Similarity — Exercise 15(A)

In the given diagram OC = 1.5 × OA, then OB is equal to :

1. 3 × OD

2. 1.5 × OD

3. $\dfrac{2}{3}$ × OD

4. OD

Answer

From figure,

In △ OAB and △ OCD,

⇒ ∠AOB = ∠COD (Vertically opposite angles are equal)

⇒ ∠OAB = ∠OCD (Alternate angles are equal)

⇒ ∠OBA = ∠ODC (Alternate angles are equal)

∴ △ OAB ~ △ OCD (By A.A.A. postulate)

Given,

⇒ OC = 1.5 × OA

⇒ $\dfrac{OA}{OC} = \dfrac{1}{1.5} = \dfrac{2}{3}$.

We know that,

Corresponding sides of similar triangle are in proportion.

$\therefore \dfrac{OB}{OD} = \dfrac{OA}{OC} \\[1em] \Rightarrow \dfrac{OB}{OD} = \dfrac{2}{3} \\[1em] \Rightarrow OB = \dfrac{2}{3} \times OD.$

Hence, Option 3 is the correct option.

Are the given triangles similar?

1. Yes

2. No

3. None of these

Answer

As we know that sum of all angles of a triangle is 180°.

In first triangle,

⇒ 65° + 55° + third angle = 180°

⇒ 120° + third angle = 180°

⇒ third angle = 180° - 120°

⇒ third angle = 60°

In second triangle,

⇒ 65° + 70° + third angle = 180°

⇒ 135° + third angle = 180°

⇒ third angle = 180° - 135°

⇒ third angle = 45°

Since, the in the given triangles neither corresponding angles are equal nor their corresponding sides are in proportion.

Hence, option 2 is the correct option.

In the given figure, OA = 5, OB = 6, OC = 3 and OD = 10, then

1. Δ AOB ∼ Δ AOD

2. Δ AOB ∼ Δ BOC

3. Δ BOC ∼ Δ COD

4. Δ AOD ∼ Δ COB

Answer

Given,

OA = 5, OB = 6, OC = 3 and OD = 10.

$\Rightarrow \dfrac{OA}{OC} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{OD}{OB} = \dfrac{10}{6} = \dfrac{5}{3} \\[1em] \therefore \dfrac{OA}{OC} = \dfrac{OD}{OB}$

From figure,

∠AOD = ∠BOC (Vertically opposite angles are equal)

∴ △ AOD ∼ △ COB (By S.A.S. axiom)

Hence, option 4 is the correct option.

In the given figure, the value of x is:

1. 15

2. 30

3. 36

4. 40

Answer

In Δ AOB and Δ DOC

⇒ ∠AOB = ∠DOC (Vertically opposite angles are equal)

⇒ ∠BAO = ∠DCO (Alternate angles are equal)

⇒ ∠ABO = ∠CDO (Alternate angles are equal)

∴ Δ AOB ∼ Δ COD (By AAA postulate)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AO}{CO} = \dfrac{OB}{OD}\\[1em] \Rightarrow \dfrac{18}{12} = \dfrac{x}{20}\\[1em] \Rightarrow x = \dfrac{18 \times 20}{12}\\[1em] \Rightarrow x = \dfrac{360}{12}\\[1em] \Rightarrow x = 30$

Hence, option 2 is the correct option.

In triangle ABC, ∠BAC = 90° and AD is perpendicular to side BC. Triangle ABD is similar to triangle CBA by :

1. SAS

2. ASA

3. AAA

4. RHS

Answer

In △ CBA and △ ABD,

⇒ ∠CAB = ∠ADB (Both equal to 90°)

⇒ ∠CBA = ∠DBA (Common angles)

Since, two angles of two triangles are equal so third angle of both the triangle will also be equal.

⇒ ∠ACB = ∠DAB.

∴ △ CBA ~ △ ABD (By A.A.A. postulate)

Hence, Option 3 is the correct option.

If AE = 10 cm, BD = 8 cm and BC = 10 cm, then AB is equal to :

1. 5 cm

2. 25 cm

3. 12.5 cm

4. 2.5 cm

Answer

From figure,

In △ ACE and △ BCD,

⇒ ∠CAE = ∠CBD (Both equal to 90°)

⇒ ∠ACE = ∠BCD (Common angles)

Since, two angles of two triangles are equal so third angle of both the triangle will also be equal.

⇒ ∠CEA = ∠CDB.

