Similarity

(i) All circles are similar.

Let there be two circles of radii r₁ and r₂.

Now, shifting the centre of smaller circle to the bigger circle.

If we slowly increase the radius of smaller circle, it will coincide with bigger circle when r₂ = r₁. Thus, the circles are similar.

(ii) All squares are similar.

Let there be two squares ABCD and EFGH. When the smaller square is kept at the centre of the square ABCD, then on increasing the side of EFGH both of them will coincide. Thus, they are similar.

Two polygons are similar if their corresponding are equal. The corresponding angles of the squares are 90°. Thus, they are similar.

(iii) All equiangular triangles are similar.

Two equiangular triangles have equal corresponding angles. Thus, they are similar by AA or AAA Similarity Rule.

(iv) Two polygons with same number of sides are similar if

(a) their all the corresponding angles are equal

(b) their corresponding sides are in the same ratio

(i) True

Two figures are said to be congruent if their corresponding sides are equal and their corresponding angles are also equal.

For figures to be similar, the corresponding angles should be equal which is true in case of congruent figures. Thus, the congruent figures are also similar.

(ii) False

For figures to be similar, the corresponding angles are equal.

Two figures are said to be congruent if their corresponding sides are equal and their corresponding angles are also equal.

Thus, two similar figures are congruent only iftheir corresponding sides are equal.

(iii) False

Two polygons have proportional sides are similar if and only if the corresponding angles are also equal.

(iv) True

(v) True

Two polygons are similar if their corresponding sides are proportional and the corresponding angles are equal.

All circles are similar.

Let there be two circles of radii r₁ and r₂.

Now, shifting the centre of smaller circle to the bigger circle

If we slowly increase the radius of smaller circle, it will coincide with bigger circle when r₂ = r₁. Thus, the circles are similar.

All squares are similar.

Let there be two squares ABCD and EFGH. When the smaller square is kept at the centre of the square ABCD, then on increasing the side of EFGH both of them will coincide. Thus, they are similar.

In ΔABC,

∵DE|| BC

∴|(By Thales Theorem)

Given,

In ΔABC,

∵DE|| BC

∴|(By Thales Theorem)

⇒

Given,

In ΔABC,

∵DE|| BC

∴|(By Thales Theorem)

⇒

AC = AE + EC = 6.3 + 3.6 = 9.9 cm

Given,

AD = 4x – 3

BD = 3x – 1

AE = 8x – 7

CE = 5x – 3

In ΔABC,

∵DE|| BC

∴|(By Thales Theorem)

⇒ 20x² – 27x + 9 = 24x² – 29x + 7

⇒ 4x² – 2x – 2 = 0

⇒ x = 1, – 1/2

Suppose DE||BC,

In ΔADE & ΔABC,

∠A = ∠A |common angle

∠ADE = ∠ABC |corresponding angles

ΔADE~ΔABC by AA Similarity Rule

By Thales Theorem,

should be true for DE||BC.

Here,

Hence, DE||BC is true.

Here,

Hence, DE||BC is false.

Here,

Hence, DE||BC is false.

Given,

AD = 5.7 cm

BD = 9.5 cm

AE = 3.3 cm

EC = 5.5 cm.

AB = AD + BD = 5.7 + 9.5 = 15.2 cm

Here,

Hence, DE||BC is true.

In OAB & OLM

LM||AB

…(1) |By Basic Proportionality Theorem(BPT)

In ΔOMN & ΔOBC,

MN||BC

…(2) |By BPT

From (1)&(2)

…(3)

In ΔOLN & ΔOAC

⇒ LN||AC by Converse of BPT

Hence, proved.

In ΔABC & ΔADE

∠B = ∠C |Given

⇒ AC = AB |sides opposite to equal angles are equal

⇒ AB = AC …(1)

Given,

BD = CE

Subtracting BD from AB and CE from AC,

⇒ AB – BD = AC – CE

⇒ AD = AE |...(2)

In ΔABC & ΔADE,

From (1) and (2)

∠B = ∠C

⇒ ABC~ADE by SAS Similarity Rule

⇒ ∠ADE = ∠ABC

⇒ DE||BC as all corresponding angles are equal.

In ΔADE & ΔABC,

∠A = ∠A |common angle

∠ADE = ∠ABC |corresponding angles

ΔADE~ ΔABC by AA Similarity Rule

…(1)

In ΔAFE & ΔADC,

∠A = ∠A |common angle

∠AFE = ∠ADC |corresponding angles

ΔAFE~ ΔADC by AA Similarity Rule

…(2)

From (1) & (2),

⇒ AD² = AB × AF

Hence, proved.

