# RS Aggarwal Solutions Chapter 4 Triangles Exercise - 4A Class 10 Maths

 Chapter Name RS Aggarwal Chapter 4 Triangles Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 4BExercise 4CExercise 4DExercise 4E Related Study NCERT Solutions for Class 10 Maths

### Exercise 4A Solutions

1. D and E are points on the sides AB and AC respectively of a ∆ABC such that DE ∥BC.

(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.

(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.

(iii) If AD/DB = 4/7 and AC = 6.6cm, find AE.

(iv) If AD/AB = 8/15 and EC = 3.5 cm, find AE.

Solution

(i) In ∆ ABC, it is given that DE ∥ BC.

Applying Thales’ theorem, we get:

∵ AD = 3.6 cm, AB = 10 cm, AE = 4.5cm

∴ DB = 10 − 3.6 = 6.4cm

Or, 3.6/6.4 = 4.5/EC

Or, EC = (6.4×4.5)/3.6

Or, EC = 8 cm

Thus, AC = AE + EC

= 4.5 + 8 = 12.5 cm

(ii) In ∆ ABC, it is given that DE || BC.

Applying Thales’ Theorem, we get:

Adding 1 to both sides, we get:

AD/DB + 1 = AE/EC + 1

⇒ 13.3/DB = 11.9/5.1

⇒ DB = (13.3×5.1)/11.9 = 5.7 cm

Therefore, AD = AB – DB = 13.5 - 5.7 = 7.6 cm

(iii) In ∆ ABC, it is given that DE || BC.

Applying Thales’ theorem, we get:

⇒ 4/7 = AE/EC

Adding 1 to both the sides, we get:

11/7 = AC/EC

⇒ EC = (66×7)11 = 4.2 cm

Therefore,

AE = AC – EC = 6.6 – 4.2 = 2.4 cm

(iv) In ∆ ABC, it is given that DE ‖ BC.

Applying Thales’ theorem, we get:

⇒ 8/15 = AE/(AE + EC)

⇒ 8/15 = AE/(AE + 3.5)

⇒ 8AE + 28 = 15AE

⇒ 7AE = 28 cm

⇒ AE = 4 cm

2. D and E are points on the sides AB and AC respectively of a ∆ABC such that DE ∥ BC. Find the value of x, when

(i) AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

(ii) AD = 4cm, DB = (x – 4) cm, AE = 8cm and EC = (3x – 19) cm.

(iii) AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm.

Solution

(i) In ∆ ABC, it is given that DE || BC.

Applying Thales’ theorem, we have:

⇒ x/(x-2) = (x + 2)/(x – 1)

⇒ x = 4 cm

(ii) In ∆ ABC, it is given that DE ‖ BC.

Applying Thales’ theorem, we have:

⇒ 4/(x – 4) = 8/(3x – 19)

⇒ 4(3x – 19) = 8(x – 4)

⇒ 12x – 76 = 8x – 32

⇒ 4x = 44

⇒ x = 11 cm

(iii) In ∆ ABC, it is given that DE || BC.

Applying Thales’ theorem, we have:

⇒ (7x – 4)/(3x + 4) = (5x – 2)/3x

⇒ 3x(7x – 4) = (5x – 2)(3x + 4)

⇒ 21x2 – 12x = 15x2 + 14x – 8

⇒ 6x2 – 26x + 8 = 0

⇒ (x – 4) (6x – 2) = 0

⇒ x = 4, 1/3

∵ x ≠ 1/3 (as if x = 1/3 then AE will become negative)

∴ x = 4 cm

3. D and E are points on the sides AB and AC respectively of a ∆ABC. In each of the following cases, determine whether DE║BC or not.

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

(ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm.

(iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.

(iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.

Solution

(i) We have,

AD/DE = 5.7/9.5 = 0.6 cm

AE/EC = 4.8/8 = 0.6 cm

Applying the converse of Thales’ theorem,

We conclude that DE || BC.

(ii) We have,

AB = 11.7 cm, DB = 6.5 cm

Therefore,

AD = 11.7 -6.5 = 5.2 cm

Similarly,

AC = 11.2 cm, AE = 4.2 cm

Therefore,

EC = 11.2 – 4.2 = 7 cm

Now,

AE/EC = 4.2/7

Applying the converse of Thales’ theorem,

We conclude that DE is not parallel to BC.

