Request a call back

# 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

D = D   … (Common angle)

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(iii)

Given that XY || BC

So, AXY ABC

### Solution 5

Triangle ABC is similar to triangle PQR

### 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

(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 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:

### 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)

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

### 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

(iii)

### Solution 17

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)

We have:

### Solution 21

(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),

### Solution 1(a)

Correct option: (iii) AAA

In ΔABD and ΔCBA,

B = B    (Common angle)

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)

DAE = BAC    (Given as angle A is common)

AED = ACB   (Given)

∗∗ 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)

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

Similarly

### Solution 1(a)

Correct option: (i) 3 : 8

Since, DE || BC, we have

… By Basic Proportionality Theorem

… From (I) and (II)

A = A   (Common angle)

DE : BC = 3 : 8

### Solution 1(b)

AB = AD + DB = AE + CE = AC   … (I)

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

A = A    (Common angle)

### 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)

A = A    (Common angle)

### Solution 1(e)

Correct option: (ii) similar

Two congruent triangles are similar.

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

(i) AP =PB

(ii)

Let

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 9

(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 1(a)

Correct option: (iii) 25 : 144

Since, DE || BC

… By Basic Proportionality Theorem

… (I)

A = A   (Common angles)

### Solution 1(b)

Correct option: (iv) 25 : 24

Since, DE || BC

… By Basic Proportionality Theorem

… (I)

A = A   (Common angles)

### Solution 1(c)

Correct option: (i) 4 : 49

Since, DE || BC

… By Basic Proportionality Theorem

… (I)

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)

A = A   (Common angles)

### 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)

DAC = (90 x)° = DBA    From (II)

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

### 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