All these RS Aggarwal class 10 solutions Chapter 4 Triangles are solved by Byju's top ranked professors as per CBSE guidelines.
| Triangles Exercise 4.1 |
| Triangles Exercise 4.2 |
| Triangles Exercise 4.3 |
Question 1: D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC such that DE || BC.
(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.
(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.
(iii) If \(\frac{AD}{DB}=\frac{4}{7}\) and AC = 6.6 cm, find AE.
(iv) If \(\frac{AD}{AB}=\frac{8}{15}\) and EC = 3.5 cm, find AE.
Solution:
(i) In \(\Delta\)ABC, DE || BC, AD = 3.6 cm, AB = 10 cm, AE = 4.5 cm
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)
\(\Rightarrow\) \(\frac{3.6}{AB-AD}=\frac{4.5}{AC-AE}\)
\(\Rightarrow\) \(\frac{3.6}{10-3.6}=\frac{4.5}{AC-4.5}\)
\(\Rightarrow\) \(\frac{3.6}{6.4}=\frac{4.5}{AC-4.5}\)
\(\Rightarrow\) 3.6AC ??? 16.2 = 28.8
\(\Rightarrow\) 3.6AC = 45
\(\Rightarrow\) AC = 12.5 cm
Therefore, EC = AC ??? AE = 12.5 ??? 4.5 = 8 cm
Hence, AC = 12.5 cm and EC = 8 cm
(ii) In \(\Delta\)ABC, DE || BC, AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)
\(\Rightarrow\) \(\frac{AD}{AB-AD}=\frac{AC-EC}{EC}\)
\(\Rightarrow\) \(\frac{AD}{13.5-AD}=\frac{11.9-5.1}{5.1}\)
\(\Rightarrow\) \(\frac{AD}{13.5-AD}=\frac{6.8}{5.1}\)
\(\Rightarrow\) 5.1AD = 91.8 ??? 6.8AD
\(\Rightarrow\) 11.9AD = 91.8
\(\Rightarrow\) AD = \(\frac{91.8}{11.9}\) = 7.7
Hence, AD = 7.7 cm
(iii) In \(\Delta\)ABC, DE || BC, AC = 6.6 cm, \(\frac{AD}{DB}=\frac{4}{7}\)
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)
\(\Rightarrow\) \(\frac{4}{7}=\frac{AE}{AC-AE}\)
\(\Rightarrow\) \(\frac{4}{7}=\frac{AE}{6.6-AE}\)
\(\Rightarrow\) 26.4 ??? 4AE = 7AE
\(\Rightarrow\) 26.4 = 11AE
\(\Rightarrow\) AE = \(\frac{26.4}{11}\) = 2.4 cm
Hence, AE = 2.4 cm
(iv) In \(\Delta\)ABC, DE || BC, Given \(\frac{AD}{AB}=\frac{8}{15}\), EC = 3.5 cm
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)
And AD = \(\frac{8}{15}\)AB [Given]
Therefore, \(\frac{AD}{AB-AD}=\frac{AE}{3.5}\) [AD = \(\frac{8}{15}\)AB]
\(\frac{\frac{8}{15}AB}{AB-\frac{8}{15}AB}=\frac{AE}{3.5}\)\(\Rightarrow\) \(\frac{\frac{8}{15}AB}{\frac{7}{15}AB}=\frac{AE}{3.5}\)
\(\Rightarrow\) \(\frac{8}{7}=\frac{AE}{3.5}\)
\(\Rightarrow\) AE = \(\frac{3.5\times 8}{7}\)
\(\Rightarrow\) AE = 4 cm
Hence, AE = 4 cm
Question 2: D and E are points on the sides AB and AC respectively of a \(\Delta\)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 = 4 cm, DB = (x ??? 4) cm, AE = 8 cm and EC = (3x ??? 19) cm.
(iii) AD = (7x ??? 4) cm, AE = (5x ??? 2)cm, DB = (3x + 4) cm and EC = 3x cm.
