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

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

### Solution 1(i)

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

### 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

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## 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

_{}

## 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

## 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) cm^{2}

= 15x cm^{2}

### 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 km^{2}

### 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

### 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.

## 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 = 90^{o}

QPM = RPQ (Common)

(ii) In QMR and PQR,

QMR = PQR = 90^{o}

QRM = QRP (Common)

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

### Solution 19

(i) In CDB,

1 + 2 +3 = 180^{o}

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

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

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, 2 = 5 = 90^{o}

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, AD^{2}
= 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 = 90^{o}

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),