∴ △ ACE ~ △ BCD (By A.A.A. postulate)

We know that,

Corresponding sides of similar triangle are in proportion.

$\therefore \dfrac{AC}{BC} = \dfrac{AE}{BD} \\[1em] \Rightarrow \dfrac{AC}{10} = \dfrac{10}{8} \\[1em] \Rightarrow AC = \dfrac{100}{8} = 12.5 \text{ cm}.$

From figure,

AB = AC - BC = 12.5 - 10 = 2.5 cm.

Hence, Option 4 is the correct option.

In the given figure :

1. △ ABE ~ △ ADE

2. △ ADE ~ △ ABC

3. △ ADE ~ △ BAC

4. △ ADE ~ △ CAB

Answer

From figure,

⇒ ∠BAE = x (let)

⇒ ∠DAB = ∠EAC = y (let)

⇒ ∠AED = ∠ACB = z (let)

⇒ ∠DAE = ∠DAB + ∠BAE = y + x

⇒ ∠BAC = ∠BAE + ∠EAC = x + y

⇒ ∠DAE = ∠BAC = x + y

In △ ABC and △ ADE,

⇒ ∠DAE = ∠BAC (Proved above)

⇒ ∠AED = ∠ACB (Both equal to z)

∴ △ ADE ~ △ ABC (By A.A. axiom)

Hence, Option 2 is the correct option.

The value of x is :

1. 2

2. 3

3. 1

4. none of these

Answer

In Δ AOB and Δ DOC

⇒ ∠AOB = ∠DOC (Vertically opposite angles are equal)

⇒ ∠BAO = ∠DCO (Corresponding angles are equal)

⇒ ∠ABO = ∠CDO (Corresponding angles since AB ∥ CD and BD is a transversal)

∴ Δ AOB ∼ Δ COD (By AAA postulates)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{OB}{OD} = \dfrac{AB}{CD}\\[1em] \Rightarrow \dfrac{3 }{4} = \dfrac{x + 1}{x + 2}\\[1em] \Rightarrow 3(x + 2) = 4(x + 1)\\[1em] \Rightarrow 3x + 6 = 4x + 4\\[1em] \Rightarrow 4x - 3x = 6 - 4\\[1em] \Rightarrow x = 2\\[1em]$

Hence, option 1 is the correct option.

In the given figure, OD = 2 × OB, OC = 2 × OA and CD = 2 × AB then △ AOB ~ △ COD by :

1. AA

2. SS

3. SAS

4. SSS

Answer

Given,

⇒ OD = 2 × OB

⇒ $\dfrac{OD}{OB}$ = 2 .......(1)

⇒ OC = 2 × OA

⇒ $\dfrac{OC}{OA}$ = 2 .......(2)

⇒ CD = 2 × AB

⇒ $\dfrac{CD}{AB}$ = 2 .......(3)

From equation (1), (2) and (3), we get :

⇒ $\dfrac{OD}{OB} = \dfrac{OC}{OA} = \dfrac{CD}{AB}$

∴ △ AOB ~ △ COD (By S.S.S. postulate)

Hence, Option 4 is the correct option.

Are the two congruent triangles always similar?

1. yes

2. no

3. none of these

Answer

Two congruent triangles have equal corresponding angles, and equal corresponding sides (same lengths).

Two similar triangles only require equal corresponding angles and proportional corresponding sides (not necessarily equal).

Hence, option 1 is the correct option.

If $\dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AC}}{\text{DF}}$, then :

1. Δ ABC ∼ Δ EDF

2. Δ ABC ∼ Δ DEF

3. Δ ABC ∼ Δ FDE

4. none of these

Answer

In Δ ABC and Δ DEF,

$\dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AC}}{\text{DF}}$

From above equation, we can conclude

AB corresponds to DE, BC corresponds to EF and AC corresponds to DF.

∴ Δ ABC ∼ Δ DEF (By SSS postulates)

Hence, option 2 is the correct option.

If Δ ABC ∼ Δ DEF, then

1. $\dfrac{\text{AB}}{\text{EF}} = \dfrac{\text{AC}}{\text{DF}}$

2. $\dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{BC}}{\text{DF}}$

3. $\dfrac{\text{AC}}{\text{DF}} = \dfrac{\text{AB}}{\text{DE}}$

4. $\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AC}}{\text{DE}}$

Answer

$\dfrac{\text{AC}}{\text{DF}} = \dfrac{\text{AB}}{\text{DE}}$

Reason

In Δ ABC ∼ Δ DEF, we can conclude :

AB corresponds to DE, BC corresponds to EF and AC corresponds to DF.