Solution:
(i) D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC such that DE || BC, AD = x cm, DB = (x ??? 2) cm, AE = (x + 2) cm, EC = (x ??? 1) cm
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (by thales theorem)
\(\Rightarrow\) \(\frac{x}{x-2}=\frac{x+2}{x-1}\)
\(\Rightarrow\) x(x ??? 1) = x2 ??? 4 \(\Rightarrow\) x2 ??? x = x2 ??? 4
\(\Rightarrow\) x = 4
(ii) In \(\Delta\)ABC, DE || BC, AD = 4 cm, DB = (x ??? 4) cm, AE = 8 cm, EC = (3x ??? 19) cm
\(\frac{AD}{AB}=\frac{AE}{EC}\) (by thales theorem)
\(\Rightarrow\) \(\frac{4}{x-4}=\frac{8}{3x-19}\)
\(\Rightarrow\) 4(3x ??? 19) = 8(x ??? 4)
\(\Rightarrow\) 12x ??? 76 = 8x ??? 32
\(\Rightarrow\) 4x = 44
\(\Rightarrow\) x = 11
\(\frac{AD}{DB}=\frac{AE}{EC}\) (by thales theorem)
\(\frac{7x-4}{3x+4}=\frac{5x-2}{3x}\)Hence, x = 11
(iii) In \(\Delta\)ABC, DE || BC, AD = (7x ??? 4) cm, AE = (5x ??? 2) cm, DB = (3x + 4) cm, EC = 3x cm
\(\Rightarrow\) 21x2 ??? 12x = 15x2 ??? 6x + 20x ??? 8
\(\Rightarrow\) 6x2 ??? 26x + 8 = 0
3x2 ??? 13x + 4 = 0
3x2 ??? 12x ??? x + 4 = 0
3x(x ??? 4) ??? 1(x ??? 4)
(x ??? 4)(3x ??? 1) = 0
x = 4 or x = \(\frac{1}{3}\)
If x = \(\frac{1}{3}\), length of sides become negative
Hence, x = 4
Question 3: D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC. In each of the following cases, determine whether DE || BC or not.
(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm.
(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm and AE = 4.2 cm.
(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm and EC = 4 cm.
(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm.
Solution:
Given: A \(\Delta\)ABC, in which D and E are points on the sides AB and AC respectively.
To prove: DE || BC
Proof:
(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm
Since D and E are the points on AB and AC repectively.
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\)
\(\Rightarrow\) \(\frac{5.7}{9.5}=\frac{4.8}{8}\)
\(\Rightarrow\) 0.6 = 0.6
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (each equal to 0.6)
Hence, by the converse of Thales theorem DE || BC
(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm, AE = 4.2 cm
Since D and E are points on AB and AC respectively.
\(\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{AB-DB}{DB}=\frac{AE}{AC-AE}\)\(\Rightarrow\) \(\frac{11.7-6.5}{6.5}=\frac{4.2}{11.2-4.2}\Rightarrow \frac{5.2}{6.5}\neq \frac{4.2}{7}\)
Hence, \(\frac{AD}{DB}\neq \frac{AE}{EC}\)
Hence, by the converse of Thales theorem DE is not parallel to BC.
(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm, EC = 4 cm
Since D and E are the points on AB and AC respectively.
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) [by thales theorem]
\(\Rightarrow\) \(\frac{AD}{AB-AD}=\frac{AC-EC}{EC}\)
\(\Rightarrow\) \(\frac{6.3}{(10.8-6.3)}=\frac{(9.6-4.0)}{4}\)
\(\Rightarrow\) \(\frac{6.3}{4.5}=\frac{5.6}{4}\)
\(\Rightarrow\) 1.4 = 1.4
Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (each is equal to 1.4)
Hence by the converse of Thales theorem DE || BC
(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm, AC = 10 cm
Since D and E are points on the side AB and AC respectively.
