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Class 10 SELINA Solutions Maths Chapter 15 - Similarity (With Applications to Maps and Models)

Similarity (With Applications to Maps and Models) Exercise TEST YOURSELF

Solution 1(a)

Correct option: (iii) B and B’

Triangles ABC and A’B’C’ are similar to each other.

So, the corresponding vertices are:

A and A’

B and B’

C and C’

Solution 1(b)

Correct option: (i) AB and RS

Triangles ABC and RSP are similar to each other. So, their corresponding sides are:

AB and RS

BC and SP

AC and RP

Solution 1(c)

Correct option: (ii) A is false, B is true

Two congruent triangles are always similar

But, two similar triangles need not to be congruent.

Solution 1(d)

Correct option: (ii)

As triangles ABC and PQR are similar to each other.

Solution 1(e)

Correct option: (iii) 256 cm2

In ΔABC and ΔAEF,

A = A    (Common angle)

AEF = ABC    (Corresponding angles)

ΔABC ΔAEF     (By AA Postulate)

Area of ΔAEF = 256 cm2

Solution 1(f)

Correct option: (iii) Yes, AA

In ΔBAD and ΔACD,

D = D   … (Common angle)

ACD = BAD   (Given)

ΔBAD ΔACD    … (By AA Postulate)

Solution 1(g)

Correct option: (iv) 250 m

Scale = 3 : 500    … (Given)

Length of the ship = 250 m

Solution 2(i)

  

Solution 2(ii)

YC over AC equals fraction numerator 4.5 over denominator 13.5 end fraction.... open curly brackets table attributes columnalign left end attributes row cell AY over AC equals fraction numerator AC minus YC over denominator AC end fraction equals 1 minus YC over AC end cell row cell 1 minus YC over AC equals fraction numerator 9 over denominator 13.5 end fraction end cell end table close
YC over AC equals 1 third

Solution 2(iii)

Given that XY || BC

So, AXY ABC

  

Solution 3(i)

  

Solution 3(ii)

  

Solution 3(iii)

Solution 4

  

Solution 5

  

Triangle ABC is similar to triangle PQR

Solution 6

  

  

Solution 7

 

  

Solution 8

  

Solution 9

  

Solution 10(i)

  

Solution 10(ii)

  

Solution 11(i)

  

Solution 11(ii)

  

Solution 11(iii)

  

Solution 12

BF + FE = 2BF

Hence,

FE = BF 

 

(ii) In AFD, EG || FD. Using Basic Proportionality theorem,

… (1)

Now, AE = EB (as E is the mid-point of AB)

AE = 2EF (Since, EF = FB, by (i))

From (1),

Hence, AG: GD = 2: 1.

Solution 13

Let us assume two similar triangles as ABC PQR

Solution 14

The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Required ratio = (3)2 : (5)2 = 9: 25

Solution 15

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

So, the ratio between the sides of the two triangles = 4: 5

(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

Solution 16

Scale :- 1 : 20000

1 cm represents 20000 cm= = 0.2 km

(i)

=

= 576 + 1024 = 1600

AC = 40 cm

 

Actual length of diagonal = 40 0.2 km = 8 km

(ii)

1 cm represents 0.2 km

1 cm2 represents 0.2 0.2

The area of the rectangle ABCD = AB BC

= 24 32 = 768

Actual area of the plot = 0.2 0.2 768 = 30.72 km2

Solution 17

The dimensions of the building are calculated as below.

Length = 1 50 m = 50 m

Breadth = 0.60 50 m = 30 m

Height = 1.20 50 m = 60 m

Thus, the actual dimensions of the building are 50 m 30 m 60 m.

(i)

Floor area of the room of the building

(ii)

Volume of the model of the building

Solution 18

left parenthesis straight i right parenthesis
In space increment PQL space space and space increment RMP
angle LPQ space equals space angle QRP
angle RPM space equals space angle RQP
increment PQL space space tilde space increment RPM space... space left parenthesis AA space test right parenthesis

left parenthesis ii right parenthesis
As comma
increment PQL space space tilde space increment RPM
PQ over RP equals PL over RM equals QL over PM
rightwards double arrow QL cross times RM equals PL cross times PM

 

(iii)

Solution 19

Triangle ABC is enlarged to DEF. So, the two triangles will be similar.

Longest side in ABC = BC = 6 cm

Corresponding longest side in DEF = EF = 9 cm

Scale factor = = 1.5

Solution 20

Let ABC and PQR be two isosceles triangles.

Then,

Also, A = P (Given)

Let AD and PS be the altitude in the respective triangles.