We know that,

Corresponding sides of similar triangles are equal.

$\therefore \dfrac{\text{AB}}{\text{DE}} = \dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AC}}{\text{DF}} \\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{AC}{DF}.$

Hence, option 3 is the correct option.

In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Also, AL and DM are medians. Prove that $\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$.

Answer

In Δ ABC and Δ DEF,

⇒ ∠A = ∠D (Given)

⇒ ∠B = ∠E (Given)

⇒ ∠C = ∠F (Given)

∴ Δ ABC ∼ Δ DEF (By AAA postulate)

Since, AL and DM are medians of triangles ABC and DEF respectively.

∴ BL = $\dfrac{1}{2}BC$ and EM = $\dfrac{1}{2}EF$

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{BC}{EF}....................(1)\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{\dfrac{1}{2}BC}{\dfrac{1}{2}EF}\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{BL}{EM} .........................(2)$

In Δ ABL and Δ DEM,

⇒ ∠B = ∠E (Given)

⇒ $\dfrac{AB}{DE} = \dfrac{BL}{EM}$ [From equation (2)]

∴ Δ ABL ∼ Δ DEM (By SAS postulates)

As, corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AL}{DM}$ .........(3)

From equation (1) and (3), we get :

$\Rightarrow \dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$

Hence, proved that $\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{AL}}{\text{DM}}$

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.

Show that angle BAC = 90°.

Answer

Triangle ABC is shown in the figure below:

Given :

AD² = BD × DC

⇒ $\dfrac{AD}{DC} = \dfrac{BD}{AD}$

∠ADB = ∠ADC [Both = 90°]

∴ △DBA ~ △DAC (By SAS).

Since, triangles are similar they will be equiangular.

∴ ∠1 = ∠C and ∠2 = ∠B

⇒ ∠1 + ∠2 = ∠B + ∠C

⇒ ∠A = ∠B + ∠C

By angle sum property :

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°.

From figure,

⇒ ∠BAC = ∠A = 90°.

Hence, proved that ∠BAC = 90°.

In the given figure, Δ ABC and Δ DEF are similar, BM and EN are their medians. If Δ ABC is similar to Δ DEF, prove that :

(i) Δ AMB ∼ Δ DNE

(ii) Δ CMB ∼ Δ FNE

(iii) $\dfrac{BM}{EN} = \dfrac{AC}{DF}$

Answer

(i) Given,

Since, BM and EN are medians of triangles ABC and DEF respectively.

∴ AM = $\dfrac{1}{2}AC$ and DN = $\dfrac{1}{2}DF$

Given,

Δ ABC ∼ Δ DEF

∴ ∠A = ∠D (Corresponding angles of similar triangles are equal)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{DF} ....................(1)\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{\dfrac{1}{2}AC}{\dfrac{1}{2}DF}\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{AM}{DN} .........................(2)$

In Δ AMB and Δ DNE,

⇒ ∠A = ∠D (Proved above)

⇒ $\dfrac{AB}{DE} = \dfrac{AM}{DN}$ [From equation (2)]

∴ Δ AMB ∼ Δ DNE (By SAS postulate)

Hence, proved that Δ AMB ∼ Δ DNE.

(ii) Given,

Since, BM and EN are medians of triangles ABC and DEF respectively.

∴ MC = $\dfrac{1}{2}AC$ and NF = $\dfrac{1}{2}DF$

Given,

Δ ABC ∼ Δ DEF

⇒ ∠C = ∠F (Corresponding angles of similar triangles are equal)

Since, corresponding sides of similar triangles are proportional.

$\therefore \dfrac{BC}{EF} = \dfrac{AC}{DF} ....................(3)\\[1em] \Rightarrow \dfrac{BC}{EF} = \dfrac{\dfrac{1}{2}AC}{\dfrac{1}{2}DF}\\[1em] \Rightarrow \dfrac{BC}{EF} = \dfrac{MC}{NF} .........................(4)$

In Δ CMB and Δ FNE,

⇒ ∠C = ∠F (Proved above)

⇒ $\dfrac{BC}{EF} = \dfrac{MC}{NF}$ [From equation (4)]

∴ Δ CMB ∼ Δ FNE (By SAS postulate)

Hence, proved that Δ CMB ∼ Δ FNE.

