Hint In △ ABC,AC= BC ⇒ \(\angle A=\angle B\)

⇒ \(180^{\circ}-\angle A=180^{\circ}-\angle B\)

⇒ \(\angle P A C=\angle Q B C\)

Also, we have AP x BQ = AC²

⇒ \(\frac{A P}{A C}=\frac{A C}{B Q} \Rightarrow \frac{A P}{B C}=\frac{A C}{B Q}\) [∵ AC= BC]

⇒ △ PAC~ △ CBQ [by SAS similarity criterion]

Hint Prove that △APB-△AQC [by AA similarly criterion]

(i) Hint Use AA similarity criterion to prove

△ ABC ~ △ ADE

(ii) Hint \(\frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E} .\)

Ans. \(A E=\frac{15}{13} \mathrm{~cm} \text { and } D E=\frac{36}{13} \mathrm{~cm}\)

Hint Given the two quadrilaterals are not similar because the corresponding sides of the quadrilaterals are not in the same ratio. Ans. No

Given △ NSQ ≅ △ MTR and \(\angle 1=\angle 2\)

To prove △PTS ~ △PRQ

Proof Since, △ NSQ ≅ △ MTR

∴ SQ = TR ...(i)

Also, \(\angle 1=\angle 2\)

⇒ PT = PS ...(ii)

[∵ sides opposite to equal angles are also equal]

From Eqs. (i) and (ii),

\(\frac{P S}{S Q}=\frac{P T}{T R}\)

⇒ \(S T \| Q R\)

[by converse of basic proportionality theorem]

∴ \(\angle 1=\angle P Q R\)

and \(\angle 2=\angle P R Q\) [corresponding angles]

In △ PTS and △ PRQ,

\(\angle T P S=\angle Q P R\) [common angle]

\(\angle 1=\angle P Q R\) [proved above]

and \(\angle 2=\angle P R Q\)

∴ △ PTS - △ PRQ [by AA similarity criterion]

Hence proved

Let AB be a pole of length6 m, BC=4 m be the length of its shadow and0be the angle which sun makes with ground.

Also, at the same time, another tower casts shadow NM of length 28 m. Let PM =h m be the height of the tower and θ be the angle which sun ray makes with ground.

In △ ABC and △ PMN,

\(\angle A B C=\angle P M N\) [each 90°]

and \(\angle A C B=\angle P N M\) [each θ]

∴ △ ABC~△ PMN [by AA similarity criterion]

Now, \(\frac{A B}{P M}=\frac{B C}{M N}\)

⇒ \(\frac{A B}{B C}=\frac{P M}{M N}\)

⇒ \(\frac{6}{4}=\frac{h}{28}\) [∵ AB = 6m, BC= 4m and MN=28m]

⇒ \(h=\frac{6 \times 28}{4}=42 \mathrm{~m}\)

Hence, the height of the tower is 42 m.

Draw △ ARC such that Dis a point on BC and join AD.

In △ ABC and △ DAC we have

\(\angle B A C=\angle A D C\) [given]

and \(\angle A C B=\angle D C A\) [common angle]

∴ △ ABC - △ DAC [by AA similarity criterion]

⇒ \(\frac{A C}{D C}=\frac{C B}{C A}\)

[∵ corresponding sides of two similar triangles are proportional]

Or \(\frac{C A}{C D}=\frac{C B}{C A}\)

⇒ CA x CA = CB × CD

⇒ CA² = CB x CD Hence proved.

Given, △ ABC and △ PQR in which AD and PM are their medians, respectively.

So, \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\)

To prove △ ABC ~ △ PQR

Proof We have, \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\) [given]

⇒ \(\frac{A B}{P Q}=\frac{1 / 2 B C}{1 / 2 Q R}=\frac{A D}{P M}\) [multiplying numerator and denominator of middle tern by 1/2)

⇒ \(\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\) [∵ medians AD and PM bisect the lines BC and QR,
respectively, i.e. BD=1/2BCand QM = 1/2QR]

⇒ △ ADB- △ PMQ [by SSS similarity criterion]

∴ \(\angle B=\angle Q\) ...(i)

[∵ corresponding angles of similar triangles are equal]

In △ ABC and △ PQR,

\(\frac{A B}{P Q}=\frac{B C}{Q R}\) [given]

and \(\angle B=\angle Q\) [from Eq. (i)]

∴ △ ABC~ △ PQR [by SAS similarity criterion] Hence proved.