Let us drop a perpendicular AG and BH to CD cutting EG at I and J and CD.

In ΔADG & ΔAEI,

∠AGD = ∠AIE |Right Angle

∠AEI = ∠ADG |corr. ∠s

ΔADG~ΔAEI by AA Similarity Rule

In ΔBJF & ΔBHC,

∠BJF = ∠BHC |Right angle

∠BFJ = ∠BCH |corr. ∠s

ΔBJF~ΔBHC by AA Similarity Rule

In rectangle ABHG & ABJI,

AI = BJ …(a) |opposite sides of rectangle are equal

AG = BH …(b) |opp. sides of rectangle

From eqn. (b) – (a)

AG – AI = BH – BJ

⇒ GI = HJ

…(2)

From (1) & (2),

Hence, proved.

(i) In ΔDPC & ΔBPL,

∠DPC = ∠BPL |vertically opposite ∠s

In ||gm ABCD,

DC||AB or DC||AL,

⇒ ∠DCP = ∠LBP

ΔDPC~ΔBPL by AA Similarity Rule

Hence,proved.

(ii)In ΔPLB & ΔDLA,

∠L = ∠L |common angle

In ||gm ABCD, AD||BC or AD||BP,

⇒ ∠LPB = ∠LDC |corresponding angles

ΔPLB~ΔDLA by AA Similarity Rule

…(1)

Subtracting 1 from both sides of the above equation,

…(2)

Multiplying (1) & (2),

Or,

Hence, proved.

In ΔABC,

EQ||AC

By Basic Proportionality Theorem,

AD = BE |Given

AE = AD + DE = BE + ED = BD

…(1)

In ΔABC,

DP||BC

By Basic Proportionality Theorem,

…(2)

From (1) & (2),

By Converse of BPT, PQ||AB.

Hence, proved.

In ΔAOB & ΔCOD,

∠AOB = ∠COD

∠ABO = ∠ODC

∠OAB = ∠OCD

ΔAOB~ΔCOD by AAA Similarity Rule

⇒

⇒

Hence,proved.

In ΔADE & ΔABC,

∠ADE = ∠ABC

∠A = ∠A

ΔADE~ΔABC by AA Similarity Rule

BD = CE

⇒ AD = AE

Now,

AD + BD = AE + CE

AB = AC

Thus, the triangle ABC is isosceles.

As per SAS rule of similarity, the angle between sides AB & BC of ΔABC and PQ & QR of ΔPQR should be equal.

∠ABC = ∠PQR

Thus,

ΔABC is not similar to ΔDEF as the order should be ΔDFE to be similar as given conditions.

For similar triangles, the ratio of corresponding sides are equal. The order of sides should be such that the similar sides are written at the same position of naming.

In the given question, the ratio cannot be written as the order of sides is not same. The order should be

The triangles may not be similar. The triangles should have corresponding sides proportional and the angle between them to be similar by SAS Similarity Rule.

If the corresponding angles in two triangles are equal, then they are said to be equiangular. The equiangular triangles are always similar by AAA Similarity Rule.

In ΔABC & ΔQ^’P^’R,

∠A = ∠Q^’

∠B = ∠P^’

∠C = ∠R^’

Thus, ΔABC~ΔQ^’P^’R^’ |AAA Similarity Rule

In ΔMNP & ΔZXY,

∠M = ∠Z

∠P = ∠Y

Thus, ΔMNP~ΔZXY |AA Similarity Rule

In ΔPQR & ΔEFG,

Thus, the corresponding sides are proportional.

Thus, ΔPQR~ΔEFG |SSS Similarity Rule

In ΔE^’G^’F^’ & ΔNML,

∠E^’G^’F^’ = ∠NML

Thus, the corresponding sides are proportional with the angle between them equal.

Thus, ΔE^’G^’F^’~ΔNML |SAS Similarity Rule

Given,

ΔPRQ ~ ΔTRS

The mutually equal angles are

∠PRQ = ∠TRS

∠RPQ = ∠RTS

∠RQP = ∠RST

Given,

∠CBE = ∠CAD

In ΔCAD & ΔCBE,

∠ACD = ∠BCE |Common Angle

∠CBE = ∠CAD |Given

⇒ ΔCAD~ΔCBE by AA Similarity Rule

Given,

PQ||RS

In ΔPOQ & ΔSOR,

⇒ ∠PQO = ∠ORS |Alternate angles

⇒ ∠QPO = ∠OSR |Alternate angles

⇒ ∠POQ = ∠ROS |Vertically opposite angles

Thus, ΔPOQ~ΔSOR by AAA Similarity Rule

The speed of give = 1.2 ms ^{– 1}

Time taken = 4 seconds

Distance = speed × time taken

In 4 seconds, the distance walked is 1.2×4 = 4.8 m.

