ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity are given here. The solutions are comprehensive and prepared by our expert faculty team to help students clear all their doubts and help in their exam preparation to obtain good marks in Maths. The ML Aggarwal Solutions for Class 10 Maths Chapter 13 given here are available for free. Students can download these PDFs and start practising offline.

Chapter 13 – Similarity. Similarity is an idea in geometry. It means that two polygons, line segments, or other figures have the same shape. Similar objects do not need to have the same size. Two shapes are similar if their angles have the same measure and their sides are proportional. Some of the important topics discussed in this ML Aggarwal Solutions for Class 10 Maths Chapter 13 are,

  • Similarity of Triangles
    • Axioms of Similarity of Triangles
    • Basic Theorem of Proportionality
  • Relation Between Areas of Similar Triangles

ML Aggarwal Solutions for Class 10 Maths Chapter 13 :-Click Here to Download PDF

 

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Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity

Exercise 13.1

1. State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 1

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 2

Solution:-

(i) From the ΔABC and ΔPQR

AB/PQ = 3.2/4

= 32/40

Divide both numerator and denominator by 8 we get,

= 4/5

AC/PR = 3.6/4.5

= 36/45

Divide both numerator and denominator by 9 we get,

= 4/5

BC/QR = 3/5.4

= 30/54

Divide both numerator and denominator by 6 we get,

= 5/9

By comparing all the results, the side are not equal.

Therefore, the triangles are not equal.

(ii) From the ΔDEF and ΔLMN

∠E = ∠N = 40o

Then, DE/LN = 4/2

Divide both numerator and denominator by 2 we get,

= 2

EF/MN = 4.8/2.4

= 48/24

Divide both numerator and denominator by 24 we get,

= 2

Therefore, ΔDEF ~ ΔLMN

2. It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?

Solution:-

From the question is given that, ∆DEF ~ ∆RPQ

∠D = ∠R and ∠F = ∠Q not ∠P

No, ∠F ≠ ∠P

3. If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 3

Solution:-

From the figure, two line segments are intersecting each other at P.

In ΔBCP and ΔDPE

5/10 = 6/12

Dividing LHS and RHS by 2 we get,

½ = ½

Therefore, ΔBCD ~ ΔDEP

5. It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.
Find the lengths of the remaining sides of the triangles.

Solution:-

As per the dimensions give in the questions,

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 4

From the question it is given that,

ΔDEF ~ ΔLMN

So, AB/ED = AC/EF = BC/DF

Consider AB/ED = AC/EF

5/12 = 7/EF

By cross multiplication,

EF = (7 × 12)/5

EF = 16.8 cm

Now, consider AB/ED = BC/DF

5/12 = BC/15

BC = (5 × 15)/12

BC = 75/12

BC = 6.25

6.

(a) If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.

Solution:-

As per the dimensions give in the questions,

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 5

Now, we have to find out the perimeter of ΔABC

Let ΔABC ~ ΔDEF

So, AB/DE = AC/DF = BC/EF

Consider, AB/DE = AC/DE

4/6 = AC/12

By cross multiplication we get,

AC = (4 × 12)/6

AC = 48/6

AC = 8 cm

Then, consider AB/DE = BC/EF

4/6 = BC/9

BC = (4 × 9)/6

BC = 36/6

BC = 6 cm

Therefore, the perimeter of ΔABC = AB + BC + AC

= 4 + 6 + 8

= 18 cm


(b) If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.

Solution:-

From the question it is given that,

∆ABC ~ ∆PQR

Perimeter of ∆ABC = 32 cm

Perimeter of ∆PQR = 48 cm

So, AB/PQ = AC/PR = BC/QR

Then, perimeter of ∆ABC/perimeter of ∆PQR = AC/PR

32/48 = AC/6

AC = (32 × 6)/48

AC = 4

Therefore, the length of AC = 4 cm.

7. Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.

Solution:-

Let us assume that, ∆ABC ~ ∆DEF

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 6

∆ABC is BC = 6cm

∆ABC ~ ∆DEF

So, AB/DE = BC/EF = AC/DF

Consider AB/DE = BC/EF

AB/8 = 6/4

AB = (6 × 8)/4

AB = 48/4

AB = 12

Now, consider BC/EF = AC/DF

6/4 = AC/7

AC = (6 × 7)/4

AC = 42/4

AC = 21/2

AC = 10.5 cm

8.

(a) In the figure given below, AB || DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 7

Solution:-

From the question it is given that,

AB||DE

AC = 3 cm

CE = 7.5 cm

BD = 14 cm

From the figure,

∠ACB = ∠DCE [because vertically opposite angles]

∠BAC = ∠CED [alternate angles]

Then, ∆ABC ~ ∆CDE

So, AC/CE = BC/CD

3/7.5 = BC/CD

By cross multiplication we get,

7.5BC = 3CD

Let us assume BC = x and CD = 14 – x

7.5 × x = 3 × (14 – x)

7.5x = 42 – 3x

7.5x + 3x = 42

10.5x = 42

x = 42/10.5

x = 4

Therefore, BC = x = 4 cm

CD = 14 – x

= 14 – 4

= 10 cm

(b) In the figure (2) given below, CA || BD, the lines AB and CD meet at G.

(i) Prove that ∆ACO ~ ∆BDO.

(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 8

Solution:-

(i) We have to prove that, ∆ACO ~ ∆BDO.

So, from the figure

Consider ∆ACO and ∆BDO

Then,

∠AOC = ∠BOD [from vertically opposite angles]

∠A = ∠B

Therefore, ∆ACO = ∆BDO

Given, BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm, AC = 3.6 cm,

∆ACO ~ ∆BOD

So, AO/OB = CO/OD = AC/BD

Consider AC/BD = AO/OB

3.6/2.4 = AO/3.2

AO = (3.6 × 3.2)/2.4

AO = 4.8 cm

Now, consider AC/BD = CO/OD

3.6/2.4 = CO/4

CO = (3.6 × 4)/2.4

CO = 6 cm

9. (a) In the figure
(i) given below, ∠P = ∠RTS.
Prove that ∆RPQ ~ ∆RTS.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 9

Solution:-

From the given figure, ∠P = ∠RTS

So we have to prove that ∆RPQ ~ ∆RTS

In ∆RPQ and ∆RTS

∠R = ∠R (common angle for both triangle)

∠P = ∠RTS (from the question)

∆RPQ ~ ∆RTS


(b) In the figure (ii) given below, ∠ADC = ∠BAC. Prove that CA² = DC x BC

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 10

Solution:-

From the figure, ∠ADC = ∠BAC

So, we have to prove that, CA² = DC x BC

In ∆ABC and ∆ADC

∠C = ∠C (common angle for both triangle)

∠BAC = ∠ADC (from the question)

∆ABC ~ ∆ADC

Therefore, CA/DC = BC/CA

We know that, corresponding sides are proportional,

Therefore, CA2 = DC × BC

10. (a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 11

Solution:-

From the question it is give that,

AP = 2PB, CP = 2PD

(i) We have to prove that, ∆ACP is similar to ∆BDP and AC || BD

AP = 2PB

AP/PB = 2/1

Then, CP = 2PD

CP/PD = 2/1

∠APC = ∠BPD [from vertically opposite angles]

So, ∆ACP ~ ∆BDP

Therefore, ∠CAP = ∠PBD [alternate angles]

Hence, AC || BD

(ii) AP/PB = AC/BD = 2/1

AC = 2BD

2BD = 4.5 cm

BD = 4.5/2

BD = 2.25 cm

(b) In the figure (2) given below,

∠ADE = ∠ACB.

(i) Prove that ∆s ABC and AED are similar.

(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 12

Solution:-

From the given figure,

(i) ∠A = ∠A (common angle for both triangles)

∠ACB = ∠ADE [given]

Therefore, ∆ABC ~ ∆AED

(ii) from (i) proved that, ∆ABC ~ ∆AED

So, BC/DE = AB/AE = AC/AD

AD = AB – BD

= 6 – 1 = 5

Consider, AB/AE = AC/AD

6/3 = AC/5

AC = (6 × 5)/3

AC = 30/3

AC = 10 cm

(c) In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 13

Solution:-

From the figure,

∠P = ∠P (common angle for both triangles)

∠PQR = ∠PRS [from the question]

So, ∆PQR ~ ∆PRS

Then, PQ/PR = PR/PS = QR/SR

Consider PQ/PR = PR/PS

PQ/8 = 8/4

PQ = (8 × 8)/4

PQ = 64/4

PQ = 16 cm

11. In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prove that BM x NP = CN x MP.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 14

Solution:-

From the question it is given that, ABC is a triangle in which AB = AC.

