Ncert Solutions For Class 10 Maths Ex 6.3

Ncert Solutions For Class 10 Maths Chapter 6 Ex 6.3

 Question 1:

Which of the following triangle pairs are similar? State the similarity criterion you used to determine the similarity of the triangles.

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Solution:

(i)For  ΔABC and ΔPQR:

∠A=∠P = 60o (Given)

∠B =∠Q = 80o(Given)

∠C =∠R = 40o(given)

∴ Δ ABC ~ ΔPQR (AAA similarity criteria)

(ii)For ΔJKL and ΔZXC

JK/XC = KL/CZ = JL/XZ
∴  ΔJKL~ ΔZXC (SSS similarity criterion)

 

(iii) For ΔJKL and ΔZXC:
JK = 2.7, KL = 2, LJ = 3, ZX = 5, XC = 4, CZ = 6
KL/ZX = 2/4 = 1/2
JL/ZC = 3/6 = 1/2
JK/XC= 2.7/5 = 27/50
Here, KL/ZX = LJ/ZC ≠ JK/XC
Thus, ΔJKL and ΔZXC are not similar.

(iv) For ΔJKL and ΔZXC
JK = 2.5, KL = 3, ∠J = 80°, XC = 6, ZC = 5, ∠C = 80°
Here, JK/ZC = 2.5/5 = 1/2
And, KL/XC = 3/6 = 1/2
⇒ ∠K ≠ ∠C
Thus, ΔJKL and ΔZXC are not similar.

(v)  For ΔJKL, we have
∠J + ∠K + ∠L = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠L = 180°
⇒ ∠L = 180° – 70° – 80°
⇒ ∠L = 30°
In ZXC, we have
∠Z + ∠X + ∠C = 180 (Sum of angles of Δ)
⇒ ∠Z + 80° + 30° = 180°
⇒ ∠Z = 180° – 80° -30°
⇒ ∠Z = 70°
In ΔJKL and ΔZXC, we have
∠J = ∠Z = 70°
∠K = ∠X = 80°
∠L = ∠C = 30°
Thus, ΔJKL ~ ΔZXC (AAA similarity criterion)

 

 

Question 2:

In the figure below, ΔJKL ¼ ΔZXL, KJL=70O and ZLX = 70O. Find JLK, JKL AND LZX

13

Solution:

 JLX is a straight line.
Thus, ∠JLK + ∠ KLX = 180°
⇒ ∠JLK = 180° – 125°
= 55°

In ΔJLK,
∠JKL+ ∠ KJL + ∠ JLK = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠JKL + 70º + 55º = 180°
⇒ ∠JKL = 55°
Given that ΔLJK ~ ΔLXZ.

∴ ∠LZX = ∠LKJ  (Corresponding angles are equal in similar triangles)
⇒ ∠ LZX = 55°
∴ ∠LZX= ∠LKJ   (Corresponding angles are equal in similar triangles)

⇒ ∠LZX= 55°

 

Question 3:

Diagonals JX and KZ of a trapezium JKZX with JK || ZX intersect each other at the point L. With the help of similarity criterion for two triangles, prove that JL/LX = KL/KZ.

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Solution:

In ΔZLX and ΔJLK,
∠XZL = ∠JKL (Alternate interior angles as JK || XZ)
∠ZXL = ∠KJL (Alternate interior angles as JK || XZ)
∠ZLX = ∠KLJ (Vertically opposite angles)
∴ ΔZLX ~ ΔKLJ (AAA similarity criterion)

∴ ZL/KL = LX/LJ  (Corresponding sides are proportional)

⇒ LJ/LX = LK/LZ

 

 

Question 4:

In the given figure, LX/LZ = LJ/KX and 1 = 2. Prove that ΔKXZ ~ ΔJLX.

15

Solution:

In ΔJLX, ∠KLX = ∠KXL
∴ KL = KX … (i)
Given, LX /LZ = LJ/KX
Using (i), we get
LX/LZ = LJ/LK ..(ii)

In ΔKLZ and ΔJLX,
LX/LZ = LJ/LK (using(ii))
∠L = ∠L
∴ ΔKZX ~ ΔJLX [SAS similarity criterion]

 

Question 5:

In the given figure, K and L are points on sides JZ and JX of ΔZXJ such that Z = JLK. Prove that ΔJZX ~ ΔJLK.

16

Solution:

In ΔJZX and ΔJKL,
∠JLK = ∠XZK (Given)
∠J = ∠J (Common angle)
∴ ΔJZX ~ ΔJLK (By AA similarity criterion)

 

Question 6:

In the figure given, if ΔJZL ΔJXK, prove that ΔJKL ~ ΔJZX.