Given, △ ABC is an isosceles triangle with AB = AC. Also

we have AD ⊥ BC and EF⊥AC.

To prove △ ABD ~ △ ECF

Proof In △ ABC,AB = AC

∴ \(\angle B=\angle C\) [∵ angle opposite to equal sides are equal]

Now, consider △ ABD and △ ECE : In this we have

\(\angle A B D=\angle E C F\) \(\text { ( } \because \angle B=\angle C \text { proved above }\)]

and \(\angle A D B=\angle E F C\) [each 90°]

∴ △ ABD~ △ECF (by AA similarly criterion) Hence proved

Given, AD and CE are altitudes which intersect each other at the point P

(i) In △ AEP and △ CDP

\(\angle A E P=\angle C D P\) [each 90°]

and \(\angle A P E=\angle C P D\) [vertically opposite angles]

∴ △ AEP ~ △ CDP [by AA similarity criterion]

(ii) In △ ABD and △ CBE,

\(\angle A D B=\angle C E B\) [each 90°]

and \(\angle A B D=\angle C B E\) [common angle]

∴ △ ABD ~ △ CBE [by AA similarity criterion]

(iii) In △ AEP and △ ADB,

\(\angle A E P=\angle A D B\) [each 90°]

and \(\angle P A E=\angle B A D\) [common angle]

∴ △ AEP ~ △ ADB [by AA similarity criterion]

(iv) In △ PDC and △ BEC,

\(\angle P D C=\angle B E C\) [each 90°]

and \(\angle P C D=\angle B C E\) [common angle]

∴ △ PDC~ △ BEC [by AA similarity criterion]

Given, △ ABE ≅ △ ACD

⇒ AB =AC and AE=AD [by CPCT]

⇒ \(\frac{A B}{A C}=1\)

and \(\frac{A D}{A E}=1\)

⇒ \(\frac{A B}{A C}=\frac{A D}{A E}\) ...(i)

In △ ADE and △ ABC, we have

\(\frac{A D}{A E}=\frac{A B}{A C}\)

⇒ \(\frac{A D}{A B}=\frac{A E}{A C}\)

and \(\angle D A E=\angle B A C\) [common angle]

∴ △ ADE ~ △ ABC [by SAS similarity criterion] Hence proved.

Draw △ PQR, such that S and T are points on sides PR and QR, respectively. We join the points S and T.

From the above figure, we have \(\triangle R P Q \text { and } \triangle R T S\) in which

\(\angle R P Q=\angle R T S\) [given]

⇒ \(\angle P R Q=\angle T R S\) [common angle]

∴ △ RPQ ~ △ RTS [by AA similarity criterion] Hence proved.

In △ PQR \(\angle 1=\angle 2\) [given]

⇒ PR=PQ [since sides opposite to equal angles of a triangle are also equal]

Given that \(\frac{Q R}{Q S}=\frac{Q T}{P R} \Rightarrow \frac{Q R}{Q S}=\frac{Q T}{P Q}\) [∵ PQ= PR, proved above]

⇒ \(\frac{Q S}{Q R}=\frac{P Q}{Q T}\) [on taking reciprocal of the terms] ...(i)

In △ PQS and △ TQR, we have

\(\angle P Q S=\angle T Q R\) [common]

and \(\frac{Q S}{Q R}=\frac{Q P}{Q T}\) [from Eq. (i))

∴ △ PQS ~ △ TQR [by SAS similarity criterion]

Hence proved.

From the given figure, it is clear that DOB is a straight line.

∴ \(\angle D O C+\angle C O B=180^{\circ}\) [by linear pair axiom]

⇒ \(\angle D O C+125^{\circ}=180^{\circ}\) [given, \(\left.\angle C O B=125^{\circ}\right]\)

⇒ \(\angle D O C=180^{\circ}-125^{\circ}=55^{\circ}\)

In △ DOC, \(\angle D C O+\angle C D O+\angle D O C=180^{\circ}\) [by angle sum property of a triangle]

⇒ \(\angle D C O+70^{\circ}+55^{\circ}=180^{\circ}\) \(\left[\because \angle C D O=70^{\circ} \text { and } \angle D O C=55^{\circ}\right]\)

⇒ \(\angle D C O+125^{\circ}=180^{\circ}\)

⇒ \(\angle D C O=180^{\circ}-125^{\circ}=55^{\circ}\) ...(i)

Also, △ ODC ~ △ OBA [given]

⇒ \(\angle O A B=\angle O C D=\angle D C O\)

⇒ \(\angle O A B=55^{\circ}\) [from Eq. (i)]

Hence, \(\angle D O C=55^{\circ}, \angle D C O=55^{\circ} \text { and } \angle O A B=55^{\circ}\)

(i) Yes; The pair of triangles is similar.

