# ML Aggarwal Solutions for Chapter 13 Similarity Class 10 Maths ICSE

Here, we are providing the solutions for Chapter 13 Similarity from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of twelfth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 13 Similarity of ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on rules for similarity of figures, pairs of similar triangles, finding sides and angles of similar triangles and prove if the triangles are similar or not. We have also added chapter test and multiple choice questions.

### Exercise 13.1

1. State which pairs of triangles in the figure 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):

Given

(i) In ∆ABC and PQR

AB/PQ = 3.2/4 = 4/5

AC/PR = (3.6/4.5) = 4/5

BC/QR = 3/5.4 = 5/9

∵ All the sides are not proportional.

∴ The triangles are not similar.

(ii) In ∆DEF and ∆LMN

∠E = ∠N = 40°

DE/LN = 4/2 = 2/1 and

EF/MN = 4.8/2.4

= 2/1

∴ ∆DEF ~ ∆LMN (SAS axiom)

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

∆DEF ~ ∆RPQ

∠D = ∠R and ∠F = ∠Q not ∠P

Not ∠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?

In two right triangles,

One of the acute angles of the one triangle is equal to an acute angle of the other triangle.

The triangles are similar. (AAA axiom)

4. In the adjoining figure, BD and CE intersect each other at the point P. Is ∆PBC ~ ∆PDE? Give reasons for your answer.

In the given figure, two line segments intersect each other at P.

In ∆BCP and ∆DEP

∠BPC = ∠DPE

5/10 = 6/12 (each = 1/2)

∴ ∆BCD ~ ∆DEP (SAS axiom)

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

∆ABC ~ ∆EDF

AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm

AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm

∵ ∆ABC ~ ∆EDF

∴ AB/ED = AC/EF = BC/DF

⇒ 5/12 = 7/EF = BC/15

⇒ 7/EF = 5/12

⇒ EF = (7×12)/5

= 84/5

= 16.8 cm

And 5/12 = BC/15

⇒ BC = (5×15)/12

= 25/4

= 6.25 cm

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

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

(a) ∆ABC ~ ∆DEF

AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm

Perimeter of ∆ABC

∵ ∆ABC ~ ∆DEF

∴ AB/DE = AC/DF = BC/EF

⇒ 4/6 = AC/12 = BC/9

AC/12 = 4/6

⇒ AC = (12×4)/6 = 8 cm

And BC/EF = 4/6

⇒ BC/9 = 4/6

⇒ BC = (9×4)/6 = 6 cm

∴ Perimeter of ∆ABC = AB + BC + AC

= 4 + 6 + 8

= 18 cm

(b)

∆ABC ~ ∆PQR

Given Perimeter of ∆ABC = 32 cm and

Perimeter of ∆PQR = 48 cm

Side PR = 6 cm

∵ ∆ABC ~ ∆PQR

∴ AB/PQ = AC/PR = BC/QR

⇒ (Perimeter of ∆ABC)/(Perimeter of ∆PQR) = AC/PR

⇒ 32/48 = AC/6

⇒ AC = (32×6)/48

= 4 cm

Hence, 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 triangle whose sides are 4 cm, 7 cm and 8 cm.

Let ∆ABG ~ ∆DEF in which shortest side of

∆ABC is BC = 6 cm

In ∆DEF, DE = 8 cm, EF = 4 cm and EF = 4 cm and DF = 7 cm

∵ ∆ABC ~ ∆DEF

∴ AB/DE = BC/EF = AC/DF

⇒ AB/8 = 6/4 = AC/7

Now, AB/8 = 6/4

⇒ AB = (8×6)/4

= 12 cm

AC/7 = 6/4

⇒ AC = (7×6)/4

= 21/2

= 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.

(b) In the figure (2) given below, CA || BD, 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.

(a) In the given figure,

AB || DE, AC = 3 cm, CE = 7.5 cm, BD = 14 cm

To calculate CB and DC

In ∆ABC and ∆CDE

∠ACB = ∠DCE (Vertically opposite angles)

∠BAC = ∠CED (Alternate angles)

∴ ∆ABC ~ ∆CDE (AA axiom)

∴ AC/CE = BC/CD

⇒ 3/(7.5) = BC/CD

⇒ BC×7.5 = 3CD

CB = 14 – DC {∵ BD = 14 cm}

Let BC = x, then CD = (14–x) cm

x × 7.5 = 3 × (14 – x)

⇒ 7.5x = 42 – 3x

⇒ 7.5x + 3x = 42

⇒ 10.5x = 42

⇒ x = 42/(10.5) = 4

∴ BC = 4 cm and DC = 14 – 4 = 10 cm

(b) In the given figure, CA ||BD

AB and CD intersect at O

To prove :

(i) ∆ACO ~ ∆BDO

(ii) BD = 2.4 cm, OD = 4 cm

OB = 3.2 cm, AC = 3.6 cm, then calculate OA and OC

Proof:

(i) In ∆ACO and ∆BDO

∠AOC = ∠BOD (Vertically opposite angles)

∠A = ∠B

∴ ∆ACO ~ ∆BDO (AA axiom)

(ii) BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm

AC = 3.6 cm

∵ ∆ACO ~ ∆BOD

∴ AO/OB = CO/OD = AC/BD

⇒ AO/3.2 = CO/4 = 3.6/2.4

⇒ AO/3.2 = 3.6/2.4

⇒ AO = (3.6×3.2)/2.4 = 4.8 cm

CO/4 = 3.6/2.4

⇒ CO = (3.6×4)/2.4 = 6 cm

9. (a) In the figure(i) given below, P = RTS.

Prove that ∆RPQ ~ ∆RTS.

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

ADC = BAC. Prove that CA2 = DC × BC.

(a) In the given figure,

∠P = ∠RTS

To prove : ∆RPQ ~ ∆RTS

Proof: In ∆RPQ and ∆RTS

∠R = ∠R (common)

∠P = ∠RTS (given)

∆RPQ ~ ∆RTS (AA axiom)

(b) In the given figure, ∠ADC = ∠BAC

To prove : CA2 = DC × BC

∠C = ∠C (common)

∴ ∆ABC ~ ∆ADC (AA axiom)

∴ CA/DC = BC/CA (Corresponding sides are proportional)

∴ CA×CA = DC×BC

⇒ CA2 = DC×BC

10. (a) In the figure (i) 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.

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

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

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

(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.

