Similarity

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

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

We know that,

Corresponding sides of similar triangle are in proportion.

$$\begin{aligned}\therefore \frac{OB}{OD} = \frac{OA}{OC} \\[1em] \Rightarrow \frac{OB}{OD} = \frac{2}{3} \\[1em] \Rightarrow OB = \frac{2}{3} \times OD.\end{aligned}$$

Hence, Option 3 is the correct option.

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.

Given,

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

$$\begin{aligned}\Rightarrow \frac{OA}{OC} = \frac{5}{3} \\[1em] \Rightarrow \frac{OD}{OB} = \frac{10}{6} = \frac{5}{3} \\[1em] \therefore \frac{OA}{OC} = \frac{OD}{OB}\end{aligned}$$

From figure,

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

∴ △ AOD ∼ △ COB (By S.A.S. axiom)

Hence, option 4 is the correct option.

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.

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

Hence, option 2 is the correct option.

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.

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.

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

From figure,

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

Hence, Option 4 is the correct option.

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.

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.

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

Hence, option 1 is the correct option.

Given,

⇒ OD = 2 × OB

⇒ $\frac{OD}{OB}$ = 2 .......(1)

⇒ OC = 2 × OA

⇒ $\frac{OC}{OA}$ = 2 .......(2)

⇒ CD = 2 × AB

⇒ $\frac{CD}{AB}$ = 2 .......(3)

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

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

∴ △ AOB ~ △ COD (By S.S.S. postulate)

Hence, Option 4 is the correct option.

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.

In Δ ABC and Δ DEF,

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

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.

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

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.

$$\begin{aligned}\therefore \frac{\text{AB}}{\text{DE}} = \frac{\text{BC}}{\text{EF}} = \frac{\text{AC}}{\text{DF}} \\[1em] \Rightarrow \frac{AB}{DE} = \frac{AC}{DF}.\end{aligned}$$

Hence, option 3 is the correct option.

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 = $\frac{1}{2}BC$ and EM = $\frac{1}{2}EF$

We know that,

Corresponding sides of similar triangles are proportional.

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

In Δ ABL and Δ DEM,

⇒ ∠B = ∠E (Given)

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

∴ Δ ABL ∼ Δ DEM (By SAS postulates)

As, corresponding sides of similar triangles are proportional.

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

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

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

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

Triangle ABC is shown in the figure below:

Given :

AD² = BD × DC

⇒ $\frac{AD}{DC} = \frac{BD}{AD}$

∠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°.

(i) Given,

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

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

Given,

Δ ABC ∼ Δ DEF

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

We know that,

Corresponding sides of similar triangles are proportional.

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

In Δ AMB and Δ DNE,

⇒ ∠A = ∠D (Proved above)

⇒ $\frac{AB}{DE} = \frac{AM}{DN}$ [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 = $\frac{1}{2}AC$ and NF = $\frac{1}{2}DF$

Given,

Δ ABC ∼ Δ DEF

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

Since, corresponding sides of similar triangles are proportional.

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

In Δ CMB and Δ FNE,

⇒ ∠C = ∠F (Proved above)

⇒ $\frac{BC}{EF} = \frac{MC}{NF}$ [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.

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

Δ AMB ∼ Δ DNE

We know that,

Corresponding sides of similar triangles are proportional.

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

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

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

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

Given,

⇒ AP x BQ = AC²

⇒ AP x BQ = AC x AC

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

⇒ $\frac{AP}{BC} = \frac{AC}{BQ}$ ........................(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)]

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

∴ Δ ACP ∼ Δ BCQ (By SAS postulates)

Hence, proved that Δ ACP ∼ Δ BCQ.

(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.

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

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

Solving (1) we get,

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

Solving (2) we get,

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

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

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 :

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

Hence, BP x DO = 48 cm².

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 :

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

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

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

Hence, DG = 20 and DE = 30.

△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 :

⇒ $\frac{PQ}{PR} = \frac{PM}{PQ}$

⇒ 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 :

⇒ $\frac{QR}{PR} = \frac{MR}{QR}$

⇒ 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.

△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 :

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

Substituting values we get :

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

Hence, AD = 6.4 cm.

(ii) From figure,

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

Substituting values in (3) we get :

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

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 :

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

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

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.

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

Hence, PR = 6 cm.

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.

$$\begin{aligned}\therefore \frac{\text{PA}}{\text{PC}} = \frac{\text{PB}}{\text{PD}} \\[1em] \Rightarrow \text{PA} \times \text{PD} = \text{PB} \times \text{PC}.\end{aligned}$$

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

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.

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

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

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

(ii) From part (i) we get,

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

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

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.

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

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,

⇒ $\frac{CA}{AB} = \frac{BC}{BP}$

⇒ AB x BC = BP x CA

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

(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.

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

Since, ∆ADE ~ ∆ABC [Proved above]

We know that,

Corresponding sides of similar triangles are proportional.

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

Also, ∆ADM ~ ∆ABN [Proved above]

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

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

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

⇒ $\frac{AD}{EC} = \frac{BD}{AE}$

∴ △ABD ~ △CAE [By SAS]

Hence, proved that △ABD ~ △CAE.

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

In ΔADC and ΔBAC,

⇒ ∠ADC = ∠BAC [Given]

⇒ ∠ACD = ∠ACB [Common]

∴ ∆ADC ~ ∆BAC [By AA]

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

⇒ $\frac{CA}{CB} = \frac{CD}{CA}$

⇒ CA² = CB x CD.

Hence, proved that CA² = CB x CD.

(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 :

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

Also,

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

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

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 :

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

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 :

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

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

(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,

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

Since, △CEF and △CAB are similar,

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

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

(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.

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

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.

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

Hence, AR = 14 cm.

(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 = $\frac{1}{2}$BC

$\frac{ED}{BC} = \frac{1}{2}$ .....(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.

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

Hence, proved that BG = 2GD.

From figure,

In △ DAE and △ BAC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠EDA = ∠CBA (Corresponding angles are equal)

∴ △ DAE ~ △ BAC (By A.A. postulate)

Given,

AD : BD = 3 : 5

Let AD = 3x and BD = 5x.

From figure,

AB = AD + BD = 3x + 5x = 8x.

We know that,

Corresponding sides of similar triangle are in proportion.

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{3x}{8x} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{3}{8} \\[1em] \Rightarrow DE : BC = 3 : 8.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

⇒ AD = AE = x (let)

⇒ BD = CE = y (let)

From figure,

⇒ AB = AD + BD = x + y

⇒ AC = AE + EC = x + y

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC

⇒ $\frac{AD}{AB} = \frac{AE}{AC} = \frac{x}{x + y}$

∴ △ ADE ~ △ ABC (By S.A.S. postulate)

Hence, Option 4 is the correct option.

From figure,

In △ AOB and △ COD,

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

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

∴ △ AOB ~ △ COD (By A.A. postulate)

We know that,

Corresponding sides of similar triangle are in proportion.

$$\begin{aligned}\therefore \frac{OB}{OD} = \frac{AB}{CD} \\[1em] \Rightarrow \frac{8}{OD} = \frac{10}{8} \\[1em] \Rightarrow OD = \frac{8 \times 8}{10} \\[1em] \Rightarrow OD = \frac{64}{10} = 6.4 \text{ cm}\end{aligned}$$

Hence, Option 3 is the correct option.

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

Given,

AE : EC = 2 : 3

Let AE = 2x and EC = 3x.

From figure,

AC = AE + EC = 2x + 3x = 5x.

We know that,

Corresponding sides of similar triangle are in proportion.

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AE}{AC} \\[1em] \Rightarrow \frac{DE}{20} = \frac{2x}{5x} \\[1em] \Rightarrow DE = 20 \times \frac{2}{5} \\[1em] \Rightarrow DE = \frac{40}{5} = 8\text{ cm}.\end{aligned}$$

Hence, Option 3 is the correct option.

We know that,

Two congruent triangles are equal in area as well are similar.

Hence, Option 2 is the correct option.

(i) Given,

$\frac{AD}{DB} = \frac{3}{5}$ and DE || BC.

By basic proportionality theorem we have :

A line drawn parallel to one side of triangle divides the other two sides proportionally.

$\therefore \frac{AE}{EC} = \frac{AD}{DB} = \frac{3}{5}.$

Hence, AE : EC = 3 : 5.

(ii) Given,

$\frac{AD}{DB} = \frac{3}{5}$

Let AD = 3x and DB = 5x.

AB = AD + DB = 3x + 5x = 8x.

$\frac{AD}{AB} = \frac{3x}{8x} = \frac{3}{8}$ = 3 : 8.

Hence, AD : AB = 3 : 8.

(iii) Given,

$$\begin{aligned}\Rightarrow \frac{AE}{EC} = \frac{3}{5} \\[1em] \Rightarrow \frac{EC}{AE} = \frac{5}{3} \\[1em] \Rightarrow \frac{EC}{AE} + 1 = \frac{5}{3} + 1 \\[1em] \Rightarrow \frac{EC + AE}{AE} = \frac{5 + 3}{3} \\[1em] \Rightarrow \frac{AC}{AE} = \frac{8}{3} \\[1em] \Rightarrow \frac{AE}{AC} = \frac{3}{8}.\end{aligned}$$

Hence, AE : AC = 3 : 8.