(iii) Given,

Δ ABC ∼ Δ DEF

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{DF}$ .......(5)

Δ AMB ∼ Δ DNE

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{BM}{EN} = \dfrac{AB}{DE}$ ......(6)

From equation (5) and (6), we get :

$\Rightarrow \dfrac{BM}{EN} = \dfrac{AC}{DF}$

Hence, proved that $\dfrac{BM}{EN} = \dfrac{AC}{DF}$.

In the given figure, Δ ABC is isosceles and AP x BQ = AC², prove that Δ ACP ∼ Δ BCQ.

Answer

Given,

⇒ AP x BQ = AC²

⇒ AP x BQ = AC x AC

⇒ AP x BQ = AC x BC (From figure, AC = BC)

⇒ $\dfrac{AP}{BC} = \dfrac{AC}{BQ}$ ........................(1)

Since, AC = BC

⇒ ∠CAB = ∠CBA ...............(2) [Angles opposite to equal sides are equal]

⇒ 180° - ∠CAB = 180° - ∠CBA

⇒ ∠CAP = ∠CBQ ...................(3)

In Δ ACP and Δ BCQ,

⇒ ∠CAP = ∠CBQ [From equation (3)]

⇒ $\dfrac{AP}{BC} = \dfrac{AC}{BQ}$ [From equation (1)]

∴ Δ ACP ∼ Δ BCQ (By SAS postulates)

Hence, proved that Δ ACP ∼ Δ BCQ.

In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:

(i) ∆APC and ∆BPD are similar.

(ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

Answer

(i) In ∆APC and ∆BPD, we have

∠APC = ∠BPD [Vertically opposite angles are equal]

∠ACP = ∠BDP [Alternate angles (as, AC || BD) are equal]

∴ ∆APC ~ ∆BPD [By A.A.]

Hence, proved that ∆APC ~ ∆BPD.

(ii) In similar triangles the ratio of corresponding sides are equal.

$\dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PA}}{\text{PB}}$ ..............(1) and,

$\dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PC}}{\text{PD}}$ ...............(2)

Solving (1) we get,

$\Rightarrow \dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PA}}{\text{PB}} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{\text{PA}}{3.2} \\[1em] \Rightarrow \text{PA} = \dfrac{3.6}{2.4} \times 3.2 \\[1em] \Rightarrow \text{PA} = \dfrac{3}{2} \times 3.2 \\[1em] \Rightarrow \text{PA} = 4.8 \text{ cm}.$

Solving (2) we get,

$\Rightarrow \dfrac{\text{AC}}{\text{BD}} = \dfrac{\text{PC}}{\text{PD}} \\[1em] \Rightarrow \dfrac{3.6}{2.4} = \dfrac{\text{PC}}{4} \\[1em] \Rightarrow \text{PC} = \dfrac{3.6}{2.4} \times 4 \\[1em] \Rightarrow \text{PC} = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow \text{PC} = 6 \text{ cm}.$

Hence, PA = 4.8 cm and PC = 6 cm.

In the given figure, AB || DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.

Answer

In ΔDOQ and ΔBOP,

As AB || DC so, PB || DQ and BD is transversal.

∴ ∠QDO = ∠PBO [Alternate angles]

∠DOQ = ∠BOP [Vertically opposite angles are equal]

Hence, ∆DOQ ~ ∆BOP [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{DO}{BO} = \dfrac{DQ}{BP} \\[1em] \Rightarrow \dfrac{DO}{6} = \dfrac{8}{BP} \\[1em] \Rightarrow BP \times DO = 8 \times 6 = 48 \text{ cm}^2. \\[1em]$

Hence, BP x DO = 48 cm².

Given: ∠GHE = ∠DFE = 90°, DH = 8, DF = 12, DG = 3x – 1 and DE = 4x + 2.

Find: the lengths of segments DG and DE.

Answer

In ΔDHG and ΔDFE,

⇒ ∠GHD = ∠DFE = 90°

⇒ ∠D = ∠D [Common]

Thus, ∆DHG ~ ∆DFE [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{DH}{DF} = \dfrac{DG}{DE} \\[1em] \Rightarrow \dfrac{8}{12} = \dfrac{(3x – 1)}{(4x + 2)} \\[1em] \Rightarrow 8(4x + 2) = 12(3x – 1) \\[1em] \Rightarrow 32x + 16 = 36x - 12 \\[1em] \Rightarrow 36x - 32x = 12 + 16 \\[1em] \Rightarrow 4x = 28 \\[1em] \Rightarrow x = 7.$

DG = 3x - 1 = 3(7) - 1 = 21 - 1 = 20,

DE = 4x + 2 = 4(7) + 2 = 28 + 2 = 30.