\(\frac{AD}{DB}=\frac{AE}{EC}\) [by thales theorem]
\(\frac{7.2}{AB-AD}=\frac{6.4}{AC-AE}\)\(\Rightarrow\) \(\frac{7.2}{12-7.2}=\frac{6.4}{10-6.4}\)
\(\Rightarrow\) \(\frac{7.2}{4.8}=\frac{3.4}{3.6}\)
but \(\frac{3}{2}\neq \frac{16}{9}\)
Hence, by the converse of Thales theorem DB is not parallel to BC
Question 4: In a \(\Delta\)ABC, AD is the bisector of \(\angle\)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) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm
Let BC = x
Now, DC = (BC ??? BD) = (x ??? 5.6) cm
In \(\Delta\)ABC, AD is the base of \(\angle\)A
So, by the angle bisector theorem, we have
\(\frac{BD}{DC}=\frac{AB}{AC}\)\(\Rightarrow\) \(\frac{5.6}{x-5.6}=\frac{6.4}{8}\)
\(\Rightarrow\) 6.4x ??? 35.84 = 44.8
\(\Rightarrow\) 6.4x = 80.64
\(\Rightarrow\) x = 12.6
Hence, BC = 12.6 cm and DC = (12.6 ??? 5.6) cm = 7 cm
(ii) AB = 10 cm, AC = 14 cm, BC = 6 cm
Let BD = x,
DC = (BC ??? BD) = (6 ??? x) cm
In \(\Delta\)ABC, AD is the bisector of \(\angle\)A
So, By angle bisector theorem,
We have, \(\frac{BD}{DC}=\frac{AB}{AC}\)
\(\Rightarrow\) \(\frac{x}{6-x}=\frac{10}{14}\)
\(\Rightarrow\) 14x = 60 ??? 10x
\(\Rightarrow\) 24x = 60
\(\Rightarrow\) x = 2.5
Hence, BD = 2.5 cm and DC = (6 ??? 2.5) cm = 3.5 cm
(iii) AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm
DC = BC ??? BD = (6 ??? 3.2) cm = 2.8 cm
Let, AC = x,
In \(\Delta\)ABC, AD is the bisector of \(\angle\)A
So, by the angle bisector theorem we have
Therefore, \(\frac{BD}{DC}=\frac{AB}{AC}\)
\(\Rightarrow\) \(\frac{3.2}{2.8}=\frac{5.6}{x}\)
\(\Rightarrow\) x = \(\frac{5.6\times 2.8}{3.2}=4.9\) cm
Hence, AC = 4.9 cm
(iv) AB = 5.6 cm, AC = 4 cm, DC = 3 cm
Let BD = x,
In \(\Delta\)ABC, AD is the bisector of \(\angle\)A
So, by the angle bisector theorem we have
Therefore, \(\frac{BD}{DC}=\frac{AB}{AC}\)
\(\Rightarrow\) \(\frac{x}{3}=\frac{5.6}{4}\)
\(\Rightarrow\) x = \(\frac{5.6\times 3}{4}=4.2\) cm
Hence, BD = 4.2 cm
So BC = BD + AC = (4.2 + 3) cm
BC = 7.2 cm
Question 5: In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals interact at O. If AO = (5x ??? 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm, find the value of x.
Solution:
We know that CD || AB in trap ABCD and its diagonals interact at O.