P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC.

We have to prove that, BM x NP = CN x MP

Consider the ∆ABC

AB = AC … [from the question]

∠B = ∠C … [angles opposite to equal sides]

Then, consider ∆BMP and ∆CNP

∠M = ∠N

Therefore, ∆BMP ~ ∆CNP

So, BM/CN = MP/NP

By cross multiplication we get,

BM x NP = CN x MP

Hence it is proved.

12. Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Solution:-

Consider the two triangles, ∆MNO and ∆XYZ

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 15

From the question it is given that, two triangles are similar triangles

So, ∆MNO ~ ∆XYZ

If two triangles are similar, the corresponding angles are equal and their corresponding sides are proportional.

MN/XY = NO/YZ = MO/XZ

Perimeter of ∆MNO = MN + NO + MO

Perimeter of ∆XYZ = XY + YZ + XZ

Therefore, (MN/XY = NO/YZ = MO/XZ) = (MN/XY + NO/YZ + MO/XZ)

= Perimeter of ∆MNO/perimeter of ∆XYZ

13. In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/OD

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 16

Using the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

Solution:-

From the given figure, ABCD is a trapezium in which AB || DC,

The diagonals AC and BD intersect at O.

So we have to prove that, AO/OC = BO/OD

Consider the ∆AOB and ∆COD,

∠AOB = ∠COD … [vertically opposite angles]

∠OAB = ∠OCD

Therefore, ∆AOB ~ ∆COD

So, OA/OC = OB/OD

Now by using above result we have to find the value of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

OA/OC = OB/OD

(3x – 19)/(x – 3) = (x – 4)/4

By cross multiplication we get,

(x – 3) (x – 4) = 4(3x – 19)

X2 – 4x – 3x + 12 = 12x – 76

X2 – 7x + 12 – 12x + 76 = 0

X2 – 19x + 88 = 0

X2 – 8x – 11x + 88 = 0

X(x – 8) – 11(x – 8) = 0

(x – 8) (x – 11) = 0

Take x – 8 = 0

X = 8

Then, x – 11= 0

X = 11

Therefore, the value of x is 8 and 11.

14. In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.

Solution:-

Consider the ∆ABC,

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 17

So, we have to prove that, AB × AE = AC × AD

Now, consider the ∆ADB and ∆AEC,

∠A = ∠A [common angle for both triangles]

∠ADB = ∠AEC [both angles are equal to 90o]

∆ADB ~ ∆AEC

So, AB/AC = AD/AE

By cross multiplication we get,

AB × AE = AC × AD

15. In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that BE/DE = AC/BC

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 18

Solution:-

From the figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC

We have to prove that, BE/DE = AC/BC

Consider the ∆ABC and ∆DEB,

∠C = 90o

∠A + ∠ABC = 90o [from the figure equation (i)]

Now in ∆DEB

∠DBE + ∠ABC = 90o [from the figure equation (ii)]

From equation (i), we get

∠A = ∠DBE

Then, in ∆ABC and ∆DBE

∠C = ∠E [both angles are equal to 90o]

So, ∆ABC ~ ∆DBE

Therefore, AC/BE = BC/DE

By cross multiplication, we get

AC/BC = BE/DE

16.

(a) In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 19

Solution:-

From the figure, ABCD is a parallelogram,

Then, E is a point on AD and produced and BE intersects CD at F.

We have to prove that ∆ABE ~ ∆CFB

Consider ∆ABE and ∆CFB

∠A = ∠C [opposite angles of a parallelogram]
∠ABE = ∠BFC [alternate angles are equal]
∆ABE ~ ∆CFB


(b) In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.

(i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN.

Solution:-

From the question it is give that,

Consider the ∆RLQ and ∆PLN,

∠RLQ = ∠NLP [vertically opposite angles are equal]

∠RQL = ∠LNP [alternate angle are equal]

Therefore, ∆RLQ ~ ∆PLN

So, QR/PN = RL/LP = 2/3

QR/PN = 2/3

10/PN = 2/3

PN = (10 × 3)/2

PN = 30/2

PN = 15 cm

Therefore, PN = 15 cm


(ii) Name a triangle similar to triangle RLM. Evaluate RM.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 20

Solution:-

From the figure,

Consider ∆RLM and ∆QLP

Then, ∠RLM = ∠QLP [vertically opposite angles are equal]

∠LRM = ∠LPQ [alternate angles are equal]

Therefore, ∆RLM ~ ∆QLP

Then, RM/PQ = RL/LP = 2/3

So, RM/16 = 2/3

RM = (16 × 2)/3

RM = 32/3

RM =
ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 21

17. The altitude BN and CM of ∆ABC meet at H. Prove that

(i) CN × HM = BM × HN

(ii) HC/HB = √[(CN × HN)/(BM × HM)]

(iii) ∆MHN ~ ∆BHC

Solution:-

Consider the ∆ABC,

Where, the altitude BN and CM of ∆ABC meet at H.

and construction: join MN

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 22

(i) We have to prove that, CN × HM = BM × HN

In ∆BHM and ∆CHN

∠BHM = ∠CHN [because vertically opposite angles are equal]

∠M = ∠N [both angles are equal to 90o]

Therefore, ∆BHM ~ ∆CHN

So, HM/HN = BM/CN = HB/HC

Then, by cross multiplication we get

CN × HM = BM × HN

(ii) Now, HC/HB = √(HN × CN)/(HM × BM)

= √(CN × HN)/(BM × HM)

Because, M and N divide AB and AC in the same ratio.

(iii) Now consider ∆MHN and ∆BHC

∠MHN = ∠BHC [because vertically opposite angles are equal]

∠MNH = ∠HBC [because alternate angles are equal]

Therefore, ∆MHN ~ ∆BHC

18. In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that:

(i) ∆AMC ~ ∆PQR

(ii) CM/RN = AB/PQ

(iii) ∆CMB ~ ∆RNQ

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 23

Solution:-

From the given figure it is given that, CM and RN are respectively the medians of ∆ABC and ∆PQR.

(i) We have to prove that, ∆AMC ~ ∆PQR

Consider the ∆ABC and ∆PQR

As ∆ABC ~ ∆PQR

∠A = ∠P, ∠B = ∠Q and ∠C = ∠R

And also corresponding sides are proportional

AB/PQ = BC/QR = CA/RP

Then, consider the ∆AMC and ∆PNR,

∠A = ∠P

AC/PR = AM/PN

Because, AB/PQ = ½ AB/½PQ

AB/PQ = AM/PN

Therefore, ∆AMC ~ ∆PNR

(ii) From solution(i) CM/RN = AM/PN

CM/RN = 2AM/2PN

CM/RN = AB/PQ

(iii)Now consider the ∆CMB and ∆RNQ

∠B = ∠Q

BC/QP = BM/QN

Therefore, ∆CMB ~ ∆RNQ

19. In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 24

Solution:-

From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE.

(i) We have to prove that, EF = FC

From the figure, D is the midpoint of BC and also DF parallel to BE.

So, F is the midpoint of EC

Therefore, EF = FC

= ½ EC

EF = ½ AE

(ii) Now consider the ∆AGE and ∆ADF

Then, (BG or GE) ||DF

Therefore, ∆AGE ~ ∆ADF

So, AG/GD = AE/EF

AG/GD = 1/½

AG/GD = 1 × (2/1)

Therefore, AG: GD = 2: 1

20.

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 25

Solution:-

From the figure it is given that, AB, EF and CD are parallel lines.

(i) Consider the ∆EFG and ∆CGD

∠EGF = ∠CGD [Because vertically opposite angles are equal]

∠FEG = ∠GCD [alternate angles are equal]

Therefore, ∆EFG ~ ∆CGD

Then, EG/GC = EF/CD

5/10 = EF/18

EF = (5 × 18)/10

Therefore, EF = 9 cm

(ii) Now, consider the ∆ABC and ∆EFC

EF ||AB

So, ∆ABC ~ ∆EFC

Then, AC/EC = AB/EF

AC/(5 + 10) = 15/9

AC/15 = 15/9

AC = (15 × 15)/9

Therefore, AC = 25 cm

(b) In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 26

Solution:-

From the figure, AF, BE and CD are parallel lines.