(iv) In ∆ADE and ∆ABC,

∠ADE = ∠ABC [As DE || BC, Corresponding angles are equal.]

∠A = ∠A [Common angles]

Hence, ∆ADE ~ ∆ABC by AA criterion for similarity.

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

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{3}{8} = \frac{2.4}{BC} \\[1em] \Rightarrow BC = \frac{8 \times 2.4}{3} \\[1em] \Rightarrow BC = 6.4 \text{ cm}.\end{aligned}$$

Hence, BC = 6.4 cm.

(v) Since, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{3}{8} = \frac{DE}{4.8} \\[1em] \Rightarrow DE = \frac{3 \times 4.8}{8} \\[1em] \Rightarrow DE = 1.8 \text{ cm}.\end{aligned}$$

Hence, DE = 1.8 cm.

(i) Given PQ || AB,

By basic proportionality theorem :

$$\begin{aligned}\therefore \frac{CP}{PA} = \frac{CQ}{BQ} \\[1em] \Rightarrow \frac{CP}{PA} = \frac{4.8}{3.6} \\[1em] \Rightarrow \frac{CP}{PA} = \frac{4}{3}.\end{aligned}$$

Hence, ratio = 4 : 3.

(ii) In ∆CPQ and ∆CAB,

∠CPQ = ∠CAB [As PQ || AB, corresponding angles are equal.]

∠PCQ = ∠ACB [Common angle]

∴ ∆CPQ ~ ∆CAB [By AA].

From figure,

CB = CQ + QB = 4.8 + 3.6 = 8.4

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

$$\begin{aligned}\therefore \frac{PQ}{AB} = \frac{CQ}{CB} \\[1em] \Rightarrow \frac{PQ}{6.3} = \frac{4.8}{8.4} \\[1em] \Rightarrow PQ = 6.3 \times \frac{4}{7} = 3.6 \text{ cm}.\end{aligned}$$

Hence, PQ = 3.6 cm

(iii) As, ∆CPQ ~ ∆CAB.

We have,

$$\begin{aligned}\frac{CP}{AC} = \frac{CQ}{CB} \\[1em] \frac{CP}{AC} = \frac{4.8}{8.4} \\[1em] \frac{CP}{AC} = \frac{4}{7}.\end{aligned}$$

So, if AC is 7 parts and CP is 4 parts, then PA is 3 parts.

Given, AP = x

or, 3 parts = x

⇒ 1 part = $\frac{x}{3}$

⇒ 7 parts = $\frac{7x}{3}$.

Hence, AC = $\frac{7x}{3}$.

Let AP = x cm.

From figure,

PB = AB - AP = 9 - x cm.

QC = AC - AQ = 6 - 4.2 = 1.8 cm.

Given PQ || AB,

By basic proportionality theorem :

$$\begin{aligned}\Rightarrow \frac{AP}{PB} = \frac{AQ}{QC} \\[1em] \Rightarrow \frac{x}{9 - x} = \frac{4.2}{1.8} \\[1em] \Rightarrow 1.8x = 4.2(9 - x) \\[1em] \Rightarrow 1.8x = 37.8 - 4.2x \\[1em] \Rightarrow 1.8x + 4.2x = 37.8 \\[1em] \Rightarrow 6x = 37.8 \\[1em] \Rightarrow x = 6.3 \text{ cm}.\end{aligned}$$

Hence, AP = 6.3 cm.

(i) From figure,

BD = AB - AD = 9 - 4 = 5 cm

In ∆ADE and ∆ABC,

$$\begin{aligned}\frac{AE}{EC} = \frac{6}{7.5} = \frac{4}{5} \\[1em] \frac{AD}{BD} = \frac{4}{5} \\[1em] \text{Since } \frac{AE}{EC} = \frac{AD}{BD}.\end{aligned}$$

Hence, DE || BC by the converse of Basic proportionality theorem.

(ii) From figure,

BD = AB - AD = 6.3 - 0.8 = 5.5 cm

In ∆ADE and ∆ABC,

$$\begin{aligned}\frac{AE}{EC} = \frac{1.6}{11} = \frac{0.8}{5.5} \\[1em] \frac{AD}{BD} = \frac{0.8}{5.5} \\[1em] \text{Since } \frac{AE}{EC} = \frac{AD}{BD}.\end{aligned}$$

Hence, DE || BC by the converse of Basic proportionality theorem.

Given,

ΔABC ~ ΔADE

Given,

AE : EC = 4 : 7

Let AE = 4y and EC = 7y.

So, AC = 4y + 7y = 11y.

From figure,

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

$$\begin{aligned}\Rightarrow \frac{AE}{AC} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{4y}{11y} = \frac{6.6}{BC} \\[1em] \Rightarrow \frac{4}{11} = \frac{6.6}{BC} \\[1em] \Rightarrow BC = \frac{11 \times 6.6}{4} = 18.15 \text{ cm}\end{aligned}$$

As ΔABC ~ ΔADE, we have :

∠ABC = ∠ADE and ∠ACB = ∠AED

So, DE || BC as ∠ADE and ∠ABC are corresponding angles.

$\therefore \frac{AB}{AD} = \frac{AC}{AE} = \frac{11y}{4y} = \frac{11}{4}$

Let perpendicular from A to DE meet DE at point P. Then,

AP = x

Let perpendicular from A to BC meet BC at point Q.

In ∆ADP and ∆ABQ,

∠ADP = ∠ABQ [Corresponding angles are equal.]

∠APD = ∠AQB [Both = 90°]

∴ ∆ADP ~ ∆ABQ [By AA]

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{AP}{AQ} \\[1em] \Rightarrow \frac{4}{11} = \frac{x}{AQ} \\[1em] \Rightarrow AQ = \frac{11}{4}x.\end{aligned}$$

Hence, BC = 18.15 cm and AQ = $\frac{11}{4}x.$

Given,

⇒ AB = 5BD

⇒ AD + BD = 5BD

⇒ AD = 5BD - BD

⇒ AD = 4BD

⇒ $\frac{\text{AD}}{\text{BD}} = \frac{4}{1}$.

Given DE || BC,

by basic proportionality theorem :

$$\begin{aligned}\therefore \frac{AD}{BD} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{4}{1} = \frac{AE}{3.2} \\[1em] \Rightarrow AE = 4 \times 3.2 = 12.8 \text{ cm}.\end{aligned}$$

Hence, AE = 12.8 cm

In ∆BEF,

DC || FE

So, by basic proportionality theorem,

$\frac{BD}{DF} = \frac{BC}{CE} = \frac{x}{3}$

$\frac{BD}{DF} = \frac{x}{3}$ .....(1)

Let BD = ax and DF = 3a.

From figure,

BF = BD + DF = ax + 3a = a(x + 3).

$\frac{BF}{DF} = \frac{a(x + 3)}{3a} = \frac{x + 3}{3}$ .....(2)

In ∆AFB and CDF,

∠AFB = ∠CFD [Common angles]

∠ABF = ∠CDF [Corresponding angles are equal]

∴ ∆AFB ~ ∆CDF [By AA]

Since, corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{BF}{DF} = \frac{AB}{CD} \\[1em] \Rightarrow \frac{BF}{DF} = \frac{7.5}{y} .....(3) \\[1em] \text{From (2) and (3) we get}, \\[1em] \Rightarrow \frac{x + 3}{3} = \frac{7.5}{y} \\[1em] \Rightarrow y = \frac{22.5}{x + 3} \space .....(4)\end{aligned}$$

In ∆BCD and ∆BEF ,

∠FBE = ∠DBC [Common angles]

∠CDB = ∠EFB [Corresponding angles are equal]

∴ ∆BEF ~ ∆BCD [By AA]

Since, corresponding sides of similar triangles are proportional.

$\frac{BD}{BF} = \frac{CD}{FE} = \frac{y}{4.5}$

$\frac{BD}{BF} = \frac{y}{4.5}$

Let BD = ay and BF = 4.5a

From figure,

DF = BF - BD = 4.5a - ay = a(4.5 - y).

$\therefore \frac{BD}{DF} = \frac{ay}{a(4.5 - y)} = \frac{y}{4.5 - y}$ .....(5)

From (1) and (5) we get,

$\Rightarrow \frac{x}{3} = \frac{y}{4.5 - y}$

Substituting value of y from (4) in above equation we get,

$$\begin{aligned}\Rightarrow \frac{x}{3} = \frac{\frac{22.5}{x + 3}}{4.5 - \frac{22.5}{x + 3}} \\[1em] \Rightarrow \frac{x}{3} = \frac{\frac{22.5}{x + 3}}{\frac{4.5x + 13.5 - 22.5}{x + 3}} \\[1em] \Rightarrow \frac{x}{3} = \frac{22.5}{4.5x - 9} \\[1em] \Rightarrow \frac{x}{3} = \frac{22.5}{3(1.5x - 3)} \\[1em] \Rightarrow x = \frac{22.5}{1.5x - 3} \\[1em] \Rightarrow x = \frac{22.5}{1.5(x - 2)} \\[1em] \Rightarrow x = \frac{15}{x - 2} \\[1em] \Rightarrow x(x - 2) = 15 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = - 3 \text{ or } x = 5.\end{aligned}$$

Since, side of triangle cannot be negative. So, x = 5 cm.