Since the diagonals of a trapezium divide each other proportionally therefore, we have
\(\frac{AO}{OC}=\frac{BO}{OD}\) (by thales theorem)
\(\Rightarrow\) \(\frac{5x-7}{2x+1}=\frac{7x+1}{7x-5}\)
\(\Rightarrow\) (5x ??? 7)()7x ??? 5) = (7x + 1)(2x + 1)
\(\Rightarrow\) 35x2 ??? 25x ??? 49x + 35 = 14x2 + 7x + 2x + 1
\(\Rightarrow\) 35x2 ??? 14x2 ??? 25x ??? 49x ??? 7x ??? 2x + 35 ??? 1 = 0
\(\Rightarrow\) 21x2 ??? 83x + 34 = 0
Therefore, \(x=\frac{83\pm \sqrt{(83)^{2}-4\times 21\times 34}}{21\times 2}\)
= \(\frac{83\pm \sqrt{4033}}{42}\)
= \(\frac{83\pm 63.51}{42}\)
= \(\frac{146.51}{42}\) or \(\frac{19.49}{42}\) = 3.49 or 0.46
\(\Rightarrow\) x = 0.46 or 3.49,
But x \(\neq\) 0.46 as all sides should be positive.
Therefore, x = 3.49 cm
Question 6: \(\Delta\)ABC and \(\Delta\)DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD is drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.
Solution:
Given: \(\Delta\)ABC and \(\Delta\)DBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.
To Prove: QR || AD
Proof: In \(\Delta\)ABC
PQ || AB
\(\Rightarrow\) \(\frac{CP}{PB}=\frac{CQ}{QA}\) ——– (1)(by thales theorem)
In \(\Delta\)BCD, PR || BD
Therefore, \(\frac{CP}{PB}=\frac{CR}{RD}\) ——– (2)(by thales theorem)
From (1) & (2), we get:
\(\frac{CQ}{QA}=\frac{CR}{RD}\)Hence, in \(\Delta\)ACD, Q and R the points in AC and CD such that
Therefore, \(\frac{CQ}{QA}=\frac{CR}{RD}\)
QR || AD (by the converse of Thales theorem)
Hence proved
Question 7: In the given figure, side BC of \(\Delta\)ABC is bisected at D and O is any point in 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 mid-point of OX. Prove that AO: AX = AF: AB and show that EF || BC.
Solution:
Given BD = CD and OD = DX
Join BX and CX
Thus, the diagonals of quad OBXC bisect each other
OBXC is a parallelogram
BX || CF and so, OF || BX
Similarly, CX || OE
In \(\Delta\)ABX, OF || BX
Therefore, \(\frac{AO}{AX}=\frac{AF}{AB}\) —– (1)
In \(\Delta\)ACX, OE || XC
Therefore, \(\frac{AO}{AX}=\frac{AE}{AC}\) —– (2)
From (1) & (2) we get:
\(\frac{AF}{AB}=\frac{AE}{AC}\)Hence, EF || BC
Question 8: ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = \(\frac{1}{4}\)AC. If PQ produced meets BC at R, prove that R is the mid-point of BC.
Solution:
Given: ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac{1}{4}\)AC. PQ produced meets BC at R.
To prove: R is the mid-point of BC
Construction: Join BD
Proof: Since the diagonals of a parallelogram bisect each other at S such that
CS = ?? AC
Now, CS = ?? AC and CQ = \(\frac{1}{4}\)AC \(\Rightarrow\) CQ = ?? CS
Therefore, Q is the mid-point of CS
So, PQ || DS
Therefore, QR || SB
In \(\Delta\)CSB, Q is the mid-point of CS and QR || SB.
So R is the midpoint of BC.
Question 9: 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: ABC is a triangle in which AB = AC. D and E are points on AB and AC respectively such that AD = AE
To prove: AB = AC (given)
\(\Rightarrow\) (AB ??? AD) = (AC ??? AE)
\(\Rightarrow\) DE = EC
\(\Rightarrow\) \(\frac{AD}{AE}=\frac{DB}{EC}\) (each equal to 1)
\(\Rightarrow\) DE || BC (by the converse of Thales???s theorem)
\(\Rightarrow\) \(\angle\)DEC + \(\angle\)ECB = 180o
\(\Rightarrow\) \(\angle\)DEC + \(\angle\)CBD = 180o [AB = AC \(\Rightarrow\) \(\angle\)C = \(\angle\)B]
Quadrilateral BCEA is cyclic
Hence, the point B, C, E, D are concyclic
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