Consider the ∆AEF and ∆CED

∠AEF and ∠CED [because vertically opposite angles are equal]

∠F = ∠C [alternate angles are equal]

Therefore, ∆AEF ~ ∆CED

So, AF/CD = AE/ED

7.5/4.5 = y/3

By cross multiplication,

y = (7.5 × 3)/4.5

y = 5 cm

So, similarly in ∆ACD, BE ||CD

Therefore, ∆ABE ~ ∆ACD

EB/CD = AE/AD

x/CD = y/y + 3

x/4.5 = 5/(5 + 3)

x/4.5 = 5/8

x = (4.5 × 5)/8

x = 22.5/8

x = 225/80

x = 45/16

x =
ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 27

21. In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 28

Solution:-

From the figure, consider ∆ABC,

So, ∠A = 90o

And AD ⊥ BC

∠BAC = 90o

Then, ∠BAD + ∠DAC = 90o … [equation (i)]

Now, consider ∆ADC

∠ADC = 90o

So, ∠DCA + ∠DAC = 90o … [equation (ii)]

From equation (i) and equation (ii)

We have,

∠BAD + ∠DAC = ∠DCA + ∠DAC

∠BAD = ∠DCA … [equation (iii)]

So, from ∆BDA and ∆ADC

∠BDA = ∠ADC … [both the angles are equal to 90o]

∠BAD = ∠DCA … [from equation (iii)]

Therefore, ∆BDA ~ ∆ADC

BD/AD = AD/DC = AB/AC

Because, corresponding sides of similar triangles are proportional

BD/AD = AD/DC

By cross multiplication we get,

AD2 = BD × CD

AD2 = 2 × 8 = 16

AD = √16

AD = 4

22. A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Solution:-

From the question it is given that,

Height of a tower PQ = 15m

It’s shadow QR = 24 m

Let us assume the height of a telephone pole MN = x

It’s shadow NO = 16 m

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 29

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 30

Given, at the same time,

∆PQR ~ ∆MNO

Therefore, PQ/MN = ON/RQ

15/x = 24/16

By cross multiplication we get,

x = (15 × 16)/24

x = 240/24

x = 10

Therefore, height of pole = 10 m.

23. A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?

Solution:-

From the question it is given that,

Height of pole (PQ) = 6m

Height of a woman (MN) = 1.5m

So, shadow NR = 3m

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 31

Therefore, pole and woman are standing in the same line

PM ||MR

∆PRQ ~ ∆MNR

So, RQ/RN = PQ/MN

(3 + x)/3 = 6/1.5

(3 + x)/3 = 60/15

(3 + x)/3 = 4/1

(3 + x) = 12

X = 12 – 3

X = 9m

Therefore, women is 9m away from the pole.

Exercise 13.2

1. (a) In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find (i) AE : EC (ii) DE.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 32

Solution:-

From the figure,

DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm

(i) AE: EC

So, AD/BD = AE/EC

AE/EC = AD/BD

AE/EC = ¾

AE: EC = 3: 4

(ii) consider ∆ADE and ∆ABC

∠D = ∠B

∠E = ∠C

Therefore, ∆ADE ~ ∆ABC

Then, DE/BC = AD/AB

DE/5 = 3/(3 + 4)

DE/5 = 3/7

DE = (3 × 5)/7

DE = 15/7

DE =
https://latex.codecogs.com/gif.latex?2%5Ctfrac%7B1%7D%7B7%7D


(b) In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 33

Solution:-

From the figure,

PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm

∠BQP = ∠BCA … [because alternate angles are equal]

Also, ∠B = ∠B … [common for both the triangles]

Therefore, ∆ABC ~ ∆BPQ

Then, BQ/BC = BP/AB = PQ/AC

BQ/BC = 6/(6 + 4) = PQ/AC

BQ/BC = 6/10 = PQ/AC

BQ/8 = 6/10 = PQ/AC … [because BC = 8 cm given]

Now, BQ/8 = 6/10

BQ = (6/10) ×8

BQ = 48/10

BQ = 4.8 cm

Also, CQ = BC – BQ

CQ = (8 – 4.8) cm

CQ = 3.2cm

Therefore, CQ = 3.2 cm and BQ = 4.8 cm


(c) In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 34

Solution:-

From the figure,

XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm,

So, PX/QX = PY/YR

1/3 = PY/4.5

By cross multiplication we get,

(4.5 × 1)/3 = PY

PY = 45/30

PY = 1.5

Then, ∠X = ∠Q

∠Y = ∠R

So, ∆PXY ~ ∆PQR

Therefore, XY/QR = PX/PQ

XY/9 = 1/(1 + 3)

XY/9 = ¼

XY = 9/4

XY = 2.25

2. In the given figure, DE || BC.

(i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
(ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 35

Solution:-

(i) From the figure, it is given that,

Consider the ∆ABC,

AD/DB = AE/EC

x/(x – 2) = (x + 2)/(x – 1)

By cross multiplication we get,

X(x – 1) = (x – 2) (x + 2)

x2 – x = x2 – 4

-x = -4

x = 4

(ii) From the question it is given that,

DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3

Consider the ∆ABC,

AD/DB = AE/EC

2x/(x – 2) = (2x + 3)/(x – 3)

By cross multiplication we get,

2x(x – 2) = (2x + 3) (x – 3)

2x2 – 4x = 2x2 – 6x + 3x – 9

2x2 – 4x – 2x2 + 6x – 3x = -9

-7x + 6x = -9

-x = – 9

x = 9

3. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.

Solution:-

From the given dimensions,

Consider the ∆PQR

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 36

So, PE/EQ = 3.9/3

= 39/30

= 13/10

Then, PF/FR = 8/9

By comparing both the results,

13/10 ≠ 8/9

Therefore, PE/EQ ≠ PF/FR

So, EF is not parallel to QR

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

Solution:-

From the dimensions given in the question,

Consider the ∆PQR

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 37

So, PQ/PE = 1.28/0.18

= 128/18

= 64/9

Then, PR/PF = 2.56/0.36

= 256/36

= 64/9

By comparing both the results,

64/9 = 64/9

Therefore, PQ/PE = PR/PF

So, EF is parallel to QR.

4. A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Give reasons for your answer.

Solution:-

From the dimensions given in the question,

Consider the ∆PQR

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 38

So, PQ/PA = 12.5/5

= 2.5/1

PR/PB = (PB + BR)/PB

= (4 + 6)/4

= 10/4

= 2.5

By comparing both the results,

2.5 = 2.5

Therefore, PQ/PA = PR/PB

So, AB is parallel to QR.

5.

(a) In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 39

Solution:-

From the question it is given that,

DE || BC and BD = CE

So, we have to prove that ABC is an isosceles triangle.

Consider the triangle ABC,

AD/DB = AE/EC

Given, DB = EC … [equation (i)]

Then, AD = AE … [equation (ii)]

By adding equation (i) and equation (ii) we get,

AD + DB = AE + EC

So, AB = AC

Therefore, ∆ABC is an isosceles triangle.


(b) In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 40

Solution:-

From the figure it is given that, AB || DE and BD || EF.

We have to prove that, DC² = CF x AC

Consider the ∆ABC,

DC/CA = CE/CB … [equation (i)]

Now, consider ∆CDE

CF/CD = CE/CB … [equation (ii)]

From equation (i) and equation (ii),

DC/CA = CF/CD

DC/AC = CF/DC

By cross multiplication we get,

DC2 = CF x AC

6.

(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 41

Solution:-

From the given figure, CD || LA and DE || AC,

Consider the ∆BCA,

BE/BC = BD/BA

By using the corollary of basic proportionality theorem,

BE/(BE + EC) = BD/AB

4/(4 + 2) = BD/AB … [equation (i)]

Then, consider the ∆BLA

BC/BL = BD/AB

By using the corollary of basic proportionality theorem,

6/(6 + CL) = BD/AB … [equation (ii)]

Now, combining the equation (i) and equation (ii), we get

6/(6 + CL) = 4/6

By cross multiplication we get,

6 x 6 = 4 x (6 + CL)

24 + 4CL = 36

4CL = 36 – 24

CL = 12/4

CL = 3 cm

Therefore, the length of CL is 3 cm.