We know that the ratio of areas of two similar triangles is equal to the square of their corresponding altitudes.

Solution 21

In triangle ABC, PO || BC. Using Basic proportionality theorem,

(i)

(ii)

Solution 22

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

(by AA- similarity)

(i) We have,

(ii)

Solution 23

left parenthesis i right parenthesis space L e t space angle C A D equals x
rightwards double arrow m angle D A B equals 90 degree minus x
rightwards double arrow m angle D B A equals 180 degree minus open parentheses 90 degree plus 90 degree minus x close parentheses equals x
rightwards double arrow angle C A D equals angle D B A space space space space space space space.... left parenthesis 1 right parenthesis
I n space triangle A D B space a n d space triangle C D A comma
angle A D B equals angle C D A space space space space space.... left square bracket E a c h space 90 degree right square bracket
angle A B D equals angle C A D space space space space space.... left square bracket F R o m space left parenthesis 1 right parenthesis right square bracket
therefore space triangle A D B tilde triangle C D A space space space space.... left square bracket B y space A. A. right square bracket

left parenthesis i i right parenthesis space S i n c e space t h e space c o r r e s p o n d i n g space s i d e s space o f space s i m i l a r space t r i a n g l e s space a r e space p r o p o r t i o n a l comma space w e space h a v e
fraction numerator B D over denominator A D end fraction equals fraction numerator A D over denominator C D end fraction
rightwards double arrow fraction numerator 18 over denominator A D end fraction equals fraction numerator A D over denominator 8 end fraction
rightwards double arrow space A D squared equals 18 cross times 8 equals 144
rightwards double arrow A D equals 12 space c m

left parenthesis i i i right parenthesis space T h e space r a t i o space o f space t h e space a r e a s space o f space t w o space s i m i l a r space t r i a n g l e s space i s space e q u a l space t o space t h e space r a t i o space o f space t h e space s q u a r e s
o f space t h e i r space c o r r e s p o n d i n g space s i d e s.
rightwards double arrow fraction numerator A r left parenthesis triangle A D B right parenthesis over denominator A r left parenthesis triangle C D A right parenthesis end fraction equals fraction numerator A D squared over denominator C D squared end fraction equals 12 squared over 8 squared equals 144 over 64 equals 9 over 4 equals 9 space colon space 4 space

Solution 24(i)

  

Solution 24(ii)

  

Solution 24(iii)

Solution 25

(i) In AGB, DE || AB , by Basic proportionality theorem,

.... (1)


 

In GBC, EF || BC, by Basic proportionality theorem,

.... (2)


 

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


 

(ii)

From (i), we have:

AD over DG equals CF over FG
therefore fraction numerator AD plus DG over denominator DG end fraction equals fraction numerator CF plus FG over denominator FG end fraction
therefore AG over DG equals CG over FG
Also comma
angle DGF space equals angle AGC space... open parentheses common space angle close parentheses
therefore increment DFG space tilde increment ACG space... open parentheses SAS space test close parentheses

Solution 26

i.

In PQR and SPR,

PSR = QPR … given

PRQ = PRS … common angle

 PQR SPR  (AA Test)

ii. Find the lengths of QR and PS.

Since ∆PQR SPR … from (i)

iii.

  

Similarity (With Applications to Maps and Models) Exercise Ex. 15(A)

Solution 2(i)

  

Solution 2(ii)

Solution 3(i)

  

 

  

Solution 3(ii)

 

  

Solution 4(i)

  

 

 

  

Solution 4(ii)

  

 

 

  

Solution 5(i)

 

  

 

 

 

  

Solution 5(ii)

  

 

angle BAP space equals space angle CAB space... space left parenthesis common right parenthesis

Solution 6

 

  

Solution 7(i)

  

Solution 7(ii)

In ΔADE and ΔABC

 

Solution 8

  

Solution 9

  

Solution 10

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

Solution 11

Solution 12

 

Solution 13

(i)  In ∆ ABC and ∆ AMP,

 BAC=  PAM [Common]

 ABC=  PMA [Each = 90°]

∆ ABC ~ ∆ AMP [AA Similarity]

(ii)

 

 

 

 

 

 

 

 

Solution 14

 

(i) In PQM and PQR,

PMQ = PQR = 90o

QPM = RPQ (Common)

 

 

(ii) In QMR and PQR,

QMR = PQR = 90o

QRM = QRP (Common)

 

 

(iii) Adding the relations obtained in (i) and (ii), we get,

Solution 15

(i) In CDB,

1 + 2 +3 = 180o

1 + 3 = 90o ..... (1)(Since, 2 = 90o)

3 + 4 = 90o .....(2) (Since, ABC = 90o)

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, 2 = 5 = 90o

Hence, AD = 6.4 cm

 

 

(iii)

Solution 16

Solution 17

 

Given, AD2 = BD DC

So, these two triangles will be equiangular.

