# ICSE Solutions for Selina Concise Chapter 15 Similarity (with Applications to Maps and Models) Class 10 Maths

### Exercise 15(A)

**1. In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:**

**(i) ∆APC and ∆BPD are similar.**

**(ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.**

**Solution**

**(i)** In ∆APC and ∆BPD, we have

∠APC = ∠BPD **[Vertically opposite angles]**

∠ACP = ∠BDP **[Alternate angles as, AC || BD]**

Thus, ∆APC ~ ∆BPD by AA similarity criterion

**(ii)** So, by corresponding parts of similar triangles, we have

PA/PB = PC/PD = AC/BD

Given, BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm

PA/(3.2) = PC/4 = 3.6/2.4

PA/3.2 = 3.6/2.4 and PC/4 = 3.6/2.4

Thus,

PA = (3.6×3.2)/2.4 = 4.8 cm and

PC = (3.6×4)/2.4 = 6 cm

**2. In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:**

**(i) Î” APB is similar to Î” CPD.**

**(ii) PA x PD = PB×PC.**

**Solution**

**(i)** In ∆APB and ∆CPD, we have

∠APB = ∠CPD **[Vertically opposite angles]**

∠ABP = ∠CDP **[Alternate angles as, AB||DC]**

Thus, ∆APB ~ ∆CPD by AA similarity criterion.

**(ii)** As ∆APB ~ ∆CPD

Since the corresponding sides of similar triangles are proportional, we have

PA/PC = PB/PD

Thus,

PA×PD = PB×PC

**3. P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:**

**(i) DP : PL = DC : BL.**

**(ii) DL : DP = AL : DC.**

**Solution**

**(i) **As AD||BC, we have AD|| BP also.

So, by BPT

DP/PL = AB/BL

And, since ABCD is a parallelogram, AB = DC

Hence,

DP/PL = DC/BL

i.e., DP : PL = DC : BL

**(ii)** As AD||BC, we have AD|| BP also.

So, by BPT

DL/DP = AL/AB

And, since ABCD is a parallelogram, AB = DC

Hence,

DL/DP = AL/AB

i.e., DL : DP = AL : DC

**4. In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:**

**(i) Î” AOB is similar to Î” COD.**

**(ii) OA x OD = OB****×****OC.**

**Solution**

**(i)** Given,

AO = 2CO and BO = 2DO,

⇒ AO/CO = 2/1 = BO/DO

And,

∠AOB = ∠DOC **[Vertically opposite angles]**

Hence, ∆AOB ~ ∆COD **[SAS criterion for similarity]**

**(ii)** As, AO/CO = 2/1 = BO/DO **[Given]**

Thus, OA×OD = OB×OC

**5. In Î” ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :**

**(i) CB : BA = CP : PA**

**(ii) AB****×****BC = BP****×****CA **

**Solution**

**(i)** In Î” ABC, we have

∠ABC = 2 ∠ACB **[Given]**

Now, let ∠ACB = x

So, ∠ABC = 2x

Also given, BP is bisector of ∠ABC

Thus, ∠ABP = ∠PBC = x

By using the angle bisector theorem,

i.e. the bisector of an angle divides the side opposite to it in the ratio of other two sides.

∴ CB: BA = CP: PA.