17

Solution:

Given, that ΔJZL ≅ ΔJXK.
∴ JZ = JX [By cpct] … (i)
Also, JK = JL [By cpct] … (ii)
In ΔJKL and ΔJZX,

JK/JZ = JL/JX [Dividing equation (ii) by (i)]

∠J = ∠J [Common angle]
∴ ΔJKL ~ ΔJZX [By SAS similarity criterion]

 

Question 7:

  In the figure given below, lines ZK and JX of ΔJZC intersect each other at the point L. Prove that:

(i) ΔZXL ~ ΔJKL
(ii) ΔZCK ~ ΔJCX
(iii) ΔZXL ~ ΔZKC
(iv) ΔLKJ ~ ΔCXJ

18

Solution:

(i) In ΔZXL and ΔJKL,
∠ZXL = ∠JKL (Each 90°)
∠ZLX = ∠JLK (Vertically opposite angles)
Thus, by using AA similarity criterion,
ΔZXL ~ ΔJKL

(ii) In ΔZCK and ΔJCX,
∠ZKC = ∠JXC (Each 90°)

∠ZCK = ∠JCX (Common)
Hence, by using AA similarity criterion,
ΔZCK ~ ΔJCX
(iii) In ΔZXL and ΔZKC,
∠ZXL = ∠ZKC (Each 90°)

∠LZX = ∠KZC (Common)
Hence, by using AA similarity criterion,
ΔZXL ~ ΔZKC
(iv) In ΔLKJ and ΔCXJ,
∠LKJ = ∠CXJ (Each 90°)
∠LJK = ∠CJX (Common angle)
Hence, by using AA similarity criterion,
ΔLJK ~ ΔCXJ

 

Question 8:

In the given figure X is a point on the side JZ of a  parallelogram JKCZ and KX  intersect ZC at L. Prove that ΔJKX ~ ΔCLK.

19

Solution:

In ΔJKX and ΔCLX,
∠J = ∠C (Opposite angles of a parallelogram)
∠JXK = ∠CKL (Alternate interior angles as JX || KC)
∴ ΔJKX ~ ΔCLX (By AA similarity criterion)

 

Question 9:

In the given figure, JKC and AMP are two right triangles, right angled at B and M respectively, show that:

i) ΔJZC ~ ΔJKX

(ii) CJ/XJ = ZC/KX

Solution:

20

(i) In ΔJZC and ΔJKX, we have

∠J= ∠J (common angle)

∠JZC = ∠JKX = 90° (each 90°)

∴ ΔJZC ~ ΔJKX (By AA similarity criterion)

 

(ii) Since, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CJ/XJ = ZC/KX

 

Question 10:

In the given figure CD and KH are the bisectors of ZCX and JKL respectively, such that D and H lie on sides ZX and JL of ΔZXC and ΔJKL respectively. If ΔZXC ~ ΔJKL. Prove that:

(i) CD/KH = ZC/JK
(ii) ΔCXD ~ ΔKLH
(iii) ΔDZC ~ ΔHKJ

21

Solution:

(i) Given that ΔZXC ~ ΔJKL
∴ ∠Z = ∠J, ∠X = ∠L, and ∠ZCX = ∠JKL
∠ZCX = ∠JKL
∴ ∠ZCD = ∠JGH (Angle bisector)
And, ∠DCX = ∠HKL (Angle bisector)
In ΔDZC and ΔJKH,
∠Z = ∠J (Proved above)
∠DCZ = ∠JKH (Proved above)
∴ ΔDZC ~ ΔJKH (By AA similarity criterion)

⇒ CD/KH = ZC/JK

 

(ii) In ΔXCD and ΔHKL,
∠XCD = ∠HK; (Proved above)
∠X = ∠L (Proved above)
∴ ΔXCD ~ ΔHKL (By AA similarity criterion)

 

(iii) In ΔZCD and ΔHJK
∠DCZ = ∠JKH (Proved above)
∠Z = ∠J (Proved above)
∴ ΔZCD ~ ΔHJK (By AA similarity criterion)

 

Question 11:

In the following figure, K is a point on side CL  of an isosceles triangle JCL, where JL= JC. If JZ LC and KF CJ. Show that ΔJLZ ~ ΔFCK.

22

Solution:

It is given that JLC is an isosceles triangle.
∴ JL = JC
⇒ ∠JLZ = ∠FCJ
In ΔJLZ and ΔFKC,
∠JDL = ∠CFK (Each 90°)
∠LJZ = ∠FKC (Proved above)
∴ ΔJLZ ~ ΔFCK (By using AA similarity criterion)

 

Question 12:

In the given figure sides ZX and XC and median ZV of a triangle ZXC are proportional to sides HJ and JL  and median HK of ΔPQR, respectively. Prove that ΔZCX ~ ΔHJL.