(iii) We have,

AB = 10.8 cm, AD = 6.3 cm

Therefore,

DB = 10.8 – 6.3 = 4.5 cm

Similarly,

AC = 9.6 cm, EC = 4cm

Therefore,

AE = 9.6 – 4 = 5.6 cm

Now,

AE/EC = 5.6/4 = 7/5

Applying the converse of Thales' theorem,

We conclude that DE ‖ BC.

(iv) We have,

AD = 7.2 cm, AB = 12 cm

Therefore,

DB = 12 – 7.2 = 4.8 cm

Similarly,

AE = 6.4 cm, AC = 10 cm

Therefore,

EC = 10 – 6.4 = 3.6 cm

Now,

AE/EC = 6.4/3.6 = 16/9

Applying the converse of Thales’ theorem,

We conclude that DE is not parallel to BC.

4. In a ∆ABC, AD is the bisector of A.

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

(ii) IF AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

Solution

(i) It is given that AD bisects ∠A.

Applying angle – bisector theorem in ∆ ABC, we get:

BD/DC = AB/AC

⇒ 5.6/DC = 6.4/8

⇒ DC = (8×5.6)/6.4 = 7 cm

(ii) It is given that AD bisects ∠A.

Applying angle – bisector theorem in ∆ ABC, we get:

BD/DC = AB/AC

Let BD be x cm.

Therefore, DC = (6 - x) cm

⇒ x/(6 – x) = 10/14

⇒ 14x = 60-10x

⇒ 24x = 60

⇒ x = 2.5 cm

Thus, BD = 2.5 cm

DC = 6 -2.5 = 3.5 cm

(iii) It is given that AD bisector ∠A.

Applying angle – bisector theorem in ∆ ABC, we get:

BD/DC = AB/AC

BD = 3.2 cm, BC = 6 cm

Therefore, DC = 6- 3.2 = 2.8 cm

⇒ 3.2/2.8 = 5.6/AC

⇒ AC = (5.6×2.8)/3.2

= 4.9 cm

(iv) It is given that AD bisects ∠A

Applying angle – bisector theorem in ∆ ABC, we get:

BD/DC = AB/AC

⇒ BD/3 = 5.6/4

⇒ BD = (5.6×3)/4

⇒ BD/3 = 4.2 cm

Hence, BC = 3 + 4.2 = 7.2 cm

5. M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that

(i) DM/MN = DC/BN

(ii) DN/DM = AN/DC

Solution

(i) Given: ABCD is a parallelogram

To prove:

(i) DM/MN = DC/BN

(ii) DN/DM = AN/DC

Proof:

In ∆DMC and ∆NMB

∠DMC = ∠NMB (Vertically opposite angle)

∠DCM = ∠NBM (Alternate angles)

By AAA-Similarity

∆DMC ~ ∆NMB

∴ DM/MN = DC/BN

Now, MN/DM = BN/DC

Adding 1 to both sides, we get

MN/DM + 1 = BN/DC + 1

⇒ (MN + DM)/DM = (BN + DC)/DC

⇒ (MN + DM)/DM = (BN + AB)/DC  [∵ ABCD is a parallelogram]

⇒ DN/DM = AN/DC

6. Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel sides

Solution

Let the trapezium be ABCD with E and F as the mid Points of AD and BC, Respectively Produce AD and BC to Meet at P.

In ∆ PAB, DC || AB.

Applying Thales’ theorem, we get

PD/DA = PC/CB

Now, E and F are the midpoints of AD and BC, respectively.

⇒ PD/2DE = PC/2CF

⇒ PD/DE = PC/CF

Applying the converse of Thales’ theorem in ∆PEF, we get that DC

Hence, EF || AB.

Thus. EF is parallel to both AB and DC.

This completes the proof.

7. In the given figure, ABCD is a trapezium in which AB║DC and its diagonals intersect at O. If AO = (5x – 7), OC = (2x + 1) , BO = (7x – 5) and OD = (7x + 1), find the value of x.