Hence, DG = 20 and DE = 30.

In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that :

(i) PQ² = PM × PR

(ii) QR² = PR × MR

(iii) PQ² + QR² = PR²

Answer

△PQR is shown in the figure below:

(i) In △PQR and △PMQ,

⇒ ∠PMQ = ∠PQR [Both = 90°]

⇒ ∠QPM = ∠RPQ [Common]

∴ △PQR ~ △PMQ [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\dfrac{PQ}{PR} = \dfrac{PM}{PQ}$

⇒ PQ² = PM × PR

Hence, proved that PQ² = PM × PR.

(ii) In △QRM and △PRQ,

⇒ ∠QMR = ∠PQR [Both = 90°]

⇒ ∠QRM = ∠QRP [Common]

∴ △QRM ~ △PRQ [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\dfrac{QR}{PR} = \dfrac{MR}{QR}$

⇒ QR² = PR × MR

Hence, proved that QR² = PR × MR.

(iii) Adding equations from (i) and (ii) we get,

⇒ PQ² + QR² = PM × PR + PR × MR .........(1)

⇒ PQ² + QR² = PR(PM + MR)

From figure,

PM + MR = PR

⇒ PQ² + QR² = PR².

Hence, proved that PQ² + QR² = PR².

In △ABC, ∠B = 90° and BD ⊥ AC.

(i) If CD = 10 cm and BD = 8 cm; find AD.

(ii) If AC = 18 cm and AD = 6 cm; find BD.

(iii) If AC = 9 cm and AB = 7 cm; find AD.

Answer

△ABC is shown in the figure below:

(i) In △CDB,

⇒ ∠1 + ∠2 + ∠3 = 180° (Sum of angles of triangle = 180°)

⇒ ∠1 + ∠3 + 90° = 180°

⇒ ∠1 + ∠3 = 90° ..........(1)

From figure,

⇒ ∠B = 90°

⇒ ∠3 + ∠4 = 90° ..........(2)

From (1) and (2) we get,

⇒ ∠1 + ∠3 = ∠3 + ∠4

⇒ ∠1 = ∠4.

From figure,

⇒ ∠2 = ∠5 [Both = 90°]

∴ △CDB ~ △BDA [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\dfrac{CD}{BD} = \dfrac{BD}{AD}$ ..........(3)

Substituting values we get :

$\Rightarrow \dfrac{10}{8} = \dfrac{8}{AD} \\[1em] \Rightarrow AD = \dfrac{8^2}{10} \\[1em] \Rightarrow AD = \dfrac{64}{10} = 6.4 \text{ cm}.$

Hence, AD = 6.4 cm.

(ii) From figure,

CD = AC - AD = 18 - 6 = 12 cm.

Substituting values in (3) we get :

$\Rightarrow \dfrac{12}{BD} = \dfrac{BD}{6} \\[1em] \Rightarrow BD^2 = 12 \times 6 \\[1em] \Rightarrow BD^2 = 72 \\[1em] \Rightarrow BD = \sqrt{72} = 6\sqrt{2} = 8.5\text{ cm}.$

Hence, BD = 8.5 cm.

(iii) In △ABC and △ABD,

⇒ ∠ADB = ∠ABC [Both = 90°]

⇒ ∠ABD = ∠ACB [As ∠1 = ∠4]

∴ △ABC ~ △ABD [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{AD}{AB} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{AD}{7} = \dfrac{7}{9} \\[1em] \Rightarrow AD = 7 \times \dfrac{7}{9} = \dfrac{49}{9} = 5\dfrac{4}{9} \text{ cm}.$

Hence, AD = $5\dfrac{4}{9}$ cm.

In the right-angled triangle QPR. PM is an altitude.

Given that QR = 8 cm and MQ = 3.5 cm, calculate the value of PR.

Answer

In △PQR and △MPR,

∠QPR = ∠PMR = 90°

∠PRQ = ∠PRM (Common)

∴ △PQR ~ △MPR [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{QR}{PR} = \dfrac{PR}{MR} \\[1em] \Rightarrow PR^2 = QR \times MR \\[1em] \Rightarrow PR^2 = 8 \times (8 - 3.5) \\[1em] \Rightarrow PR^2 = 8 \times 4.5 = 36 \\[1em] \Rightarrow PR = \sqrt{36} = 6 \text{ cm}.$

Hence, PR = 6 cm.

