In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4B Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 4 |

Chapter Name | Linear equations in two variables |

Exercise | 4B |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4B**

**Question 1.****Solution:**

We know that two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.

(i) In ∆ABC and ∆PQR

∠A = ∠Q = 50°

∠B = ∠P = 60° and ∠C = ∠R = 70°

∆ABC ~ ∆QPR (AAA axiom)

(ii) In ∆ABC and ∆DEF

In ∆ABC,

AB = 3 cm, BC = 4.5

and in ∆DEF

DF = 6 cm, DE = 9 cm

∆ABC is not similar to ∆DEF

As in ∆ABC, ∠A is not included of two sides AB and BC.

(iii) In ∆ABC and ∆PQR

In ∆ABC,

AC = 8 cm BC = 6 cm

Included ∠C = 80°

In ∆PQR,

PQ = 4.5 cm, QR = 6 cm

and included ∠Q = 80°

(v) In ∆ABC,

∠A = 80°, ∠C = 70°

and third angle

∠B = 180° – (80° + 70°)

⇒ ∠B = 180° – 150° = 30°

In ∆MNR,

∠M = 80°, ∠N = 30°, and ∠R = [180° – (80° + 30°)]

∠R = 180° – 110° = 70°

Now, in ∆ABC

∠A = ∠M – 80°, ∠B = ∠N = 30°

and ∠C = ∠R = 70°

∆ABC ~ ∆MNR (AAA or AA axiom)

**Question 2.****Solution:**

In the given figure, ∆ODC ~ ∆OBA and ∠BOC =115°, ∠CDO = 70°

To find :

(i) ∠DOC

(ii) ∠DCO

(iii) ∠OAB

(iv) ∠OBA

∆ODC ~ ∆OBA

∠D = ∠B = 70°

∠C = ∠A

∠COD = ∠AOB

(i) But ∠DOC + ∠BOC = 180° (Linear pair)

⇒ ∠DOC + 115°= 180°

⇒ ∠DOC = 180° – 115° = 65°

(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)

⇒ 65° + 70° + ∠DCO = 180°

⇒ 135° + ∠DCO = 180°

⇒ ∠DCO = 180° – 135°

∠DCO = 45°

(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)

∠OAB = ∠DCO = 45° (∆ODC ~ ∆OBA)

(iv) ∠OBA = ∠CDO = 70° (∆ODC ~ ∆OBA)

**Question 3.****Solution:**

In the given figure, ∆OAB ~ ∆OCD

AB = 8 cm, BO = 6.4 cm OC = 3.5 cm, CD = 5 cm

**Question 4.****Solution:**

Given : In the given figure,

∠ADE = ∠B

To prove:

(i) ∆ADE ~ ∆ABC

(ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE

Proof: (i) In ∆ADE and ∆ABC

∠ADE = ∠B (given)

∠A = ∠A (common)

∆ADE ~ ∆ABC (AA axiom)

**Question 5.****Solution:**

∆ABC ~ ∆PQR,

PQ = 12 cm

To find AB.

Perimeter of ∆ABC = AB + BC + CA = 32 cm

Perimeter of ∆PQR = PQ + QR + RP = 24 cm

Now,

∆ABC ~ ∆PQR

**Question 6.****Solution:**

∆ABC ~ ∆DEF

BC = 9.1 cm, EF = 6.5 cm

Perimeter of ∆DEF = 25 cm

**Question 7.****Solution:**

Given : In the given figure,

∠CAB = 90° and AD ⊥ BC

To prove : ∆BDA ~ ∆BAC

If AC = 75 cm, AB = 1 m or 100 cm,

BC = 1.25 m or 125 cm

Find AD.

∆BDA ~ ∆BAC (corresponding sides and proportional)

**Question 8.****Solution:**

In the given figure,

∠ABC = 90°, BD ⊥ AC.

AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm

To find BC,

In ∆ABC and ∆BDC,

∠ABC = ∠BDC (each 90°)

∠BCA = ∠BCD (common)

∆ABC ~ ∆BDC (AA axiom)

Corresponding sides are proportional

**Question 9.****Solution:**

In the given figure, ”

∠ABC = 90°, BD ⊥ AC

BD = 8 cm, AD = 4cm

To find CD,

Let CD = x

Now in ∆DBC and ∆BDA

∠BDC = ∠BDA (each 90°)

∠C = ∠ABD

∆DBC ~ ∆BDA

Sides are proportional

**Question 10.****Solution:**

In ∆ABC, P and Q are points on the sides AB and AC respectively such that

AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm.