As both the lamp post and the give are perpendicular to the ground, they make the following similar triangles.

In ΔBDE & ΔBAC,

∠B = ∠B |common angle

DE||AC

∠BDE = ∠BAC

ΔBDE~ΔBAC by AA Similarity Rule

4x = x + 4.8

⇒ x = 1.6 m

Given,

length of the pillar = A₁B₁ = 12 m

length of the shadow of pillar = A₁C₁ = 8 m

length of the shadow of the tower = AC = 56 m

Let the length of the tower = AB be x m.

As seen from the figure,

ΔABC is similar to ΔA₁B₁C₁

Given,

AD² = BD × DC

|(1)

∠ADB = 90°

In ΔADC & ΔADB,

∠ADB = ∠CDA = 90°

ΔADC~ΔADB by SAS Similarity Rule

⇒ ∠CAD = ∠ABD

⇒ ∠ACD = ∠BAD …(1)

In ΔADC,

∠CAD + ∠ACD + ∠ADC = 180°

⇒ ∠CAD + ∠ACD = 180° – 90° = 90°

From (1),

⇒ ∠CAD + ∠BAD = 90°

⇒ ∠BAC = 90°

Thus, ABC is right angled triangle.

Let there be a ΔABC with the mid points D,E and F of sides AB, AC and BC.

In ΔADE and ΔABC,

D is the mid point of AB and E is the mid point of AC.

By Midpoint theorem,

⇒

⇒

⇒ DE = BF |(1)

Similarly in ΔBFD & ΔBCA,

DF = EC = AE |(2)

Similarly in ΔCFE & ΔCBA

EF = AD = DB |(3)

In ΔADE & ΔBDF,

AD = DB |D is mid point

BF = DE |From (1)

DF = EA |From (2)

Thus, ΔADE & ΔBDF are similar to each other by SSS Similarity Rule.

⇒ ΔADE~ΔBDF

Similarly,

ΔADE~ΔEFC

ΔDBF~ΔEFC

In ΔADE & ΔDEF,

AD = EF |From (3)

DE = DE

EA = DF |From (2)

⇒ ΔADE~ΔDEF

Thus, all the smaller triangles are similar to each other.

In ΔABC and ΔCED,

DC ⊥ BC

⇒ ∠ACB + ∠DCE = 90°

Let ∠ACB = x

⇒ ∠DCE = 90° – x

In ΔABC,

∠ACB + ∠ABC + ∠BAC = 180°

⇒ x + 90° + ∠BAC = 180°

⇒ ∠BAC = 90° – x

In ΔABC and ΔCED,

∠ABC = ∠DEC = 90°

∠ACB = ∠CDE = x

∠BAC = ∠DCE = 90° – x

Thus, the ΔABC and ΔCED are similar by AAA Similarity Rule.

Let a point F on AC such that DF||BX.

By Converse of Mid Point Theorem, as D is mid point of BC, F is the mid point of AC.

⇒ CF = XF

In ΔCFD & ΔCXB,

|By Mid Point Theorem

⇒ BX = 2DF

In ΔAXE & ΔAFD

E is the mid point of AD

and EX | | DF

By Mid Point Theorem,

AX = XF

Hence, proved.

(i) False

The ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, in the given question the ratio of areas should be 16:81.

(ii) False

The ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

In two triangles respectively ΔABC and ΔDEF of then ΔABC~ΔDEF.

(iii) True

(iv) True

In the ΔABO & ΔOCD,

∠AOB = ∠DOC |vertically opp. angles

As AB||CD

∠ABO = ∠DCO |alternate angles

ΔABO~ΔOCD

The sides BC and XY may or may not be coincident.

The ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

In two triangles respectively ΔABC~ΔDEF then

…(1)

In ΔAOF & ΔDOG,

∠AOF = ∠DOG |vertically opp. angles

∠AFO = ∠DGO |both right angles

ΔAOF~ΔDOF by AA Similarity Rule

…(2)

From (1) & (2),

Hence, proved.

Let AD and DB be 2k and 3k.