**(ii)** In Î” ABC and Î” APB,

∠ABC = ∠APB **[Exterior angle property]**

∠BCP = ∠ABP **[Given]**

Thus, ∆ABC ~ ∆APB by AA criterion for similarity

Now, since corresponding sides of similar triangles are proportional we have

CA/AB = BC/BP

∴ AB×BC = BP×CA

**6. In Î” ABC, BM ⊥ AC and CN ⊥ AB; show that:**

**AB/AC = BM/CN = AM/AN**

**Solution**

In Î” ABM and Î” ACN,

∠AMB = ∠ANC **[Since, BM ⊥ AC and CN ⊥ AB]**

∠BAM = ∠CAN **[Common angle]**

Hence, ∆ABM ~ ∆ACN by AA criterion for similarity

So, by corresponding sides of similar triangles we have

**7. In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.**

**(i) Write all possible pairs of similar triangles.**

**(ii) Find the lengths of ME and DM.**

**Solution**

**(i)** In Î” AME and Î” ANC,

∠AME = ∠ANC **[Since DE || BC so, ME || NC]**

∠MAE = ∠NAC **[Common angle]**

Hence, ∆AME ~ ∆ANC by AA criterion for similarity

In Î” ADM and Î” ABN,

∠ADM = ∠ABN **[Since DE || BC so, DM || BN]**

∠DAM = ∠BAN **[Common angle]**

Hence, ∆ADM ~ ∆ABN by AA criterion for similarity

In Î” ADE and Î” ABC,

∠ADE = ∠ABC **[Since DE || BC so, ME || NC]**

∠AED = ∠ACB **[Since DE || BC]**

Hence, ∆ADE ~ ∆ABC by AA criterion for similarity

**(ii)** Proved above that, ∆AME ~ ∆ANC

So as corresponding sides of similar triangles are proportional, we have

ME/NC = AE/AC

⇒ ME/ 6 = 15/ 24

⇒ ME = 3.75 cm

And, ∆ADE ~ ∆ABC **[Proved above]**

So as corresponding sides of similar triangles are proportional, we have

AD/AB = AE/AC = 15/24 **…. (1)**

Also, ∆ADM ~ ∆ABN **[Proved above]**

So as corresponding sides of similar triangles are proportional, we have

DM/BN = AD/AB = 15/24 **…. From (1)**

⇒ DM/ 24 = 15/ 24

⇒ DM = 15 cm

**8. In the given figure, AD = AE and AD ^{2} = BD x EC. Prove that: triangles ABD and CAE are similar. **

**Solution**

In Î” ABD and Î” CAE,

∠ADE = ∠AED [Angles opposite to equal sides are equal.]

So, ∠ADB = ∠AEC **[As ∠ADB + ∠ADE = 180 ^{o }and ∠AEC + ∠AED = 180^{o}]**

And, AD^{2} = BD×EC **[Given]**

AD/BD = EC/AD

AD/BD = EC/AE

Thus, ∆ABD ~ ∆CAE by SAS criterion for similarity.

**9. In the given figure, AB ‖ DC, BO = 6 cm and DQ = 8 cm; find: BP****×****DO.**

**Solution**

In Î” DOQ and Î” BOP,

∠QDO = ∠PBO **[As AB || DC so, PB || DQ]**

So, ∠DOQ = ∠BOP **[Vertically opposite angles]**

Hence, ∆DOQ ~ ∆BOP by AA criterion for similarity

Since, corresponding sides of similar triangles are proportional we have

DO/BO = DQ/BP

DO/6 = 8/BP

BP×DO = 48 cm^{2}

^{}

**10. Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.**

**Solution**

In Î” ABC,

AC = AB **[Given]**

So, ∠ABC = ∠ACB **[Angles opposite to equal sides are equal.]**

In Î” PRC and Î” PQB,

∠ABC = ∠ACB

∠PRC = ∠PQB **[Both are right angles.]**

Hence, ∆PRC ~ ∆PQB by AA criterion for similarity

Since, corresponding sides of similar triangles are proportional we have

PR/PQ = RC/QB = PC/PB

⇒ PR/PQ = PC/PB

⇒ 9/15 = 12/PB

Thus,

PB = 20 cm

**11. State, true or false:**

**(i) Two similar polygons are necessarily congruent.**

**(ii) Two congruent polygons are necessarily similar.**

**(iii) All equiangular triangles are similar.**

**(iv) All isosceles triangles are similar.**

**(v) Two isosceles-right triangles are similar.**

**(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.**

**(vii) The diagonals of a trapezium, divide each other into proportional segments.**

**Solution**

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

**12. Given: **∠**GHE = **∠**DFE = 90 ^{o}, DH = 8, DF = 12, DG = 3x – 1 and DE = 4x + 2.**

**Find: the lengths of segments DG and DE.**

**Solution**

In Î” DHG and Î” DFE,

∠GHD = ∠DFE = 90^{o}

∠D = ∠D **[Common]**

Thus, ∆DHG ~ ∆DFE by AA criterion for similarity

So, we have

DH/DF = DG/DE

⇒ 8/12 = (3x – 1)/ (4x + 2)

