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

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

Solution 1(i)

  

Solution 1(ii)

Solution 2(i)

  

 

  

Solution 2(ii)

 

  

Solution 3(i)

  

 

 

  

Solution 3(ii)

  

 

 

  

Solution 4(i)

 

 

  

Solution 4(ii)

 

 

 

 

  

Solution 5(i)

 

  

 

 

 

  

Solution 5(ii)

  

 

 

  

Solution 6

 

  

Solution 7(i)

  

Solution 7(ii)

  

Solution 8

  

Solution 9

  

Solution 10

  

 

 

  

Solution 11

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

Solution 12

Solution 13

Solution 14

(i)  In ∆ ABC and ∆ AMP,

 BAC=  PAM [Common]

 ABC=  PMA [Each = 90°]

∆ ABC ~ ∆ AMP [AA Similarity]

(ii)

 

 

 

 

 

 

 

 

Solution 15

(i)

(ii)

Since, triangles PQT and RQS are similar.

Solution 16

Hence, DP CR = DC PR

Solution 17

Solution 18

(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 19

(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 20

Solution 21

Given, AE: EC = BE: ED

Draw EF || AB

In ABD, EF || AB

Using Basic Proportionality theorem,

Thus, in DCA, E and F are points on CA and DA respectively such that

Thus, by converse of Basic proportionality theorem, FE || DC.

But, FE || AB.

Hence, AB || DC.

Thus, ABCD is a trapezium.

Solution 22

Given, AD2 = BD DC

So, these two triangles will be equiangular.

Solution 23

(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 24

Given, QR is parallel to AB. Using Basic proportionality theorem,

Also, DR is parallel to QB. Using Basic proportionality theorem,

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

Solution 25

1 = 6 (Alternate interior angles)

2 = 3 (Vertically opposite angles)

DM = MC (M is the mid-point of CD)

So, DE = BC (Corresponding parts of congruent triangles)

Also, AD = BC (Opposite sides of a parallelogram)

AE = AD + DE = 2BC

Now, 1 = 6 and 4 = 5

Solution 26

(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 27

We have:

Solution 28

(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),

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

Solution 1(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 5AD over DB equals AE over EC
rightwards double arrow AE over EC equals 3 over 5 

Solution 1(ii)

  

Solution 1(iii)

  

Solution 1(iv)

  

Solution 1(v)

  

Solution 2(i)

  

Solution 2(ii)

  

Solution 2(iii)

  

Solution 3

  

 

 

  

Solution 4(i)

  

 

 

 

  

Solution 4(ii)

 

 

 

 

  

 

Solution 5

  

 

 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 6

  

 

 

 

 

  

Solution 7

 

Solution 8

  

Solution 9

  

Solution 10

  

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

Solution 1

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

 

(i) Required ratio =

(ii) Required ratio =

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

(i)

(ii)

Solution 8

 

Solution 9

Solution 10

(i) Since  APB and  CPB have common vertex at B and their bases AP and PC are along the same straight line

(ii) Since  DPC and  BPA are similar

(iii) Since  ADP and  APB have common vertex at A and their bases DP and PB are along the same straight line

(iv) Since  APB and  ADB have common vertex at A and their bases BP and BD are along the same straight line

Solution 11

 

(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 12

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

Solution 1(i)

  

Solution 1(ii)

  

Solution 2(i)

  

Solution 2(ii)

  

Solution 3(i)

  

Solution 3(ii)

  

Solution 3(iii)

  

Solution 3(iv)

Given that triangle ABC is enlarged and the scale factor is m = 3 to the triangle A'B'C'.

OC = 21 cm

So, (OC)3 = OC'

i.e. 21 x 3 = OC'

i.e. OC' = 63 cm


  

Solution 4(i)

  

Solution 4(ii)

  

Solution 5

  

Solution 6(i)

  

Solution 6(ii)

  

Solution 7

 

  

Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(E)

Solution 1(i)

  

Solution 1(ii)

  

Solution 1(iii)

Given that XY || BC

So, AXY ABC

  

Solution 2(i)

  

Solution 2(ii)

  

Solution 2(iii)

Solution 3

  

Solution 4

  

  

Solution 5

  

  

Solution 6

 

  

Solution 7

  

Solution 8

Join AR.

In  ACR, BX || CR. By Basic Proportionality theorem,

In  APR, XQ || AP. By Basic Proportionality theorem,

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

Solution 9

  

Solution 10(i)

  

Solution 10(ii)

  

Solution 11(i)

  

Solution 11(ii)

  

Solution 11(iii)

  

Solution 12

In ABC, PR || BC. By Basic proportionality theorem,

Also, in PAR and ABC,

Similarly,

Solution 13

(i)

(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 14

Let us assume two similar triangles as ABC PQR

Solution 15

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 16

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 17

In PXY and PQR, XY is parallel to QR, so corresponding angles are equal.

Hence, (By AA similarity criterion)

 (i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

 (ii) Ar (trapezium XQRY) = Ar (PQR) - Ar (PXY)

= (16x - x) cm2

= 15x cm2

Solution 18

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 19

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 20

(i)

(ii)

(iii)

Solution 21

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 22

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 23

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

(i)

(ii)

Solution 24

Solution 25

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

(by AA- similarity)

(i) We have,

(ii)

Solution 26

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 27(i)

  

Solution 27(ii)

  

Solution 27(iii)

 

 

  

 

 

 

Solution 28(i)

  

Solution 28(ii)

  

Solution 28(iii)

Solution 29

(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:

Solution 30

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.