SOLUTION:

Soln. : (i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar.
(A) Their corresponding angles are equal and
(B) Their corresponding sides are proportional.

SOLUTION:

Soln. : (i) (a) Any two circles are similar figures.
(b) Any two squares are similar figures.
(ii) (a) A circle and a triangle are non-similar figures.
(b) An isosceles triangle and a scalene triangle are non-similar figures.

SOLUTION:

Soln.: On observing the given figures, we find that
Their corresponding sides are proportional but their corresponding angles are not equal.
∴ The given figures are not similar.

SOLUTION:

Soln. : (i) Since DE ∥ BC [Given]
∴ Using the Basic proportionality theorem,
We have
Since, AD = 1.5 cm, DB = 3 cm and AE = 1 cm,
∴
By cross-multiplication, we have EC × 1.5 = 1 × 3
⇒
∴ EC = 2 cm
(ii) In ∆ABC, DE ∥ BC
∴ Using the Basic proportionality theorem, we have
∴
⇒ AD × 5.4 = 1.8 × 7.2
⇒
∴ AD = 2.4 cm.

SOLUTION:

Soln. : (i) We have, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

∴
And
∵
∴
⇒ EF is not parallel to QR.
(ii) We have, PE = 4 cm, QE = 4.5 cm
PF = 8 cm and RF = 9 cm

∴
And
Since,
⇒ EF is parallel to QR.
(iii) We have, PE = 0.18 cm, PQ = 1.28 cm
PF = 0.36 cm and PR = 2.56 cm

∴
And
Since,
⇒ EF is parallel to QR.

SOLUTION:

Soln.: In ∆ABC, LM ∥ CB [given]
∴ Using the Basic proportionality theorem, we have
.....(1)
∵
⇒
⇒
Similarly, in ∆ACD ⇒ LN ∥ CD
∴ Using the Basic proportionality theorem, we have
.....(2)
From (1) and (2),

⇒ (Proved)

SOLUTION:

Soln.: In ∆ABC,
∵ DE ∥ AC [given]
∴ .....(1)
[By the basic proportionality theorem]
In ∆ABE,
∴ DF ∥ AE [given]
∴ Using the basic proportionality theorem, we have
.....(2)
From (1) and (2)

⇒

SOLUTION:

Soln.: In ∆PQO,
∵ DE ∥ OQ [given]
∴ Using the Basic proportionality theorem, we have
.....(1)
Similarly, in ∆POR, (in which DF ∥ OR) [given]
∴ Using the Basic proportionality theorem, we have
.....(2)
From (1) and (2),

⇒
Now, in ∆PQR,
∵ E and F are two distinct points on PQ and PR respectively and , ie., E and F dividing the two sides PQ and PR in the same ratio in ∆PQR.
∴ By converse of Basic proportionality theorem, EF ∥ QR.

SOLUTION:

Soln.: In ∆PQR, O is a point and OP, OQ and OR are joined. We have points A, B, and C on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR.
Now, in ∆OPQ,
∵ AB ∥ PQ [Given]
∴ .....(1)
[By the Basic proportionality theorem]
Again, in ∆OPR,
∵ AC ∥ PR [Given]
∴ Using the Basic proportionality theorem, we have
.....(2)
From (1) and (2),

⇒
Now, in ∆OQR,
∵ B is a point on OQ, C is a point on OR
[Proved above]
i.e., B and C divide the sides OQ and OR in the same ratio in ∆OQR.
∴ BC ∥ QR.

SOLUTION:

Soln. : We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE ∥ BC.

∵ DE ∥ BC [given]
∴ Using the Basic proportionality theorem, we get
.....(1)
But D is the mid-point of AB
∴ AD = DB
⇒ .....(2)
From (1) and (2),

⇒ EC = AE
⇒ E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

SOLUTION:

Soln. : We have ∆ABC, in which D and E are the mid-points of sides AB and AC respectively.
∴ AD = DB and AE = EC
From (1) and (2), we have


⇒
⇒ DE ∥ BC (By converse of Basic proportionality theorem).