To prove : BC = 3 PQ

BC = 3 PQ

Hence proved.

**Question 11.****Solution:**

Given: ABCD is a parallelogram.

E is a point on BC and diagonal BD intersects AE at F.

To prove : AF x FB = EF x FD

Proof: In ∆AFD and ∆BFE

∠AFD = ∠BFE (vertically opposite angles)

∠ADF = ∠FBE (alternate angles)

∆AFD ~ ∆BFE (AA axiom)

AFEF = FDFB

By cross multiplication,

⇒ AF x FB = EF x FD

Hence proved.

**Question 12.****Solution:**

Given : In the given figure,

DB ⊥ BC, DE ⊥ AB and AC ⊥ BC

**Question 13.****Solution:**

In ∆ABC and ∆DEF

AC is stick and BC is its shadow.

DF is the tower and EF is its shadow

AC = 7.5 m, BC = 5 m EF = 24 m,

let DF = x m

**Question 14.****Solution:**

Given : In isosceles ∆ABC,

CA = CB, base AB and BA are produced in P and Q such that

AP x BQ = AC²

To prove : ∆ACP ~ ∆BCQ

Proof: In ∆ABC,

CA = CB

∠CAB = ∠CBA (Angles opposite to equal sides)

⇒ 180° – ∠CAB = 180° – ∠CBA (Subtracting each from 180°)

⇒ ∠CAP = ∠CBQ

AP x BQ = AC² (given)

**Question 15.****Solution:**

Given : In the given figure, ∠1 = ∠2

**Question 16.****Solution:**

Given : In quadrilateral ABCD,

AD = BC

P, Q, R and S are midpoints of AB, AC, CD and BD respectively.

To proof : PQRS is a rhombus.

Proof: In ∆ABC,

P and Q are the midpoints of sides AB and AC respectively.

PQ || BC ……(i)

**Question 17.****Solution:**

In the given circle, two chords AB and CD intersect each other at P inside the circle.

To prove :

(a) ∆PAC ~ ∆PDB

(b) PA . PB = PC . PD

Proof:

(a) In ∆PAC and ∆PDB

∠APC = ∠BPD (Vertically opposite angles)

∠A = ∠D (Angles in the same segment)

∆PAC ~ ∆PDB (AA axiom)

PAPD = PCPB

⇒ PA x PB = PC x PD

Hence proved.

**Question 18.****Solution:**

In a circle, two chords AB and CD intersect each other at the point P outside the circle.

AC and BD are joined.

To prove :

(a) ∆PAC ~ ∆PDB

(b) PA . PB = PC . PD

Proof: In the circle, quadrilateral ABDC is a cyclic.

Ext. ∠PAC = ∠D

Now, in ∆PDB and ∆PAC

∠P = ∠P (common)

∠PAC = ∠D (proved)

∆PDB ~ ∆PAC (AA axiom)

or ∆PAC ~ ∆PBD

PAPD = PCPB

⇒ PA . PB = PC . PD

Hence proved

**Question 19.****Solution:**

Given : In right ∆ABC, ∠B = 90°

D is a point on hypotenuse AC such that

BD ⊥ AC and DP ⊥ AB and DQ ⊥ BC.

To prove :

(a) DQ² = DP . QC

(b) DP² = DQ . AP

Proof: AB ⊥ BC and DQ ⊥ BC

AB || DQ

DP ⊥ AB

DP || BC

Now, AB or PB || DQ and BC or BQ || DP BQDP is a rectangle

BQ = DP and BP = DQ

Now, in right ∆BQD

∠1 + ∠2 = 90° …..(i)

Similarly in rt. ∆DQC,

∠3 + ∠4 = 90° (DQ ⊥ BC) …(ii)

and in right ∆BDC,

∠2 + ∠3 = 90° …(iii)

∠BDC = 90° (BD ⊥ AC)

From (i) and (ii),

∠1 = ∠3

and from (ii) and (iii),

∠2 = ∠4

Now, in ∆BQD and ∆DQC

∠1 = ∠3

∠2 = ∠4 (proved)

∆BQD ~ ∆DQC (AA axiom)

BQDQ = DQQC

⇒ DQ² = BQ x QC

⇒ DQ² = DP x QC

(b) Similarly, we can prove that

∆PDA ~ ∆PBD

PDPB = APDP

⇒ DP² = BP x AP

⇒ DP² = DQ . AP (BP = DQ)

Hence proved.

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.