⇒ 32x + 16 = 36x – 12

⇒ 28 = 4x

⇒ x = 7

Hence,

DG = 3×7 – 1 = 20

DE = 4×7 + 2 = 30

**13. D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA ^{2} = CB x CD.**

**Solution**

In Î” ADC and Î” BAC,

∠ADC = ∠BAC **[Given]**

∠ACD = ∠ACB **[Common]**

Thus, ∆ADC ~ ∆BAC by AA criterion for similarity

So, we have

CA/CB = CD/CA

∴ CA^{2} = CB×CD

**14. In the given figure, ∆ ABC and ∆ AMP are right angled at B and M respectively.**

**Given AC = 10 cm, AP = 15 cm and PM = 12 cm.**

**(i) ∆ ABC ~ ∆ AMP.**

**(ii) Find AB and BC.**

**Solution**

**(i)** In ∆ ABC and ∆ AMP, we have

∠BAC = ∠PAM** [Common]**

∠ABC = ∠PMA **[Each = 90 ^{o}]**

Hence, ∆ABC ~ ∆AMP by AA criterion for similarity

**(ii)** Now, in right triangle AMP

By using Pythagoras theorem, we have

AM = √(AP^{2} – PM^{2}) = √(15^{2} – 12^{2}) = 9

As ∆ABC ~ ∆AMP,

AB/AM = BC/PM = AC/AP

⇒ AB/9 = BC/12 = 10/15

So,

AB/9 = 10/15

⇒ AB = (10×9)/15 = 6 cm

⇒ BC/12 = 10/ 15

⇒ BC = 8 cm

### Exercise 15(B)

**1. In the following figure, point D divides AB in the ratio 3: 5. Find:**

**(i) AE/EC**

**(ii) AD/AB**

**(iii) AE/AC**

**Also if,**

**(iv) DE = 2.4 cm, find the length of BC.**

**(v) BC = 4.8 cm, find the length of DE.**

**Solution**

**(i)** Given, AD/DB = 3/5

And DE || BC.

So, by Basic Proportionality theorem, we have

AD/DB = AE/EC

AE/EC = 3/5

**(ii)** Given, AD/DB = 3/5

So, DB/AD = 5/3

Adding 1 both sides, we get

DB/AD + 1 = 5/3 + 1

⇒ (DB + AD)/ AD = (5 + 3)/3

⇒ AB/AD = 8/3

∴ AD/AB = 3/8

**(iii)** In ∆ABC, as DE || BC

By BPT, we have

AD/DB = AE/ EC

So, AD/AB = AE/AC

From above, we have AD/AB = 3/8

∴ AE/AC = 3/8

**(iv)** In ∆ADE and ∆ABC,

∠ADE = ∠ABC **[As DE || BC, corresponding angles are equal.]**

∠A = ∠A **[Common angle]**

Hence, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

AD/AB = DE/BC

⇒ 3/8 = 2.4/BC

⇒ BC = 6.4 cm

**(v)** Since, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

AD/AB = DE/BC

⇒ 3/8 = DE/4.8

⇒ DE = 1.8 cm

**2. In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:**

**(i) CP/PA**

**(ii) PQ**

**(iii) If AP = x, then the value of AC in terms of x.**

**Solution**

**(i)** In ∆CPQ and ∆CAB,

∠PCQ = ∠APQ **[As PQ || AB, corresponding angles are equal.]**

∠C = ∠C **[Common angle]**

Hence, ∆CPQ ~ ∆CAB by AA criterion for similarity

So, we have

CP/CA = CQ/CB

⇒ CP/CA = 4.8/ 8.4 = 4/7

Thus, CP/PA = 4/3

**(ii)** As, ∆CPQ ~ ∆CAB by AA criterion for similarity

We have,

PQ/AB = CQ/CB

⇒ PQ/6.3 = 4.8/8.4

⇒ PQ = 3.6 cm

**(iii)** As, ∆CPQ ~ ∆CAB by AA criterion for similarity

We have,

CP/AC = CQ/CB

⇒ CP/AC = 4.8/8.4 = 4/7

So, if AC is 7 parts and CP is 4 parts, then PA is 3 parts.