(b) In the give figure, ∠D = ∠E and AD/BD = AE/EC. Prove that BAC is an isosceles triangle.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 42

Solution:-

From the given figure, ∠D = ∠E and AD/BD = AE/EC,

We have to prove that, BAC is an isosceles triangle

So, consider the ∆ADE

∠D = ∠E … [from the question]

AD = AE … [sides opposite to equal angles]

Consider the ∆ABC,

Then, AD/DB = AE/EC … [equation (i)]

Therefore, DE parallel to BC

Because AD = AE

DB = EC … [equation (ii)]

By adding equation (i) and equation (ii) we get,

AD + DB = AE + EC

AB = AC

Therefore, ∆ABC is an isosceles triangle.

7. In the adjoining given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 43

Solution:-

Consider the ∆POQ

AB || PQ … [given]

So, OA/AP = OB/BQ … [equation (i)]

Then, consider the ∆OPR

AC || PR

OA/AP = OC/CR … [equation (ii)]

Now by comparing both equation (i) and equation (ii),

OB/BQ = OC/CR

Then, in ∆OQR

OB/BQ = OC/CR

Therefore, BC || QR

8. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that AO/BO = CO/DO

Solution:-

From the question it is given that,

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 44

Now consider the ∆OAB and ∆OCD,

∠AOB = ∠COD [because vertically opposite angles are equal]

∠OBA = ∠ODC [because alternate angles are equal]

∠OAB = ∠OCD [because alternate angles are equal]

Therefore, ∆OAB ~ ∆OCD

Then, OA/OC = OB/OD

AO/OB = CO/DO … [by alternate angles]

9.

(a) In the figure (1) given below, AB || CR and LM || QR.

(i) Prove that BM/MC = AL/LQ

(ii) Calculate LM : QR, given that BM : MC = 1 : 2.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 45

Solution:-

From the question it is given that, AB || CR and LM || QR

(i) We have to prove that, BM/MC = AL/LQ

Consider the ∆ARQ

LM || QR … [from the question]

So, AM/MR = AL/LQ … [equation (i)]

Now, consider the ∆AMB and ∆MCR

∠AMB = ∠CMR … [because vertically opposite angles are equal]

∠MBA = ∠MCR … [because alternate angles are equal]

Therefore, AM/MR = BM/MC … [equation (ii)]

From equation (i) and equation (ii) we get,

BM/MR = AL/LQ

(ii) Given, BM : MC = 1 : 2

AM/MR = BM/MC

AM/MR = ½ … [equation (iii)]

LM || QR … [given from equation]

AM/MR = LM/QR … [equation (iv)]

AR/AM = QR/LM

(AM + MR)/AM = QR/LM

1 + MR/AM = QR/LM

1 + (2/1) = QR/LM

3/1 = QR/LM

LM/QR = 1/3

Therefore, the ratio of LM: QR is 1: 3.


(b) In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 46

Solution:-

From the question it is given that,

AD is bisector of ∠BAC

AB = 6 cm, AC = 4 cm and BD = 3cm

Construction, from C draw a straight line CE parallel to DA and join AE

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 47

∠1 = ∠2 … [equation (i)]

By construction CE || DE

So, ∠2 = ∠4 … [because alternate angles are equal] [equation (ii)]

Again by construction CE || DE

∠1 = ∠3 … [because corresponding angles are equal] [equation (iii)]

By comparing equation (i), equation (ii) and equation(iii) we get,

∠3 = ∠4

So, AC = AE … [equation (iv)]

Now, consider the ∆BCE,

CE || DE

BD/DC = AB/AE

BD/DC = AB/AC

3/DC = 6/4

By cross multiplication we get,

3 × 4 = 6 × DC

DC = (3 × 4)/6

DC = 12/6

DC = 2

Therefore, BC = BD + DC

= 3 + 2

= 5 cm

Exercise 13.3

1. Given that ∆s ABC and PQR are similar.
Find:
(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.
(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.

Solution:-

From the question it is given that,

(i) The area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3

Then, ∆ABC ~ ∆PQR

area of ∆ABC/area of ∆PQR = BC2/QR2

So, BC : QR = 1 : 3

Therefore, ∆ABC/area of ∆PQR = 12/32

= 1/9

Hence the ratio of the area of ∆ABC to the area of ∆PQR is 1: 9

(ii) The area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 25 : 36

Then, ∆ABC ~ ∆PQR

area of ∆ABC/area of ∆PQR = BC2/QR2

area of ∆ABC/area of ∆PQR = BC2/QR2 = 25/36

= (BC/QR)2 = (5/6)2

BC/QR = 5/6

Hence the ratio of their corresponding sides is 5 : 6

2. ∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

Solution:-

From the question it is given that,

∆ABC ~ DEF

Area of ∆ABC = 9 sq. cm

Area of ∆DEF =16 sq. cm

We know that,

area of ∆ABC/area of ∆DEF = BC2/EF2

area of ∆ABC/area of ∆DEF = BC2/EF2

9/16 = BC2/EF2

9/16 = (2.1)2/x2

2.1/x = √9/√16

2.1/x = ¾

By cross multiplication we get,

2.1 × 4 = 3 × x

8.4 = 3x

x = 8.4/3

x = 2.8

Therefore, EF = 2.8 cm

3. ∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.

Solution:-

From the question it is given that,

∆ABC ~ ∆DEF

BC = 3 cm, EF = 4 cm

Area of ∆ABC = 54 sq. cm.

We know that,

Area of ∆ABC/ area of ∆DEF = BC2/EF2

54/area of ∆DEF = 32/42

54/area of ∆DEF = 9/16

By cross multiplication we get,

Area of ∆DEF = (54 × 16)/9

= 6 × 16

= 96 cm

4. The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Solution:-

From the question it is given that,

The area of two similar triangles are 36 cm² and 25 cm².

Let us assume ∆PQR ~ ∆XYZ, PM and XN are their altitudes.

So, area of ∆PQR = 36 cm2

Area of ∆XYZ = 25 cm2

PM = 2.4 cm

Assume XN = a

We know that,

area of ∆PQR/area of ∆XYZ = PM2/XN2

36/25 = (2.4)2/a2

By cross multiplication we get,

36a2 = 25 (2.4)2

a2 = 5.76 × 25/36

a2 = 144/36

a2 = 4

a = √4

a = 2 cm

Therefore, altitude of the other triangle XN = 2 cm.

5.

(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 48

Solution:-

From the question it is given that, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm²

From the figure,

Consider the ∆AOQ and ∆BOP,

∠OAQ = ∠OBP … [both angles are equal to 90o]

∠AOQ = ∠BOP … [because vertically opposite angles are equal]

Therefore, ∆AOQ ~ ∆BOP

Then, area of ∆AOQ/area of ∆BOP = OQ2/PO2

Area of ∆AOQ/120 = 92/62

Area of ∆AOQ/120 = 81/36

Area of ∆AOQ = (81 × 120)/36

Area of ∆AOQ = 270 cm2

b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.

(i) Prove that ∆OAB ~ ∆OCD.

(ii) Find CD and OB.

(iii) Find the ratio of areas of ∆OAB and ∆OCD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 49

Solution:-

From the question it is given that,

AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm

(i) We have to prove that, ∆OAB ~ ∆OCD

So, consider the ∆OAB and ∆OCD

∠AOB = ∠COD … [because vertically opposite angles are equal]

∠OBA = ∠OCD … [because alternate angles are equal]

Therefore, ∆OAB ~ ∆OCD … [from AAA axiom]

(ii) Consider the ∆OAB and ∆OCD

OA/OC = OB/OD = AB/CD

Now consider OA/OC = OB/OD

10/5 = OB/2.8

OB = (10 × 2.8)/5

OB = 2 × 2.8

OB = 5.6 cm

Then, consider OA/OC = AB/CD

10/5 = 6.5/CD

CD = (6.5 × 5)/10

CD = 32.5/10

CD = 3.25 cm

(iii) We have to find the ratio of areas of ∆OAB and ∆OCD.

From (i) we proved that, ∆OAB ~ ∆OCD

Then, area of (∆OAB)/area of ∆OCD

AB2/CD2 = (6.5)2/(3.25)2

= (6.5 × 6.5)/(3.25 × 3.25)

= 2 × 2/1

= 4/1

Therefore, the ratio of areas of ∆OAB and ∆OCD = 4: 1.

