ML Aggarwal Similarity Exe-13.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-13.3 Questions for Similarity as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Similarity Exe-13.3 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-13 | Similarity |

Writer / Book | Understanding |

Topics | Solutions of Exe-13.3 |

Academic Session | 2024-2025 |

### Similarity Exe-13.3

ML Aggarwal Class 10 ICSE Maths Solutions

**Question 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.**

**Answer 1**

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 = BC^{2}/QR^{2}

BC : QR = 1 : 3

Hence, ∆ABC/area of ∆PQR = 1^{2}/3^{2}

= 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 = BC^{2}/QR^{2}

area of ∆ABC/area of ∆PQR = BC^{2}/QR^{2} = 25/36

= (BC/QR)^{2} = (5/6)^{2}

BC/QR = 5/6

Hence the ratio of their corresponding sides is 5 : 6

**Question 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.**

**Answer 2**

Let EF = x

Given that

∆ABC ~ ∆DEF,

Area of ∆ABC = 9 sq. cm

Area of ∆DEF =16 sq. cm

area of ∆ABC/area of ∆DEF = BC^{2}/EF^{2}

area of ∆ABC/area of ∆DEF = BC^{2}/EF^{2}

9/16 = BC^{2}/EF^{2}

9/16 = (2.1)^{2}/x^{2}

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

Hence, EF = 2.8 cm

**Question 3 **

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

**Answer 3**

∆ABC ~ ∆DEF

BC = 3 cm, EF = 4 cm

Area of ∆ABC = 54 sq. cm.

We know that,

Area of ∆ABC/ area of ∆DEF = BC^{2}/EF^{2}

54/area of ∆DEF = 3^{2}/4^{2}

54/area of ∆DEF = 9/16

By cross multiplication we get,

Area of ∆DEF = (54 × 16)/9

= 6 × 16

= 96 cm

**Question 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.**

**Answer 4**

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 cm^{2}

Area of ∆XYZ = 25 cm^{2}

PM = 2.4 cm

Assume XN = a

area of ∆PQR/area of ∆XYZ = PM^{2}/XN^{2}

36/25 = (2.4)^{2}/a^{2}

By cross multiplication we get,

36a^{2} = 25 (2.4)^{2}

a^{2} = 5.76 × 25/36

a^{2} = 144/36

a^{2} = 4

a = √4

a = 2 cm

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

**Question 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. (2006)**

**(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.**

**Answer 5**

**(a) **From the question, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm²

Consider the ∆AOQ and ∆BOP,

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

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

Hence, ∆AOQ ~ ∆BOP

area of ∆AOQ/area of ∆BOP = OQ^{2}/PO^{2}

Area of ∆AOQ/120 = 9^{2}/6^{2}

Area of ∆AOQ/120 = 81/36

Area of ∆AOQ = (81 × 120)/36

Area of ∆AOQ = 270 cm^{2}

**(b) **From the question

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

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

consider the ∆OAB and ∆OCD

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

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

Hence, ∆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

AB^{2}/CD^{2} = (6.5)^{2}/(3.25)^{2}

= (6.5 × 6.5)/(3.25 × 3.25)

= 2 × 2/1

= 4/1

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

### ML Aggarwal Similarity Exe-13.3 Class 10 ICSE Maths Solutions

Page 287

**Question 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.**

**Answer 6**

(a) In the figure,

DE || BC

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

Now consider the ∆ADE and ∆ABC,

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

Hence, ∆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 cm^{2}

(b)

From the question,

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]

Hence, ∆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 = AD^{2}/AB^{2}

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

**Question 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 = 18cm², find the area of trapezium DBCE**

**Answer 7**

**(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]

Hence, ∆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

Hence, DE = 4.5/3

DE = 1.5 cm

**(iii) From the question it is given that, area of ∆ABC = 18 cm ^{2}**

Then, area of ∆ADE/area of ∆ABC = DE^{2}/BC^{2}

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

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

= 18 – 2

= 16 cm^{2}

**Question 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.**

**Answer 8**

(i) Consider the ∆ABC and ∆DEC,

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

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

Hence, ∆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) area of ∆ABC/area of ∆DEC = AB^{2}/DE^{2}

area of ∆ABC/area of ∆DEC = 6^{2}/4^{2}

area of ∆ABC/area of ∆DEC = 36/16

area of ∆ABC/area of ∆DEC = 9/4

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

**Question 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 ? (2007)**

**Answer 9**

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

From the question

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]