SOLUTION:

Soln. : We have, a trapezium ABCD such that
AB ∥ DC. The diagonals AC and BD intersect each other at O.

Let us draw OE parallel to either AB or DC.
In ∆ADC
OE ∥ DC [By construction]
∴ Using the Basic proportionality theorem, we get
.....(1)
In ∆ABD
OE ∥ AB [By construction]
∴ Using the Basic proportionality theorem, we get

⇒ .....(2)
From (1) and (2),

⇒

SOLUTION:

Soln. : It is given that
From
(Through O, draw OE ∥ BA)
In ∆ADB
OE ∥ AB [By construction]

∴ Using the Basic proportionality theorem, we get

⇒ .....(1)
Also, .....(2)
From (1) and (2), we have

i.e., the points O and E on the sides AC and AD (of ∆ADC) respectively are in the same ratio.
∴ Using the Basic proportionality theorem, we get
OE ∥ DC and OE ∥ AB
⇒ AB ∥ DC
⇒ ABCD is a trapezium.

SOLUTION:

Soln. : (i) In ∆ABC and ∆PQR,
We have : ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
∴ The corresponding angles are equal.
∴ Using the AAA similarity rule,
∆ABC ~ ∆PQR
(ii) In ∆ABC and ∆QRP,



i.e., ,
i.e., Using the SSS similarity, ∆ABC ~ ∆QRP
(iii) In ∆LMP and ∆DEF,
∵
⇒
⇒ ∆s are not similar.
(iv) In ∆MNL and ∆QPR,
and ∠NML = ∠PQR
∴ Using SAS criteria of similarity, we have
∆MNL ~ ∆QPR.
(v) In ∆ABC and ∆FDE,
∠A = ∠F = 80°
∴ are unknown.
∴ The ∆s cannot said to be similar.
(vi) In ∆DEF and ∆PQR,
∠D = ∠P = 70°
[∵ ∠P = 180° - (80° + 30°) = 180° - 110° = 70°]
∠E = ∠Q = 80°
∠F = ∠R = 30° [( ∠F = 180°]
∴ Using the AAA similarity rule, ∆DFE ~ ∆PRQ.

SOLUTION:

Soln.: We have, ∠BOC = 125° and ∠CDO = 70°
since, ∠DOC + ∠BOC = 180° [Linear pair]
⇒ ∠DOC = 180° - 125° = 55° .....(1)
In ∆DOC,
Using the angle sum property for DODC,
we get
∠DOC + ∠ODC + ∠DCO = 180°
⇒ 55° + 70° + ∠DCO = 180°
⇒ ∠DCO = 180° - 55° - 70° = 55°
Again,
∠DOC = ∠BOA .....(2)
[vertically opposite angles]
and ∠OCD = ∠OAB = 55° .....(3)
(corresponding angles of similar triangles)
Thus, from (1), (2) and (3)
∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

SOLUTION:

Soln.: We have a trapezium ABCD in which AB ∥ DC. The diagonals AC and BD intersect at O.
In ∆OAB and ∆OCD,
∵ AB ∥ DC and AC and BD intersect each other at point O.
∴ ∠OBA = ∠ODC (Alternate angles)
and ∠OAB = ∠OCD (Alternate angles)
∴ Using AA similarity rule, ∆OAB ~ ∆OCD
So, (Ratios of corresponding sides of the similar triangles)
⇒

SOLUTION:

Soln.: In ∆PQR
∵ ∠1 = ∠2 [Given]
∴ PR = QP .....(1)
[( In a ∆, sides opposite to equal angles are equal]
∵ [Given] .....(2)
From (1) and (2),
.....(3)
(By taking reciprocals)
Now, in ∆PQS and ∆TQR,
[From (3)]
and ∠SQP = ∠RQT = ∠1
Now, using SAS similarity rule,
∆PQS ~ ∆TQR.

SOLUTION:

Soln. : In ∆PQR,
T is a point on QR and S is a point on PR such that ∠RTS = ∠P.