6.

(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 50

Solution:-

From the question it is given that,

DE || BC, DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm

From the fig, ∠D = ∠B and ∠E = ∠C … [corresponding angles are equal]

Now consider the ∆ADE and ∆ABC,

∠A = ∠A … [common angles for both triangles]

Therefore, ∆ADE ~ ∆ABC

Then, area of ∆ADE/area of ∆ABC = (DE)2/(BC)2

28/area of ∆ABC = (6)2/(9)2

28/area of ∆ABC = 36/81

area of ∆ABC = (28 × 81)/36

area of ∆ABC = 2268/36

area of ∆ABC = 63 cm2

(b) In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 51

Solution:-

From the question it is given that, DE || BC and AD : DB = 1 : 2,

∠D = ∠B, ∠E = ∠C … [corresponding angles are equal]

Consider the ∆ADE and ∆ABC,

∠A = ∠A … [common angles for both triangles]

Therefore, ∆ADE ~ ∆ABC

But, AD/DB = ½

Then, DB/AD = 2/1

Now, adding 1 for both side LHS and RHS,

(DB/AD) + 1 = (2/1) + 1

(DB + AD)/AD = (2 + 1)

Therefore, ∆ADE ~ ∆ABC

Then, area of ∆ADE/area of ∆ABC = AD2/AB2

Area of ∆ADE/area of ∆ABC = (1/3)2

Area of ∆ADE/area of ∆ABC = 1/9

Area of ∆ABC = 9 area of ∆ADE

Area of trapezium DBCE

Area of ∆ABC – area of ∆ADE

9 area of ∆ADE – area of ∆ADE

8 area of ∆ADE

Therefore, area of ∆ADE/area of trapezium = 1/8

Then area of ∆ADE : area of trapezium DBCE = 1: 8

7.

In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.

(ii) Given that AD = ½ BD, calculate DE if BC = 4.5 cm.

(iii) If area of ∆ABC = 18cm2, find the area of trapezium DBCE

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 52

Solution:-

(i) From the question it is given that, DE || BC

We have to prove that, ∆ADE and ∆ABC are similar

∠A = ∠A … [common angle for both triangles]

∠ADE = ∠ABC … [because corresponding angles are equal]

Therefore, ∆ADE ~ ∆ABC … [AA axiom]

(ii) From (i) we proved that, ∆ADE ~ ∆ABC

Then, AD/AB = AB/AC = DE/BC

So, AD/(AD + BD) = DE/BC

(½ BD)/ ((½BD) + BD) = DE/4.5

(½ BD)/ ((3/2)BD) = DE/4.5

½ × (2/3) = DE/4.5

1/3 = DE/4.5

Therefore, DE = 4.5/3

DE = 1.5 cm

(iii) From the question it is given that, area of ∆ABC = 18 cm2

Then, area of ∆ADE/area of ∆ABC = DE2/BC2

area of ∆ADE/18 = (DE/BC)2

area of ∆ADE/18 = (AD/AB)2

area of ∆ADE/18 = (1/3)2 = 1/9

area of ∆ADE = 18 × 1/9

area of ∆ADE = 2

So, area of trapezium DBCE = area of ∆ABC – area of ∆ADE

= 18 – 2

= 16 cm2

8. In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 53

Solution:-

(i) Consider the ∆ABC and ∆DEC,

∠ABC = ∠DEC … [both angles are equal to 90o]

∠C = ∠C … [common angle for both triangles]

Therefore, ∆ABC ~ ∆DEC … [by AA axiom]

(ii) AC/CD = AB/DE

Corresponding sides of similar triangles are proportional

15/CD = 6/4

CD = (15 × 4)/6

CD = 60/6

CD = 10 cm

(iii) we know that, area of ∆ABC/area of ∆DEC = AB2/DE2

area of ∆ABC/area of ∆DEC = 62/42

area of ∆ABC/area of ∆DEC = 36/16

area of ∆ABC/area of ∆DEC = 9/4

Therefore, the ratio of the area of ∆ABC : area of ∆DEC is 9 : 4.

9. In the adjoining figure, ABC is a triangle. DE is parallel to BC and AD/DB = 3/2,

(i) Determine the ratios AD/AB, DE/BC

(ii) Prove that ∆DEF is similar to ∆CBF. Hence, find EF/FB.

(iii) What is the ratio of the areas of ∆DEF and ∆CBF?

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 54

Solution:-

(i) We have to find the ratios AD/AB, DE/BC,

From the question it is given that, AD/DB = 3/2

Then, DB/AD = 2/3

Now add 1 for both LHS and RHS we get,

(DB/AD) + 1 = (2/3) + 1

(DB + AD)/AD = (2 + 3)/3

From the figure (DB + AD) = AB

So, AB/AD = 5/3

Now, consider the ∆ADE and ∆ABC,

∠ADE = ∠B … [corresponding angles are equal]

∠AED = ∠C … [corresponding angles are equal]

Therefore, ∆ADE ~ ∆ABC … [by AA similarity]

Then, AD/AB = DE/BC = 3/5

(ii) Now consider the ∆DEF and ∆CBF

∠EDF = ∠BCF … [because alternate angles are equal]

∠DEF = ∠FBC … [because alternate angles are equal]

∠DFE = ∠ABFC … [because vertically opposite angles are equal]

Therefore, ∆DEF ~ ∆CBF

So, EF/FB = DE/BC = 3/5

(iii) we have to find the ratio of the areas of ∆DEF and ∆CBF,

We know that, Area of ∆DFE/Area of ∆BFC = DE2/BC2

Area of ∆DFE/Area of ∆BFC = (DE/BC)2

Area of ∆DFE/Area of ∆BFC = (3/5)2

Area of ∆DFE/Area of ∆BFC = 9/25

Therefore, the ratio of the areas of ∆DEF and ∆CBF is 9: 25.

10. In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 55

Solution:-

From the question it is given that,

PB = 2: 3

PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

(i) we have to find the area ∆APO: area ∆ABC,

Then,

∠A = ∠A … [common angles for both triangles]

∠APO = ∠ABC … [because corresponding angles are equal]

Then, ∆APO ~ ∆ABC … [AA axiom]

We know that, area of ∆APO/area of ∆ABC = AP2/AB2

= AP2/(AP + PB)2

= 22/(2 + 3)2

= 4/52

= 4/25

Therefore, area ∆APO: area ∆ABC is 4: 25

(ii) we have to find the area ∆APO: area ∆CQO

Then, ∠AOP = ∠COQ … [because vertically opposite angles are equal]

∠APQ = ∠OQC … [because alternate angles are equal]

Therefore, area of ∆APO/area of ∆CQO = AP2/CQ2

area of ∆APO/area of ∆CQO = AP2/PB2

area of ∆APO/area of ∆CQO = 22/32

area of ∆APO/area of ∆CQO = 4/9

Therefore, area ∆APO: area ∆CQO is 4: 9

11.

(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 56

Solution:-

From the question it is given that,

ABCD is a trapezium in which AB || DC and AB = 2 CD,

Then, ∠OAB = ∠OCD … [because alternate angles are equal]

∠OBA = ∠ODC

Then, ∆AOB ~ ∆COD

So, area of ∆AOB/area of ∆COD = AB2/CD2

= (2CD)2/CD2 … [because AB = 2 CD]

= 4CD2/CD2

= 4/1

Therefore, the ratio of the areas of ∆AOB and ∆COD is 4: 1.


(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 57

Solution:-

From the question it is given that,

ABCD is a parallelogram, AM ⊥ DC and AN ⊥ CB

AM = 6 cm

AN = 10 cm

The area of parallelogram ABCD is 45 cm²

Then, area of parallelogram ABCD = DC × AM = BC × AN

45 = DC × 6 = BC × 10

(i) DC = 45/6

Divide both numerator and denominator by 3 we get,

= 15/2

= 7.5 cm

Therefore, AB = DC = 7.5 cm

(ii) BC × 10 = 45

BC = 45/10

BC = 4.5 cm

(iii) Now, consider ∆ADM and ∆ABN

∠D = ∠B … [because opposite angles of a parallelogram]

∠M = ∠N … [both angles are equal to 90o]

Therefore, ∆ADM ~ ∆ABN

Therefore, area of ∆ADM/area of ∆ABN = AD2/AB2

= BC2/AB2

= 4.52/7.52

= 20.25/56.25

= 2025/5625

= 81/225

= 9/25

Therefore, area of ∆ADM : area of ∆ANB is 9: 25


(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD
(iii) area of ∆ABD : area of trapezium AFED
(iv) area of ∆FEO : area of ∆OBC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 58

Solution:-

From the question it is given that, ABCD is a parallelogram.