Hence, ∆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]

Hence, ∆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 = DE^{2}/BC^{2}

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

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

Page 288

**Question 10 **

In ∆PQR, MN is parallel to QR and PM/MQ = 2/3

(i) Find MN/QR

(ii) Prove that ∆OMN and ∆PRQ are similar

(iii) Find area of ∆OMN : area of ∆ORQ

**Answer 10**

(i) In ΔPMN and ΔPQR, MN || QR

⇒∠PMN=∠PQR (alternate angles)

⇒∠PNM=∠PRQ (alternate angles)

=> ΔPMN ~ ΔPQR (AA postulate)

(ii) In ΔOMN and ΔORQ,

∠OMN=∠ORQ (alternate angles)

∠MNO=∠OQR (alternate angles)

=> ΔOMN ~ ΔORQ (AA postulate)

(iii)

## ML Aggarwal Similarity Exe-13.3 Class 10 ICSE Maths Solutions

**Question 11**

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. ** (2008)**

**Answer 11 :**

**Question 12 **

**(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**

**(i) EF : AD**

**(ii) area of ∆BEF : area of ∆ABD**

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

**(iv) area of ∆FEO : area of ∆OBC.**

**Answer 12**

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

∠OAB = ∠OCD [alternate angles] ∠OBA = ∠ODC

∆AOB ~ ∆COD

So, area of ∆AOB/area of ∆COD = AB^{2}/CD^{2}

= (2CD)^{2}/CD^{2} … [because AB = 2 CD]

= 4CD^{2}/CD^{2}

= 4/1

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

**(b) **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

Hence, 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 90^{o}]

Hence, ∆ADM ~ ∆ABN

Therefore, area of ∆ADM/area of ∆ABN = AD^{2}/AB^{2}

= BC^{2}/AB^{2}

= 4.5^{2}/7.5^{2}

= 20.25/56.25

= 2025/5625

= 81/225

= 9/25

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

**(c) **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) 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

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 = 3^{2}/5^{2}

= 9/25

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

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 = EF^{2}/BC^{2}

EF^{2}/AD^{2} = 9/25

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

**Question 13 ****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 || gm ABCD.**

**Answer 13**

In the figure, ABCD is a parallelogram.

P is a point on BC such that BP : PC = 1 : 2

and DP is produced to meet ABC produced at Q.

Area ∆CPQ = 20 cm²

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

= BP/PC = ½

**(i) So, area ∆BPQ = ½ area of ∆CPQ**

= ½ × 20

area of ∆BPQ = 10 cm^{2}

**(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]

∆CDP ~ ∆BQP … [AA axiom]

area of ∆CDP/area of ∆BQP = PC^{2}/BP^{2}

area of ∆CDP/area of ∆BQP = 2^{2}/1^{2}

area of ∆CDP/area of ∆BQP = 4/1

area of ∆CDP = 4 × area ∆BQP

area of ∆CDP = 4 × 10

= 40 cm^{2}

**(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) cm^{2}

= 2 × 60 cm^{2}

= 120 cm^{2}

Hence, the area of parallelogram ABCD is 120 cm^{2}.

### Similarity Exe-13.3

ML Aggarwal Class 10 ICSE Maths Solutions

Page 289

**Question 14**

**(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.**

**Answer 14**

(a) In ∆ABC, DE || BC

Now in ∆ABC and ∆ADE

∠A = ∠A (common)

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

∆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} = 4^{2}/9^{2}

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

Hence, DE: BC = 2: 3

**(b) 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 = DC^{2}/AB^{2}

area of ∆EDC/area of ∆ABE = DC^{2}/(2DC)^{2}

area of ∆EDC/area of ∆ABE = DC^{2}/4DC^{2}

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

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

#### Question 15

**(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.**

**Answer 15**

(a) In trapezium ABCD, DC || 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

BP = 7.2 cm

(ii) area of ∆APB/area of ∆CPD = AB^{2}/CD^{2}

area of ∆APB/area of ∆CPD = 9^{2}/6^{2}

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

Hence, the ratio of areas of ∆APB and ∆DPC is 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

Consider AC/BC = AD/AB

4/BC = 5/8

BC = (4 × 8)/5

BC = 32/5

BC = 6.4 cm

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 = AC^{2}/AB^{2}

= 4^{2}/8^{2}

= 16/64

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

= ¼

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

**Question 16 **

(a) In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm.