Now in ∆RPQ and ∆RTS,
∠RPQ = ∠RTS [Given]
∠PRQ = ∠TRS [Common]
∴ Using AA similarity, we have ∆RPQ ~ ∆RTS.

SOLUTION:

Soln. : We have, ∆ABE ≅ ∆ACD
∴ Their corresponding parts are equal, i.e.,
AB = AC, AE = AD
⇒
∴
⇒ .....(1) [∵ AE = AD]
Now in ∆ADE and ∆ABC, [from (1)]
and ∠DAE = ∠BAC (each = ∠A)
∴ Using the SAS similarity, we have
∆ADE ~ ∆ABC.

SOLUTION:

Soln.: We have a ∆ABC in which altitude AD and CE intersect each other at P.
⇒ ∠D = ∠E = 90° .....(1)
(i) In ∆AEP and ∆CDP,
∠AEP = ∠CDP [From (1)]
∠EPA = ∠DPC [Vertically opp. angles]
∴ Using AA similarity, we get ∆AEP ~ ∆CDP
(ii) In ∆ABD and ∆CBE,
∠ADB = ∠CEB [From (1)]
Also, ∠ABD = ∠CBE [Common]
∴ Using AA similarity, we have
∆ABD ~ ∆CBE
(iii) In ∆AEP and ∆ADB,
∵ ∠AEP = ∠ADB [From (1)]
Also, ∠EAP = ∠DAB [Common]
∴ Using AA similarity, we have
∆AEP ~ ∆ADB
(iv) In ∆PDC and ∆BEC,
∵ ∠PDC = ∠BEC [From (1)]
Also, ∠DCP = ∠ECB [Common]
∴ Using AA similarity, we have
∆PDC ~ ∆BEC

SOLUTION:

Soln.: We have a parallelogram ABCD in which AD is produced to E and BE is joined such that BE intersect CD at F.
Now, in ∆ABE and ∆CFB
∠BAE = ∠FCB [Opp. angles of a ∥ gm are always equal]
∠AEB = ∠CBF [∵ Parallel sides are intersected by the transversal BE]
Now, using AA similarity, we have ∆ABE ~ ∆CFB.

SOLUTION:

Soln. : We have ∆ABC, right angled at B and ∆AMP, right angled at M.
∴ ∠B = ∠M = 90° .....(1)
(i) In ∆ABC and ∆AMP,
∵ ∠ABC = ∠AMP [From (1)]
and ∠BAC = ∠MAP [Common]
∴ Using AA similarity, we get ∆ABC ~ ∆AMP
(ii) ( ∆ABC ~ ∆AMP [As proved above]
∴ Their corresponding sides are proportional.
⇒

SOLUTION:

Soln.: We have, two similar ∆ABC and ∆FEG such that CD and GH are the bisectors of ∠ACB and ∠FGE respectively.
(i) In ∆ACD and ∆FGH,
∠A = ∠F [∵ ∆ABC ~ ∆FEG] .....(1)
since ∆ABC ~ ∆FEG
∴ ∠C = ∠G
⇒ ∠ACD = ∠FGH .....(2)
From (1) and (2),
∆ACD ~ ∆FGH [AA similarity]
∴ Their corresponding sides are proportional,
∴
(ii) In ∆DCB and ∆HGE,
∠B = ∠E [∵ ∆ABC ~ ∆FEG] .....(1)
Again, ∆ABC ~ ∆FEG ⇒ ∠ACB = ∠FGE
∴
⇒ ∠DCB = ∠HGE .....(2)
From (1) and (2),
∆DCB ~ ∆HGE [AA similarity]
(iii) In ∆DCA and ∆HGF,
∵ ∆ABC ~ ∆FEG ⇒ ∠CAB = ∠GFE
⇒ ∠CAD = ∠GFH ⇒ ∠DAC = ∠HFG .....(1)
Also, ∆ABC ~ ∆FEG ⇒ ∠ACB = ∠FGE
∴
⇒ ∠DCA = ∠HGF .....(2)
From (1) and (2), we have
∆DCA ~ ∆HGF [AA similarity]