In △ ABC and △ PQR,

\(\angle A=\angle P=60^{\circ} \text { and } \angle C=\angle R=40^{\circ}\) [given]

Then, \(\angle B=\angle Q=180^{\circ}-60^{\circ}-40^{\circ}=80^{\circ}\)

∴ △ ABC~△PQR [by AAA similarity criterion]

(ii) Yes, the pair of triangles is similar.

In △ MNL and △ QPR,

\(\angle N M L=\angle P Q R=70^{\circ}\)

\(\frac{M N}{P Q}=\frac{2.5}{5}=\frac{1}{2}\)

and \(\frac{M L}{Q R}=\frac{5}{10}=\frac{1}{2}\)

Then, \(\frac{M N}{P Q}=\frac{M L}{Q R}\)

∴ △ MNL~ △ QPR [by SAS similarity criterion]

(iii) No, because in △ ABC, \(\angle B A C\) is given but the measure of included side AC is not given.

Let AB be the lamp-post, CD be the girl

and D be the position of girl after 4 s.

Again, let DE = x m be the length of shadow of the girl.

Given, CD = 90cm= 0.9 m and AB =3.6 m

and speed of the girl = 1.2m/s

∴ Distance of the girl from lamp-post after 4 s,

BD = 1.2 x 4= 4.8 m [∵ distance = speed x time]

In △ ABE and △ CDE,

\(\angle B=\angle D\) [each 90°]

\(\angle E=\angle E\) [common angle]

∴ \(\triangle A B E \sim \triangle C D E\) [by AA similarity criterion]

⇒ \(\frac{B E}{D E}=\frac{A B}{C D}\) ...(i)

[since corresponding sides of similar triangles are proportional)

On substituting all the values in Eq. (i),we get

\(\frac{4.8+x}{x}=\frac{3.6}{0.9}\) [∵ BE= BD + DE= 4.8 + x]

⇒ \(\frac{4.8+x}{x}=4\)

⇒ 48 +x =4 ⇒ 3x = 4.8

⇒ \(x=\frac{4.8}{3}=16 \mathrm{~m}\)

Hence, the length of her shadow after 4 s is 1.6 m.

(i) △ ABC~ △ PQR [given]

⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\) ..(i)

and \(\angle A=\angle P, \angle B=\angle Q \text { and } \angle C=\angle R\) ...(ii)

In △ AMC and △ PNR,

2AM = AB and 2PN= PQ

(∵ CM and RN are medians)

⇒ \(\frac{2 A M}{2 P N}=\frac{C A}{R P}\) [from Eq. (i)

⇒ \(\frac{A M}{P N}=\frac{C A}{R P}\)

Also, \(\angle M A C=\angle N P R\) [from Eq. (ii)]

∴ \(\triangle M A C \sim \triangle N P R\) [SAS similarity criterion] Hence proved.

(ii) We have, △ AMC~ △ PNR

⇒ \(\frac{A M}{P N}=\frac{A C}{P R}=\frac{C M}{R N}\)

[since triangles are similar, hence corresponding sides will be proportional]

∴ \(\frac{C M}{R N}=\frac{A C}{P R}\)

⇒ \(\frac{C M}{R N}=\frac{A B}{P Q}\) [from Eq. (i)] Hence proved.

In △ ABC and △ RQP,

\(\frac{A B}{R Q}=\frac{3.8}{7.6}=\frac{1}{2},\)

\(\frac{B C}{Q P}=\frac{6}{12}=\frac{1}{2}\)

and \(\frac{C A}{P R}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}\)

Since, \(\frac{A B}{R Q}=\frac{B C}{Q P}=\frac{C A}{P R}=\frac{1}{2}\)

∴ △ ABC - △ RQP [by SSS similarity criterion]

So, \(\angle P=\angle C\) ...(i)

[since corresponding angles of similar triangles are equal]

In △ ABC

\(\angle A+\angle B+\angle C=180^{\circ}\) [by angle sum property of a triangle]

⇒ 80°+ 60° + \(\angle C=180^{\circ}\) [∵ \(\angle A=\) 80° and \(\angle B =\) 60°, given]

⇒ \(\angle C=180^{\circ}-140^{\circ}=40^{\circ}\)

From Eq. (i) \(\angle P=40^{\circ}\)

(i) Examples of similar figures:
(a) Two squares of different sizes.
(b) Two regular hexagons of different sizes.
(ii) Examples of non-similar figures:
(a) Two isosceles triangles with different angle measures.

