ML Aggarwal Similarity Exe-13.2 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-13.2 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-13.2 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-13 | Similarity |

Writer / Book | Understanding |

Topics | Solutions of Exe-13.2 |

Academic Session | 2024-2025 |

**Similarity Exe-13.2 Questions**

ML Aggarwal Class 10 ICSE Maths Solutions

**Question 1**

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

**(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.**

**Answer 1**

**(a) **In the figure (i)

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

(i) AE: EC

So, AD/BD = AE/EC

AE/EC = AD/BD

AE/EC = ¾

AE: EC = 3: 4

#### (ii) consider ∆ADE and ∆ABC

∠D = ∠B

∠E = ∠C

Therefore, ∆ADE ~ ∆ABC

Then, DE/BC = AD/AB

DE/5 = 3/(3 + 4)

DE/5 = 3/7

DE = (3 × 5)/7

DE = 15/7

DE = 2 (1/7)

**(b) **In the figure

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

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

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

Therefore, ∆ABC ~ ∆BPQ

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

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

BQ/BC = 6/10 = PQ/AC

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

Now, BQ/8 = 6/10

BQ = (6/10) ×8

BQ = 48/10

BQ = 4.8 cm

Also, CQ = BC – BQ

CQ = (8 – 4.8) cm

CQ = 3.2cm

Hence, CQ = 3.2 cm and BQ = 4.8 cm

**(c) **In the figure

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

So, PX/QX = PY/YR

1/3 = PY/4.5

By cross multiplication we get,

(4.5 × 1)/3 = PY

PY = 45/30

PY = 1.5

Then, ∠X = ∠Q

∠Y = ∠R

So, ∆PXY ~ ∆PQR

Hence, XY/QR = PX/PQ

XY/9 = 1/(1 + 3)

XY/9 = ¼

XY = 9/4

XY = 2.25

#### ML Aggarwal Similarity Exe-13.2 Class 10 ICSE Maths Solutions

Page 273

**Question 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.**

**Answer 2 **In the given figure, DE || BC

**(i) **Consider the ∆ABC,

AD/DB = AE/EC

x/(x – 2) = (x + 2)/(x – 1)

By cross multiplication we get,

X(x – 1) = (x – 2) (x + 2)

x^{2} – x = x^{2} – 4

-x = -4

x = 4

**(ii) **From the question it is given that,

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

Consider the ∆ABC,

AD/DB = AE/EC

2x/(x – 2) = (2x + 3)/(x – 3)

By cross multiplication we get,

2x(x – 2) = (2x + 3) (x – 3)

2x^{2} – 4x = 2x^{2} – 6x + 3x – 9

2x^{2} – 4x – 2x^{2} + 6x – 3x = -9

-7x + 6x = -9

-x = – 9

x = 9

**Question 3 ****E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:**

**(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.**

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

**Answer 3**

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

So, PE/EQ = 3.9/3

= 39/30

= 13/10

Then, PF/FR = 8/9

By comparing both the results,

13/10 ≠ 8/9

Therefore, PE/EQ ≠ PF/FR

So, EF is not parallel to QR

(ii) From the dimensions given in the question,

Consider the ∆PQR

So, PQ/PE = 1.28/0.18

= 128/18

= 64/9

PR/PF = 2.56/0.36

= 256/36

= 64/9

By comparing both the results,

64/9 = 64/9

Hence, PQ/PE = PR/PF

EF is parallel to QR.

**Question 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.**

**Answer 4**

In ∆PQR, A and B are points on the sides PQ and PR such that

PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

PQ/PA = 12.5/5

= 2.5/1

PR/PB = (PB + BR)/PB

= (4 + 6)/4

= 10/4, = 2.5

By comparing both the results,

2.5 = 2.5

Hence, PQ/PA = PR/PB

AB is parallel to QR.

**Question 5 **

**(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.**

**Answer 5**

(a) From the figure, CD || LA and DE || AC,

Consider the ∆BCA,

BE/BC = BD/BA

By using the corollary of basic proportionality theorem,

BE/(BE + EC) = BD/AB

4/(4 + 2) = BD/AB … [equation (i)]

consider the ∆BLA

BC/BL = BD/AB

By using the corollary of basic proportionality theorem,

6/(6 + CL) = BD/AB … [equation (ii)]

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

6/(6 + CL) = 4/6

By cross multiplication we get,

6 x 6 = 4 x (6 + CL)

24 + 4CL = 36

4CL = 36 – 24

CL = 12/4

CL = 3 cm

Hence, the length of CL is 3 cm.

(b) From the figure, ∠D = ∠E and AD/BD = AE/EC,

We have to prove that, BAC is an isosceles triangle

So, consider the ∆ADE

∠D = ∠E … [from the question]

AD = AE … [sides opposite to equal angles]

Consider the ∆ABC,

AD/DB = AE/EC … [equation (i)]

Hence, DE parallel to BC

Because AD = AE

DB = EC … [equation (ii)]

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

AD + DB = AE + EC

AB = AC

Hence, ∆ABC is an isosceles triangle.

### ML Aggarwal Similarity Exe-13.2 Class 10 ICSE Maths Solutions

**Question 6 ****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.**

**Answer 6**

Consider the ∆POQ

AB || PQ … [given]

So, OA/AP = OB/BQ … [equation (i)]

Then, consider the ∆OPR

AC || PR

OA/AP = OC/CR … [equation (ii)]

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

OB/BQ = OC/CR

Then, in ∆OQR

OB/BQ = OC/CR

Hence, BC || QR

**Question 7 ****ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that AO/BO = CO/DO**

**Answer 7**

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

Its diagonals AC and BD intersect each other at O

Now consider the ∆OAB and ∆OCD,

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

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

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

Therefore, ∆OAB ~ ∆OCD

Then, OA/OC = OB/OD

AO/OB = CO/DO … [by alternate angles]

**Question 8 ****In the figure given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC**

**Answer 8**

From the question it is given that,

AD is bisector of ∠BAC

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

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

∠1 = ∠2 … [equation (i)]

By construction CE || DE

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

Again by construction CE || DE

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

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

∠3 = ∠4

So, AC = AE … [equation (iv)]

Now, consider the ∆BCE,

CE || DE

BD/DC = AB/AE

BD/DC = AB/AC

3/DC = 6/4

By cross multiplication we get,

3 × 4 = 6 × DC

DC = (3 × 4)/6

DC = 12/6

DC = 2

Hence, BC = BD + DC

= 3 + 2

= 5 cm

— : End of ML Aggarwal Similarity Exe-13.2 Class 10 ICSE Maths Solutions : –

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