Substituting value of x in (4) we get,

$$\begin{aligned}\Rightarrow y = \frac{22.5}{x + 3} \\[1em] \Rightarrow y = \frac{22.5}{5 + 3} \\[1em] \Rightarrow y = \frac{22.5}{8} = 2.8125 \text{ cm}\end{aligned}$$

Hence, x = 5 cm and y = 2.8125 cm

Given XY || QR,

By basic proportionality theorem :

$$\begin{aligned}\therefore \frac{PX}{XQ} = \frac{PY}{YR} \\[1em] \Rightarrow \frac{1}{2} = \frac{4}{YR} \\[1em] \Rightarrow YR = 4 \times 2 = 8 \text{ cm}.\end{aligned}$$

From figure,

PR = PY + YR = 4 + 8 = 12 cm.

Since, PQR is a right-angled triangle.

By pythagoras theorem we get,

$$\begin{aligned}\Rightarrow PR^2 = PQ^2 + QR^2 \\[1em] \Rightarrow 12^2 = 6^2 + QR^2 \\[1em] \Rightarrow 144 = 36 + QR^2 \\[1em] \Rightarrow QR^2 = 144 - 36 \\[1em] \Rightarrow QR^2 = 108 \\[1em] \Rightarrow QR = \sqrt{108} = 10.392 \text{ cm}.\end{aligned}$$

Hence, PR = 12 cm and QR = 10.392 cm.

In ∆BME and ∆DMC,

∠BME = ∠CMD [Vertically opposite angles are equal.]

∠MCD = ∠MBE [Alternate angles are equal]

BM = MC [M is mid-point of BC]

∴ ∆BME ≅ ∆DMC [By AAS congruence rule]

∴ BE = DC [By C.P.C.T]

Since, opposite sides of parallelogram are equal.

∴ AB = DC

or, AB = DC = BE. ...........(1)

In ∆DCP and ∆EPA,

∠DPC = ∠EPA [Vertically opposite angles are equal.]

∠CDP = ∠AEP [Alternate angles are equal]

∴ ∆DCP ~ ∆EAP [By AA]

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

$$\begin{aligned}\Rightarrow \frac{DC}{EA} = \frac{PD}{PE} \\[1em] \Rightarrow \frac{EA}{DC} = \frac{PE}{PD} \\[1em] \Rightarrow \frac{PE}{PD} = \frac{AB + BE}{DC} \\[1em] \Rightarrow \frac{PE}{PD} = \frac{2DC}{DC} \\[1em] \Rightarrow \frac{PE}{PD} = 2 \\[1em] \Rightarrow PE = 2PD.\end{aligned}$$

Hence proved that PE = 2PD.

From figure,

FB = AF + AB = 8 + 12 = 20 cm.

In △DEC and △EAF

⇒ ∠DEC = ∠FEA [Vertically opposite angles are equal]

⇒ ∠EDC = ∠EAF [Alternate angles are equal]

∴ △DEC ~ △EAF [By AA]

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

$$\begin{aligned}\Rightarrow \frac{DE}{AE} = \frac{DC}{AF} \\[1em] \Rightarrow \frac{DE}{AE} = \frac{AB}{AF} \space \Big[\because\text{ AB = CD}\Big] \\[1em] \Rightarrow \frac{DE}{4} = \frac{12}{8} \\[1em] \Rightarrow DE = \frac{48}{8} = 6 \text{ cm}.\end{aligned}$$

Since, ABCD is a ||gm.

AB = CD and BC = AD.

From figure,

AD = AE + ED = 4 + 6 = 10 cm.

Perimeter of ||gm ABCD = AB + BC + CD + AD

= 12 + 10 + 12 + 10

= 44 cm.

Hence, perimeter of ||gm ABCD = 44 cm.

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

From figure,

AC = AE + EC = 5 + 7 = 12 cm.

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ ADE}}{\text{Area of △ ABC}} = \frac{AE^2}{AC^2} \\[1em] = \frac{5^2}{12^2} \\[1em] = \frac{25}{144} \\[1em] = 25 : 144.\end{aligned}$$

Hence, Option 3 is the correct option.

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

From figure,

AB = AD + DB = 5 + 2 = 7 cm.

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ ADE}}{\text{Area of △ ABC}} = \frac{AD^2}{AB^2} \\[1em] = \frac{5^2}{7^2} \\[1em] = \frac{25}{49} \\[1em] = 25 : 49.\end{aligned}$$

Let area of △ ADE = 25x and area of △ ABC = 49x.

From figure,

Area of trapezium DBCE = Area of △ ABC - Area of △ ADE = 49x - 25x = 24x.

$\therefore \frac{\text{Area of △ ADE}}{\text{Area of trapezium DBCE}} = \frac{25x}{24x} = \frac{25}{24}$

⇒ Area of △ ADE : Area of trapezium DBCE = 25 : 24.

Hence, Option 4 is the correct option.

Given,

AD : DB = 2 : 5

Let AD = 2x and DB = 5x

From figure,

AB = AD + DB = 2x + 5x = 7x.

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

∴ △ ADE ~ △ ABC (By A.A. postulate)

We know that,

Corresponding sides of similar triangles are in proportion.

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{2x}{7x} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{2}{7}.\end{aligned}$$

In △ ODE and △ OCB,

⇒ ∠DOE = ∠BOC (Vertically opposite angle are equal)

⇒ ∠ODE = ∠OCB (Alternate angles are equal)

∴ △ ODE ~ △ OCB (By A.A. postulate)

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of △ ODE}}{\text{Area of △ OCB}} = \frac{DE^2}{BC^2} \\[1em] = \frac{2^2}{7^2} \\[1em] = \frac{4}{49} \\[1em] = 4 : 49.\end{aligned}$$

Hence, Option 1 is the correct option.

In △ ADE and △ ABC,

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

⇒ ∠DAE = ∠BAC (Common angle)

∴ △ ADE ~ △ ABC (By A.A. postulate)

Given,

Area of △ ADE : Area of trapezium BCED = 25 : 39

Let area of △ ADE = 25x and area of trapezium BCED = 39x.

From figure,

Area of △ ABC = Area of △ ADE + Area of trapezium BCED = 25x + 39x = 64x.

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ ADE}}{\text{Area of △ ABC}} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{25x}{64x} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{25}{64} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{AD}{AB} = \sqrt{\frac{25}{64}} = \frac{5}{8}.\end{aligned}$$

Let AD = 5y and AB = 8y.

From figure,

⇒ BD = AB - AD = 8y - 5y = 3y.

⇒ AD : BD = 5y : 3y = 5 : 3.

Hence, Option 4 is the correct option.

From figure,

In Δ BAC and Δ ADC,

⇒ ∠BAC = ∠ADC (Both equal to 90°)

⇒ ∠ACB = ∠ACD (Common angle)

∴ Δ BAC ~ Δ ADC (By A.A. postulate)

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of Δ BAC}}{\text{Area of Δ ADC}} = \frac{BC^2}{AC^2} \\[1em] \Rightarrow \frac{\text{Area of Δ BAC}}{\text{Area of Δ ADC}} = \frac{13^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of Δ BAC}}{\text{Area of Δ ADC}} = \frac{169}{25}.\end{aligned}$$

Let, area of Δ BAC = 169x and area of Δ ADC = 25x.

From figure,

Area of Δ DBA = Area of Δ BAC - Area of Δ ADC = 169x - 25x = 144x.

Substituting values we get :

area of △ ADC : area of △ DBA = 25x : 144x = 25 : 144.

Hence, Option 3 is the correct option.

Given, AP = $\frac{1}{3}$PB

So, $\frac{AP}{PB} = \frac{1}{3}$

Let AP = x and PB = 3x.

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

$\therefore \frac{AP}{AB} = \frac{x}{4x} = \frac{1}{4}$.

In ∆APQ and ∆ABC,

∠APQ = ∠ABC and ∠AQP = ∠ACB [Corresponding angles are equal]

Hence, ∆APQ ~ ∆ABC by AA criterion for similarity

(i) We know that,

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

$$\begin{aligned}\frac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}} = \frac{\text{AB}^2}{\text{AP}^2} \\[1em] = \frac{4^2}{1^2} \\[1em] = \frac{16}{1} \\[1em] = 16 : 1\end{aligned}$$

Hence, $\frac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}}$ = 16 : 1.

(ii) From figure,

Area of Trapezium PBCQ = Area of ΔABC – Area of ΔAPQ

From part (i) above we get,

$\frac{\text{Area of ΔABC}}{\text{Area of ΔAPQ}} = \frac{16}{1}$

Let Area of ΔABC = 16a and Area of ΔAPQ = a

Area of trapezium PBCQ = 16a - a = 15a.

$\frac{\text{Area of ΔAPQ}}{\text{Area of trap. PBCQ}} = \frac{a}{15a} = \frac{1}{15}$ = 1 : 15.

Hence, $\frac{\text{Area of ΔAPQ}}{\text{Area of trap. PBCQ}}$ = 1 : 15.

Let the triangles be ∆ABC and ∆DEF.

