
DEBCTherefore, by B.P.T.,ADDB = AEEC = 12

 DBAD = 21

 DBAD  +1 = 3

 ABAD = 3   ADAB = 13

 ar(ADE)ar(ABC) = AD2AB2 = 19

III.
We have:
AP = 3 cm
PB = 6 cm
AQ =  5 cm
QC =  10 cm
BC =  3PQ

BCPQ = ABAP = ACAQ = 31 BC = 3PQ
Thus, I and III  are true, II is false.



Page No 257:

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 257:

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 257:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 258:

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 259:

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 259:

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 259:

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 259:

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 259:

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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