Hence, AC = 7/3 ×PA = (7/3)x

**3. A line PQ is drawn parallel to the side BC of Î” ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.**

**Solution**

In ∆ APQ and ∆ ABC,

∠APQ = ∠ABC **[As PQ || BC, corresponding angles are equal.]**

∠PAQ = ∠BAC **[Common angle]**

Hence, ∆APQ ~ ∆ABC by AA criterion for similarity

So, we have

AP/AB = AQ/AC

⇒ AP/9 = 4.2/6

Thus, AP = 6.3 cm

**4. In Î” ABC, D and E are the points on sides AB and AC respectively.**

**Find whether DE ‖ BC, if**

**(i) AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm. **

**(ii) AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.**

**Solution**

**(i)** In ∆ ADE and ∆ ABC,

AE/EC = 6/7.5 = 4/5

⇒ AD/BD = 4/5 **[BD = AB – AD = 9 – 4 = 5 cm]**

So, AE/EC = AD/BD

∴ DE || BC by the converse of BPT.

**(ii)** In ∆ ADE and ∆ ABC,

AE/EC = 1.6/11 = 0.8/5.5

⇒ AD/BD = 0.8/5.5 **[BD = AB – AD = 6.3 – 0.8 = 5.5 cm]**

So, AE/EC = AD/BD

∴ DE || BC by the converse of BPT.

**5. In the given figure, Î” ABC ~ Î” ADE. If AE: EC = 4: 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.**

**Solution**

Given,

Î” ABC ~ Î” ADE

So, we have

AE/AC = DE/BC

⇒ 4/11 = 6.6/BC

⇒ BC = (11×6.6)/4 = 18.15 cm

And, also

As, Î” ABC ~ Î” ADE, we have

∠ABC = ∠ADE and ∠ACB = ∠AED

So, DE || BC

And, AB/AD = AC/AE = 11/4 [Since, AE/EC = 4/7]

In ∆ ADP and ∆ ABQ,

∠ADP = ∠ABQ **[As DP || BQ, corresponding angles are equal.]**

∠APD = ∠AQB **[As DP || BQ, corresponding angles are equal.]**

Hence, ∆ADP ~ ∆ABQ by AA criterion for similarity

AD/AB = AP/AQ

⇒ 4/11 = x/AQ

Thus, AQ = (11/4)x

### Exercise 15(C)

**1. (i) The ratio between the corresponding sides of two similar triangles is 2: 5. Find the ratio between the areas of these triangles.**

**(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.**

**Solution**

We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

So,

**(i) **The required ratio is given by,

**(ii)** The required ratio is given by,

**2. A line PQ is drawn parallel to the base BC of Î” ABC which meets sides AB and AC at points P and Q respectively. If AP = 1/3 PB; find the value of:**

**(i) Area of Î” ABC/Area of Î” APQ**

**(ii) Area of Î” APQ/ Area of Trapezium PBCQ**

**Solution**

Given, AP = (1/3) PB

So, AP/PB = 1/3

In ∆ APQ and ∆ ABC,

As PQ || BC, corresponding angles are equal

∠APQ = ∠ABC and ∠AQP = ∠ACB

Hence, ∆APQ ~ ∆ABC by AA criterion for similarity

So,

**(i)** Area of ∆ABC/ Area of ∆APQ

= AB^{2}/ AP^{2}

= 4^{2}/1^{2}

= 16: 1

**[AP/PB = 1/3 so, AB/AP = 4/1]**

**(ii)** Area of Î” APQ/Area of Trapezium PBCQ

= Area of Î” APQ/(Area of Î” ABC – Area of Î” APQ)

= 1/(16/1)