(a) In the given figure,

AP = 2PB, CP = 2PD

To prove:

(i) ∆ACP ~ ∆BDP and AC ||BD

(ii) If AC = 4.5 cm, find length of BD

Proof:

(i) AP = 2PB

⇒ AP/PB = 2/1

And CP = 2PD

⇒ CP/PD = 2/1

And ∠APC = ∠BPD (Vertically opposite angles)

∴ ∆ACP ~ ∆BDP (SAS axiom)

∴ ∠CAP = ∠PBD

But these are alternate angles

∴ AC || BD

(ii) ∴ AP/PB = AC/BD = 2/1

∴ AC = 2BD

⇒ 2BD = 4.5 cm

∴ BD = (4.5)/2

= 2.25 cm

(b) In the given figure, ∠ADE = ∠ACB

To prove:

(i) ∆ABC ~ ∆AED

(ii) If AE = 3 cm, BD = 1 cm

And AB = 6 cm, find AC

Proof:

(i) In ∆ABC and ∆AED

∠A = ∠A (common)

∴ ∆ABC ~ ∆AED (AA axiom)

∴ BC/DE = AB/AE = AC/AD

(ii) BC/DE = 6/3 = AC/5

{∴ AD = AB – BD = 6 – 1 = 5 cm}

6/3 = AC/5

⇒ AC = (6×5)/3

= 10 cm

(c) In the given figure, ∠PQR = ∠PRS

To prove:

(i) ∆PQR ~ ∆PRS

(ii) If PR = 8 cm, PS = 4 cm, find PQ

Proof:
(i) In ∆PQR and ∆PRS

∠P = ∠P (common)

∠PQR = ∠PRS (given)

∴ ∆PQR ~ ∆PRS (AA axiom)

∴ PQ/PR = PR/PS = QR/SR (sides are proportional)

(ii) PQ/8 = 8/4

⇒ PQ = (8×8)/4

= 16 cm

∴ PQ = 16 cm

11. In 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 × NP = CN × MP.

In the given figure, ABC in which AB = AC.

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

To prove: BM×NP = CN×MP

Proof : In ∆ABC, AB = AC

∴ ∠B = ∠C (Angles opposite to equal sides)

Now in ∆BMP and ∆CNP

∠M = ∠N (each 90°)

∴ ∠B = ∠C (proved)

∴ ∆BMP ~ ∆CNP (AA axiom)

∴ BM/CN = MP/NP (sides are proportional)

∴ BM×NP = CN×MP (By cross multiplication)

Hence, proved

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

Given: ∆ABC ~ ∆PQR

To prove: Ratio in their perimeters k the same as the ratio in their corresponding sides.

Proof :

∵ ∆ABC ~ ∆PQR

∴ AB/PQ = BC/QR = CA/RP

= (AB + BC + CA)/(PQ + QR + RP)

= (Perimeter of ∆ABC)/(Perimeter of ∆PQR)

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.

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

ABCD is a trapezium in which AB ||DC

Diagonals AC and BD intersect each other at O.

To prove:

(i) AO/OC = BO/OD

(ii) If OA = 3x – 19, OB = x – 4, OC = x – 3,

OD = 4

Find the value of x

Proof:

(i) In ∆AOB and COD

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD

∴ ∆AOB ~ ∆COD

∴ OA/OC = OB/OD

(ii) ∵ OA = 3x – 19, OB = x – 4, OC = x – 3,

OD = 4

Then (3x – 19)/(x – 3) = (x – 4)/x

By cross multiplication,

(x – 3)(x – 4) = 4(3x – 19)

x2 – 4x – 3x + 12 = 12x – 76

⇒ x2 – 7x – 12x + 12 + 76 = 0

⇒ x2 – 19x + 88 = 0

⇒ x2 – 8x – 11x + 88 = 0

⇒ x(x – 8) – 11(x – 8) = 0

(x – 8)(x – 11) = 0

Either x – 8 = 0, then x = 8

Or x – 11 = 0, then x = 11

∴ x = 8, 11

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

In ∆ABC, ∠A is acute

BD and CE are perpendicular on AC and AB respectively

To prove : AB×AE = AC×AD

Proof : In ∆ADB and ∆AEC

∠A = ∠A (common)

∴ ∆ADB ~ ∆AEC (AA axiom)

⇒ AB×AE = AC×AD (By cross multiplication)

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

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC

To prove: BE/DE = AC/BC

Proof : In ∆ABC = ∆DEB

∠C = 90°

∴ ∠A + ∠ABC = 90° …(i)

And In ∆DEB

∠DBE + ∠ABC = 90° …(ii)

From (i),

∠A = ∠DBE

Now in ∆ABC and ∆DBE

∠C = ∠E (each 90°)

∴ ∆ABC ~ ∆DBE (AA axiom)

∴ AC/BE = BC/DE

⇒ 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.

(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.

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

(a) In the given figure, ABCD is a ||gm

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

To prove : ∆ABE ~ ∆CFB

Proof : In ∆ABE and ∆CFB

∠A = ∠C (opposite angles of a ||gm)

∠ABE = ∠BFC (alternate angles)

∆ABE ~ ∆CFB (AA axiom)

(b) In the given figure, PQRS is a ||gm PQ = 16 cm,

QR = 10 cm

L is a point on PR such that

RL : LP = 2:3

QL is produced to meet RS at M and PS produced at N.

To prove:

(i) ∆RLQ ~ ∆PLN and find PN

(ii) Name triangle similar to ∆RLM and evaluate RM

Proof :

(i) In ∆RLQ and ∆PLN

∠RLQ = ∠NLP (vertically opposite angles)

∠RQL = ∠LNP (alternate angles)

∴ ∆RLQ ~ ∆PLN

∴ QR/PN = RL/LP = 2/3

⇒ 10/PN = 2/3

⇒ PN = (10 × 3)/2

= 15 cm

(ii) In ∆RLM and ∆QLP

∠RLM = ∠QLP (vertically opposite angles)

∠LRM = ∠LPQ (alternate angles)

∴ ∆RLM ~ ∆QLP

∴ RM/PQ = RL/LP = 2/3

RM/16 = 2/3

⇒ RM = (16 × 2)/3

= 32/2

= 10.2/3 cm

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

(i) CN. HM = BM. HN.

(ii)

(iii) ∆MHN ~ ∆BHC.

In the given figure, BN ⊥ AC and CM ⊥ AB of ∆ABC which intersect each other at H.

To prove:

(i) CN.HM = BM. HN

(ii)

(iii) ∆MHN ~ ∆BHC.