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:

(i) ΔAPB is similar to ΔCPD.

(ii) PA x PD = PB x PC.

Answer

Trapezium ABCD is shown in the figure below:

(i) In ∆APB and ∆CPD, we have

∠APB = ∠CPD [Vertically opposite angles]

∠ABP = ∠CDP [Alternate angles (as AB||DC) are equal]

∴ ∆APB ~ ∆CPD [By A.A.]

Hence, proved that ∆APB ~ ∆CPD.

(ii) We know that,

In similar triangles the ratio of corresponding sides are equal.

$\therefore \dfrac{\text{PA}}{\text{PC}} = \dfrac{\text{PB}}{\text{PD}} \\[1em] \Rightarrow \text{PA} \times \text{PD} = \text{PB} \times \text{PC}.$

Hence, proved that PA x PD = PB x PC.

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:

(i) DP : PL = DC : BL.

(ii) DL : DP = AL : DC.

Answer

Parallelogram ABCD is shown in the figure below:

(i) In ∆DPC and ∆BPL, we have

∠DPC = ∠BPL [Vertically opposite angles area equal]

∠DCP = ∠PBL [Alternate angles (as AB || DC) are equal]

∴ ∆DPC ~ ∆BPL [By A.A.]

Since, corresponding sides of similar triangles are proportional.

$\therefore \dfrac{DP}{PL} = \dfrac{DC}{BL}$.

i.e., DP : PL = DC : BL.

Hence, proved that DP : PL = DC : BL.

(ii) From part (i) we get,

$\Rightarrow \dfrac{DP}{PL} = \dfrac{DC}{BL} \\[1em] \Rightarrow \dfrac{PL}{DP} = \dfrac{BL}{DC} \\[1em] \Rightarrow \dfrac{PL}{DP} + 1 = \dfrac{BL}{DC} + 1 \\[1em] \Rightarrow \dfrac{PL + DP}{DP} = \dfrac{BL + DC}{DC} \\[1em] \Rightarrow \text{Since, AB = DC as ABCD is a || gm} \\[1em] \Rightarrow \dfrac{PL + DP}{DP} = \dfrac{BL + AB}{DC} \\[1em] \Rightarrow \dfrac{DL}{DP} = \dfrac{AL}{DC}.$

Hence, proved that DL : DP = AL : DC.

In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :

(i) CB : BA = CP : PA

(ii) AB x BC = BP x CA

Answer

ΔABC is shown in the figure below:

(i) Let ∠ACB = x, so ∠ABC = 2x.

Since, BP is the bisector of ∠ABC.

So, ∠ABP = ∠PBC = x.

By angle bisector theorem,

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

$\therefore \dfrac{CB}{BA} = \dfrac{CP}{PA}$

i.e. CB : BA = CP : PA.

Hence, proved that CB : BA = CP : PA.

(ii) From figure,

∠APB = ∠PBC + ∠PCB = 2x. [Exterior angle is equal to the sum of opposite two interior angles].

∴ ∠APB = ∠ABC

∠BCP = ∠ABP [Both = x]

∴ △ABC ~ △APB [By A.A.]

Since corresponding sides of similar triangles are proportional we have,

⇒ $\dfrac{CA}{AB} = \dfrac{BC}{BP}$

⇒ AB x BC = BP x CA

Hence, proved that AB x BC = BP x CA.

In the given figure, DE || BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.

(i) Write all possible pairs of similar triangles.

(ii) Find the lengths of ME and DM.

Answer

(i) In ΔAME and ΔANC,

⇒ ∠AME = ∠ANC [Since DE || BC so, ME || NC and AN is transversal]

⇒ ∠MAE = ∠NAC [Common angle]

∴ ∆AME ~ ∆ANC [By AA]

In ΔADM and ΔABN,

⇒ ∠ADM = ∠ABN [Since DE || BC so, DM || BN and AB is transversal]

⇒ ∠DAM = ∠BAN [Common angle]

∴ ∆ADM ~ ∆ABN [By AA]

In ΔADE and ΔABC,

⇒ ∠ADE = ∠ABC [Since DE || BC and AB is transversal]

⇒ ∠AED = ∠ACB [Since DE || BC and AC is transversal]

∴ ∆ADE ~ ∆ABC [By AA]

Hence, ∆ADM ~ ∆ABN, ∆AME ~ ∆ANC and ∆ADE ~ ∆ABC.

