Fill in the blanks using the correct word given in brackets :

(i) All circles are .....(congruent, similar)

(ii) All squares are .....(similar, congruent)

(iii) All .......... triangles are similar (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (A) their corresponding angles are ....... and (B) their corresponding sides are .....(equal, proportional).

**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.

Give two different examples of pair of

(i) similar figures

(ii) non-similar figures

**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.

State whether the following quadrilaterals are similar or not :

**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.

In figures (i) and (ii), *DE* ∥ *BC*. Find *EC* in (i) and *AD* in (ii).

**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.

*E* and *F* are points on the sides *PQ* and *PR* respectively of a ∆*PQR*. For each of the following cases, state whether *EF* ∥ *QR*;

(i) *PE* = 3.9 cm, *EQ* = 3 cm, *PF* = 3.6 cm and *FR* = 2.4 cm

(ii) *PE* = 4 cm, *QE* = 4.5 cm, *PF* = 8 cm and *RF* = 9 cm

(iii) *PQ* = 1.28 cm, *PR* = 2.56 cm, *PE* = 0.18 cm and *PF* = 0.36 cm

**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*.

In the figure, if *LM* ∥ *CB* and *LN* ∥ *CD*, prove that

**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)

In the figure, *DE* ∥ *AC* and *DF* ∥ *AE*. Prove that

**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)

⇒

In the figure, *DE* ∥ *OQ* and *DF* ∥ *OR*. Show that *EF* ∥ *QR*.

**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*.

In the figure *A*, *B* and *C* are points on *OP*, *OQ* and *OR* respectively such that *AB* ∥ *PQ* and
*AC* ∥ *PR*. Show that *BC* ∥ *QR*.

**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*.

Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

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.

Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to 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).

*ABCD* is a trapezium in which *AB* ∥ *DC* and its diagonals intersect each other at the point *O*. Show that

**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),

⇒

The diagonals of a quadrilateral *ABCD* intersect each other at the point *O* such that

Show that *ABCD* is a trapezium.

**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.

State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

**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*.

2. In the figure, ∆*ODC* ~ ∆*OBA*, ∠*BOC* = 125° and ∠*CDO* = 70°. Find ∠*DOC*, ∠*DCO* and ∠*OAB*.

**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 D*ODC*,

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°.

Diagonals *AC* and *BD* of a trapezium *ABCD* with *AB* ∥ *DC* intersect each other at the point *O*. Using a similarity criterion for two triangles, show that

**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)

⇒

In the figure, and ∠1 = ∠2. Show that ∆*PQS* ~ ∆*TQR*.

**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*.

*S* and *T* are points on sides *PR* and *QR* of ∆*PQR* such that ∠*P* = ∠*RTS*. Show that

∆*RPQ* ~ ∆*RTS*.

**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*.

In the figure, if ∆*ABE * ≅ ∆*ACD*, show that ∆*ADE* ~ ∆*ABC*.

**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*.

In the figure, altitudes *AD* and *CE* of ∆*ABC* intersect each other at the point *P*. Show that:

(i) ∆*AEP* ~ ∆*CDP*

(ii) ∆*ABD* ~ ∆*CBE*

(iii) ∆*AEP* ~ ∆*ADB*

(iv) ∆*PDC* ~ ∆*BEC*

**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*

*E* is a point on the side *AD* produced of a parallelogram *ABCD* and *BE* intersects *CD* at *F*. Show that ∆*ABE* ~ ∆*CFB*.

**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*.

In the figure, *ABC* and *AMP* are two right triangles, right angled at *B* and *M* respectively.

Prove that : (i) ∆*ABC* ~ ∆*AMP*

(ii)

**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.

⇒

*CD* and *GH* are respectively the bisectors of ∠*ACB* and ∠*EGF* such that *D* and *H* lie on sides *AB* and *FE* of ∆*ABC* and ∆*EFG* respectively. If ∆*ABC* ~ ∆*FEG*, show that :

(i)

(ii) ∆*DCB* ~ ∆*HGE*

(iii) ∆*DCA* ~ ∆*HGF*

**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]

In the figure, *E* is a point on side *CB* produced of an isosceles triangle *ABC* with *AB* = *AC*. If *AD* ⊥ *BC* and *EF* ⊥ *AC*, prove that ∆*ABD* ~ ∆*ECF*.

**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]

Side *AB* and *BC* and median *AD* of a triangle *ABC* are respectively proportional to sides *PQ* and *QR* and median *PM* of ∆*PQR* (see figure). Show that ∆*ABC* ~ ∆*PQR*.

**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*.

*D* is a point on the side *BC* of a triangle *ABC* such that ∠*ADC* = ∠*BAC*. Show that *CA*2 = *CB*⋅*CD*.

**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*

⇒ *CA*^{2} = *CB* × *CD*

Sides *AB* and *AC* and median *AD* of a triangle *ABC* are respectively proportional to sides *PQ* and *PR* and median *PM* of another triangle *PQR*. Show that ∆*ABC* ~ ∆*PQR*.

**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)

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

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.

16. If *AD* and *PM* are medians of triangles *ABC* and *PQR*, respectively where, ∆*ABC* ~ ∆*PQR*, prove that

**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* = 2*BD* and *QR* = 2*QM*

∴ From (1), .....(3)

And ∠*B* = ∠*Q* ⇒ ∠*ABD* = ∠*PQM* .....(4)

∴ From (3) and (4), we have

∆*ABD* ~ ∆*PQM*

∴ Their corresponding sides are proportional.

⇒