Construction: Join MN

Proof :

(i) In ∆BHM and ∆CHN

∠BHM = ∠CHN (vertically opposite angles)

∠M = ∠N (each 90°)

∴ ∆BHM ~ ∆CHN (AA axiom)

∴ HM/HN = BM/CN = HB/HC

By cross multiplication,

CN.HM = BM.HN

(ii) Now, HC/HB

∵ M and N divide AB and AC in the same ratio

∴ MN || BC

(iii) Now in ∆MHN and ∆BHC

∠MHN = ∠BHC (vertically opposite angles)

∠MNH = ∠HBC (alternate angles)

∴ ∆MHN ~ ∆BHC (AA axiom)

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

(i) ∆AMC ~ ∆PNR

(ii) CM/RN = AB/PQ

(iii) ∆CMB ~ ∆RNQ.

In the given figure, CM and RN are medians of ∆ABC and ∆PQR respectively and ∆ABC ~ ∆PQR.

To prove:

(i) ∆AMC ~ ∆PNR

(ii) CM/RN = AB/PQ

(iii) ∆CMB ~ ∆RNQ

Proof:

(i) ∵ ∆ABC ~ ∆PQR

∠A = ∠P, ∠B = ∠Q and ∠C = ∠R and corresponding sides are also proportional

i.e., AB/PQ = BC/QR = CA/RP

Now in ∆AMC and ∆PNR,

∠A = ∠P

AC/PR = AM/PN

{ ∵ AB/PQ = (1/2.AB)/(1/2.PQ) = AM/PN}

∴ ∆AMC ~ ∆PNR (SAS axiom)

(ii) CM/RN = AM/PN

⇒ 2AM/2PN = AB/PQ

(iii) Now in ∆CMB and ∆RNQ

∠B = ∠Q

BC/QP = BM/QN

{∵ BM/QN = (1/2.AB)/(1/2.PQ) = AB/PQ}

∴ ∆CMB ~ ∆RNQ (SAS axiom)

19. In the adjoining figure, medians AD and BE of ∆ABC meet at point G, and DF is drawn parallel to BE. Prove that

(i) EF = FC

(ii) AG : GD = 2 : 1

In the given figure,

AD and BE are the medians of ∆ABC intersecting each other at G.

DF || BE is drawn

To prove :

(ii) EF = FC

(iii) AG : GD = 2 : 1

Proof:

(i) D is the mid-point of BC and DF ||BE

∴ F is the midpoint of CE

⇒ EF = FC = (1/2)EC

⇒ EF = 1/2.AE

∵ GE or BG || DF

(ii) Now,

AG/GD = AE/EF = (1/1)/2

= 1 × 2/1

= 2/1

∴ 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

(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.

(a) In the given figure,

AB ||EF ||CD

AB = 15 m, EG = 5 cm. GC = 10 cm and DC = 18 cm

Calculate :

(i) EF

(ii) AC

Proof : In ∆EFG and ∆CGD

(i) ∠EGF = ∠CGD (vertically opposite angles)

∠FEG = ∠GDC (alternate angles)

∴ EG/GC = EF/CD

⇒ 5/10 = EF/18

⇒ EF = (5×18)/10

= 9 cm

∴ EF = 9 cm

(ii) In ∆ABC and ∆EFC

EF || AB

∴ ∆ABC ~ ∆EFC

∴ AC/EC = AB/EF

⇒ AC/(5 + 10)

= 15/9

⇒ AC/15 = 15/9

⇒ AC = (15×15)/9

= 25 cm

∴ AC = 25 cm

(b) In the given figure, AF ||BE ||CD

AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm

BE = x and AE = y

To find the value of x and y

In ∆AEF and ∆CED

∠AEF = ∠CED (vertically opposite angles)

∠F = ∠C (alternate angles)

∴ ∆AEF ~ ∆CED

∴ AF/CD = AE/ED

⇒ 7.5/4.5 = y/3

⇒ y = (7.5×3)/4.5 = 5.0 cm

Similarly in ∆ACD, BE||CD

∴ ∆ABE ~ ∆ACD

∴ x/CD = y/(y + 3)

⇒ x/(4.5) = 5/(5 + 3) = 5/8

x = (4.5×5)/8

= (22.5)/8

= 225/(10×8)

= 45/16 cm

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

In ∆ABC, we have ∠A = 90°

Now,

In ∆ABC, we have,

∠BAC = 90°

⇒ ∠BAD + ∠DAC = 90° …(i)

So, ∠DCA + ∠DAC = 90° ...(ii)

From (i) and (ii), we have

∠BAD + ∠DAC = ∠DCA + ∠DAC

So, ∆BDA ~ ∆ADC (AA similarity)

⇒ BD/AD = AD/DC = AB/AC (Corresponding sides of similar ∆’s are proportional)

⇒ AD2 = BD × CD

⇒ AD2 = 2×8 = 16

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.

Height of a tower AB = 15 m

And its shadow BC = 24 m

At the same time and position

Let the height of a telephone pole DE = x m

And its shadow EF = 16 m

∵ The time is same

∴ ∆ABC ~ ∆DEF

∴ AB/DE = CD/EF

⇒ 15/x = 24/16

⇒ x = (15×16)/24

= 10

∴ 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?

Height of height pole (AB) = 6 m

And height of a woman (DE) = 1.5 m

Here shadow EF = 3 m

∵ Pole and woman are standing in the same line

∴ ∆AFB ~ ∆DFE

∴ FB/EF = AB/DE

⇒ (3 + x)/3 = 6/(1.5)

= 4/1

⇒ 3 + x = 12

⇒ x = 12 – 3 = 9 m

∴ Woman is 9 m away from the pole.

### Exercise 13.2

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

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

(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.

(a) In the figure (i)

Given: DE || BC, AD = 3 cm, BD = 4 cm and BC = 5 cm.

To find:

(i) AE:EC and

(ii) DE since DE || BC of ∆ABC

= 3/4 [∵ AD = 3 cm and BD = 4 cm]

AE : EC = 3 : 4

∠D = ∠B and ∠E = ∠C [∵ DE || BC given]

⇒ DE/5 = 3/(3 + 4) [∵ AB = AD + BD = 3 cm + 4 cm]

⇒ DE/5 = 3/7

⇒ DE = (3×5)/7

DE = 15/7 cm

(b) In the figure

Given: PQ ||AC, AP = 4 cm,

To find: CQ and BQ

Now PQ || AC (Given)

∠BQP = ∠BCA (Alternate angles)

Also, ∠B = ∠B (common)

∴ ∆ABC ~ ∆BPQ

∴ 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 [∵ BC = 8 cm given]

Now, BQ/8 = 6/10

⇒ BQ = 6/10 × 8

= 48/10

= 4.8 cm

Also, CQ = BC – BQ

CQ = (8 – 4.8) cm

⇒ CQ = 3.2 cm

Hence, CQ = 3.2 cm and BQ = 4.8 cm

(c) In the given figure,

Given : XY || QR,

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

To find: PY and XY

Now , XY ||QR (Given)

∴ PX/QX = PY/YR

⇒ 1/3 = PY/4.5

⇒ (4.5×1)/3 = PY

⇒ 1.5 = PY

⇒ PY = 1.5 cm

Also, ∠X = ∠Q

And ∠Y = ∠R (XY || QR given)

∵ ∆PXY ~ ∆PQR

∴ XY/QR = PX/PQ

⇒ XY/9 = 1/(1 + 3) [PQ = 1 + 3 = 4 cm]

⇒ XY/9 = 1/4

⇒ XY = 9/4

= 2.25 cm

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.