SOLUTION:

Soln.: We have an isosceles ∆ABC in which AB = AC.
In ∆ABD and ∆ECF,
AB = AC [Given]
⇒ Angles opposite to them are equal
∴ ∠ACB = ∠ABC
⇒ ∠ECF = ∠ABD .....(1)
Again, AD ⊥ BC and EF ⊥ AC
⇒ ∠ADB = ∠EFC = 90° .....(2)
From (1) and (2), we have
∆ABD ~ ∆ECF [AA similarity]

SOLUTION:

Soln.: We have ∆ABC and ∆PQR in which AD and PM are medians corresponding to sides BC and QR
respectively such that
⇒
⇒
∴ Using SSS similarity, we have
∆ABD ~ ∆PQM
∴ Their corresponding angles are equal.
⇒ ∠ABD = ∠PQM ⇒ ∠ABC = ∠PQR
Now, in ∆ABC and ∆PQR, .....(1)
Also, ∠ABC = ∠PQR .....(2)
∴ Using SAS similarity from (1) and (2), we get ∆ABC ~ ∆PQR.

SOLUTION:

Soln.: We have a ∆ABC and a point D on its side BC such that ∠ADC = ∠BAC.
In ∆BAC and ∆ADC,
∵ ∠BAC = ∠ADC [Given]
and ∠BCA = ∠DCA [Common]
∴ Using AA similarity, we have ∆BAC ~ ∆ADC.
∴ Their corresponding sides are proportional.
⇒
⇒ CA × CA = CB × CD
⇒ CA² = CB × CD

SOLUTION:

Soln. : Given : ∆ABC and ∆PQR in which AD and PM are medians.
Also, .....(1)
To prove : ∆ABC ~ ∆PQR
Construction : Produce AD to E and PM to N such that AD = DE and PM = MN. Join BE, CE, QN and RN.

Proof : Quadrilaterals ABEC and PQNR are parallelograms, since their diagonals bisect each other at point D and M respectively.
⇒ BE = AC and QN = PR
⇒
[By (1)]
i.e., .....(2)
From (1),
⇒ .....(3)
From (2) and (3), we have

⇒ ∆ABE ~ ∆PQN ⇒ ∠1 = ∠3 .....(4)
Similarly, we can prove
∆ACE ~ ∆PRN ⇒ ∠2 = ∠4 .....(5)
From (4) and (5)
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P .....(6)
Now, in ∆ABC and ∆PQR, we have
[From (1)]
and ∠A = ∠P [From (6)]
∴ ∆ABC ~ ∆PQR (SAS similarity criterion)

SOLUTION:

Soln. : Let AB = 6m be the pole and BC = 4 m be its shadow (in right ∆ABC), whereas DE and EF denote the tower and its shadow respectively.
∵ EF = Length of the shadow of the tower = 28 m
and DE = h = Height of the tower

In ∆ABC and ∆DEF, we have ∠B = ∠E = 90°
∠A = ∠D [∵ Angular elevation of the
sun at the same time is equal]
∴ Using AA criteria of similarity, we have
∆ABC ~ ∆DEF
∴ Their sides are proportional i.e.,
⇒
Thus, the required height of the tower is 42 m.

SOLUTION:

Soln. : We have ∆ABC ~ ∆PQR such that AD and PM are the medians corresponding to the sides BC and QR respectively.

∵ ∆ABC ~ ∆PQR
And the corresponding sides of similar triangles are proportional.
∴ .....(1)
∵ Corresponding angles are also equal in two similar triangles.
∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R .....(2)
Since, AD and PM are medians.
∴ BC = 2BD and QR = 2QM
∴ From (1), .....(3)
And ∠B = ∠Q ⇒ ∠ABD = ∠PQM .....(4)
∴ From (3) and (4), we have
∆ABD ~ ∆PQM
∴ Their corresponding sides are proportional.
⇒