* **Question 1:*

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

**Solution: **

(i)For ΔABC and ΔPQR:

∠A=∠P = 60^{o }(Given)

∠B =∠Q = 80^{o}(Given)

∠C =∠R = 40^{o}(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=70 ^{O} and *

*∠*

*ZLX = 70*^{O}. Find

*∠*

*JLK,*

*∠*

*JKL*

*AND*

*∠*

*LZX***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.*

**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.*

**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.*

**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.*

**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

**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.*

**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:**

(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

**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.*

**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.*

**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 CJ ^{2} = CK.LC*

**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

⇒ CJ^{2} = 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*

**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.*

**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

⇒ *h *= 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.*

**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.