In the given figure, DE || BC

(i) AD = x, DB = x – 2, AE = x + 2, EC = x – 1

In ∆ABC,

∵ DE || BC

⇒ x/(x – 2) = (x + 2)/(x – 1) (By cross multiplication)

x(x – 1) = (x – 2)(x + 2)

x2 – x = x2 – 4

- x = - 4

⇒ x = 4

(ii) DB = x – 3, AB = 2x

EC = x – 2, AC = 2x + 3

In ∆ABC

∴ DE || BC

∴ AB/DB = AC/EC

⇒ 2x/(x – 3) = (2x + 3)/(x – 2)

By cross multiplication,

2x(x – 2) = (2x + 3)(x – 3)

⇒ 2x2 – 4x = 2x2 – 6x + 3x – 9

⇒ 2x2 – 4x – 2x2 + 6x – 3x = - 9

⇒ - x = - 9

∴ x = 9

PE/EQ = (3.9)/3

= 39/30

= 13/10

PF/FR = 8/9

∴ PE/EQ ≠ PF/FR

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 RF = 9 cm.

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0. 18 cm and PF = 0.36 cm.

(i) In ∆PQR, E and F are the points on the sides PQ and PR respectively

PE = 3.9 cm, EQ = 3 cm, PF = 8 cm,

RF = 9 cm

Is EF || QR ?

∴ EF is not parallel to QR

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

PQ/PE = (1.28)/(0.18)

= 128/18

= 64/9

PR/PF = (2.56)/(0.36)

= 256/36

= 64/9

∵ PQ/PE = PR/PF

∴ EF || 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.

In ∆PQR, A and B are points on the points PQ and PR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

Now PQ/PA = (12.5)/5 = (2.5)/1

And PR/PB = (PB + BR)/PB = (4 + 6)/4 = 10/4

= 2.5/1

∵ PQ/PA = PR/PB

∴ AB || QR

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

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

(a) Given: In the figure,

DE || BC and BD = CE

To prove: ∆ABC is an isosceles triangle

Proof: In ∆ABC, DE || BC

But DB = EC (Given)…(i)

AD + DB = AE + EC

⇒ AB = AC

∴ ∆ABC is an isosceles triangle

(b) Given: In the given figure,

AB || DE, BD|| EF

To Prove: DC2 = CF × AC

Proof : In ∆ABC, DE || AB

∴ DC/CA = CE/CB …..(i)

In ∆CDE

EF || DB

CF/CD = CE/CB …(ii)

From (i) and (ii),

DC/CA = CF/CD

⇒ DC/AC = CF/DC

By cross multiplication,

DC2 = CF × 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.

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

(a) Given: CD || LA and DE || AC

Length of BE = 4 cm

Length of EC = 2 cm

Now, In ∆BCA

DE || AC

BE/BC = BD/BA (Corollary of basic proportionality theorem)

⇒ BE/(BE + EC) = BD/AB

⇒ 4/(4+2) = BD/AB …(i)

Now, In ∆BLA

CD || LA

∴ BC/BL = BD/AB (Corollary of basic proportionality theorem)

⇒ BC/(BC + CL) = BD/AB

6/(6 + CL) = BD/AB …(ii)

Combining eq. (i) and (ii), we get

6/(6 + CL) = 4/6

⇒ 6×6 = 4×(6 + CL)

⇒ 24 + 4CL = 36

⇒ 4CL = 36 – 24

⇒ CL = 12/4 = 3 cm

∴ The length of CL = 3 cm

(b) Given : In the given figure, ∠D = ∠E

To prove: ∆BAC is an isosceles triangle

∠D = ∠F (Given)

∴ AD = AE (Sides opposite to equal angles)

In ∆ABC,

∴ DE || BC

∴ DB = EC [From (i)]

AD + DB = AE + EC

⇒ AB = AC

∴ ∆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.

In the given figure, A, B, C are points on OP, OQ and OR respectively and AB || PQ and AC || PR

To prove: BC || QR

Proof : In POQ,

AB || PQ

∴ OA/AP = OB/BQ …(i)

Similarly in ∆OPR

AC || PR

∴ OA/AP = OC/CR ….(ii)

From (i) and (ii),

OB/BQ = OC/CR

Now in ∆OQR

∵ OB/BQ = OC/CR

∴ BC || QR

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

Given : ABCD is a trapezium in which AB || DC

Its diagonals AC and BD intersect each other at O.

To prove: AO/BO = CO/DO

Proof : In ∆OAB and ∆OCD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate angles)

And ∠OBA = ∠ODC (Alternate angles)

∴ ∆OAB ~ ∆OCD

∴ OA/OC = OB/OD

⇒ AO/OB = CO/DO (By alternedo)

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.

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

(a) Given : AB || CR and LM || QR.

Also BM : MC = 1 : 2

To prove:

(i) BM/MC = AL/LQ

(ii) to calculate LM : QR

Proof:

(i) In ∆ARQ

∵ LM || QR

∴ AM/MR = AL/LQ …(i)

Now, in ∆AMB and ∆MCR

∠AMB = ∠CMR (Vertically opposite angles)

∠MBA = MCR (Alternate angles)

[∵ AB || CR (given)]

∴ ∆AMB and ∆MCR

∴ AM/MR = BM/MC …(ii)

From (i) and (ii), we get

BM/MC = AL/LQ

(ii) from (2),

AM/MR = BM/MC

⇒ AM/MR = 1/2 …(iii)

[∵ BM : MC = 1 : 2, ∴ BM/MC = 1/2]

∵ LM ||QR (given)

∵ AM/MR = LM/QR

Or, AR/AM = QR/LM

Or, (AM + MR)/AM = QR/LM

Or, 1 + MR/AM = QR/LM

Or, 1 + 2/1 = QR/LM [∵ From (iii), AM/MR = 1/2 ∴ MR/AM = 2/1]

Or, 3/1 = QR/LM

Or, LM/QR = 3/1

∴ LM/QR = 1 : 3

(b) Given : ∆ABC, AD is (internal) bisector of ∠BAC.

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

To calculate: The value of BC.