Solution 18

(i) The three pair of similar triangles are:

BEF and BDC

CEF and CAB

ABE and CDE

 

(ii) Since, ABE and CDE are similar,

 

Since, CEF and CAB are similar,

Solution 19

(i) Given, AP: PB = 4: 3.

Since, PQ || AC. Using Basic Proportionality theorem,

 

Now, PQB = ACB (Corresponding angles)

QPB = CAB (Corresponding angles)

 

(ii) ARC = QSP = 90o

ACR = SPQ (Alternate angles)

Solution 20

We have:

Solution 21

(i) Since, BD and CE are medians.

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

 

In EGD and CGB,

 

(ii) Since,

 

In AED and ABC,

From (1),

Solution 1(a)

Correct option: (iii) AAA

In ΔABD and ΔCBA,

B = B    (Common angle)

ADB = CAB   (Each 90°)

Thus, ΔABD ΔCBA   … (By AA or AAA postulate)

Solution 1(b)

Correct option: (iii)

In ΔOCD and ΔOAB,

COD = AOB    … (Vertically opposite angles)

OCD = OAB   (Internally opposite angles)

ODC = OBA   (Internally opposite angles)

ΔOCD ΔOAB   AA postulate

Solution 1(c)

Correct option: (iv) 2.5 cm

In ΔACE and ΔBCD,

CAE = CBD    (Each 90°)

C = C   … (Common angles)

ΔACE ΔBCD   … AA postulate

Now, AB = AC – BC = 2.5 cm

Solution 1(d)

Correct option: (ii) ΔADE ΔABC

In ΔADE and ΔABC

DAE = BAC    (Given as angle A is common)

AED = ACB   (Given)

ΔADE ΔABC    By AA postulate

∗∗ Answer in the book is given as (iii) ΔADE ΔBAC, which is not correct. ∗∗

Solution 1(e)

Correct option: (iv) SSS

Since, OD = 2 × OB, OC = 2 × OA and CD = 2 × AB

ΔAOB ΔCOD   … By SSS postulate

Similarity (With Applications to Maps and Models) Exercise Ex. 15(B)

Solution 2(i)

Given space that space AD over DB equals 3 over 5

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

AD over DB equals AE over EC
rightwards double arrow AE over EC equals 3 over 5


Solution 2(ii)

  

Solution 2(iii)

  

Solution 2(iv)

  

Solution 2(v)

  

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4

  

Solution 5(i)

  

 

 

 

  

Solution 5(ii)

 

Solution 6

  

 

 begin mathsize 14px style increment ABC tilde increment ADE comma
rightwards double arrow AE over AC equals DE over BC
rightwards double arrow 4 over 11 equals fraction numerator 6.6 over denominator BC end fraction
rightwards double arrow BC equals fraction numerator 11 cross times 6.6 over denominator 4 end fraction equals 18.15 space cm end style

  

Solution 7

  

 

 

 

 

  

Solution 8

Hence comma
BD space equals space straight x over 3 cross times FD
In space increment AFB comma space DC space vertical line vertical line space AB
rightwards double arrow FD over FB equals FC over FA equals CD over AB
rightwards double arrow FD over FB equals CD over AB
rightwards double arrow FD over CD equals FB over AB
rightwards double arrow FD over CD equals fraction numerator FD plus BD over denominator AB end fraction
rightwards double arrow FD over CD equals fraction numerator FD plus straight x over 3 cross times FD over denominator AB end fraction
rightwards double arrow 3 over straight y equals fraction numerator straight x plus 3 over denominator 7.5 end fraction space... space left parenthesis straight i right parenthesis

 

Similarly

 

Solution 9

  

Solution 10

Solution 11

Solution 1(a)

Correct option: (i) 3 : 8

Since, DE || BC, we have

   … By Basic Proportionality Theorem

In ΔADE and ΔABC,

  … From (I) and (II)

A = A   (Common angle)

ΔADE ΔABC    By SAS Postulate

DE : BC = 3 : 8

Solution 1(b)

Correct option: (iv) ΔADE ΔABC

AB = AD + DB = AE + CE = AC   … (I)

DE || BC    … (By Converse of Basic Proportionality Theorem)

In ΔADE and ΔABC,

A = A    (Common angle)

ADE = ABC   (Corresponding angles)

ΔADE ΔABC   By AA Postulate

Solution 1(c)

Correct option: (iii) 6.4 cm

In ΔAOB and ΔCOD,

AOB = COD   (Vertically opposite angles)

BAO = DCO   (Alternate interior angles)

ΔAOB ΔCOD   By AA postulate

Solution 1(d)

Correct option: (iii) 8 cm

Since, DE || BC

   … By Basic Proportionality Theorem

    … (I)

In ΔADE and ΔABC,

A = A    (Common angle)

ADE = ABC   (Corresponding angles)

ΔADE ΔABC   By AA Postulate

Solution 1(e)

Correct option: (ii) similar

Two congruent triangles are similar.