= 1: 16

**3. The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.**

**Solution**

Let ∆ABC ~ ∆DEF

So, AB/DE = BC/EF = AC/DF = (AB + BC + AC)/(DE + EF + DF)

= Perimeter of Î” ABC/Perimeter of Î” DEF

Perimeter of Î” ABC/Perimeter of Î” DEF = AB/DE

30/24 = 12/DE

⇒ DE = 9.6 cm

**4. In the given figure, AX: XB = 3: 5.**

**Find:**

**(i) the length of BC, if the length of XY is 18 cm.**

**(ii) the ratio between the areas of trapezium XBCY and triangle ABC.**

**Solution**

Given, AX/XB = 3/5 ⇒ AX/AB = 3/8 **…. (1)**

**(i)** In Î” AXY and Î” ABC,

As XY || BC, corresponding angles are equal.

∠AXY = ∠ABC and ∠AYX = ∠ACB

Hence, ∆AXY ~ ∆ABC by AA criterion for similarity.

So, we have

AX/AB = XY/BC

⇒ 3/8 = 18/BC

⇒ BC = 48 cm

**(ii)** Area of Î” AXY/ Area of Î” ABC = AX^{2}/AB^{2} = 9/64

(Area of Î” ABC – Area of Î” AXY)/Area of Î” ABC = (64 – 9)/64 = 55/64

Area of trapezium XBCY/ Area of Î” ABC = 55/64

**5. ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.**

**Solution**

It’s given that,

Ar(Î” APQ) = ½ Ar(Î” ABC)

⇒ Ar(Î” APQ)/Ar(Î” ABC) = ½

⇒ AP^{2}/AB^{2} = ½

⇒ AP/AB = 1/√2

⇒ (AB – BP)/ AB = 1/√2

⇒ 1 – (BP/AB) = 1/√2

⇒ BP/AB = 1 – 1/√2

Thus,

**6. In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4.**

**Calculate the value of ratio:**

**(i) PL/PQ and then LM/QR**

**(ii) Area of Î” LMN/ Area of Î” MNR**

**(iii) Area of Î” LQM/ Area of Î” LQN**

**Solution**

**(i)** In Î” PLM and Î” PQR,

As LM || QR, corresponding angles are equal.

∠PLM = ∠PQR

∠PML = ∠PRQ

Hence, ∆PLM ~ ∆PQR by AA criterion for similarity.

So, we have

PM/ PR = LM/ QR

3/7 = LM/QR [Since, PM/MR = ¾ ⇒ PM/PR = 3/7]

And, by BPT we have

PL/LQ = PM/MR = ¾

⇒ LQ/PL = 4/3

⇒ 1 + (LQ/PL) = 1 + 4/3

⇒ (PL + LQ)/PL = (3 + 4)/3

⇒ PQ/PL = 7/3

Hence, PL/PQ = 3/7

**(ii) **As Î” LMN and Î” MNR have common vertex at M and their bases LN and NR are along the same straight line

Hence, Ar (Î” LMN)/ Ar (Î” RNQ) = LN/NR

Now, in Î” LMN and Î” RNQ we have,

∠NLM = ∠NRQ **[Alternate angles]**

∠LMN = ∠NQR **[Alternate angles]**

Thus, ∆LNM ~ ∆RNQ by AA criterion for similarity.

So,

MN/QN = LN/NR = LM/QR = 3/7

∴ Ar (Î” LMN)/ Ar (Î” MNR) = LN/NR = 3/7

**(iii)** As Î” LQM and Î” LQN have common vertex at L and their bases QM and QN are along the same straight line.

Area of Î” LQM/Area of Î” LQN = QM/QN = 10/7

**[Since, MN/QN = 3/7 ⇒ QM/QN = 10/7]**

### Exercise 15(D)

**1. A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’ Calculate:**

**(i) the length of AB, if A’ B’ = 6 cm. **

**(ii) the length of C’ A’ if CA = 4 cm. **

**Solution**

Given that, Î” ABC has been enlarged by scale factor m of 2.5 to Î” A’B’C’.