(ii) Since, ∆AME ~ ∆ANC

We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{ME}{NC} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{ME}{6} = \dfrac{15}{15 + 9} \\[1em] \Rightarrow ME = \dfrac{15}{24} \times 6 \\[1em] \Rightarrow ME = 3.75 \text{ cm}.$

Since, ∆ADE ~ ∆ABC [Proved above]

We know that,

Corresponding sides of similar triangles are proportional.

$\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{15}{24}$ ......... (1)

Also, ∆ADM ~ ∆ABN [Proved above]

$\therefore \dfrac{DM}{BN} = \dfrac{AD}{AB} = \dfrac{15}{24} ........[From (1)] \\[1em] \therefore \dfrac{DM}{BN} = \dfrac{15}{24} \\[1em] DM = \dfrac{15}{24} \times BN = \dfrac{15}{24} \times 24 = 15 \text{ cm}.$

Hence, ME = 3.75 cm and DM = 15 cm.

In the given figure, AD = AE and AD² = BD x EC. Prove that: triangles ABD and CAE are similar.

Answer

From figure,

⇒ ∠ADE = ∠AED [Angles opposite to equal sides of a triangle are equal].

⇒ 180° - ∠ADE = 180° - ∠AED

⇒ ∠ADB = ∠AEC

Given,

⇒ AD² = BD x EC

⇒ AD x AD = BD x EC

⇒ AD x AE = BD x EC

⇒ $\dfrac{AD}{EC} = \dfrac{BD}{AE}$

∴ △ABD ~ △CAE [By SAS]

Hence, proved that △ABD ~ △CAE.

State, true or false:

(i) Two similar polygons are necessarily congruent.

(ii) Two congruent polygons are necessarily similar.

(iii) All equiangular triangles are similar.

(iv) All isosceles triangles are similar.

(v) Two isosceles-right triangles are similar.

(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.

(vii) The diagonals of a trapezium, divide each other into proportional segments.

Answer

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA² = CB x CD.

Answer

In ΔADC and ΔBAC,

⇒ ∠ADC = ∠BAC [Given]

⇒ ∠ACD = ∠ACB [Common]

∴ ∆ADC ~ ∆BAC [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\dfrac{CA}{CB} = \dfrac{CD}{CA}$

⇒ CA² = CB x CD.

Hence, proved that CA² = CB x CD.

In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.

Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

(i) Prove that : ∆ABC ~ ∆AMP.

(ii) Find AB and BC.

Answer

(i) In ∆ABC and ∆AMP, we have

⇒ ∠BAC = ∠PAM [Common]

⇒ ∠ABC = ∠PMA [Each = 90°]

∴ ∆ABC ~ ∆AMP [By AA]

Hence, proved that, ∆ABC ~ ∆AMP.

(ii) In right angle triangle AMP,

By pythagoras theorem,

⇒ AP² = AM² + MP²

⇒ AM² = AP² - MP²

⇒ AM² = 15² - 12²

⇒ AM² = 225 - 144

⇒ AM² = 81

⇒ AM = $\sqrt{81}$ = 9 cm.

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{AB}{AM} = \dfrac{AC}{AP} \\[1em] \Rightarrow \dfrac{AB}{9} = \dfrac{10}{15} \\[1em] \Rightarrow AB = 9 \times \dfrac{10}{15} \\[1em] \Rightarrow AB = 6 \text{ cm}.$

Also,

$\Rightarrow \dfrac{BC}{MP} = \dfrac{AC}{AP} \\[1em] \Rightarrow \dfrac{BC}{12} = \dfrac{10}{15} \\[1em] \Rightarrow BC = 12 \times \dfrac{10}{15} \\[1em] \Rightarrow BC = 8 \text{ cm}.$

Hence, AB = 6 cm and BC = 8 cm.

In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

Answer

In △RLQ and △PLN,

⇒ ∠RLQ = ∠PLN [Vertically opposite angles are equal]

⇒ ∠LRQ = ∠LPN [Alternate angles are equal]

∴ △RLQ ~ △PLN [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{RL}{LP} = \dfrac{RQ}{PN} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{10}{PN} \\[1em] \Rightarrow PN = \dfrac{30}{2} = 15 \text{ cm}.$

In △RLM and △PLQ,

⇒ ∠RLM = ∠PLQ [Vertically opposite angles are equal]

⇒ ∠LRM = ∠LPQ [Alternate angles are equal]

∴ △RLM ~ △PLQ [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{RM}{PQ} = \dfrac{RL}{LP} \\[1em] \Rightarrow \dfrac{RM}{16} = \dfrac{2}{3} \\[1em] \Rightarrow RM = \dfrac{32}{3} = 10\dfrac{2}{3} \text{ cm}.$

Hence, PN = 15 cm and RM = $10 \dfrac{2}{3}$ cm.