Construction: Through C, Draw a straight line, CE||DA meeting BA produced in E

Now, In ∆ABC,

As AD is bisector of ∠A,

∠1 = ∠2 …(i)

∵ CE || DE (By construction) and AC cuts them,

∠2 = ∠4 (Alternate angles) …(ii)

Again CE || DA and BE cuts them,

∠1 = ∠3 (corresponding angles) …(iii)

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

∠3 = ∠4

⇒ AC = AE …(iv)

In ∆BCE, DA || CE,

∴ BD/DC = AB/AE

⇒ BD/DC = AB/AC (∵ From (4), AC = AE)

⇒ 3/DC = 6/4

⇒ 3×4 = 6×DC

⇒ DC = (3×4)/6 = 2

∴ BC = BD + DC

= 3 cm + 2 cm

= 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.

(i) ∴ ∆ABC ~ ∆PQR

(Area of ∆ABC)/(area of ∆PQR) = BC2/QR2

(By theorem 15.1)

But BC = QR = 1 : 3

∴ (Area of ∆ABC)/(Area of ∆PQR)

= (1)2/(3)2 = 1/9

Hence area of ∆ABC : area of ∆PQR

= 1 : 9

(ii) ∵ ∆ABC ~ ∆PQR

(area of ∆ABC)/(area of ∆PQR) = (BC)2/(QR)2

(By theorem 15.1)

But area of ∆ABC = Area of ∆PQR

= 25 : 36

∴ BC2/QR2 = 25/36

⇒ (BC/QR)2 = (5/6)2

⇒ BC/QR = 5/6

⇒ BC : QR = 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.

Let EF = x

Given that

∆ABC ~ ∆DEF,

∴ (area of ∆ABC)/(area of ∆DEF) = BC2/EF2

⇒ 9/16 = BC2/EF2

⇒ (2.1)2/x2 = 9/16

⇒ 2.1/x = 3/4 (Taking square root)

⇒ 3x = 4×2.1

⇒ x = (4×2.1)/3

∴ x = 2.8

Hence 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.

∆ABC ~ ∆DEF

(Area of ∆ABC)/(Area of ∆DEF) = BC2/EF2

⇒ 54/(area of ∆DEF) = (3)2/(4)2

⇒ 54/(area of ∆DEF) = 9/16

⇒ 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.

Let ∆ABC ~ ∆DEF, AL and DM are their altitudes then area of ∆ABC = 36 cm2

Area of ∆DEF = 25 cm2 and AL = 2.4 cm

Let DM = x

Now ∆ABC ~ ∆DEF

∴ (Area of ∆ABC)/(Area of ∆DEF) = AL2/DM2

⇒ 36/25 = (2.4)2/x2

⇒ 36 x2 = (2.4)2×25

⇒ x2 = {(2.4)2×2.5}/36

= (576×25)/(100×36)

= 16/4

∴ x = 2 cm

Hence altitude of the other triangle = 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.

(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5 cm, AB = 6.6 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.

(a) In ∆AOQ and ∆BOP, we have

∠OAQ = ∠OBP [Each = 90°]

∠AOQ = ∠BOP [Vertically opposite angles]

∆AOQ ~ ∆BOP [AA similarity]

(Area of ∆AOQ)/(Area of ∆BOP) = OQ2/PQ2

[∵ Area’s of similar triangles are proportional to the squares of their corresponding sides]

(Area of ∆AOQ)/120 = (9)2/(6)2

⇒ (Area of ∆AOQ)/120 = 81/36

⇒ Area of ∆AOQ = (81 × 120)/36

= 9×30 cm2

= 270 cm2

(b) Given : In the figure AB || CD

AO = 10 cm, OC = 5 cm,

AB = 6.5 cm and OD = 2.8 cm

To prove: (i) ∆OAB ~ ∆OCD

(ii) Find CD and OB

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

Proof: In the ∆OAB and ∆OCD,

(i) ∴ ∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate angles)

∠OBA = ODC (Alternate angles)

∴ ∆OAB ~ ∆OCD (AAA axiom)

(ii) ∴ OA/OC = OB/OD = AB/CD

⇒ 10/5 = OB/2.8 = 6.5/CD

OB/2.8 = 10/5

⇒ OB = 10/5×2.8

= 5.6 cm

∴ OB = 5.6 cm

Also,

6.5/CD = 10/5

⇒ CD = (6.5 × 5)/10

CD = (32.5)/10

= 3.25 cm

(iii) ∵ ∆OAB ~ ∆OCD (Proved)

∴ {ar(∆OAB)/ar(∆OCD)} = AB2/CD2

= (6.5)2/(3.25)2

= (6.5×6.5)/(3.25×3.25)

= (2×2)/1

= 4/1

ar(∆OAB) : ar(∆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.

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

(a) In the figure,

DE || BC

∠D = ∠B and ∠E = ∠C (Corresponding angles)

∠D = ∠B, ∠E = ∠C (proved)

∠A = ∠A (common)

∴ ∆ADE ~ ∆ABC (AAA postulate)

∴ (Area of ∆ADE)/(Area of ∆ABC)

= (DE)2/(BC)2

⇒ 28/(area of ∆ABC)

= (6)2/(9)2

= 36/81

⇒ area of ∆ABC = (28 × 81)/36

= 63

∴ Area of ∆ABC = (28 × 81)/36

= 63 cm2

(b) In the figure, DE || BC

∴ ∠D = ∠B and ∠E = C (Corresponding angles)

∠D = ∠B, ∠E = ∠C (Proved)

∴ ∠A = ∠A (Common)

∴ ∆ADE ~ ∆ABC (AAA postulate)

= 2/1 + 1

∴ (Area of ∆ADE)/(Area of ∆ABC)

= 1/9

⇒ area of ∆ABC – area of ∆Ade

∴ (Area of ∆ADE)/(Area of trapezium DBCE) = 1/8

∴ 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 = (1/2).BD, calculate DE if BC = 4.5 cm.

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

(i) Given: In ∆ABC, DE || BC.

To Prove : In ∆ADE ~ ∆ABC

Proof : In ∆ADE and ∆ABC,

∠A = ∠A (Common)

∴ ∆ADE ~ ∆ABC (AA axiom)

∴ AD/AB = AB/AC = DE/BC

⇒ (1/2.BD)/(1/2.BD +BD) = DE/4.5 (∵ BC = 4.5)

⇒ {(1/2).BD}/{(3/2).BD} = DE/4.5

⇒ 1/2 × 2/3 = DE/(4.5)

⇒ 1/3 = DE/(4.5)

∴ DE = (4.5)/3

= 1.5 cm.