(i) Prove that △PQR ∼ APST.

(ii) Find Area of △PQR : Area of quadrilateral SRQT.

**Answer 16**

**Question 17 ****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.**

**Answer 17**

(i) Consider DADE and DACB

∠A = ∠A (Common)

m ∠B = m ∠E = 90°

Thus by angle-angle similarity, triangles,

∆ACB ~ ∆ADE

(ii) Consider ∆ADE and ∆ACB

Since they are similar triangles,

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

Consider the ∆ABC, is a right angle triangle

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

13^{2} = AB^{2} + 5^{2}

169 = AB^{2} + 25

AB^{2} = 169 – 25

AB^{2} = 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,

Area of ∆ADE = ½ × AE × DE

= ½ × 4 × (5/3)

= 10/3 cm^{2}

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 cm^{2}

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

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

### ML Aggarwal Class 10 ICSE Maths Solutions

Similarity Exe-13.3

Page 289

**Question 18 ****Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.**

**Answer 18**

In two isosceles ∆s ABC and DEF

∠P = ∠X … [from the question]

So, ∠Q + ∠R = ∠Y + ∠Z

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

Hence, ∠Q = ∠Y and ∠R = ∠Z

∆PQR ~ ∆XYZ

Then, area of ∆PQR/area of ∆XYZ = PM^{2}/XN^{2} … [from corollary of theorem]

PM^{2}/XN^{2} = 7/16

PM/XN = √7/√16

PM/XN = √7/4

ratio of PM: DM = √7: 4

**Question 19 ****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:**

**Answer 19**

Scale factor k = 1 : 250000 = 1/250000

Length on map,

AB = 3 cm, BC = 4 cm

(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 cm^{2}

Then, area of actual plot = 1/k^{2} × 6 cm^{2}

= 250000^{2} × 6

To covert cm into km divide by (100000)^{2}

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

= (25/4) × 6

= 75/2

= 37.5 km^{2}

**Question 20 **On a map drawn to a scale of 1 : 50,000, a rectangular plot of land ABCD has the following dimensions AB = 6 cm; BC = 8 cm and all angles are right angles. Find :

(i) the actual length of the diagonal distance AC of the plot in km.

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

**Answer 20**

Scale used on the map is 1 : 50,000

Dimensions of a rectangular plot ABCD are AB = 6 cm, BC = 8 cm Since each angle is right angle

∴ By using Pythagoras theorem, we have

(i) Actual length of the diagonal AC = 10 × 50000 cm

= (500000/100000) km

= 5 km

(ii) Area of the rectangular field ABCD on map

= 6 × 8 = 48 cm^{2
}Actual area of the field = 48 × 500000 × 500000

= 12(10)^{10} sq. cm.

= 12 sq. km.

**Question 21 ****A map of a square plot of land is drawn to a scale 1 : 25000. If the area of the plot in the map is 72 cm², find:**

**(i) the actual area of the plot of the land**

**(ii) the length of the diagonal in the actual plot of land**

**Answer 21**

**Question 22 ****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. (2009)**

**Answer 22**

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) 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 m^{3}

Hence, Volume of model = 1000 cm^{3}

**Question 23 ****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 litres, find the volume of the ship in m³.**

**(100 litres = m³)**

**Answer 23**

Scale = 1 : 200

(i) Length of a model of ship = 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 m^{2}

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

Then, Volume of ship = 200 × (200/1)^{3}

= 200 × 8000000

= (200 × 8000000)/100

= 1600000 m^{3}

— : End of ML Aggarwal Similarity Exe-13.3 Class 10 ICSE Maths Solutions : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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