Given ∆ABC ~ ∆DEF

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

So, $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$

Adding numerator and denominator we get :

$$\begin{aligned}\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \\[1em] = \frac{AB + BC + AC}{DE + EF + DF} \\[1em] = \frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}}\end{aligned}$$

So,

$$\begin{aligned}\frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF }} = \frac{AB}{DE} \\[1em] \frac{30}{24} = \frac{12}{DE} \\[1em] DE = \frac{24 \times 12}{30} \\[1em] DE = 9.6 \text{ cm}.\end{aligned}$$

Hence, the length of corresponding side of second triangle is 9.6 cm.

(i) Given, AX : XB = 3 : 5

Let AX = 3a and XB = 5a.

From figure,

AB = AX + XB = 3a + 5a = 8a.

(i) In ΔAXY and ΔABC,

As XY || BC, corresponding angles are equal.

∠AXY = ∠ABC and ∠AYX = ∠ACB

∴ ∆AXY ~ ∆ABC [By AA]

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

$$\begin{aligned}\therefore \frac{AX}{AB} = \frac{XY}{BC} \\[1em] \Rightarrow \frac{3a}{8a} = \frac{18}{BC} \\[1em] \Rightarrow BC = \frac{18 \times 8a}{3a} = 48 \text{ cm}.\end{aligned}$$

Hence, BC = 48 cm.

(ii) We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of ∆AXY}}{\text{Area of ∆ABC}} = \frac{AX^2}{AB^2} \\[1em] = \frac{(3a)^2}{(8a)^2} \\[1em] = \frac{9a^2}{64a^2} \\[1em] = \frac{9}{64}. \\[1em]\end{aligned}$$

Let area of ∆AXY = 9b and area of ∆ABC = 64b.

From figure,

Area of trapezium XBCY = Area of ∆ABC - Area of ∆AXY

= 64b - 9b = 55b.

$\therefore \frac{\text{Area of trap. XBCY}}{\text{Area of ∆ABC}} = \frac{55b}{64b} = \frac{55}{64}$

Hence, ratio of area of trapezium XBCY and triangle ABC = 55 : 64.

Triangle ABC is shown in the figure below:

In ΔAPQ and ΔABC,

∠PAQ = ∠BAC [Common]

∠APQ = ∠ABC [Corresponding angles are equal]

∴ ΔAPQ ~ ΔABC [By AA].

According to question,

Area of ΔAPQ = $\frac{1}{2}$ Area of ΔABC

$\frac{\text{Area of ΔAPQ}}{\text{Area of ΔABC}} = \frac{1}{2}$

We know that,

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

$$\begin{aligned}\Rightarrow \frac{AP^2}{AB^2} = \frac{1}{2} \\[1em] \Rightarrow \frac{AP}{AB} = \frac{1}{\sqrt{2}}.\end{aligned}$$

Let AP = x and AB = $\sqrt{2}$x

From figure,

BP = AB - AP = $\sqrt{2}x - x$

$$\begin{aligned}\therefore \frac{BP}{AB} = \frac{\sqrt{2}x - x}{\sqrt{2}x} \\[1em] = \frac{\sqrt{2} - 1}{\sqrt{2}} \\[1em] = \frac{\sqrt{2} - 1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\[1em] = \frac{2 - \sqrt{2}}{2}.\end{aligned}$$

Hence, BP : AB = $(\sqrt{2} - 1) : \sqrt{2}$ = $(2 - \sqrt{2}) : 2$.

(i) Given,

$\frac{PM}{MR} = \frac{3}{4}$

Let PM = 3x and MR = 4x.

From figure,

PR = PM + MR = 3x + 4x = 7x.

$\frac{PM}{PR} = \frac{3x}{7x} = \frac{3}{7}$.

In ΔPLM and ΔPQR,

As LM || QR, corresponding angles are equal.

∠PLM = ∠PQR

∠PML = ∠PRQ

∴ ∆PLM ~ ∆PQR [By AA]

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

$$\begin{aligned}\therefore \frac{PL}{PQ} = \frac{PM}{PR} = \frac{LM}{QR} \\[1em] \therefore \frac{PL}{PQ} = \frac{LM}{QR} = \frac{3}{7}.\end{aligned}$$

Hence, PL : PQ = 3 : 7 and LM : QR = 3 : 7.

(ii) As ΔLMN and ΔMNR have common vertex at M and their bases LN and NR are along the same straight line.

$\therefore \frac{\text{Area of ΔLMN}}{\text{Area of ΔMNR}} = \frac{\text{LN}}{\text{NR}}$ .....(1)

Now, in ΔLMN and ΔRNQ we have,

⇒ ∠NLM = ∠NRQ [Alternate angles are equal]

⇒ ∠LMN = ∠NQR [Alternate angles are equal]

∴ ∆LNM ~ ∆RNQ [By AA]

Since corresponding sides of similar triangle are proportional to each other, we have :

$\frac{MN}{QN} = \frac{LN}{NR} = \frac{LM}{QR} = \frac{3}{7}$

Substituting value in (1) we get :

$\frac{\text{Area of ΔLMN}}{\text{Area of ΔMNR}} = \frac{3}{7}$.

(iii) From part (ii) we get :

$\frac{MN}{QN} = \frac{3}{7}$

Let MN = 3a and QN = 7a

From figure,

MQ = MN + QN = 3a + 7a = 10a.

As ΔLQM and ΔLQN have common vertex at L and their bases QM and QN are along the same straight line.

$\frac{\text{ Area of ΔLQM }}{\text{ Area of ΔLQN }} = \frac{QM}{QN} = \frac{10a}{7a} = \frac{10}{7}$.

Hence, $\frac{\text{ Area of ΔLQM }}{\text{ Area of ΔLQN }} =\frac{10}{7}$

(i) In △BPQ and △CPD,

⇒ ∠BPQ = ∠CPD [Vertically opposite angles are equal]

⇒ ∠BQP = ∠PDC [Alternate angles are equal]

∴ △BPQ ~ △CPD [By AA]

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

$\therefore \frac{BP}{PC} = \frac{PQ}{PD} = \frac{BQ}{CD} = \frac{1}{2}$.

As ΔBPQ and ΔCPQ have common vertex at Q and their bases BP and CP are along the same straight line.

So,

$$\begin{aligned}\Rightarrow \frac{\text{Area of ΔBPQ}}{\text{Area of ΔCPQ}} = \frac{BP}{PC} \\[1em] \Rightarrow \frac{\text{Area of ΔBPQ}}{20} = \frac{1}{2} \\[1em] \Rightarrow \text{Area of ΔBPQ} = \frac{20}{2} \\[1em] \Rightarrow \text{Area of ΔBPQ} = 10 \text{ cm}^2\end{aligned}$$

We know that,

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

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{Area of ΔBPQ}}{\text{Area of ΔCPD}} = \frac{BP^2}{PC^2} \\[1em] \Rightarrow \frac{10}{\text{Area of ΔCPD}} = \frac{1^2}{2^2} \\[1em] \Rightarrow \text{ Area of ΔCPD} = \frac{10 \times 4}{1} \\[1em] \Rightarrow \text{ Area of ΔCPD} = 40 \text{ cm}^2.\end{aligned}$$

Hence, area of ΔCPD = 40 cm².

(ii) From part (i),

$\frac{PQ}{PD} = \frac{1}{2}$

Let PQ = x and PD = 2x.

From figure,

QD = PQ + PD = x + 2x = 3x.

$\frac{QD}{QP} = \frac{3x}{x} = 3$ ........(1)

In △BPQ and △AQD,

⇒ ∠QBP = ∠QAD [Corresponding angles are equal]

⇒ ∠BQP = ∠AQD [Common]

∴ △BPQ ~ △AQD [By AA]

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

$\therefore \frac{AQ}{BQ} = \frac{QD}{QP} = \frac{AD}{BP} = 3.$

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of △AQD}}{\text{Area of △BPQ}} = \Big(\frac{AQ}{BQ}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of △AQD}}{10} = 3^2 \\[1em] \Rightarrow \frac{\text{Area of △AQD}}{10} = 9 \\[1em] \Rightarrow \text{Area of △AQD} = 90 \text{ cm}^2.\end{aligned}$$

Area of trapezium ADPB = Area of △AQD - Area of △BPQ = 90 - 10 = 80 cm².

Area of || gm ABCD = Area of △CDP + Area of trapezium ADPB = 40 + 80 = 120 cm².

Hence, area of || gm ABCD = 120 cm².

Area of △ADE = Area of △ABC + Area of trapezium BCED = 25 + 24 = 49 cm².

Given,

BC || DE.

∠ABC = ∠ADE [Corresponding angles are equal]

∠ACB = ∠AED [Corresponding angles are equal]

∴ △ABC ~ △ADE [By AA]

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △ADE}} = \frac{BC^2}{DE^2}\\[1em] \Rightarrow \frac{25}{49} = \frac{BC^2}{14^2} \\[1em] \Rightarrow BC^2 = \frac{25}{49} \times 196 \\[1em] \Rightarrow BC^2 = 100 \\[1em] \Rightarrow BC = 10 \text{ cm}.\end{aligned}$$

Let height of trapezium BCED be h cm.

