# 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)

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

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

(i) In ∆ ABC and ∆ AMP,

BAC= PAM [Common]

ABC= PMA [Each = 90°]

∆ ABC ~ ∆ AMP [AA Similarity]

(ii)

(i)

(ii)

Since, triangles PQT and RQS are similar.

Hence, DP CR = DC PR

(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,

(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)

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.

Given, AD^{2}
= BD DC

So, these two triangles will be equiangular.

(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,

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,

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

(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)

We have:

(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),

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

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

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

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 = _{}

(i) AP =_{}PB _{}

_{}

(ii) _{}

Let _{}

_{}

Given, _{}

(i)

_{}

(ii)

_{}

From the given information, we have:

_{}

(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

_{}

(i)

_{}

(ii)

_{}

_{}

_{}

_{}

(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

_{}

(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

_{}

_{}

(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)

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

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)

Join AR.

In _{}ACR, BX || CR. By Basic Proportionality theorem,

_{}

In _{}APR, XQ || AP. By Basic Proportionality theorem,

_{}

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

_{}

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

Also, in PAR and ABC,

Similarly,

(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.

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),

Let us assume two similar triangles as ABC PQR

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

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

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}

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}

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

(i)

(ii)

(iii)

In CDB,

1 + 2 +3 = 180^{o}

1 + 3 = 90^{o} … (1) (Since, 2 = 90o)

3 + 4 = 90^{o} … (2) (Since, ACB = 90o)

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, ADC = ACB = 90^{o}

(AA similarity)

Now, BDC = ACB = 90o

CBD = ABC (Common)

(AA similarity)

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

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

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.

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

(i)

(ii)

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

(by AA- similarity)

(i) We have,

(ii)

(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:

## Selina Concise Mathematics X Class 10 Chapter Solutions

- Chapter 1 - Value Added Tax
- Chapter 2 - Banking (Recurring Deposit Accounts)
- Chapter 3 - Shares and Dividends
- Chapter 4 - Linear Inequations (in one variable)
- Chapter 5 - Quadratic Equations
- Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)
- Chapter 7 - Ratio and Proportion (Including Properties and Uses)
- Chapter 8 - Remainder And Factor Theorems
- Chapter 9 - Matrices
- Chapter 10 - Arithmetic Progression
- Chapter 11 - Geometric Progression
- Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)
- Chapter 13 - Section and Mid-Point Formula
- Chapter 14 - Equation of a Line
- Chapter 15 - Similarity (With Applications to Maps and Models)
- Chapter 16 - Loci (Locus and its Constructions)
- Chapter 17 - Circles
- Chapter 18 - Tangents and Intersecting Chords
- Chapter 19 - Constructions (Circles)
- Chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume)
- Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)
- Chapter 22 - Heights and Distances
- Chapter 23 - Graphical Representation (Histograms and Ogives)
- Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)
- Chapter 25 - Probability

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