Similarity (With Applications to Maps and Models) Exercise Ex. 15(C)

Solution 2

 

(i) AP = PB

 

(ii)

Solution 3

Let

Solution 4

Given,

(i)

(ii)

Solution 5

From the given information, we have:

Solution 6

(i)

(ii) Since  LMN and  MNR have common vertex at M and their bases LN and NR are along the same straight line

 

(iii) Since  LQM and  LQN have common vertex at L and their bases QM and QN are along the same straight line

Solution 7

 

Solution 8

Solution 9

 

(i) Given, DE || BC and

In  ADE and  ABC,

 A =  A(Corresponding Angles)

 ADE =  ABC(Corresponding Angles)

  (By AA- similarity)

  ..........(1)

Now

Using (1), we get .........(2)

(ii) In  DEF and  CBF,

 FDE =  FCB(Alternate Angle)

 DFE =  BFC(Vertically Opposite Angle)

  DEF   CBF(By AA- similarity)

 using (2)

 .

(iii) Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, therefore

Solution 10

Solution 1(a)

Correct option: (iii) 25 : 144

Since, DE || BC

   … By Basic Proportionality Theorem

   … (I)

In ΔADE and ΔABC,

ADE = ABC   (Corresponding angles)

A = A   (Common angles)

ΔADE ΔABC

Solution 1(b)

Correct option: (iv) 25 : 24

Since, DE || BC

   … By Basic Proportionality Theorem

   … (I)

In ΔADE and ΔABC,

ADE = ABC   (Corresponding angles)

A = A   (Common angles)

ΔADE ΔABC

Solution 1(c)

Correct option: (i) 4 : 49

Since, DE || BC

   … By Basic Proportionality Theorem

   … (I)

In ΔADE and ΔABC,

ADE = ABC   (Corresponding angles)

A = A   (Common angles)

ΔADE ΔABC   … By AA Postulate

In ΔODE and ΔOCB,

DOE = COB   Vertically opposite angles

ODE = OCB   Alternate interior angles

ΔODE ΔOCB   By AA Postulate

Solution 1(d)

Correct option: (iv) 5 : 3

Since, DE || BC

   … By Basic Proportionality Theorem

   … (I)

In ΔADE and ΔABC,

ADE = ABC   (Corresponding angles)

A = A   (Common angles)

ΔADE ΔABC    By AA Postulate

Solution 1(e)

Correct option: (iii) 25 : 144

In right-angled triangle BAC.

AB2 + AC2 = BC2

AB = 12 cm    … (I)

Let C = x°

B = (90 x)°   (II)

In ΔADC and ΔBDA,

DAC = (90 x)° = DBA    From (II)

ADC = BDA    (Each 90°)

ΔADC ΔBDA   By AA Postulate

Similarity (With Applications to Maps and Models) Exercise Ex. 15(D)

Solution 2(i)

  

Solution 2(ii)

Solution 3(i)

  

Solution 3(ii)

  

Solution 4(i)

Solution 4(ii)

Solution 5

Actual space dimensions space equals space open parentheses 30 cross times 1.2 close parentheses cross times open parentheses 30 cross times 0.75 close parentheses cross times open parentheses 30 cross times 2 close parentheses

Solution 6(i)

Solution 6(ii)

  

Solution 7

 

  

Solution 1(a)

Correct option: (i) 6.4 cm

Scale factor = 1:25

Solution 1(b)

Correct option: (ii) 2.4 m & 1.4 m

Length of the model of a rectangular object = 12 cm

Scale factor = 1 : 20

Solution 1(c)

Correct option: (iii) 1 : 15

Solution 1(d)

Correct option: (iv) 1 : 100

Let the scale factor = k

Area of the room = Length × breadth

Area of the model = k2 × Area of the room

Therefore, Scale factor = 1 : 100

Solution 1(e)

Correct option: (ii) 5 : 3

Let the scale factor = k

Capacity of a container = Volume of the container = length × breadth × height

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