23

Solution:

It is given that:

ΔZCX and ΔHJL, ZX, CX and median ZV of ΔZCX are proportional to sides HJ, JL and median HK of ΔHJL

Or, ZX/HJ = CX/JL= ZV/

To Prove: ΔZCX ~ ΔHJL

 

Proof: ZX/HJ = CX/JL = ZV/HK

 

⇒ ZX/HJ = CX/JL = ZV/HK (V is the mid-point of CX and K is the midpoint of JL)

⇒ ΔZXV ~ ΔHJK [SSS similarity criterion]

∴ ∠ZXV = ∠HJK [Corresponding angles of two similar triangles are equal]

⇒ ∠ZXC = ∠HJL

In ΔABC and ΔPQR

ZX/HJ = CX/JL ….(i)

∠CXZ = ∠HJL….. (ii)

Hence, from equation   (i) and (ii), we get

ΔCXZ ~ ΔHJL [By SAS similarity criterion]

 

Question 13:

J is a point on the side CK of a triangle CJK such that JLC = KJC. Prove that CJ2 = CK.LC

24

Solution:

In ΔCJL and ΔJKC,
∠CLJ = ∠CJK (Given)
∠JCL = ∠KCJ (Common angle)
∴ ΔCJL ~ ΔJKC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.

∴ CJ/KC =CL/JC

⇒ CJ2 = CK.LC.

 

Question 14:

In the given figure, sides ZX and ZC and median ZV of a triangle CXZ are proportional to sides JK and JH and median JL of another triangle JKH, respectively. Show that ΔCXZ~ ΔJKH

25

Solution:

It is given that:

Two triangles ΔCXZ and ΔJKH in which ZV and JL are medians such that ZX/JK = CZ/JH = ZV/JL

To Prove: ΔZXC ~ ΔJKH

Construction: Produce ZV  to F such that ZV = VF. Connect CF. Similarly produce JL to N so that JL = LN, also connect HN.

 

Proof:

In ΔZXV and ΔVCF, we have

 

ZV = VF                                   [By Construction]

XV = VC                                   [∴ AP is the median]

And, ∠ZVX = ∠CVF                  [Vertically opp. angles]

∴ ΔZXV ≅ ΔCVF                       [By SAS criterion of congruence]

⇒ ZX = CF                                [CPCT] ….. (i)

 

Also, in ΔJKL and ΔLNH, we have

JL = LN                                     [By Construction]

KL = LH                                    [∴ PM is the median]

and, ∠JLK = ∠NLH                   [Vertically opposite angles]

∴ ΔJKL = ΔLHN                         [By SAS criterion of congruence]

⇒ JK = HN                               [CPCT]  …. (ii)

Now, ZX/JK = ZC/JH = ZV/JL

⇒ CF/HN = ZC/JH = ZV/JL … [From (i) and (ii)]

⇒ CF/HN = ZC/JH = 2ZV/2JL

⇒ CF/HN = ZC/JH = ZF/JN     [∴ 2AD = AE and 2PM = PN]

∴ ΔZCF ~ ΔJHN                        [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠Z = ∠J … (iii)

Now, In ΔZXC and ΔJKH, we have

ZX/JK = ZC/JH       (Given)

∠Z = ∠J [From (iii)]

∴ ΔZXC ~ ΔJKH                             [By SAS similarity criterion]

 

Question 15:

A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time, a tower casts a shadow 28 m long. Find the height of the tower.

26

Solution:

Length of the vertical pole = 6m (Given)

Shadow of the pole = 4 m (Given)

Let Height of tower = h m

Length of shadow of the tower = 28 m (Given)

In ΔZXC and ΔJKL,

∠C = ∠K (angular elevation of sum)

∠X = ∠L = 90°

∴ ΔZXC ~ ΔJKL (By AA similarity criterion)

∴ ZX/JL = XC/KL (when two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒ h = 6×28/4

⇒ h = 6 × 7

⇒ = 42 m

Therefore, the height of the tower is 42 m.

 

Question 16:

In the given figure if ZV and JM are medians of triangles ZXC and JKL, respectively. Where ΔZXC ~ ΔJKL prove that ZX/JK = ZV/JM.

27

Solution:

It is given that: ΔZXC ~ ΔJKL
We know that, the corresponding sides of similar triangles are in proportion

.∴ ZX/JK = ZC/JL = KC/KL … (i)
Also, ∠ Z= ∠J, ∠X = ∠K, ∠C = ∠L …(ii)
As ZV and JM are medians, their opposite sides will be divided by them

.∴ XV = XC/2 and KM = KL/2 …(iii)
From equations (i) and (iii), we have
ZX/JK = XV/KM …(iv)
In ΔZXV and ΔJKM,
∠X = ∠K [Using equation (ii)]
ZX/JK = XV/KM [Using equation (iv)]
∴ ΔZXV ~ ΔJKM (By SAS similarity criterion)

⇒ ZX/JK = XV/KM = ZV/JL.

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