**(i)** A’B’ = 6 cm

So,

AB(2.5) = A’B’ = 6 cm

⇒ AB = 2.4 cm

**(ii)** CA = 4 cm

We know that,

CA(2.5) = C’A’

⇒ C’A’ = 4×2.5 = 10 cm

**2. A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:**

**(i) the length of M’ N’, if MN = 8 cm.**

**(ii) the length of LM, if L’ M’ = 5.4 cm.**

**Solution**

Given, Î” LMN has been reduced by a scale factor m = 0.8 to Î” L’M’N’.

**(i)** MN = 8 cm

So, MN (0.8) = M’N’

⇒ (8)(0.8) = M’N’

⇒ M’N’ = 6.4 cm

**(ii)** L’M’ = 5.4 cm

So, LM (0.8) = L’M’

⇒ LM (0.8) = 5.4

⇒ LM = 6.75 cm

**3. A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find:**

**(i) A’B’, if AB = 4 cm.**

**(ii) BC, if B’C’ = 15 cm.**

**(iii) OA, if OA’ = 6 cm**

**(iv) OC’, if OC = 21 cm**

**Also, state the value of: **

**(a) OB’/OB (b) C’A’/CA**

**Solution**

Given that, Î” ABC is enlarged and the scale factor m = 3 to the Î” A’B’C’.

**(i) **AB = 4 cm

So, AB(3) = A’B’

⇒ (4)(3) = A’B’

⇒ A’B’ = 12 cm

**(ii)** B’C’ = 15 cm

So, BC(3) = B’C’

⇒ BC(3) = 15

⇒ BC = 5 cm

**(iii)** OA’ = 6 cm

So, OA (3) = OA’

⇒ OA (3) = 6

⇒ OA = 2 cm

**(iv)** OC = 21 cm

So, OC(3) = OC’

⇒ 21×3 = OC’

⇒ OC’ = 63 cm

The ratio of the lengths of the two corresponding sides of two triangles.

Î” ABC is enlarged and the scale factor m = 3 to the Î” A’B’C’

Hence,

(a) OB’/OB = 3

(b) C’A’/CA = 3

### Exercise 15(E)

**1. In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.**

**Find:**

**(i) AY/YC**

**(ii) YC/AC**

**(iii) XY**

**Solution**

Given, XY || BC.

So, In Î” AXY and Î” ABC

∠AXY = ∠ABC [Corresponding angles]

∠AYX = ∠ACB [Corresponding angles]

Hence, ∆AXY ~ ∆ABC by AA criterion for similarity.

As corresponding sides of similar triangles are proportional, we have

**(i)** AX/AB = AY/AC

⇒ 9/13.5 = AY/AC

⇒ AY/YC = 9/4.5

⇒ AY/YC = 2

⇒ AY/YC = 2/1

**(ii)** We have,

AX/AB = AY/AC

⇒ 9/13.5 = AY/ AC

⇒ YC/AC = 4.5/13.5 = 1/3

**(iii) **As, ∆AXY ~ ∆ABC

AX/AB = XY/BC

⇒ 9/13.5 = XY/18

⇒ XY = (9×18)/13.5 = 12 cm

**2. In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm,**

**Calculate:**

**(i) EC**

**(ii) AF**

**(iii) PE**

**Solution**

**(i)** In Î” AEB and Î” FEC,

∠AEB = ∠FEC **[Vertically opposite angles]**

∠BAE = ∠CFE **[Since, AB||DC]**

Hence, ∆AEB ~ ∆FEC by AA criterion for similarity.

So, we have

AE/FE = BE/EC = AB/FC

⇒ 15/EC = 9/13.5

⇒ EC = 22.5 cm

**(ii)** In Î” APB and Î” FPD,

∠APB = ∠FPD **[Vertically opposite angles]**

∠BAP = ∠DFP **[Since, AB||DF]**

Hence, ∆APB ~ ∆FPD by AA criterion for similarity.