E is a point on AB, CE intersects the diagonal BD at O.

AE : EB = 2 : 3

(i) We have to find EF : AD

So, AB/BE = AD/EF

EF/AD = BE/AB

AE/EB = 2/3 … [given]

Now add 1 to both LHS and RHS we get,

(AE/EB) + 1 = (2/3) + 1

(AE + EB)/EB = (2 + 3)/3

AB/EB = 5/3

EB/AB = 3/5

Therefore, EF : AD is 3: 5

(ii) we have to find area of ∆BEF: area of ∆ABD,

Then, area of ∆BEF/area of ∆ABD = (EF)2/(AD)2

area of ∆BEF/area of ∆ABD = 32/52

= 9/25

Therefore, area of ∆BEF: area of ∆ABD is 9: 25

(iii) From (ii) area of ∆ABD/area of ∆BEF = 25/9

25 area of ∆BEF = 9 area of ∆ABD

25(area of ∆ABD – area of trapezium AEFD) = 9 area of ∆ABD

25 area of ∆ABD – 25 area of trapezium AEFD = 9 area of ∆ABD

25 area of trapezium AEFD = 25 area of ∆ABD – 9 area of ∆ABD

25 area of trapezium AEFD = 16 area of ∆ABD

area of ∆ABD/area of trapezium AEFD = 25/16

Therefore, area of ∆ABD : area of trapezium AFED = 25: 16

(iv) Now we have to find area of ∆FEO : area of ∆OBC

So, consider ∆FEO and ∆OBC,

∠EOF = ∠BOC … [because vertically opposite angles are equal]

∠F = ∠OBC … [because alternate angles are equal]

∆FEO ~ ∆OBC

Then, area of FEO/area of ∆OBC = EF2/BC2

EF2/AD2 = 9/25

Therefore, area of ∆FEO: area of ∆OBC = 9: 25.

12. In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 59

(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of parallelogram ABCD.

Solution:-

From the question it is given that, ABCD is a parallelogram.

BP: PC = 1: 2

area of ∆CPQ = 20 cm²

Construction: draw QN perpendicular CB and Join BN.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 60

Then, area of ∆BPQ/area of ∆CPQ = ((½BP) × QN)/((½PC) × QN)

= BP/PC = ½

(i) So, area ∆BPQ = ½ area of ∆CPQ

= ½ × 20

Therefore, area of ∆BPQ = 10 cm2

(ii) Now we have to find area of ∆CDP,

Consider the ∆CDP and ∆BQP,

Then, ∠CPD = ∠QPD … [because vertically opposite angles are equal]

∠PDC = ∠PQB … [because alternate angles are equal]

Therefore, ∆CDP ~ ∆BQP … [AA axiom]

area of ∆CDP/area of ∆BQP = PC2/BP2

area of ∆CDP/area of ∆BQP = 22/12

area of ∆CDP/area of ∆BQP = 4/1

area of ∆CDP = 4 × area ∆BQP

Therefore, area of ∆CDP = 4 × 10

= 40 cm2

(iii) We have to find the area of parallelogram ABCD,

Area of parallelogram ABCD = 2 area of ∆DCQ

= 2 area (∆DCP + ∆CPQ)

= 2 (40 + 20) cm2

= 2 × 60 cm2

= 120 cm2

Therefore, the area of parallelogram ABCD is 120 cm2.

13. (a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 61

Solution:-

From the question it is given that,

DE || BC

The ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5

Now, consider the ∆ABC and ∆ADE

∠A = ∠A … [common angle for both triangles] 

∠D = ∠B and

∠E = ∠C  … [because corresponding angles are equal] 

Therefore, ∆ADE ~ ∆ABC 

So, area of ∆ADE/area of ∆ABC = (DE)2/(BC)2 … [equation (i)]

Then, area of ∆ADE/area of trapezium DBCE = 4/5

area of trapezium DBCE/area of ∆ADE = 5/4

Add 1 for both LHS and RHS we get,

(area of trapezium DBCE/area of ∆ADE) + 1 = (5/4) + 1

(area of trapezium DBCE + area of ∆ADE)/area of ∆ADE = (5 + 4)/4

area of ∆ABC/area of ∆ADE = 9/4

area of ∆ADE/area of ∆ABC = 4/9

From equation (i),

area of ∆ADE/area of ∆ABC = (DE)2/(BC)2

area of ∆ADE/area of ∆ABC = (DE)2/(BC)2 = 42/92

area of ∆ADE/area of ∆ABC = (DE)2/(BC)2 = 2/3

Therefore, DE: BC = 2: 3

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 62

Solution:-

From the question it is given that,

AB || DC

AB = 2 DC, AD = 3 cm, BC = 4 cm

Now consider ∆EAB,

EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1

(i) EA = 2, DA = 2 × 3 = 6 cm

Then, ED = EA – DA

= 6 – 3

= 3 cm

(ii) EB/CB = 2/1

EB = 2 CB

EB = 2 × 4

EB = 8 cm

(iii) Now, consider the ∆EAB, DC || AB

So, ∆EDC ~ ∆EAB

Therefore, area of ∆EDC/area of ∆ABE = DC2/AB2

area of ∆EDC/area of ∆ABE = DC2/(2DC)2

area of ∆EDC/area of ∆ABE = DC2/4DC2

area of ∆EDC/area of ∆ABE = ¼

Therefore, area of ABE = 4 area of ∆EDC

Then, area of ∆EDC + area of trapezium ABCD = 4 area of ∆EDC

Area of trapezium ABCD = 3 area of ∆EDC

So, area of ∆EDC/area of trapezium ABCD = 1/3

Therefore, area of ∆EDC: area of trapezium ABCD = 1: 3

14. (a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find (i) BP (ii) the ratio of areas of ∆APB and ∆DPC. 

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 63

Solution:-

From the question it is given that,

DC is parallel to AB

AB = 9 cm, DC = 6 cm and BB = 12 cm

(i) Consider the ∆APB and ∆CPD

∠APB = ∠CPD … [because vertically opposite angles are equal]

∠PAB = ∠PCD … [because alternate angles are equal]

So, ∆APB ~ ∆CPD

Then, BP/PD = AB/CD

BP/(12 – BP) = 9/6

6BP = 108 – 9BP

6BP + 9BP = 108

15BP = 108

BP = 108/15

Therefore, BP = 7.2 cm

(ii) We know that, area of ∆APB/area of ∆CPD = AB2/CD2

area of ∆APB/area of ∆CPD = 92/62

area of ∆APB/area of ∆CPD = 81/36

By dividing both numerator and denominator by 9, we get,

area of ∆APB/area of ∆CPD = 9/4

Therefore, the ratio of areas of ∆APB and ∆DPC is 9: 4

(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm. (i) Prove that ∆ACD is similar to ∆BCA (ii) Find BC and CD (iii) Find the area of ∆ACD : area of ∆ABC.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 64

Solution:-

From the question it is given that,

∠ABC = ∠DAC

AB = 8 cm, AC = 4 cm, AD = 5 cm

(i) Now, consider ∆ACD and ∆BCA

∠C = ∠C … [common angle for both triangles]

∠ABC = ∠CAD … [from the question]

So, ∆ACD ~ ∆BCA … [by AA axiom]

(ii) AC/BC = CD/CA = AD/AB

Consider AC/BC = AD/AB

4/BC = 5/8

BC = (4 × 8)/5

BC = 32/5

BC = 6.4 cm

Then, consider CD/CA = AD/AB

CD/4 = 5/8

CD = (4 × 5)/8

CD = 20/8

CD = 2.5 cm

(iii) from (i) we proved that, ∆ACD ~ ∆BCA

area of ∆ACB/area of ∆BCA = AC2/AB2

= 42/82

= 16/64

By dividing both numerator and denominator by 16, we get,

= ¼

Therefore, the area of ∆ACD : area of ∆ABC is 1: 4.

15. ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that: 

(i) ∆ADE ~ ∆ACB. 

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD. 

(iii) Find, area of ∆ADE : area of quadrilateral BCED. 