In the given figure, AB || EF || DC; AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm.

(i) Name the three pairs of similar triangles.

(ii) Find the length of EC and EF.

Answer

(i) The three pairs of similar triangle are :

In △BEF and △BDC

⇒ ∠FBE = ∠CBD [Common angle]

⇒ ∠BFE = ∠BCD [Corresponding angles are equal]

∴ △BEF ~ △BDC [By AA]

In △CEF and △CAB

⇒ ∠FCE = ∠BCA [Common angle]

⇒ ∠CFE = ∠CBA [Corresponding angles are equal]

∴ △CEF ~ △CAB [By AA]

In △ABE and △CDE

⇒ ∠AEB = ∠CED [Vertically opposite angles are equal]

⇒ ∠BAE = ∠ECD [Alternate angles are equal]

∴ △ABE ~ △CDE [By AA]

(ii) Since, △ABE and △CDE are similar,

$\therefore \dfrac{AB}{CD} = \dfrac{AE}{CE} \\[1em] \Rightarrow \dfrac{67.5}{40.5} = \dfrac{52.5}{CE} \\[1em] \Rightarrow CE = \dfrac{40.5 \times 52.5}{67.5} \\[1em] \Rightarrow CE = 31.5 \text{ cm}.$

Since, △CEF and △CAB are similar,

$\therefore \dfrac{CE}{CA} = \dfrac{EF}{AB} \\[1em] \Rightarrow \dfrac{31.5}{CE + AE} = \dfrac{EF}{67.5} \\[1em] \Rightarrow \dfrac{31.5}{31.5 + 52.5} = \dfrac{EF}{67.5} \\[1em] \Rightarrow EF = \dfrac{31.5 \times 67.5}{84} \\[1em] \Rightarrow EF = \dfrac{2126.25}{84} \\[1em] \Rightarrow EF = 25\dfrac{5}{16}\text{ cm}.$

Hence, CE = 31.5 cm and EF = $25\dfrac{5}{16}$ cm.

In the given figure, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.

(i) Calculate the ratio PQ : AC, giving reason for your answer.

(ii) In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°. Given QS = 6 cm, calculate the length of AR.

Answer

(i) Given,

AP : PB = 4 : 3

Let AP = 4x and PB = 3x.

From figure,

AB = AP + PB = 4x + 3x = 7x.

PB : AB = 3x : 7x = 3 : 7.

In △PQB and △ACB,

QP || AC

∠BPQ = ∠BAC (Corresponding angles are equal)

∠BQP = ∠BCA (Corresponding angles are equal)

△PQB ~ △ACB.

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{PQ}{AC} = \dfrac{PB}{AB} = \dfrac{3}{7}$.

Hence, PQ : AC = 3 : 7.

(ii) In △ARC and △QSP,

∠ARC = ∠QSP = 90°

∠ACR = ∠SPQ (Alternate angles are equal)

∴ △ARC ~ △QSP [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{AR}{QS} = \dfrac{AC}{PQ} \\[1em] \Rightarrow \dfrac{AR}{6} = \dfrac{7}{3} \\[1em] \Rightarrow AR = 6 \times \dfrac{7}{3} \\[1em] \Rightarrow AR = 14 \text{ cm}.$

Hence, AR = 14 cm.

In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that :

(i) △EGD ~ △CGB and

(ii) BG = 2GD from (i) above.

Answer

(i) Since, BD and CE are medians.

So, E is mid-point of AB and D is mid-point of AC.

By converse of mid-point theorem,

ED || BC and ED = $\dfrac{1}{2}$BC

$\dfrac{ED}{BC} = \dfrac{1}{2}$ .....(1)

In △EGD and △CGB,

∠EGD = ∠BGC (Vertically opposite angles are equal)

∠DEG = ∠GCB (Alternate angles are equal)

∴ △EGD ~ △CGB [By AA].

Hence, proved that △EGD ~ △CGB.

(ii) Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{BG}{GD} = \dfrac{ED}{BC} \\[1em] \Rightarrow \dfrac{BG}{GD} = \dfrac{1}{2} \space .....\text{ (From 1)} \\[1em] \Rightarrow BG = 2GD.$

Hence, proved that BG = 2GD.