So, we have

AP/FP = AB/FD

⇒ 6/FP = 9/31.5

⇒ FP = 21 cm

So, AF = AP + PF = 6 + 21 = 27 cm

**(iii)** We already have, ∆AEB ~ ∆FEC

So,

AE/FE = BE/CE = AB/FC

⇒ AE/FE = 9/13.5

⇒ (AF – EF)/ FE = 9/13.5

⇒ AF/EF – 1 = 9/13.5

⇒ 27/EF = 9/13.5 + 1 = 22.5/ 13.5

⇒ EF = (27×13.5)/22.5 = 16.2 cm

Now, PE = PF – EF = 21 – 16.2 = 4.8 cm

**3. In the following figure, AB, CD and EF are perpendicular to the straight line BDF.**

**If AB = x and; CD = z unit and EF = y unit, prove that:**

**1/x + 1/y = 1/z**

**Solution**

In Î” FDC and Î” FBA,

∠FDC = ∠FBA **[As DC || AB]**

∠DFC = ∠BFA **[common angle]**

Hence, ∆FDC ~ ∆FBA by AA criterion for similarity.

So, we have

DC/AB = DF/BF

⇒ z/x = DF/BF **…. (1)**

In Î” BDC and Î” BFE,

∠BDC = ∠BFE **[As DC || FE]**

∠DBC = ∠FBE **[Common angle]**

Hence, ∆BDC ~ ∆BFE by AA criterion for similarity.

So, we have

BD/BF = z/y **... (2)**

Now, adding (1) and (2), we get

BD/BF + DF/BF = z/y + z/x

⇒ 1 = z/y + z/x

Thus,

1/z = 1/x + 1/y

Hence, Proved

**4. Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.**

**Solution**

Given, ∆ABC ~ ∆PQR

AD and PM are the medians, so BD = DC and QM = MR

So, we have

AB/PQ = BC/QR [Corresponding sides of similar triangles are proportional.]

Then,

AB/PQ = (BC/2)/ (QR/2) = BD/QM

And, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM

Hence, ∆ABD ~ ∆PQM by SAS criterion for similarity.

Thus,

AB/PQ = AD/PM.

**5.** **Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: AB/PQ = AD/PM.**

**Solution**

Given, ∆ABC ~ ∆PQR

So,

∠ABC = ∠PQR i.e. ∠ABD = ∠PQM

Also, ∠ADB = ∠PMQ **[Both are right angles]**

Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.

Thus,

AB/PQ = AD/PM

**6. Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: AB/PQ = AD/PM**

**Solution**

Given, ∆ABC ~ ∆PQR

And, AD and PM are the angle bisectors.

So,

∠BAD = ∠QPM

Also, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM

Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.

Thus,

AB/PQ = AD/PM

**7. In the following figure, ∠AXY = ∠AYX. If BX/AX = CY/AY, show that triangle ABC is isosceles.**

**Solution**

Given,

∠AXY = ∠AYX

So, AX = AY **[Sides opposite to equal angles are equal.]**

Also, from BPT we have

BX/AX = CY/AY

Thus,

AX + BX = AY + CY

So, AB = AC

∴ ∆ABC is an isosceles triangle.

**8. In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown. **

**Prove that: AB/BC = PQ/QR**

**Solution**

Let join AR such that it intersects BQ at point X.

In ∆ACR, BX || CR. By BPT, we have

AB/BC = AX/XR **…(1)**

In ∆APR, XQ || AP. By BPT, we have

PQ/QR = AX/XR** …(2)**

From (1) and (2),

AB/BC = PQ/QR

Hence, Proved

**9. In the following figure, DE || AC and DC || AP. Prove that: BE/EC = BC/CP.**

**Solution**

Given, DE || AC

So,

BE/EC = BD/DA **[By BPT]**

And, DC || AP

So,

BC/CP = BD/DA **[By BPT]**

∴ BE/EC = BC/CP

**10. In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.**

**Calculate: (i) EF (ii) AC**

**Solution**

**(i)** In ∆PCD and ∆PEF,

∠CPD = ∠EPF **[Vertically opposite angles]**

∠DCE = ∠FEP **[As DC || EF, alternate angles.]**

Hence, ∆PCD ~ ∆PEF by AA criterion for similarity.