(iii) Area of ∆ABC = 18 cm2

∴ (area of ∆ ADE)/( area of ∆ABC) = DE2/BC2

(Area of similar triangles are proportional to the square of their corresponding sides)

⇒ (Area of ∆ADE)/18 = (1/3)2 = 1/9 [Proved in (ii)]

⇒ Area of ∆ADE = 18× 1/9 = 2

∴ 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.

(i) Prove that : ∆ABC ~ ∆DEC

In ∆ABC and ∆DEC

∠ABC = ∠DEC = 90°

∠C = ∠C (common)

∴ ∆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 = 10 cm

(iii) (Area of ∆ABC)/(Area of ∆DEC) = AB2/DE2

= 62/42

= 36/16

= 9/4

= 9 : 4

9. In the adjoining figure, ABC is a triangle.

DE is parallel to BC and AD/DB = 3/2

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

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

= 2/3 + 1

∠AED = ∠C (Corresponding ∠s)

∴ By AA similarity

⇒ DE/BC = 3/5

(ii) In ∆DEF and ∆CBF

∠1 = ∠2 (Alternate ∠s)

∠3 = ∠4 (Alternate ∠s)

∠5 = ∠6 (Vertically opp. ∠s)

∴ ∆DEF ~ ∆CBF

EF/FB = DE/BC = 3/5

(iii) As the ratio of the area of two similar triangles is equal to the ratio of the square of any two corresponding sides.

∴ (Area of ∆DFE)/(Area of ∆BFC) = DE2/BC2

= (DE/BC)2

= (3/5)2

= 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.

In the figure,

PQ || BC and PO is produced to Q such that Q such that CQ || BA

And AP: PB = 2 : 3

(i) Now in ∆APO and ∆ABC

∠A = ∠A (Common)

∠APO = ∠ABC (Corresponding angle)

∴ ∆APO ~ ∆ABC (AA axiom)

∵ Areas of the similar triangles are proportional to the square of their corresponding sides

∴ {Area(∆APO)/Area(∆ABC)}

= AP2/AB2

= AP2/(AP + PB)2

= (2)2/(2 + 3)2

= 4/(5)2

= 4/25

Area (∆APO) : area (∆ABC) = 4 : 25

(ii) In ∆APO and ∆CQO

∠AOP = ∠COQ (Vertically opposite angles)

∠APQ = ∠OQC (Alternate angles)

∴ ∆APQ ~ ∆CQO

∴ {area(∆APO)/area(∆CQO)} = AP2/CQ2

= AP2/PB2 (∵ PB = CQ)

= (2)2/(3)2

= 4/9

Area (∆APO) : area (∆CQO)

= 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.

(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.

(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

(ii) area of ∆BEF : area of ∆ABD

(iii) area of ∆ABD : area of trap. AFED

(iv) area of ∆FEO : area of ∆OBC.

(i)

(ii)
(iii)

(a) In trapezium ABCD, AB || DC.

∠OAB = ∠OCD [alternate angles]

∠OBA = ∠ODC

∆AOB ~ ∆COD

∴ (area of ∆AOB)/(area of ∆COD) = AB2/CD2

= (2 CD)2/CD2 (∵ AB = 2 CD)

= (4 CD2)/(CD)2

= 4/1

∴ area of ∆AOB : area of ∆COD

= 4 : 1

(b) In || gm ABCD, AM ⊥ DC and AN ⊥ CB

Now area of ||gm ABCD = DC×AM or BC×AN

∴ DC×AM = BC×AN

= area of ||gm

⇒ DC × 6 = BC × 10

= 45

(i) ∴ DC = 45/6

= 15/2

= 7.5 cm

∴ AB = 7.5 cm (∵ AB = DC)

(ii) and BC = 45/10 = 4.5 cm

(iii) Now in ∆ADM and ∆ABN

∠D = ∠B (opposite angles of a ||gm)

∠M = ∠N (each 90°)

= BC2/AB2

= (4.5)2/(7.5)2

= (20.25)/(56.25)

= 2025/5625

= 81/225

= 9/25

∴ area of ∆ADM : area of ∆ABN

= 9 : 25

(c) In ||gm ABCD, E is a point on AB, CE intersects the diagonal BD at O EF|| BC and AE : EB = 2 : 3

In ∆ABD, EF || BC or AD

But AE/EB = 2/3

⇒ AE/EB + 1 = 2/3 + 1

⇒ (AE + EB)/EB = (2 + 3)/3

⇒ AB/EB = 5/3

⇒ BE/AB = 3/5

⇒ EF/AD = BE/AB = 3/5

⇒ EF : AD = 3 : 5

(iii) ∵ ∆BEF ~ ∆ABD

= (area of ∆BEF)/(area of ∆ABD)

= (3)2/(5)2

= 9/25

∴ area of ∆BEF : area of ∆ABD = 9 : 25

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

25 area of ∆BEF = 9 area of ∆ABD

⇒ 25(area of ∆ABD – area of trap. AEFD)

= 9 area of ∆ABD

⇒ 25 area of trap AEFD = 25 area of ∆ABD – 9 area of ∆ABD

⇒ 25 area of AEFD = 16 area of ∆ABD

⇒ (area of ∆ABD)/(area of trap AEFD) = 25/16

⇒ area of ∆ABD : area of trap. AEFD

= 25 : 16

(iv) In ∆FEO and ∆OBC

∠EOF = ∠BOC (Vertically opposite angles)

∠F = ∠OBC (alternate angles)

∴ ∆FEO ~ ∆OBC

∴ (area of FEO)/(area of ∆OBC) = EF2/BC2

= EF2/AD2 = 9/25 [form (i)]

∴ 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

(i) area of ∆BPQ.

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

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.

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.

(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.

(a) In ∆ABC, DE || BC

∠A = ∠A (common)

∠D = ∠B and ∠E = ∠C (Corresponding angles)

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

But (area of ∆ADE)/(area of trap DBCE) = 4/5

⇒ (area of trap. DBCE)/(area of ∆ADE) = 5/4

⇒ (area of trap DBCE)/(area of ∆ADE) + 1

= 5/4 + 1 (Adding 1 both sides)

⇒ (area of trap DBCE + area of ∆ADE)/(area of ∆ ADE)

= (5 + 4)/4 = 9/4

⇒ (area of ∆ABC)/(area of ∆ADE) = 9/4

⇒ (area of ∆ADE)/(area of ∆ABC) = 4/9

Now from (i) (DE)2/(BC)2 = 4/9

⇒ (2)2/(3)2

⇒ (DE)/(BC) = 2/3

⇒ DE : BC = 2 : 3

(b) 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.

(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.