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 65

Solution:-

From the question it is given that,

∠ABC = 90°

AB and DE is perpendicular to AC

(i) Consider the ∆ADE and ∆ACB,

∠A = ∠A … [common angle for both triangle]

∠B = ∠E … [both angles are equal to 90o]

Therefore, ∆ADE ~ ∆ACB

(ii) from (i) we proved that, ∆ADE ~ ∆ACB

So, AE/AB = AD/AC = DE/BC … [equation (i)]

Consider the ∆ABC, is a right angle triangle

From Pythagoras theorem, we have

AC2 = AB2 + BC2

132 = AB2 + 52

169 = AB2 + 25

AB2 = 169 – 25

AB2 = 144

AB = √144

AB = 12 cm

Consider the equation (i),

AE/AB = AD/AC = DE/BC

Take, AE/AB = AD/AC

4/12 = AD/13

1/3 = AD/13

(1 × 13)/3 = AD

AD = 4.33 cm

Now, take AE/AB = DE/BC

4/12 = DE/5

1/3 = DE/5

DE = (5 × 1)/3

DE = 5/3

DE = 1.67 cm

(iii) Now, we have to find area of ∆ADE : area of quadrilateral BCED,

We know that, Area of ∆ADE = ½ × AE × DE

= ½ × 4 × (5/3)

= 10/3 cm2

Then, area of quadrilateral BCED = area of ∆ABC – area of ∆ADE

= ½ × BC × AB – 10/3

= ½ × 5 × 12 – 10/3

= 1 × 5 × 6 – 10/3

= 30 – 10/3

= (90 – 10)/3

= 80/3 cm2

So, the ratio of area of ∆ADE : area of quadrilateral BCED = (10/3)/(80/3)

= (10/3) × (3/80)

= (10 × 3)/(3 × 80)

= (1 × 1)/(1 × 8)

= 1/8

Therefore, area of ∆ADE : area of quadrilateral BCED is 1: 8.

16. Two isosceles triangles have equal vertical angles and their areas are in the ratio

7: 16. Find the ratio of their corresponding height.

Solution:-

Consider the two isosceles triangle PQR and XYZ,

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 66

∠P = ∠X … [from the question]

So, ∠Q + ∠R = ∠Y + ∠Z

∠Q = ∠R and ∠Y = ∠Z [because opposite angles of equal sides]

Therefore, ∠Q = ∠Y and ∠R = ∠Z

∆PQR ~ ∆XYZ

Then, area of ∆PQR/area of ∆XYZ = PM2/XN2 … [from corollary of theorem]

PM2/XN2 = 7/16

PM/XN = √7/√16

PM/XN = √7/4

Therefore, ratio of PM: DM = √7: 4

17. On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate 

(i) the actual length of AB in km. 

(ii) the area of the plot in sq. km:

Solution:-

From the question it is given that,

Map drawn to a scale of 1: 250000

AB = 3 cm, BC = 4 cm and ∠ABC = 90o

(i) We have to find the actual length of AB in km. 

Let us assume scale factor K = 1: 250000

K = 1/250000

Then, length of AB of actual plot = 1/k × length of AB on the map

= (1/(1/250000)) × 3

= 250000 × 3

To covert cm into km divide by 100000

= (250000 × 3)/(100 × 1000)

= 15/2

length of AB of actual plot = 7.5 km

(ii) We have to find the area of the plot in sq. km

Area of plot on the map = ½ × AB × BC

= ½ × 3 × 4

= ½ × 12

= 1 × 6

= 6 cm2

Then, area of actual plot = 1/k2 × 6 cm2

= 2500002 × 6

To covert cm into km divide by (100000)2

= (250000 × 250000 × 6)/(100000 × 100000)

= (25/4) × 6

= 75/2

= 37.5 km2

18. On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm. Calculate: 

(i) the distance of a diagonal of the plot in km. 

(ii) the area of the plot in sq. km.

Solution:-

From the question it is given that,

Map drawn to a scale of 1: 25000

AB = 12 cm, BG = 16 cm

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 67

Consider the ∆ABC,

From the Pythagoras theorem,

AC2 = AB2 + BC2

AC = √(AB2 + BC2)

= √((12)2 + (16)2)

= √144 + 256

= √400

= 20 cm

Then, area of rectangular plot ABCD = AB × BC

= 12 × 16

= 192 cm2

(i) We have to find the distance of a diagonal of the plot in km. . 

Let us assume scale factor K = 1: 25000

K = 1/25000

Then, length of AB of actual plot = 1/k × length of diagonal of rectangular plot

= (1/(1/25000)) × 3

= 25000 × 20

To covert cm into km divide by 100000

= (25000 × 20)/(100 × 1000)

= 5 km

(ii) We have to find the area of the plot in sq. km.

Then, area of actual plot = 1/k2 × 192 cm2

= 250002 × 192

To covert cm into km divide by (100000)2

= (25000 × 25000 × 192)/(100000 × 100000)

= 12 km2

19. The model of a building is constructed with the scale factor 1 : 30. 

(i) If the height of the model is 80 cm, find the actual height of the building in metres. 

(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model.

Solution:-

From the question it is given that,

The model of a building is constructed with the scale factor 1 : 30

So, Height of the model/Height of actual building = 1/30

(i) Given, the height of the model is 80 cm

Then, 80/H = 1/30

H = (80 × 30)

H = 2400 cm

H = 2400/100

H = 24 m

(ii) Given, the actual volume of a tank at the top of the building is 27 m³

Volume of model/Volume of tank = (1/30)3

V/27 = 1/27000

V = 27/27000

V = 1/1000 m3

Therefore, Volume of model = 1000 cm3

20. A model of a ship is made to a scale of 1 : 200. 

(i) If the length of the model is 4 m, find the length of the ship. 

(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model. 

(iii) If the volume of the model is 200 liters, find the volume of the ship in m³. (100 liters = 1 m³)

Solution:-

From the question it is given that, a model of a ship is made to a scale of 1 : 200

(i) Given, the length of the model is 4 m

Then, length of the ship = (4 × 200)/1

= 800 m

(ii) Given, the area of the deck of the ship is 160000 m²

Then, area of deck of the model = 160000 × (1/200)2

= 160000 × (1/40000)

= 4 m2

(iii) Given, the volume of the model is 200 liters

Then, Volume of ship = 200 × (200/1)3

= 200 × 8000000

= (200 × 8000000)/100

= 1600000 m3

Chapter test

1. In the adjoining figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 68

Solution:-

From the question it is given that,

∠1 = ∠2 and ∠3 = ∠4

We have to prove that, PT x QR = PR x ST

Given, ∠1 = ∠2

Adding ∠6 to both LHS and RHS we get,

∠1 + ∠6 = ∠2 + ∠6

∠SPT = ∠QPR

Consider the ∆PQR and ∆PST,

From above ∠SPT = ∠QPR

∠3 = ∠4

Therefore, ∆PQR ~ ∆PST

So, PT/PR = ST/QR

By cross multiplication we get,

PT x QR = PR x ST

Hence, it is proved that PT x QR = PR x ST

2. In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 69

Solution:-

From the given figure,

AB = AC. If PM ⊥ AB and PN ⊥ AC

We have to show that, PM x PC = PN x PB

Consider the ∆ABC,

AB = AC … [given]

∠B = ∠C

Then, consider ∆CPN and ∆BPM

∠N = ∠M … [both angles are equal to 90o]

∠C = ∠B … [from above]

Therefore, ∆CPN ~ ∆BPM … [from AA axiom]

So, PC/PB = PN/PM

By cross multiplication we get,

PC x PM = PN x PB

Therefore, it is proved that, PM x PC = PN x PB

3.