So, we have

27/EF = 15/7.5

Thus,

EF = 13.5

**(ii)** And, as EF || AB

∆CEF ~ ∆CAB by AA criterion for similarity.

EC/AC = EF/AB

22.5/AC = 13.5/22.5

Thus, AC = 37.5 cm.

**11. In Î”ABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.**

**(i) Prove that Î”ACD is similar to Î”BCA.**

**(ii) Find BC and CD**

**(iii) Find the area of Î”ACD: area of Î”ABC**

**Solution**

**(i)** In ∆ACD and ∆BCA,

∠DAC = ∠ABC [Given]

∠ACD = ∠BCA **[Common angles]**

Hence, ∆ACD ~ ∆BCA by AA criterion for similarity.

**(ii)** Since, ∆ACD ~ ∆BCA

We have,

AC/BC = CD/CA = AD/AB

⇒ 4/BC = CD/4 = 5/8

⇒ 4/BC = 5/8

So, BC = 32/5 = 6.4 cm

And,

CD/4 = 5/8

Thus, CD = 20/8 = 2.5 cm

**(iii)** As, ∆ACD ~ ∆BCA

We have,

Ar(∆ACD)/ Ar(∆BCA) = AD^{2}/ AB^{2} = 5^{2}/8^{2}

⇒ Ar(∆ACD)/ Ar(∆BCA) = 25/64

**12. In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.**

**Solution**

In ∆ABC, as PR || BC by BPT we have

AP/PB = AR/RC

And, in ∆PAR and ∆BAC,

∠PAR = ∠BAC **[Common]**

∠APR = ∠ABC **[Corresponding angles]**

Hence, ∆PAR ~ ∆BAC by AA criterion for similarity

So, we have

PR/BC = AP/AB

⇒ PR/BC = ½ **[Since, P is the mid-point of AB]**

⇒ PR = ½ BC

Similarly,

PQ = ½ AC

RQ = ½ AB

So,

PR/BC = PQ/AC = RQ/AB

∴ ∆QRP ~ ∆ABC by SSS similarity.

**13. In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:**

**(i) EF = FB,**

**(ii) AG: GD = 2: 1**

**Solution**

**(i)** In ∆BFD and ∆BEC,

∠BFD = ∠BEC **[Corresponding angles]**

∠FBD = ∠EBC **[Common]**

Hence, ∆BFD ~ ∆BEC by AA criterion for similarity.

So,

BF/BE = BD/BC

⇒ BF/BE = ½ **[Since, D is the mid-point of BC]**

⇒ BE = 2BF

⇒ BF = FE = 2BF

Thus,

EF = FB

(ii) In ∆AFD, EG || FD and using BPT we have

AE/EF = AG/GD **….(1)**

Now, AE = EB **[Since, E is the mid-point of AB]**

AE = 2EF **[As, EF = FB, by (1)]**

So, from (1) we have

AG/GD = 2/1

∴ AG: GD = 2:1

**14. Two similar triangles are equal in area. Prove that the triangles are congruent.**

**Solution**

Let’s consider two similar triangles as ∆ABC ~ ∆PQR

So,

Ar(∆ABC)/Ar(∆PQR) = (AB/PQ)^{2} = (BC/QR)^{2} = (AC/PR)^{2}

Since,

Area of ∆ABC = Area of ∆PQR **[Given]**

Hence,

AB = PQ

BC = QR

AC = PR

So, as the respective sides of two similar triangles are all of same length.

We can conclude that,

∆ABC ≅ ∆PQR **[By SSS rule]**

Hence, Proved

**15. The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:**

**(i) medians. (ii) perimeters. (iii) areas.**

**Solution**

The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

So,

**(i) **The ratio between the medians of two similar triangles is same as the ratio between their sides.

Thus, the required ratio = 3: 5

**(ii)** The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Thus, the required ratio = 3: 5

**(iii)** The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Thus, the required ratio = (3)^{2}: (5)^{2} = 9: 25