Area = $\frac{1}{2}$ × (Sum of || sides) × h

⇒ 24 = $\frac{1}{2}$ × (BC + DE) × h

⇒ 24 × 2 = (BC + DE) × h

⇒ 48 = (10 + 14) × h

⇒ 24h = 48

⇒ h = 2 cm.

Area of △BCD = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × h

= $\frac{1}{2}$ × 10 × 2

= 10 cm².

Hence, BC = 10 cm and area of △BCD = 10 cm².

(i) Given,

$\frac{AD}{DB} = \frac{3}{2}$

Let AD = 3x and BD = 2x.

From figure,

AB = AD + DB = 3x + 2x = 5x.

$\frac{AD}{AB} = \frac{3x}{5x} = \frac{3}{5}$.

In △ADE and △ABC,

∠A = ∠A [Common]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ △ADE ~ △ABC [By AA]

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

$\therefore \frac{DE}{BC} = \frac{AD}{AB} = \frac{3}{5}$ ...........(1)

Hence, $\frac{AD}{AB} = \frac{DE}{BC} = \frac{3}{5}$.

(ii) In △DEF and △CBF,

∠FDE = ∠FCB (Alternate angles are equal)

∠DFE = ∠BFC (Vertically opposite angles are equal)

∴ △DEF ~ △CBF [By AA]

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

$\therefore \frac{EF}{FB} = \frac{DE}{BC} = \frac{3}{5}$

Hence, $\frac{EF}{FB} = \frac{3}{5}$.

(iii) We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of △DFE}}{\text{Area of △CBF}} = \frac{EF^2}{FB^2}\\[1em] = \frac{3^2}{5^2} = \frac{9}{25}.\end{aligned}$$

Hence, ratio of the areas of △DEF and △BFC = 9 : 25.

Given,

⇒ ∠ACD = ∠BCE

⇒ ∠ACD + ∠BCD = ∠BCE + ∠BCD

⇒ ∠ACB = ∠DCE

Also, ∠B = ∠E

∴ △ABC ~ △DEC [By AA]

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of △ABC}}{\text{Area of △DEC}} = \frac{AB^2}{DE^2}\\[1em] = \Big(\frac{10.4}{7.8}\Big)^2 = \Big(\frac{4}{3}\Big)^2 \\[1em] = \frac{16}{9}.\end{aligned}$$

Hence, ratio between areas of the △ABC and △DEC = 16 : 9.

Given,

Scale factor (k) = 1 : 25 or $\frac{1}{25}$

By formula,

⇒ Length of model = k × Length of vehicle

⇒ Length of model = $\frac{1}{25} \times 1.6$ = 0.064 m = 0.064 × 100 = 6.4 cm.

Hence, Option 1 is the correct option.

Given,

Scale factor (k) = 1 : 20 or $\frac{1}{20}$

By formula,

⇒ Length of model = k × Length of object

⇒ Length of object = $\frac{1}{k}$ × Length of model

⇒ Length of object = $\frac{1}{\frac{1}{20}} \times 12$ = 20 × 12 = 240 cm = $\frac{240}{100}$ = 2.4 m.

By formula,

⇒ Breadth of model = k × Breadth of object

⇒ Breadth of object = $\frac{1}{k}$ × Breadth of model

⇒ Breadth of object = $\frac{1}{\frac{1}{20}} \times 7$ = 20 × 7 = 140 cm = $\frac{140}{100}$ = 1.4 m.

Hence, Option 2 is the correct option.

Let scale factor be k.

Given,

Length of model = 16 cm

Length of boat = 2.4 m = 2.4 × 100 = 240 cm.

By formula,

⇒ Length of model = Scale factor × Length of boat

Substituting values we get :

⇒ 16 = k × 240

⇒ k = $\frac{16}{240} = \frac{1}{15}$ = 1 : 15.

Hence, Option 3 is the correct option.

Let scale factor be k.

Given,

Floor area of room = 96 m²

Floor area of model = 96 cm² = $\frac{96}{100^2} = \frac{96}{10000}$ = 96 × 10^-4 m²

By formula,

Floor area of model = (Scale factor)² × Floor area of room

Substituting values we get :

⇒ 96 × 10^-4 = k² × 96

⇒ k² = $\frac{96 \times 10^{-4}}{96}$

⇒ k² = 10^-4

⇒ k = $\sqrt{10^{-4}}$

⇒ k = 10^-2 = $\frac{1}{10^2} = \frac{1}{100}$ = 1 : 100.

Hence, Option 4 is the correct option.

Let scale factor of B with respect to A be k.

Given,

A and B are two similar containers with capacities 216 m³ and 1000 m³ respectively.

Volume of container B = k³ × Volume of container A

⇒ 1000 = k³ × 216

⇒ k³ = $\frac{1000}{216}$

⇒ k³ = $\Big(\frac{10}{6}\Big)^3$

⇒ k = $\frac{10}{6} = \frac{5}{3}$ = 5 : 3.

Hence, Option 2 is the correct option.

Given that, ΔABC has been enlarged by scale factor m of 2.5 to ΔA'B'C'.

(i)

∴ A'B' = m × AB

⇒ 6 = 2.5 × AB

⇒ AB = $\frac{6}{2.5} = 2.4$ cm.

Hence, AB = 2.4 cm

(ii)

∴ C'A' = m × CA

⇒ C'A' = 2.5 × 4

⇒ C'A' = 10 cm.

Hence, C'A' = 10 cm.

Given, ΔLMN has been reduced by a scale factor (k) = 0.8 to Δ L'M'N'.

(i)

∴ M'N' = k × MN

⇒ M'N' = 0.8 × 8

⇒ M'N' = 6.4 cm

Hence, M'N' = 6.4 cm.

(ii)

∴ L'M' = k × LM

⇒ 5.4 = 0.8 × LM

⇒ LM = $\frac{5.4}{0.8} = 6.75$ cm.

Hence, LM = 6.75 cm.

(i) Given,

Scale-factor (k) = $\frac{1}{400}$

⇒ Length of model = k × Length of aeroplane

⇒ Length of model = $\frac{1}{400} \times 40 = \frac{1}{10}$ = 0.1 m = 10 cm.

Hence, length of model = 10 cm.

(ii) Given,

Scale-factor (k) = $\frac{1}{400}$

⇒ Length of model = k × Length of aeroplane

⇒ 16 = $\frac{1}{400}$ × Length of aeroplane

⇒ 16 × 400 = Length of aeroplane

⇒ Length of aeroplane = 6400 cm = 64 m.

Hence, length of aeroplane = 64 m.

Dimensions of model = 1.2 m × 75 cm × 2 m = 1.2 m × 0.75 m × 2 m

So,

Length = 1.2 m

Breadth = 0.75 m

Height = 2 m.

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{Length of model}}{\text{Length of building}} = k \\[1em] \Rightarrow \frac{1.2}{\text{Length of building}} = \frac{1}{30} \\[1em] \Rightarrow \text{Length of building} = 1.2 \times 30 = 36 \text{ m}. \\[2em] \phantom{\Rightarrow} \frac{\text{Breadth of model}}{\text{Breadth of building}} = k \\[1em] \Rightarrow \frac{0.75}{\text{Breadth of building}} = \frac{1}{30} \\[1em] \Rightarrow \text{Breadth of building} = 0.75 \times 30 = 22.5 \text{ m}. \\[2em] \phantom{\Rightarrow} \frac{\text{Height of model}}{\text{Height of building}} = k \\[1em] \Rightarrow \frac{\text{2}}{\text{Height of building}} = \frac{1}{30} \\[1em] \Rightarrow \text{Height of building} = 2 \times 30 = 60 \text{ m}.\end{aligned}$$

Hence, actual dimension of building = 36 m × 22.5 m × 60 m.

(i) Given,

Scale factor for reduction (k) = $\frac{1}{250000}$

⇒ Length of AB in model = k × Length of AB

⇒ 3 = $\frac{1}{250000}$ × Length of AB

⇒ Length of AB = 3 × 250000 = 750000 cm = $\frac{750000}{100 \times 1000}$ = 7.5 km.

⇒ Length of BC in model = k × Length of BC

⇒ 4 = $\frac{1}{250000}$ × Length of BC

⇒ Length of BC = 4 × 250000 = 1000000 cm = $\frac{1000000}{100 \times 1000}$ = 10 km.

Hence, actual length of AB = 7.5 km and BC = 10 km.

(ii) Area of plot = $\frac{1}{2} \times$ base × height

= $\frac{1}{2}$ × BC × AB

= $\frac{1}{2}$ × 10 × 7.5

= 37.5 sq. km.

Hence, area of plot = 37.5 sq. km.

(i) Given,

Scale factor (k) = 1 : 300 = $\frac{1}{300}$

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{Length of model of ship}}{\text{Length of ship}} = k \\[1em] \Rightarrow \frac{2}{\text{Length of ship}} = \frac{1}{300} \\[1em] \Rightarrow \text{Length of ship} = 2 \times 300 = 600 \text{ m}.\end{aligned}$$

Hence, length of ship = 600 m.

(ii)

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{Area of deck of model}}{\text{Area of deck of ship}} = k^2 \\[1em] \Rightarrow \frac{\text{Area of deck of model}}{180000} = \frac{1}{300} \times \frac{1}{300} \\[1em] \Rightarrow \text{Area of deck of model} = 180000 \times \frac{1}{90000} = 2 \text{ m}^2.\end{aligned}$$

Hence, area of deck of model = 2 m².