AB = AB + DB = 2k + 3k = 5k

In ΔADE & ΔABC,

DE||BC

⇒ ∠ADE = ∠ABC

⇒ ∠AED = ∠ACB |alternate angles

ΔADE~ΔABC

In ΔPOB & ΔQOA

∠PBO = ∠QAO = 90°

∠POB = ∠QOA |vertically opposite angles

ΔPBO~ΔQOA by AA Similarity Rule

In ΔBCD & ΔACE

∠C = ∠C |common angle

∠CBD = ∠CAE |given

⇒ ΔBCD~ΔACE

Let ∠ABD = x

⇒ ∠BAD = 90 – x

∠DBC = 90 – x

⇒ DCB = 90 – (90 – x) = x

In ΔADB&ΔBDC,

∠ADB = ∠BDC = 90°

∠ABD = ∠DCB = x

∠BAD = ∠DBC = 90 – x

⇒ ΔADB~ΔBDC by AAA Similarity Rule

Hence, proved.

Let there be a square ABCD with diagonal AC of side ‘a’ .

For equilateral triangle drawn on one side of the square,

In ΔB₁C₁E₁,

Side = a

For the equilateral triangle formed on one diagonal of that square,

In ΔABC,

side = √2a

Thus, the area of an equilateral triangle formed an one side of a square is half of the area of the equilateral triangle formed on one diagonal of that square itself.

Hence, proved.

DE||BC

By Basic Proportionality Theorem(BPT),

Putting the values,

Given,

AB = 6 cm, BD = 8 cm, DC = 6 cm

In ΔABC, AD is the internal angle bisector of angle A.

By internal angle bisector theorem, the internal angle bisector divides the opposite side in the ratio of other two sides.

Putting given values,

Given,

AD = x + 4

DB = x + 3

AE = 2x – 1

EC = x + 1

In ΔADE & ΔABC,

DE||BC

By BPT,

Putting the values,

(x + 1) = (2x – 1)(x + 3)

⇒ x² + 5x + 4 = 2x² + 5x – 3

⇒ x² + 4 = 2x² – 3

⇒ x² = 7

⇒ x = √7

Given,

AB = 3.4 cm, BD = 4 cm, BC = 10 cm

DC = BC – BD = 10 – 4 = 6 cm

In ΔACB, AD is the internal angle bisector of angle A.

By internal angle bisector theorem,

Putting given values,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding medians.

In ΔAOB and ΔCOD,

∠O = ∠O |vertically opposite angle

As AB||CD

∠BAO = ∠OCD |alternate ∠s

∠OBA = ∠ODC |alternate ∠s

Thus, ΔAOB~ΔCOD

For two similar triangles, the ratio of their area is equal to the square of the ratio of their corresponding sides.

For similar triangles, the corresponding angles should be equal. Thus, the order of equal corresponding angles are angles A,B,C & angles F,E,D. So, ΔABC ~ ΔFED

To find: Ratio of Area of Similar triangles

Given: Their corresponding sides AB = 10 cm and DE = 8 cm

For similar triangles ABC & DEF, the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

For both triangles there corresponding sides will be AB and DE

Hence, Ratio of area of triangles ABC anf DEF is 25:16

Given,

AD = 8 cm

AB = 12 cm

DB = AB – AD = 12 – 8 = 4 cm

AE = 12 cm

In ΔADE & ΔABC,

DE||BC

Putting the values of AD,DB,AE,

The tower and the vertical rod are perpendicular to the ground and thus make a triangle like the above figure.

height of tower(AD) = 40 m

height of rod(EF) = 12 m

In ΔOEF & ΔOAD,

EF||AD

⇒ AD = 60 m

Given,

∠B = 70°

∠C = 50°

⇒ ∠A = 180° – (70 + 50)° = 60°

By Converse of Internal Angle Bisector Theorem, if the side opposite to the angle is in the ratio of the other two sides, then the side dividing side opposite to the angle is the internal angle bisector of the triangle.

Thus,

Given,

AD = 6 cm

DB = 9 cm

AE = 8 cm

In ΔADE & ΔABC,

DE||BC

Putting the values of AD,DB,AE,

AC = AE + EC

⇒ AC = 8 + 12 = 20 cm

Given,

AB = 8 cm

BD = 5 cm

DC = 4 cm

By internal angle bisector theorem, the bisector of internal angle of a triangle divides the side opposite to the angle in the ratio of the other two sides.

In ΔABD & ΔADC,

∠A has internal bisector AD

Putting the values of AB,BD and DC,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding heights.