(a) In the figure given below. ∠AED = ∠ABC. Find the values of x and y.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 70

Solution:-

From the figure it is given that,

∠AED = ∠ABC

Consider the ∆ABC and ∆ADE

∠AED = ∠ABC … [from the figure]

∠A = ∠A … [common angle for both triangles]

Therefore, ∆ABC ~ ∆ADE … [by AA axiom]

Then, AD/AC = DE/BC

3/(4 + 2) = y/10

3/6 = y/10

By cross multiplication we get,

y = (3 x 10)/6

y = 30/6

y = 5

Now, consider AB/AE = BC/DE

(3 + x)/4 = 10/y

Substitute the value of y,

(3 + x)/4 = 10/5

By cross multiplication,

5(3 + x) = 10 x 4

15 + 5x = 40

5x = 40 – 15

5x = 25

X = 25/5

x = 5

Therefore, the value of x = 5 cm and y = 5 cm

(b) In the figure given below, CD = ½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that : 

(i) CE || AG 

(ii) 3 ED = GD

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 71

Solution:-

From the question it is given that,

CD = ½ AC

BF || AG

(i) We have to prove that, CE || AG 

Consider, CD = ½ AC

AC = 2BC … [because from the figure B is mid-point of AC]

So, CD = ½ (2BC)

CD = BC

Hence, CE || BF … [equation (i)]

Given, BF || AG … [equation (ii)]

By comparing the results of equation (i) and equation (ii) we get,

CE || AG

(ii) We have to prove that, 3 ED = GD

Consider the ∆AGD,

CE || AG … [above it is proved]

So, ED/GD = DC/AD

AD = AB + BC + DC

= DC + DC + DC

= 3DC

So, ED/GD = DC/(3DC)

ED/GD = 1/(3(1))

ED/GD = 1/3

3ED = GD

Hence it is proved that, 3ED = GD

4. In the adjoining figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that: 

(i) DF || BH 

(ii) AH = 3 AF. 

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 72

Solution:-

From the question it is given that, 2 AD = BD, EC || BH

(i) Given, E is mid-point of BD

2DE = BD … [equation (i)]

2AD = BD … [equation (ii)]

From equation (i) and equation (ii) we get,

2DE = 2AD

DE = AD

Also given that, F is mid-point of AC

DF || EC … [equation (iii)]

Given, EC || BH … [equation (iv)]

By comparing equation (iii) and equation (iv) we get,

DF || BH

(ii) We have to prove that, AH = 3 AF,

Given, E is mid-point of BD and EC || BH

And c is midpoint of AH,

Then, FC = CH … [equation (v)]

Also given F is mid-point of AC

AF = FC … [equation (vi)]

By comparing both equation (v) and equation (vi) we get,

FC = AF = CH

AF = (1/3)AH

By cross multiplication we get,

3AF = AH

Therefore, it is proved that 3AF = AH

5. In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.

Solution:-

From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively.

DE || BC

AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 73

Consider the ∆ABC,

Given, DE || BC

So, AD/AB = AE/AC = DE/BC

Now, consider AD/AB = DE/BC

2.4/AB = 2/5

AB = (2.4 x 5)/2

AB = 12/2

AB = 6 cm

Then, consider AE/AC = DE/BC

3.2/AC = 2/5

AC = (3.2 x 5)/2

AC = 16/2

AC = 8 cm

Hence, BD = AB – AD

= 6 – 2.4

= 3.6 cm

CE = AC – AE

= 8 – 3.2

= 4.8 cm

6. In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.

Solution:-

From the question it is given that,

In a ∆ABC, D and E are points on the sides AB and AC respectively.

AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 74

Consider the ∆ABC,

EC = AC – AE

= 8.8 – 3.3

= 5.5 cm

Then, AD/DB = 5.7/9.5

= 57/95

By dividing both numerator and denominator by 19 we get,

= 3/5

AE/EC = 3.3/5.5

= 33/55

By dividing both numerator and denominator by 11 we get,

= 3/5

So, AD/DB = AE/EC

Therefore, DE || BC

7. If the areas of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.

Solution:-

From the question it is given that, the areas of two similar triangles are 360 cm² and 250 cm².

one side of the first triangle is 8 cm

So, PQR and XYZ are two similar triangles,

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 75

So, let us assume area of ∆PQR = 360 cm2, QR = 8 cm

And area of ∆XYZ = 250 cm2

Assume YZ = a

We know that, area of ∆PQR/area of ∆XYZ = QR2/yz2

360/250 = (8)2/a2

360/250 = 64/a2

By cross multiplication we get,

a2 = (250 x 64)/360

a2 = 400/9

a = √(400/9)

a = 20/3

a =
ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 76

Therefore, the length of the corresponding side of the second triangle YZ =
ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 77

8. In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find 

(i) BC 

(ii) DC 

(iii) area of ∆ACD : area of ∆BCA.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 78

Solution:-

From the question it is given that,

∠ABD = ∠CAD

AB = 5 cm, AC = 3 cm and AD = 4 cm

Now, consider the ∆ABC and ∆ACD

∠C = ∠C … [common angle for both triangles]

∠ABC = ∠CAD … [from the question]

So, ∆ABC ~ ∆ACD

Then, AB/AD = BC/AC = AC/DC

(i) Consider AB/AD = BC/AC

5/4 = BC/3

BC = (5 x 3)/4

BC = 15/4

BC = 3.75 cm

(ii) Consider AB/AD = AC/DC

5/4 = 3/DC

DC = (3 x 4)/5

DC = 12/5

Dc = 2.4 cm

(iii) Consider the ∆ABC and ∆ACD

∠CAD = ∠ABC … [from the question]

∠ACD = ∠ACB … [common angle for both triangle]

Therefore, ∆ACD ~ ∆ABC

Then, area of ∆ACD/area of ∆ABC = AD2/AB2

= 42/52

= 16/25

Therefore, area of ∆ACD : area of ∆BCA is 16: 25.

9. In the adjoining figure, the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of ∆AOE : area of parallelogram ABCD. 

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 79

Solution:-

From the given figure,

The diagonals of a parallelogram intersect at O.

OE is drawn parallel to CB to meet AB at E.

In the figure four triangles have equal area.

So, area of ∆OAB = ¼ area of parallelogram ABCD

Then, O is midpoint of AC of ∆ABC and DE || CB

E is also midpoint of AB

Therefore, OE is the median of ∆AOB

Area of ∆AOE = ½ area of ∆AOB

= ½ × ¼ area of parallelogram ABCD

= 1/8 area of parallelogram ABCD

So, area of ∆AOE/area of parallelogram ABCD = 1/8

Therefore, area of ∆AOE: area of parallelogram ABCD is 1: 8.

10. In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 80

Solution:-

From the question it is given that, ABCD is a trapezium in which AB || DC. If 2AB = 3DC.

So, 2AB = 3DC

AB/DC = 3/2

Now, consider ∆AOB and ∆COD

∠AOB = ∠COD … [because vertically opposite angles are equal]

∠OAB = ∠OCD … [because alternate angles are equal]

Therefore, ∆AOB ~ ∆COD … [from AA axiom]

Then, area of ∆AOB/area of ∆COD = AB2/DC2

= 32/22

= 9/4

Therefore, the ratio of the areas of ∆AOB and ∆COD is 9: 4

11. In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity Image 81

Solution:-

From the question it is given that,

ABCD is a parallelogram. E is mid-point of BC.

DE meets the diagonal AC at O.

(i) Now consider the ∆AOD and ∆EDC,

∠AOD = ∠EOC … [because Vertically opposite angles are equal]

∠OAD = ∠OCB … [because alternate angles are equal]

Therefore, ∆AOD ~ ∆EOC

Then, OA/OC = DO/OE = AD/EC = 2EC/EC

OA/OC = DO/OE = 2/1

Therefore, OA: OC = 2: 1

(ii) From (i) we proved that, ∆AOD ~ ∆EOC

So, area of ∆OEC/area of ∆AOD = OE2/DO2

area of ∆OEC/area of ∆AOD = 12/22

area of ∆OEC/area of ∆AOD = ¼

Therefore, area of ∆OEC: area of ∆AOD is 1: 4.

13. A model of a ship is made to a scale of 1: 250 calculate:

(i) The length of the ship, if the length of model is 1.6 m.

(ii) The area of the deck of the ship, if the area of the deck of model is 2.4 m2.

(iii) The volume of the model, if the volume of the ship is 1 km3.

Solution:-

From the question it is given that, a model of a ship is made to a scale of 1 : 250

(i) Given, the length of the model is 1.6 m

Then, length of the ship = (1.6 × 250)/1

= 400 m

(ii) Given, the area of the deck of the ship is 2.4 m²

Then, area of deck of the model = 2.4 × (1/250)2

= 1,50,000 m2 = 4 m2

(iii) Given, the volume of the model is 1 km3

Then, Volume of ship = (1/2503) × 1 km3

= 1/(250)3 × 10003

= 43

= 64 m3

Therefore, volume of ship is 64 m3.

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