(iii)

$$\begin{aligned}\Rightarrow \frac{\text{Volume of model of ship}}{\text{Volume of ship}} = k^3 \\[1em] \Rightarrow \frac{6.5}{\text{Volume of ship}} = \Big(\frac{1}{300}\Big)^3 \\[1em] \Rightarrow \text{Volume of ship} = 6.5 \times (300)^3 = 6.5 \times 27000000 \text{ m}^3 \\[1em] = 175500000 \text{ m}^3.\end{aligned}$$

Hence, volume of ship = 175500000 m³.

Given,

Triangles ABC and A'B'C' are similar to each other.

So, the corresponding vertices are A and A', B and B', C and C'.

Hence, Option 3 is the correct option.

Given,

Triangles ABC and RSP are similar to each other.

We know that,

Corresponding sides of similar triangle are proportional.

$\therefore \frac{AB}{RS} = \frac{BC}{SP} = \frac{CA}{PR}$

Hence, Option 1 is the correct option.

We know that,

Congruent triangles have both the same shape and the same size but it is not necessary that similar triangle are equal in size.

Hence, Option 2 is the correct option.

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

$\Rightarrow \frac{\text{Ar.(△ABC)}}{\text{Ar.(△PQR)}} = \frac{AB^2}{PQ^2}$ .......(1)

$\Rightarrow \frac{\text{Ar.(△ABC)}}{\text{Ar.(△PQR)}} = \frac{AC^2}{PR^2}$ .........(2)

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

$\Rightarrow \frac{AB^2}{PQ^2} = \frac{AC^2}{PR^2}$

Hence, Option 2 is the correct option.

From figure,

In △ AEF and △ ABC,

⇒ ∠EAF = ∠BAC (Common angle)

⇒ ∠AEF = ∠ABC (Corresponding angle)

∴ △ AEF ~ △ ABC (By A.A. axiom)

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ AEF}}{\text{Area of △ ABC}} = \frac{EF^2}{BC^2} \\[1em] \Rightarrow \frac{\text{Area of △ AEF}}{576} = \frac{8^2}{12^2} \\[1em] \Rightarrow \frac{\text{Area of △ AEF}}{576} = \frac{64}{144} \\[1em] \Rightarrow \text{Area of △ AEF} = \frac{64}{144} \times 576 \\[1em] \Rightarrow \text{Area of △ AEF} = 64 \times 4 = 256 \text{ cm}^2.\end{aligned}$$

Hence, Option 3 is the correct option.

From figure,

In △ BAD and △ ACD,

⇒ ∠BAD = ∠ACD (Given)

⇒ ∠ADB = ∠ADC (Common angle)

∴ △ BAD ~ △ ACD (By A.A. axiom)

Hence, Option 3 is the correct option.

Given,

Scale factor (k) = 3 : 500

By formula,

⇒ Length of model = Scale factor × Length of ship

Substituting values we get :

$$\begin{aligned}\Rightarrow 1.5 = \frac{3}{500} \times \text{ Length of ship} \\[1em] \Rightarrow \text{Length of ship} = \frac{1.5 \times 500}{3} \\[1em] \Rightarrow \text{Length of ship} = 0.5 \times 500 = 250 \text{ m}.\end{aligned}$$

Hence, Option 4 is the correct option.

In Δ ABC and Δ ABD,

⇒ ∠ABC = ∠ADB (Both are 90°)

⇒ ∠BAC = ∠BAD (Common angle)

∴ Δ ABC ∼ Δ ADB (By AA postulate) .......... (1)

Similarly, in Δ ABC and Δ BDC,

⇒ ∠ABC = ∠BDC (Both are 90°)

⇒ ∠BCA = ∠BCD (Common angles)

∴ Δ ABC ∼ Δ BDC (By AA postulate) .......... (2)

As,

Triangles, which are similar to the same triangle, are similar to each other.

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

Δ ADB ∼ Δ BDC

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given, Δ ABC and Δ PQR are two congruent triangles.

Corresponding angles and sides are equal.

⇒ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R

∴ Δ ABC ∼ Δ PQR (By AAA postulate)

So, assertion (A) is true.

Similar triangles have the same shape but not necessarily the same size.

Congruent triangles have the same shape and size.

Size can be different in similar triangles but it has to be same in congruent triangles.

So, reason (R) is false.

Hence, option 1 is the correct option.

Reflection in the origin means each point (x, y) becomes (-x, -y). Shape, size, angles and size length remains same in reflection.

So, Δ ABC is reflected in origin to get Δ A'B'C'.

Hence, Δ ABC is congruent to Δ A'B'C'.

So, statement 1 is true.

As we know that all congruent triangles are similar.

Hence, Δ ABC is similar to Δ A'B'C'.

So, statement 2 is true.

Hence, option 1 is the correct option.

Given, △ ABC ∼ △ DEF.

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of Δ DEF}}{\text{Area of Δ ABC}} = \frac{DE^2}{AB^2}\\[1em] \Rightarrow\frac{121}{64} = \frac{DE^2} {AB^2}\\[1em] \Rightarrow \frac{AB^2}{DE^2} = \frac{64}{121} \\[1em] \Rightarrow \frac{AB}{DE} = \sqrt{\frac{64}{121}}\\[1em] \Rightarrow \frac{AB}{DE} = \frac{8}{11}\end{aligned}$$

Since, corresponding sides of similar triangle are proportional.

$\Rightarrow \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$

We know that,

For any two or more equal ratios, each ratio is equal to the ratio between sum of their antecedents and sum of their consequents.

$$\begin{aligned}\Rightarrow \frac{AB}{DE} = \frac{AB + BC + AC}{DE + EF + DF} \\[1em] \Rightarrow \frac{\text{Perimeter of Δ ABC}}{\text{Perimeter of Δ DEF}} = \frac{AB}{DE} \\[1em] \Rightarrow \frac{\text{Perimeter of Δ DEF}}{\text{Perimeter of Δ ABC}} = \frac{DE}{AB} = \frac{11}{8}\\[1em]\end{aligned}$$

So, statement 1 is false.

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

So, statement 2 is false.

Hence, option 2 is the correct option.

Given,

AD : DB = 2 : 3

Let AD = 2x and DB = 3x.

From figure,

AB = AD + DB = 2x + 3x = 5x.

In Δ ADE and Δ ABC

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Corresponding angles are equal)

⇒ ∠AED = ∠ACB (Corresponding angles are equal)

∴ ΔADE ∼ ΔABC (By AAA postulate)

We know that,

In similar triangles, corresponding sides are proportional (or in the same ratio).

$$\begin{aligned}\therefore \frac{DE}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{DE}{BC} = \frac{2x}{5x} = \frac{2}{5}.\end{aligned}$$

∴ Assertion (A) is false, reason (R) is true.

Hence, option 2 is the correct option.

In Δ ODQ and Δ OBP

⇒ ∠DOQ = ∠BOP (Common angle)

⇒ ∠OQD = ∠OPB (Both equal to 90°)

∴ Δ ODQ ∼ Δ OBP (By A.A. postulate)

From figure,

OB = a + 2a = 3a

OD = a + 2a + 3a + 4a = 10a

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of Δ ODQ}}{\text{Area of Δ OBP}} = \frac{OD^2}{OB^2}\\[1em] = \frac{(10a)^2}{(3a)^2}\\[1em] = \frac{100a^2}{9a^2}\\[1em] = \frac{100}{9}.\end{aligned}$$

Let the area of Δ OBP be 9m and that of area of Δ ODQ be 100m.

From figure,

Area of trapezium BPQD = Area of Δ ODQ - Area of Δ OBP = 100m - 9m

Now,

$$\begin{aligned}\Rightarrow \frac{\text{Δ OBP}}{\text{Trapezium BDQP}} = \frac{9m}{100m - 9m}\\[1em] = \frac{9m}{(100 - 9)m}\\[1em] = \frac{9}{100 - 9}\end{aligned}$$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

In Δ ADE and Δ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∠ADE = ∠ABC (Both are 90°)

∴ Δ ADE ∼ Δ ABC (By AA postulate)

We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{AE}{AC}\\[1em] \Rightarrow \frac{AB}{AD} = \frac{AC}{AE}\\[1em]\end{aligned}$$

So, both the statements are true.

Hence, option 1 is the correct option.

Given,

Area of triangle ADE = 9 cm² and area of trapezium DBCE = 16 cm².

In Δ ABC and Δ ADE,

⇒ ∠BAC = ∠DAE (Common angle)

⇒ ∠ABC = ∠ADE (Corresponding angles are equal)

⇒ ∠ACB = ∠AED (Corresponding angles are equal)

∴ Δ ABC ∼ Δ ADE (By A.A.A. postulate)

From figure,

⇒ Area of triangle ABC = Area of triangle ADE + Area of trapezium DBCE = 9 + 16 = 25 cm²

We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of Δ ADE}}{\text{Area of Δ ABC}} = \frac{DE^2}{BC^2}\\[1em] \Rightarrow\frac{9}{25} = \frac{DE^2}{BC^2}\\[1em] \Rightarrow \frac{DE}{BC} = \sqrt{\frac{9}{25}} = \frac{3}{5}\end{aligned}$$

So, statement 1 is false but statement 2 is true.