(a) In trapezium ABCD, DC || AB

AB = 9 cm, DC = 6 cm, BD = 12 cm

(i) In ∆APB and ∆CPD

∠APB = ∠CPD (Vertically opposite angles)

∠PAB = ∠PCD (alt. angles)

∴ ∆APB ~ ∆CPD (AA postulate)

∴ BP/PD = AB/CD

⇒ BP/(12 – BP) = 9/6

⇒ 6 BP = 108 - 9BP

⇒ 6 BP + 9 BP = 108

⇒ 15 BP = 108

⇒ BP = 108/15

= 7.2 cm

(ii) Again

∵ ∆APB ~ ∆CPD

∴ (area of ∆APB)/(area of ∆CPD) = AB2/CD2

= (9)2/(6)2

= 81/36

= 9/4

(b) 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

4/BC = 5/8

BC = (4×8)/5

BC = 32/5

BC = 6.4 cm

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:

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

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]

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

So, AE/AB = AD/AC = DE/BC ...(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),

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.

Consider the two isosceles triangle PQR and XYZ,

∠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:

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.

From the question it is given that,

Map drawn to a scale of 1: 25000

AB = 12 cm, BG = 16 cm

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.

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)

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

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 22):

1. In the adjoining figure, ∆ABC ~ ∆QPR. Then R is

(a) 60°

(b) 50°

(c) 70°

(d) 80°

(a) 60°

In the given figure,

∆ABC ~ ∆QPR

∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R

But ∠A = 70°, ∠B = 50°

∴ ∠C = 180° - (70° + 500)

= 180° - 120°

= 60°

∠C = ∠R

∴ ∠R = ∠C = 60°

2. In the adjoining figure, ∆ABC ~ ∆QPR. The value of x is

(a) 2.25 cm

(b) 4 cm

(c) 4.5 cm

(d) 5.2 cm

(a) 2.25 cm

In the figure figure

∆ABC ~ ∆QPR

∴ AC/QR = BC/PR

⇒ 6/3 = (4.5)/x

⇒ x = (3 × 4.5)/6

= 2.25

∴ x = 2.25 cm

3. In the adjoining figure, line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, APB = 50° and CDP = 30°. Then PBA is equal to

(a) 50°

(b) 30°

(c) 60°

(d) 100°

(d) 100°

In the given figure two line segments AC and BD intersect each other at P

And PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°

In ∆APB and ∆CPD

AP/PD = 6/5 and BP/CP = 3/2.5 = 6/5

∵ AP/PD = BP/CP and ∆APB = ∠CPD (Vertically opposite angles)

∴ ∆APB ~ ∆CPD

∴ ∠A = ∠D = 30°’

and third ∠PBA = 180° - (50° + 30°)

= 180° - 80°

= 100°

4. If in two triangles ABC and PQR,

AB/QR = BC/PR = CA/PQ

then

(a) ∆PQR ~ ∆CAB

(b) ∆PQR ~ ∆ABC

(c) ∆CBA ~ ∆PQR

(d) ∆BCA ~ ∆PQR

(a) ∆PQR ~ ∆CAB

In two ∆ABC and ∆PQR

AB/QR = BC/PR = CA/PQ

⇒ PQ/CA = QR/AB = RP/BC

Then ∆PQR ~ ∆CAB

5. In triangles ABC and DEF, B = E, F = C and AB = 3DE, then the two triangles are

(a) congruent but not similar

(b) similar but not congruent

(c) neither congruent nor similar

(d) congruent as well as similar

(b) similar but not congruent.

In ∆ABC and ∆DEF

∠B = ∠E, ∠F = ∠C

AB = 3DE

∵ Two angles of the one triangles are equal to corresponding two angles of the other

But sides are equal

∵ Triangles are similar but not congruent.

6. In the adjoining figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are

(a) similar

(b) congruent

(c) both similar and congruent

(d) neither similar nor congruent

(a) similar

D, E and F are the mid-points of the sides BC, CA and AB of ∆ABC then two triangles ABC and DEF are similar.

7. The adjoining figure, AB || DE. The length of CD is

(a) 2.5 cm

(b) 2.7 cm

(c) 10/3 cm

(d) 3.5 cm

(b) 2.7 cm

In the given figure, AB || DE

∆ABC ~ ∆DCE

∴ AB/DE = BC/CD

⇒ 5/3 = (4.5)/CD

⇒ CD = (3 × 4.5)/5

= 2.7 cm

8. If in triangles ABC and DEF, AB/DE = BC/FD, then they will be similar when

(a) B = E

(b) A = D

(c) B = D

(d) A = F

(c) ∠B = ∠D

In two triangles ABC and DEF,

AB/DE = BC/FD

They will be similar if their included angles are equal

∠B = ∠D

9. If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is

(a) 20.25 cm

(b) 27 cm

(c) 48 cm

(d) 64 cm

(b) 27 cm

∆PQR ~ ∆ABC

PQ = 6 cm, AB = 8 cm

Perimeter of ∆ABC = 36 cm

Let perimeter of ∆PQR be x cm

∴ PQ/AB = (Perimeter of ∆PQR)/(Perimeter of ∆ABC)

⇒ 6/8 = x/36

⇒ x = (6×36)/8

= 27 cm

Perimeter of ∆PQR = 27 cm

10. In the adjoining figure, DE || BC and all measurements are in centimeters. The length of AE is

(a) 2 cm

(b) 2.25 cm

(c) 3.5 cm

(d) 4 cm

(b) 2.25 cm

In th.e given figure,

DE || BC

⇒ 3/4 = x/3

⇒ x = (3×3)/4

= 9/4

= 2.25 cm

11. In the given figure, PQ ||CA and all lengths are given in centimeters. The length of BC is

(a) 6.4 cm

(b) 7.5 cm

(c) 8 cm

(d) 9 cm

(c) 8 cm

In the given figure,

PQ ||CA

Let BC = x

∴ BQ/QC = BP/PA

⇒ 5/(x – 5) = 4/(2.4)

⇒ 4x – 20 = 12

⇒ 4x = 12 + 20

= 32

⇒ x = 32/4 = 8

∴ BC = 8 cm

12. In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is

(a) 4 cm

(b) 3.6 cm

(c) 2 cm

(d) 3 cm

(d) 3 cm

In the given figure, MN||QR

PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm

Let PM = x cm

∴ PM/MQ = PN/NR

⇒ x(5 – x) = (3.6)/(2.4)

⇒ x/(5 – x) = 36/24

⇒ 36(5 – x) = 24x

⇒ 180 – 36x = 24x

⇒ 180 = 24x + 36x

⇒ 60x = 180

⇒ x = 180/60

= 3

∴ PM = 3 cm

13. D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is

(a) 2.5 cm

(b) 3 cm

(c) 5 cm

(d) 6 cm

(b) 3 cm

D and E are the points on sides AB and AC of ∆ABC,

AD = 2 cm, BD = 3 cm,

BC = 7.5 cm

DE || BC

Let DE = x cm

In ∆ABC, DE || BC

⇒ 2/(2 + 3) = x/(7.5)

