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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models)

## Selina Textbook Solutions Chapter 15 - Similarity (With Applications to Maps and Models)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 15 - Similarity (With Applications to Maps and Models).

All Selina textbook questions of Chapter 15 - Similarity (With Applications to Maps and Models) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models) Page/Excercise 15(A)

Solution 1(ii)

Solution 1(i)

Solution 2(ii)

Solution 2(i)

Solution 3(i)

Solution 3(ii)

Solution 4(ii)

Solution 4(i)

Solution 5(ii)

Solution 5(i)

Solution 6

Solution 7(ii)

Solution 7(i)

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

(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

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.

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

In EGD and CGB,

(ii) Since,

In AED and ABC,

From (1),

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models) Page/Excercise 15(B)

Solution 1(v)

Solution 1(iv)

Solution 1(iii)

Solution 1(ii)

Solution 1(i)

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

Solution 2(iii)

Solution 2(ii)

Solution 2(i)

Solution 3

Solution 4(ii)

Solution 4(i)

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models) Page/Excercise 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

A = A(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

Solution 13

(i) AB = AC(Given)

(ii) Similarly, it can be proved that

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models) Page/Excercise 15(D)

Solution 1(ii)

Solution 1(i)

Solution 2(ii)

Solution 2(i)

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 3(iii)

Solution 3(ii)

Solution 3(i)

Solution 4(ii)

Solution 4(i)

Solution 5

Solution 6(ii)

Solution 6(i)

Solution 7

Solution 8

15cm represents = 30 m

1cm represents

1 represents 2m 2m = 4

Surface area of the model = 150

Actual surface area of aeroplane = 150 2 2= 600

50 is left out for windows

Area to be painted = 600 - 50 =50

Cost of painting per = Rs. 120

Cost of painting 550 = 120 550 = Rs. 66000

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 15 - Similarity (With Applications to Maps and Models) Page/Excercise 15(E)

Solution 1(ii)

Solution 1(i)

Solution 2(iii)

Solution 2(ii)

Solution 2(i)

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(ii)

Solution 10(i)

Solution 11(ii)

Solution 11(i)

Solution 12

Solution 13

Solution 14(iii)

Solution 14(ii)

Solution 14(i)

Solution 15

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

Also, in PAR and ABC,

Similarly,

Solution 16

(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 17

Since ABC PQR

So, their respective sides will be in proportion

Or,

Also, A = P, B = Q, C = R

Since, AX and PY are medians so they will divide their opposite sides.

Or,

Therefore, we have:

B = Q

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal.

Hence, ABX PQY (by SAS similarity rule)

So,

From (1) and (2),

Solution 18

Let us assume two similar triangles as ABC PQR

Solution 19

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 20

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 21

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 22

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 23

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 24

(i)

(ii)

(iii)

Solution 25

In CDB,

1 + 2 +3 = 180o

1 + 3 = 90o … (1) (Since, 2 = 90o)

3 + 4 = 90o … (2) (Since, ACB = 90o)

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, ADC = ACB = 90o

(AA similarity)

Now, BDC = ACB = 90o

CBD = ABC (Common)

(AA similarity)

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

Solution 26

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 27

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 28

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

(i)

(ii)

Solution 29

Solution 30

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

(by AA- similarity)

(i) We have,

(ii)

Solution 31

Solution 32(iii)

Solution 32(ii)

Solution 32(i)

Solution 33(iii)

Solution 33(ii)

Solution 33(i)

Solution 34

(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:

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# Text Book Solutions

ICSE X - Mathematics

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