Hence, option 4 is the correct option.

(i) By basic proportionality theorem,

A line drawn parallel to a side of triangle divides the other two sides proportionally.

$$\begin{aligned}\therefore \frac{AY}{YC} = \frac{AX}{XB} \\[1em] = \frac{9}{4.5} \\[1em] = \frac{2}{1}.\end{aligned}$$

Hence, $\frac{AY}{YC} = 2.$

(ii) Given,

$\Rightarrow \frac{AY}{YC} = \frac{2}{1}$

Let AY = 2x and YC = x.

From figure,

AC = AY + YC = 2x + x = 3x.

$\therefore \frac{YC}{AC} = \frac{x}{3x} = \frac{1}{3}$.

Hence, $\frac{YC}{AC} = \frac{1}{3}$.

(iii) In △AXY and △ABC,

∠AXY = ∠ABC [Corresponding angles are equal]

∠A = ∠A [Common]

∴ △AXY ~ △ABC.

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

$$\begin{aligned}\Rightarrow \frac{AX}{AB} = \frac{XY}{BC} \\[1em] \Rightarrow \frac{AX}{AX + XB} = \frac{XY}{BC} \\[1em] \Rightarrow \frac{9}{9 + 4.5} = \frac{XY}{18} \\[1em] \Rightarrow \frac{9}{13.5} = \frac{XY}{18} \\[1em] \Rightarrow XY = \frac{9}{13.5} \times 18 \\[1em] \Rightarrow XY = 12 \text{ cm}. \\[1em]\end{aligned}$$

Hence, XY = 12 cm.

(i) In ΔAEB and ΔFEC,

∠AEB = ∠FEC [Vertically opposite angles are equal]

∠BAE = ∠CFE [Alternate angles are equal]

∴ ∆AEB ~ ∆FEC [By AA]

Since, corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{BE}{EC} = \frac{AB}{FC} \\[1em] \Rightarrow \frac{15}{EC} = \frac{9}{13.5} \\[1em] \Rightarrow EC = \frac{15 \times 13.5}{9} = 22.5\text{ cm}\end{aligned}$$

Hence, EC = 22.5 cm.

(ii) In ΔAPB and ΔFPD,

∠APB = ∠FPD [Vertically opposite angles are equal]

∠BAP = ∠DFP [Alternate angles are equal]

∴ ∆APB ~ ∆FPD [By AA]

Since, corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{AP}{FP} = \frac{AB}{FD} \\[1em] \Rightarrow \frac{AP}{FP} = \frac{AB}{DC + CF} \\[1em] \Rightarrow \frac{6}{FP} = \frac{9}{18 + 13.5} \\[1em] \Rightarrow \frac{6}{FP} = \frac{9}{31.5} \\[1em] \Rightarrow FP = \frac{6 \times 31.5}{9} \\[1em] \Rightarrow FP = \frac{63}{3} = 21\text{ cm}.\end{aligned}$$

From figure,

AF = AP + FP = 6 + 21 = 27 cm.

Hence, AF = 27 cm.

(iii) We already have, ∆AEB ~ ∆FEC

So,

$$\begin{aligned}\Rightarrow \frac{AE}{FE} = \frac{AB}{FC} \\[1em] \Rightarrow \frac{AE}{FE} = \frac{9}{13.5} \\[1em] \Rightarrow \frac{(AF – EF)}{ FE} = \frac{9}{13.5} \\[1em] \Rightarrow \frac{AF}{EF} – 1 = \frac{9}{13.5} \\[1em] \Rightarrow \frac{27}{EF} = \frac{9}{13.5} + 1 \\[1em] \Rightarrow \frac{27}{EF} = \frac{22.5}{13.5} \\[1em] \Rightarrow EF = \frac{(27 \times 13.5)}{22.5} = 16.2 \text{ cm}.\end{aligned}$$

From figure,

PE = PF – EF = 21 – 16.2 = 4.8 cm

Hence, PE = 4.8 cm

Let BD = a and DF = b.

In ΔFDC and ΔFBA,

∠FDC = ∠FBA [Both = 90°]

∠DFC = ∠BFA [Common angle]

∴ ∆FDC ~ ∆FBA [By AA]

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

$$\begin{aligned}\Rightarrow \frac{DF}{BF} = \frac{DC}{AB} \\[1em] \Rightarrow \frac{DF}{BD + DF} = \frac{z}{x} \\[1em] \Rightarrow \frac{b}{a + b} = \frac{z}{x} ..... (1)\end{aligned}$$

In ΔBDC and ΔBFE,

∠BDC = ∠BFE [Both = 90°]

∠DBC = ∠FBE [Common angle]

∴ ∆BDC ~ ∆BFE [By AA]

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

$$\begin{aligned}\Rightarrow \frac{BD}{BF} = \frac{CD}{EF} \\[1em] \Rightarrow \frac{BD}{BD + DF} = \frac{z}{y} \\[1em] \Rightarrow \frac{a}{a + b} = \frac{z}{y} ............ (2)\end{aligned}$$

Adding (1) and (2) we get :

$$\begin{aligned}\Rightarrow \frac{b}{a + b} + \frac{a}{a + b} = \frac{z}{x} + \frac{z}{y} \\[1em] \Rightarrow \frac{a + b}{a + b} = z\Big(\frac{1}{x} + \frac{1}{y}\Big) \\[1em] \Rightarrow 1 = z\Big(\frac{1}{x} + \frac{1}{y}\Big) \\[1em] \Rightarrow \frac{1}{z} = \frac{1}{x} + \frac{1}{y}.\end{aligned}$$

Hence, proved that $\frac{1}{z} = \frac{1}{x} + \frac{1}{y}.$.

Given, ∆ABC ~ ∆PQR

AD and PM are the medians, so BD = DC and QM = MR

Since, corresponding sides of similar triangles are proportional.

$\frac{AB}{PQ} = \frac{BC}{QR}$

We can write,

$\frac{AB}{PQ} = \frac{\frac{BC}{2}}{\frac{QR}{2}} = \frac{BD}{QM}$

And, ∠ABC = ∠PQR i.e., ∠ABD = ∠PQM

∴ ∆ABD ~ ∆PQM [By SAS]

Since, corresponding sides of similar triangles are proportional.

$\frac{AB}{PQ} = \frac{AD}{PM}$.

Hence, proved that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Given, ∆ABC ~ ∆PQR

So,

∠ABC = ∠PQR i.e., ∠ABD = ∠PQM

Also, ∠ADB = ∠PMQ [Both are right angles]

∴ ∆ABD ~ ∆PQM [By AA]

Since, corresponding sides of similar triangles are proportional.

$\frac{AB}{PQ} = \frac{AD}{PM}$.

Hence, proved that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Given, ∆ABC ~ ∆PQR and AD and PM are the angle bisectors.

So,

⇒ ∠A = ∠P

⇒ $\frac{∠\text{A}}{2} = \frac{∠\text{P}}{2}$

⇒ ∠BAD = ∠QPM

Also, ∠ABC = ∠PQR i.e., ∠ABD = ∠PQM.

∴ ∆ABD ~ ∆PQM [By AA]

Since, corresponding sides of similar triangles are proportional.

$\frac{AB}{PQ} = \frac{AD}{PM}$.

Hence, proved that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Given,

∠AXY = ∠AYX

So, AX = AY = a (let) [Sides opposite to equal angles are equal.]

Given,

$$\begin{aligned}\Rightarrow \frac{BX}{AX} = \frac{CY}{AY} \\[1em] \Rightarrow \frac{BX}{a} = \frac{CY}{a} \\[1em] \Rightarrow BX = CY = b.\end{aligned}$$

From figure,

AB = AX + XB = a + b.

AC = AY + CY = a + b.

So, AB = AC.

Hence, proved that ∆ABC is an isosceles triangle.

Given, DE || AC

In △BAC,

By basic proportionality theorem,

$\frac{BE}{EC} = \frac{BD}{AD}$ .....(1)

Given, DC || AP

In △BPA,

By basic proportionality theorem,

$\frac{BC}{CP} = \frac{BD}{DA}$ .....(2)

$\frac{BE}{EC} = \frac{BC}{CP}$.

Hence, proved that $\frac{BE}{EC} = \frac{BC}{CP}$.

(i) In ∆PCD and ∆PEF,

∠CPD = ∠EPF [Vertically opposite angles are equal]

∠DCP = ∠FEP [Alternate angles are equal]

∴ ∆PCD ~ ∆PEF [By AA]

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

$$\begin{aligned}\Rightarrow \frac{CD}{EF} = \frac{PC}{EP} \\[1em] \Rightarrow \frac{27}{EF} = \frac{15}{7.5} \\[1em] \Rightarrow EF = \frac{27 \times 7.5}{15} \\[1em] \Rightarrow EF = 13.5 \text{ cm}.\end{aligned}$$

Hence, EF = 13.5 cm.

