## RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
- RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

**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)

\(\frac { AF }{ EF }\) = \(\frac { FD }{ FB }\)

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)

\(\frac { PA }{ PD }\) = \(\frac { PC }{ PB }\)

⇒ 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

\(\frac { PA }{ PD }\) = \(\frac { PC }{ PB }\)

⇒ 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)

\(\frac { BQ }{ DQ }\) = \(\frac { DQ }{ QC }\)

⇒ DQ² = BQ x QC

⇒ DQ² = DP x QC

(b) Similarly, we can prove that

∆PDA ~ ∆PBD

\(\frac { PD }{ PB }\) = \(\frac { AP }{ DP }\)

⇒ DP² = BP x AP

⇒ DP² = DQ . AP (BP = DQ)

Hence proved.

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B are helpful to complete your math homework.

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