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

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

**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:**

In ∆ABC, points D and E are on the sides AB and AC respectively such that DE || BC.

⇒ 7x = 28 ⇒ x = 4

**Question 2.**

**Solution:**

D and E are the points on the sides AB and AC of ∆ABC respectively and DE || BC,

By cross multiplication,

⇒ (7x – 4) (3x) = (5x – 2) (3x + 4)

⇒ 21x² – 12x = 15x² + 20x – 6x – 8

⇒ 21x² – 12x – 15x² – 20x + 6x = -8

⇒ 6x² – 26x + 8 = 0

⇒ 3x² – 13x + 4 = 0

⇒ 3x² – 12x – x + 4 = 0

⇒ 3x (x – 4) – 1 (x – 4) = 0

⇒ (x – 4) (3x – 1) = 0

Either x – 4 = 0, then x = 4

or 3x – 1 = 0 then 3x = 1 ⇒ x = \(\frac { 1 }{ 3 }\) which is not possible

x = 4

**Question 3.**

**Solution:**

D and E are the points on the sides AB and AC of ∆ABC

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

**Question 4.**

**Solution:**

In ∆ABC, AD is the bisector of ∠A

**Question 5.**

**Solution:**

Given : M is a point on the side BC of a parallelogram ABCD.

DM when produced meets AB produced at N.

**Question 6.**

**Solution:**

Given : In trapezium AB || DC.

M and N are the mid points of sides AD and BC respectively.

MN is joined.

To prove : MN || AB or DC.

Construction: Produce AD and BC to meet at P.

**Question 7.**

**Solution:**

ABCD is a trapezium and its diagonals AC and BD intersect each other at O.

**Question 8.**

**Solution:**

Given : In ∆ABC, M and N are points on the sides AB and AC respectively such that

BM = CN and if ∠B = ∠C.

**Question 9.**

**Solution:**

Given : ∆ABC and ∆DBC are on the same side of BC.

P is any point on BC.

PQ || AB and PR || BD are drawn, which meet AC at Q and CD at R.

AD is joined.

To prove : QR || AD.

Proof:

In ∆ABC, PQ||AB

Hence proved.

**Question 10.**

**Solution:**

Given : In the given figure, D is the mid point of BC. AD is joined. O is any point on AD. BO and CO are produced to meet AC at F and AB at E respectively. AD is produced to X such that D is the mid point of OX.

To prove: AO : AX = AF : AB and EF || BC

Construction. Join BX and CX.

D is the midpoint of BC.

Proof: BD = DC and OD = DX. (given)

OBXC is a parallelogram. (Diagonals of a ||gm bisect each other)

**Question 11.**

**Solution:**

Given : In ||gm, ABCD, P is mid point of DC and Q is the mid point of AC. Such that

CQ = \(\frac { 1 }{ 4 }\) AC.

PQ is produced to meet BC at R.

To prove : R is the mid point of BC.

Construction : Join BD which intersects AC at O.

Proof: Diagonals of a parallelogram bisect each other

CO = \(\frac { 1 }{ 2 }\) AC.

But CQ = \(\frac { 1 }{ 2 }\) CO (given)

Q is the mid point of CO.

PQ || DO

⇒ PR || DB and P is mid point of CD

R is the midpoint BC.

Hence proved.

**Question 12.**

**Solution:**

Given : In ∆ABC, AB = AC

D and E are the points on AB and AC such that

AD = AE

To prove : BCED is a cyclic quadrilateral.

Proof: In ∆ABC,

AB = AC,

D and E are the mid points of AB and AC respectively.

But these are opposite angles of quad. BCED

BCED is a cyclic quadrilateral

Hence, B, C, E and D are concyclic.

**Question 13.**

**Solution:**

Given : In ∆ABC, BD is the bisector of AB which meets AC at D.

A line PQ || AC which meets AB, BC, BD at P, Q, R respectively.

To prove : PR x BQ = QR x BP

Proof: In ∆ABC, PQ || AC

and BD is the bisector of ∠B

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

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