Solution

In trapezium ABCD, AB ‖ CD and the diagonals AC and BD intersect at O. Therefore,

AO/OC = BO/OD

⇒ (5x - 7) (7x + 1) = (7x - 5) (2x +1)

⇒ 35x2 + 5x – 49x – 7 = 14x2 – 10x + 7x – 5

⇒ 21x2 – 41x – 2 = 0

⇒ 21x2 – 42x + x - 2 = 0

⇒ 21x (x - 2) + 1(x - 2) =0

⇒ (x - 2) (21x + 1) = 0

⇒ x = 2, -(1/21)

∵ x ≠ -1/21

∴ x = 2

8. In ΔABC, M and N are points on the sides AB and AC respectively such that BM = CN. If ∠ = ∠B = C then show that MN || BC

Solution

In ∆ABC, ∠B = ∠C

∴ AB = AC (Sides opposite to equal angle are equal)

Subtracting BM from both sides, we get

AB – BM = AC – BM

AB – BM = AC – CN (∵ BM = CN)

AM = AN

∴ ∠AMN =∠ANM (Angles opposite to equal sides are equal)

Now, in ∆ABC,

∠A + ∠B + ∠C = 180° ---(1) (Angle Sum Property of triangle)

Again In ∆AMN,

∠A + ∠AMN + ∠ANM = 180° ---(2) (Angle Sum Property of triangle)

From (1) and (2), we get

∠B + ∠C = ∠ AMN + ∠ ANM

⇒ 2∠B = 2∠ AMN

⇒ ∠B = ∠AMN

Since, ∠B and ∠AMN are corresponding angles.

∴ MN ‖ BC

9. ΔABC and ΔDBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.

Solution

In ∆ CAB, PQ || AB.

Applying Thales' theorem, we get:

CP/PB = CQ/QA …(1)

Similarly, applying Thales theorem in ΔBDC, Where PR||DM we get:

CP/PB = CR/RD ...(2)

Hence, from (1) and (2), we have:

CQ/QA = CR/RD

Applying the converse of Thales’ theorem, we conclude that QR ‖ AD in ∆ ADC. This completes the proof.

10. In the given figure, side BC of a ∆ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX.

Prove that AO : AX = AF : AB and show that EF║BC.

Solution

It is given that BC is bisected at D.

∴ BD = DC

It is also given that OD = OX

The diagonals OX and BC of quadrilateral BOCX bisect each other.

Therefore, BOCX is a parallelogram.

∴ BO || CX and BX || CO

BX || CF and CX || BE

BX || OF and CX || OE

Applying Thales’ theorem in ∆ ABX, we get:

AO/AX = AF/AB …(1)

Also, in ∆ ACX, CX || OE.

Therefore by Thales’ theorem, we get:

AO/AX = AE/AC …(2)

From (1) and (2), we have:

AO/AX = AE/AC

Applying the converse of Theorem in ∆ ABC, EF || CB.

This completes the proof.

11. ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = 1/4.AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.

Solution

We know that the diagonals of a parallelogram bisect each other.

Therefore,

CS = 1/2.AC …(i)

Also, it is given that CQ = 1/4AC …(ii)

Dividing equation (ii) by (i), we get:

CQ/CS = (1/4.AC)/(1/2.AC)

Or, CQ = 1/2 CS

Hence, Q is the midpoint of CS.

Therefore, according to midpoint theorem in ∆CSD

PQ || DS

If PQ || DS, we can say that QR || SB

In ∆CSB, Q is midpoint of CS and QR ‖ SB.

Applying converse of midpoint theorem, we conclude that R is the midpoint of CB. This completes the proof.

12. In the adjoining figure, ABC is a triangle in which AB = AC. IF D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.

Solution

Given,

AB = AC …(ii)

Subtracting AD from both sides, we get:

⇒ AB – AD = AC - AE (Since, AD = AE)

⇒ BD = EC …(iii)

Dividing equation (i) by equation (iii), we get:

Applying the converse of Thales’ theorem, DE‖BC

⇒ ∠DEC + ∠ECB = 180° (Sum of interior angles on the same side of a Transversal Line is 0°)

⇒ ∠DEC + ∠CBD = 180° (Since, AB = AC ⇒ ∠B = ∠C)

Therefore, B,C,E and D are concylic points

13. In ∆ABC, the bisector of ∠B meets AC at D. A line OQ║AC meets AB, BC and BD at O, Q and R respectively. Show that BP × QR = BQ × PR .

Solution

In triangle BQO, BR bisects angle B.

Applying angle bisector theorem, we get:

QR/PR = BQ/BP

⇒ BP × QR = BQ × PR

This completes the proof.