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 

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/DB = AE/EC
∵ 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:
AD/DB = AE/EC
Adding 1 to both sides, we get:
AD/DB + 1 = AE/EC + 1
⇒ AD/DB = AC/EC
⇒ 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:
AD/DB = AE/EC
⇒ 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:
AD/AB = AE/AC
⇒ 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:
AD/DB = AE/EC
⇒ x/(x2) = (x + 2)/(x – 1)
⇒ x = 4 cm
(ii) In ∆ ABC, it is given that DE ‖ BC.
Applying Thales’ theorem, we have:
AD/DB = AE/EC
⇒ 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:
AD/DB = AE/EC
⇒ (7x – 4)/(3x + 4) = (5x – 2)/3x
⇒ 3x(7x – 4) = (5x – 2)(3x + 4)
⇒ 21x^{2} – 12x = 15x^{2} + 14x – 8
⇒ 6x^{2 }– 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
Hence, AD/DB = AE/EC
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,
AD/DB = 5.2/6.5 = 4/5
AE/EC = 4.2/7
Thus, AD/DB ≠ AE/EC
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,
AD/DB = 6.3/4.5 = 7/5
AE/EC = 5.6/4 = 7/5
⇒ AD/DB = AE/EC
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,
AD/DB = 7.2/4.8 = 3/2
AE/EC = 6.4/3.6 = 16/9
This, AD/DB ≠ AE/EC
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 = 6010x
⇒ 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 AAASimilarity
∆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.
SolutionIn 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)
⇒ 35x^{2 }+ 5x – 49x – 7 = 14x^{2 }– 10x + 7x – 5
⇒ 21x^{2 }– 41x – 2 = 0
⇒ 21x^{2 }– 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 Î”ABC, 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. Î”ABC and Î”DBC 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.
SolutionIn ∆ CAB, PQ  AB.
Applying Thales' theorem, we get:
CP/PB = CQ/QA …(1)
Similarly, applying Thales theorem in Î”BDC, Where PRDM 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.
SolutionWe 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.
SolutionGiven,
AD = AE …(i)
AB = AC …(ii)
Subtracting AD from both sides, we get:
⇒ AB – AD = AC – AD
⇒ AB – AD = AC  AE (Since, AD = AE)
⇒ BD = EC …(iii)
Dividing equation (i) by equation (iii), we get:
AD/DB = AE/EC
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)
Hence, quadrilateral BCED is cyclic.
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 .
SolutionIn triangle BQO, BR bisects angle B.
Applying angle bisector theorem, we get:
QR/PR = BQ/BP
⇒ BP × QR = BQ × PR
This completes the proof.