(b) Two rhombus with different angle measures.

Hint △ ACD ~ △ ABC [by AA similarity criterion]

∴ \(\frac{A C}{A B}=\frac{A D}{A C}\)

[∵ corresponding sides of similar triangles are proportional]

Hint △ CED ~ △ CBA [by AA similarity criterion]

⇒ \(\frac{C E}{C B}=\frac{E D}{B A}\)

In the given figure, we have

\(\angle B C A=90^{\circ}\)

and \(\angle A D E=90^{\circ}\)

Since, \(\angle A i\)s common.

So, it is clear that △ BAC is similar to △ EAD by AA similarity criterion.

Hint If \(\angle B=\angle P,\) then

△ ABC~△ QPR [by SAS similarity criterion]

Given, △ OCA ~△ OBD

Also, given \(\angle A O B=110^{\circ} \text { and } \angle B D O=58^{\circ}\)

∵ \(\angle A O B=\angle C O D\) [vertically opposite angles]

⇒ 110° = \(\angle C O D\)

Now, \(\angle A O C=180^{\circ}-\angle C O D\) [linear pair]

= 180° - 110° = 70°

\(\angle A O C=\angle B O D=70^{\circ}\)

In △ BOD,

\(\angle B O D+\angle B D O+\angle D B O=180^{\circ}\)

⇒ 70°+ 58° + \(\angle D B O=180^{\circ}\)

⇒ \(\angle D B O=180^{\circ}-128^{\circ}=52^{\circ}\)

∵ △ OCA ~ △ OBD

∴ \(\angle D B O=\angle O A C=52^{\circ}\)

(i) XY || PQ

∴ \(\angle X A P=\angle A P Q=45^{\circ}=\angle A P D\)

(ii) \(\angle Y A Q=\angle A Q D=30^{\circ}\) [alternate angles]

(iii) (a) In right angled △ ADP,

tan 45°\(=\frac{A D}{P D}\)

⇒ \(1=\frac{100}{P D} \Rightarrow P D=100 \mathrm{~m}\)

Hence, the Boat P is at a distance of 100 m from the light house.

Or

(b) In right angled △ ADQ,

\(\tan 30^{\circ}=\frac{A D}{D Q}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{A D}{D Q}\)

⇒ \(D Q=100 \sqrt{3} \mathrm{~m}\)

Hence, the Boat Q is at a distance of 100\(\sqrt{3}\) from the light house

Let CD = h m be the height of the tower, BE = 20 m be the height of Vijay's house and GF be the height of Ajay's house.

△ ACD ~ △ ABE

∴ \(\frac{A C}{A B}=\frac{C D}{E B}\)

⇒ \(\frac{50}{10}=\frac{h}{20}\)

⇒ h= 100 m

(i) Given, AB = 12 m, let AC = x

In similar △ ABE and △ ACD,

\(\frac{A B}{A C}=\frac{B E}{C D}\)

⇒ \(\frac{12}{x}=\frac{20}{100}\)

⇒ \(x=\frac{12 \times 100}{20}=12 \times 5=60 \mathrm{~m}\)

(ii) Let the height of Ajay's house be GF= h₁

Since, △ HFG ~ △ HCD

∴ \(\frac{H F}{H C}=\frac{F G}{C D}\)

⇒ \(\frac{20}{50}=\frac{h_1}{100} \Rightarrow h_1=\frac{20 \times 100}{50}=40 \mathrm{~m}\)

(iii) (a) Given, HC = 40 m

Let length of the shadow of Ajay's house be HF = l m

Since, △ HFG ~ △ HCD

∴ \(\frac{H F}{H C}=\frac{F G}{C D}\)

⇒ \(\frac{l}{40}=\frac{40}{100} \Rightarrow l=\frac{40 \times 40}{100}=16 \mathrm{~m}\)

Or

(b) Given, AC = 40 m

Let length of the shadow of Vijay's house be AB =l m

Since, △ ABE ~ △ ACD

∴ \(\frac{A B}{A C}=\frac{E B}{C D}\)

⇒ \(\frac{l}{40}=\frac{20}{100} \Rightarrow l=\frac{20 \times 40}{100}=8 \mathrm{~m}\)