# 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 cm^{2}

In ΔABC and ΔAEF,

∠A = ∠A … (Common angle)

∠AEF = ∠ABC … (Corresponding angles)

⇒ ΔABC ∼ ΔAEF … (By AA Postulate)

⇒ Area of ΔAEF = 256 cm^{2}

### 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)

### 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 km^{2}

### 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

(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

### 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:

### 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)

### 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 = 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 15

(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 16

### Solution 17

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

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 **∼** **Δ**BAC, 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

### 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

### Solution 7

### Solution 8

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.

AB^{2} + AC^{2} = BC^{2}

⇒ 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

### 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 = k^{2} × 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