⇒ 2/5 = x/(7.5)

⇒ x = (2 × 7.5)/5

= 15/5

= 3

∴ DE = 3 cm

14. It is given that ∆ABC ~ ∆PQR with BC/QR = 1/3 then (area of ∆PQR)/(area of ∆ABC) equal to

(a) 9

(b) 3

(c) 1/3

(d) 1/9

(a) 9

∆ABC ~ ∆PQR

BC/QR = 1/3

∴ (area of ∆PQR)/(area of ∆ABC) = QR2/BC2

= (3)2/(1)2

= 9/1

= 9

15. If the area of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio

(a) 9 : 4

(b) 3 : 2

(c) 2 : 3

(d) 16 : 81

(c) 2 : 3

Ratio in the areas of two similar triangles = 4 : 9

Ratio in their corresponding sides = √4 : √9

= 2 : 3

16. If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is

(a) 8:6

(b) 3:4

(c) 9:16

(d) 16:9

(d) 16:9

∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm

(area of ∆ABC)/(area of ∆PQR) = BC2/QR2

= 82/62

= 64/36

= 16/9

17. If ∆ABC ~ ∆QRP (area of ∆ABC)/(area of ∆PQR), AB = 18 cm and BC = 15 cm, then the length of PR is equal to

(a) 10 cm

(b) 12 cm

(c) 20/3

(d) 8 cm

(a) 10 cm

∆ABC ~ ∆QRP

(area of ∆ABC)/(Area of ∆PQR)

AB = 18 cm, BC = 15 cm2

Length of PR

(area of ∆ABC)/(Area of ∆PQR) = 9/4 = (BC)2/(PR)2

⇒ 9/4 = 152/PR2

⇒ 3/2 = 15/PR (Taking square root)

∴ PR = (15×2)/3

= 10 cm

18. If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm2, area of ∆PQR = 144 cm2 and QR = 6 cm, then length of BC is

(a) 4 cm

(b) 4.5 cm

(c) 9 cm

(d) 12 cm

(b) 4.5 cm

∆ABC ~ ∆PQR,

Area of ∆ABC = 81 cm2 and area of ∆PQR = 144 cm2,

QR = 6 cm, BC = ?

(Area of ∆ABC)/(area of ∆PQR) = BC2/QR2

= (BC)2/(6)2

= 81/144 (given)

⇒ BC2/(6)2 = (9)2/12

⇒ BC/6 = 9/12

∴ BC = (6×9)/12

= 9/2

= 4.5 cm

19. In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is

(a) 4:1

(b) 9:1

(c) 9:4

(d) 3:2

(c) 9 : 4

In the given figure, DE || CA,

D is a point on BC and BD : DC = 2 : 1

BD : BC = 2 : (2 + 1) = 2 : 3

In ∆ABC, DE || CA

∆ABC ~ ∆BDE

∴ (area of ∆ABC)/(Area of ∆BDE) = BC2/BD2

= (3)2/(2)2

= 9/4

∴ Ratio = 9:4

20. If ABC and BDE are two equilateral such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is

(a) 2:1

(b) 1:2

(c) 1:4

(d) 4:1

(d) 4:1

∆ABC and ∆BDE are an equilateral triangles

∴ ∆ABC ~ ∆BDE

D is mid-point of BC

∴ E is mid-point of AB

∴ DE || CA = 1/2.CA

(area of ∆ABC)/(area of ∆BDE) = AC2/ED2

= AC2/(1/2.AC)2

= (AC)2/(1/4.AC)2

= 4/1

= 4 : 1

21. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is

(a) 9 cm

(b) 7 cm

(c) 6 cm

(d) 4.5 cm

(d) 4.5 cm

Areas of two similar triangles are 81 cm2 and 49 cm2

The altitude of the smaller triangle is 3. 5 cm

Let the altitude of the bigger triangle is x, then

(area of bigger triangle)/(area of smaller triangle) = 81/49

(bigger altitude)2/(smaller altitude)2

⇒ 81/49 = x2/(3.5)2

⇒ 9/7 = x/(3.5)

⇒ x = (9×3.5)/ 7

= 4.5 cm

∴ Altitude of bigger triangle = 4.5 cm

22. Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm2 and area of ∆PQR = 24 cm2. If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is

(a) 49/3 cm

(b) 20/3 cm

(c) 15 cm

(d) 22.5 cm

(c) 15 cm

∆ABC ~ ∆PQR

area of ∆ABC = 54 cm2

and of ∆PQR = 24 cm2

AD and PM are their median respectively

PM = 10 cm

Let AD = x cm, then

(area of ∆ABC)/(area of ∆PQR) = AD2/PM2

= x2/(10)2

= 54/24

∴ (x2)/(10)2 = 54/24 = 9/4 = (3/2)2

∴ x/10 = 3/2

⇒ x = (3×10)/2

= 15 cm

### Chapter Test

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

From the question it is given that,

∠1 = ∠2 and ∠3 = ∠4

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

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

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

From the given figure,

AB = AC.

If PM ⊥ AB and PN ⊥ AC

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

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

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

(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

(a) From the figure it is given that,

∠AED = ∠ABC

∠AED = ∠ABC [from the figure]

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

Therefore, ∆ABC ~ ∆ADE [by AA axiom]

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) 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]

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.

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

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

2DE = BD [equation (i)]

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

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.

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

Consider the ∆ABC,

Given, DE || BC

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

2.4/AB = 2/5

⇒ AB = (2.4×5)/2

⇒ AB = 12/2

⇒ AB = 6 cm

Then, consider AE/AC = DE/BC

3.2/AC = 2/5

⇒ AC = (3.2×5)/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 = 9.5 cm, AE = 3.3cm and AC = 8.8 cm. Is DE || BC? Justify your answer.

From the question it is given that,

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

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

Consider the ∆ABC,

EC = AC – AE

= 8.8 – 3.3

= 5.5 cm

= 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

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.

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,

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

Therefore, the length of the corresponding side of the second triangle YZ = 20/3

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.

From the question it is given that,

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

5/4 = BC/3

BC = (5×3)/4

BC = 15/4

BC = 3.75 cm

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.

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.

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

(i) DO : OE = 2:1

(ii) area of ∆OEC : area of ∆OAD = 1:4.

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.

12. 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.

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.

The solutions provided for Chapter 13 Similarity of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 13 Similarity contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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