(ii) In ∆CEF and ∆CAB

∠FCE = ∠BCA [Common angles]

∠CFE = ∠CBA [As FE || BA, corresponding angles are equal]

∴ ∆CEF ~ ∆CAB [By AA]

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

$$\begin{aligned}\Rightarrow \frac{EC}{AC} = \frac{EF}{AB} \\[1em] \Rightarrow \frac{PC + PE}{AC} = \frac{13.5}{22.5} \\[1em] \Rightarrow \frac{15 + 7.5}{AC} = \frac{13.5}{22.5} \\[1em] \Rightarrow \frac{22.5}{AC} = \frac{13.5}{22.5} \\[1em] \Rightarrow AC = \frac{22.5 \times 22.5}{13.5} \\[1em] \Rightarrow AC = 37.5 \text{ cm}.\end{aligned}$$

Hence, AC = 37.5 cm.

(i) In ∆ACD and ∆BCA,

∠DAC = ∠ABC [Given]

∠ACD = ∠BCA [Common angles]

∴ ∆ACD ~ ∆BCA [By AA]

Hence, proved that ∆ACD ~ ∆BCA.

(ii) Since, ∆ACD ~ ∆BCA

We know that,

Corresponding sides of similar triangle are proportional.

$$\begin{aligned}\Rightarrow \frac{AC}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{4}{BC} = \frac{5}{8} \\[1em] \Rightarrow BC = \frac{4 \times 8}{5} \\[1em] \Rightarrow BC = \frac{32}{5} = 6.4 \text{ cm}.\end{aligned}$$

Also,

$$\begin{aligned}\Rightarrow \frac{CD}{AC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{CD}{4} = \frac{5}{8} \\[1em] \Rightarrow CD = \frac{4 \times 5}{8} \\[1em] \Rightarrow CD = \frac{5}{2} = 2.5 \text{ cm}.\end{aligned}$$

Hence, BC = 6.4 cm and CD = 2.5 cm.

(iii) As, ∆ACD ~ ∆BCA

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

$$\begin{aligned}\frac{\text{Area of ∆ACD}}{\text{Area of ∆ABC}} = \frac{AD^2}{AB^2} \\[1em] = \frac{5^2}{8^2} \\[1em] = \frac{25}{64}.\end{aligned}$$

Hence, area of ∆ACD : area of ∆ABC = 25 : 64.

(i) In ∆BFD and ∆BEC,

∠BFD = ∠BEC [Corresponding angles are equal]

∠FBD = ∠EBC [Common]

∴ ∆BFD ~ ∆BEC [By AA].

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \frac{BF}{BE} = \frac{BD}{BC}$

$\Rightarrow \frac{BF}{BE} = \frac{1}{2}$ [∵ AD is median so D is the mid-point of BC]

$\Rightarrow BE = 2BF$

From figure,

⇒ BE = BF + FE

⇒ 2BF = BF + FE

⇒ BF = FE.

Hence, proved that EF = FB.

(ii) In ∆AFD, EG || FD.

By basic proportionality theorem we have,

$\frac{AE}{EF} = \frac{AG}{GD}$ .....(1)

Now, AE = EB [∵ CE is median so E is the mid-point of AB]

As, AE = EB = 2EF [As, EF = FB].

Substituting value of AE in (1) we get,

$\frac{AG}{GD} = \frac{2EF}{EF} = \frac{2}{1}$.

Hence, AG : GD = 2 : 1.

Let's consider two similar triangles as ∆ABC ~ ∆PQR

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

So,

$\frac{\text{Area of ∆ABC}}{\text{Area of ∆PQR}} = \Big(\frac{AB}{PQ}\Big)^2 = \Big(\frac{BC}{QR}\Big)^2 = \Big(\frac{AC}{PR}\Big)^2$

Since,

Area of ∆ABC = Area of ∆PQR [Given]

Hence,

AB = PQ

BC = QR

AC = PR

So, as the respective sides of two similar triangles are all of same length.

We can conclude that,

∆ABC ≅ ∆PQR [By SSS rule]

Hence proved that both triangles are congruent.

Let △ABC and △PQR be two similar triangles with AD and PS as perpendiculars.

So,

∠ABD = ∠PQS [As ∠ABC = ∠PQR]

∠ADB = ∠PSQ [Both = 90°]

So, △ABD ~ △PQS.

$\therefore \frac{AB}{PQ} = \frac{AD}{PS} = \frac{3}{5}$.

(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Hence, the required ratio = 3 : 5.

(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Hence, the required ratio = 3 : 5.

(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Ratio = (3)² : (5)² = 9 : 25.

Hence, the required ratio = 9 : 25.

We know that,

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Given,

Ratio between areas = 16 : 25.

So, ratio between sides = $\sqrt{16} : \sqrt{25}$ = 4 : 5.

(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Hence, the required ratio = 4 : 5.

(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

Hence, the required ratio = 4 : 5.

(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Hence, the required ratio = 4 : 5.

The rectangular plot of land ABCD on the map is shown below:

(i) By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 24² + 32²

⇒ AC² = 576 + 1024

⇒ AC² = 1600

⇒ AC = $\sqrt{1600}$ = 40 cm.

Given,

Scale (k) = $\frac{1}{20000}$

Length of diagonal distance in map = k × Length of diagonal of plot

40 = $\frac{1}{20000}$ × Length of diagonal of plot

Length of diagonal of plot = 40 × 20000 = 800000 cm = $\frac{800000}{100000}$ km = 8 km.

Hence, diagonal distance of plot = 8 km.

(ii) Area of plot on map = AB × BC = 24 × 32 = 768 cm².

Area of plot on map = k² × Area of actual plot

768 = $\Big(\frac{1}{20000}\Big)^2$ × Area of actual plot

Area of actual plot = 768 × 20000 × 20000 cm²

= $\frac{768 \times 20000 \times 20000}{100000 \times 100000}$ km²

= 30.72 km².

Hence, area of plot = 30.72 km².

Given,

Scale factor (k) = 1 : 50.

Dimensions of model = l × b × h = 1 m × 0.6 m × 1.2 m.

By formula,

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{Length of building's model}}{\text{Length of building}} = k \\[1em] \Rightarrow \frac{1}{\text{Length of building}} = \frac{1}{50} \\[1em] \Rightarrow \text{Length of building} = 50 \text{ m}. \\[2em] \phantom{\Rightarrow} \frac{\text{Breadth of building's model}}{\text{Breadth of building}} = k \\[1em] \Rightarrow \frac{0.6}{\text{Breadth of building}} = \frac{1}{50} \\[1em] \Rightarrow \text{Breadth of building} = 50 \times 0.6 = 30 \text{ m}. \\[2em] \phantom{\Rightarrow} \frac{\text{Height of building's model}}{\text{Height of building}} = k \\[1em] \Rightarrow \frac{1.2}{\text{Height of building}} = \frac{1}{50} \\[1em] \Rightarrow \text{Height of building} = 50 \times 1.2 = 60 \text{ m}.\end{aligned}$$

Dimensions of building = 50 m × 30 m × 60 m.

(i) By formula,

Floor area of model room = k² × Floor area of building room

50 = $\frac{1}{50} \times \frac{1}{50}$ × Floor area of building room

Floor area of building room = 50 × 50 × 50 = 125000 cm² = $\frac{125000}{100 \times 100}$ m² = 12.5 m².

Hence, floor area of a room of the building = 12.5 m².

(ii) By formula,

Volume of a room of model = k³ × Volume of a room of building

Volume of a room of model = $\frac{1}{50} \times \frac{1}{50} \times \frac{1}{50} \times 90$ = 0.00072 m³ = 0.00072 x 100 x 100 x 100 cm³ = 720 cm³.

Hence, the space inside a room of the model = 720 cm³.

(i) In △PQL and △RPM

∠LPQ = ∠MRP (Given ∠LPQ = ∠QRP)

∠LQP = ∠RPM (Given ∠RPM = ∠RQP)

∴ △PQL ~ △RPM [By AA]

Hence, proved that △PQL ~ △RPM.

(ii) Since, △PQL ~ △RPM and corresponding sides of similar triangle are proportional to each other.

$$\begin{aligned}\therefore \frac{QL}{PM} = \frac{PL}{RM} \\[1em] \Rightarrow QL \times RM = PL \times PM.\end{aligned}$$

Hence, proved that QL × RM = PL × PM.

(iii) In △PQL and △RQP

∠LPQ = ∠QRP (Given)

∠Q = ∠Q [Common]

∴ △PQL ~ △RQP [By AA]

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

$$\begin{aligned}\therefore \frac{PQ}{RQ} = \frac{QL}{QP} \\[1em] \Rightarrow PQ^2 = QR \times QL.\end{aligned}$$

Hence, proved that PQ² = QR x QL.

Longest side in △ABC = BC = 6 cm

Corresponding longest side in △DEF = EF = 9 cm.

Scale factor (k) = $\frac{EF}{BC} = \frac{9}{6} = \frac{3}{2}$ = 1.5

Triangle ABC is enlarged to DEF. So, the two triangles will be similar.

$\therefore \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{2}{3}$.

So,

$$\begin{aligned}\phantom{\Rightarrow} \frac{AB}{DE} = \frac{2}{3} \\[1em] \Rightarrow DE = \frac{3}{2}AB \\[1em] = \frac{9}{2} \\[1em] = 4.5 \text{ cm}. \\[2em] \phantom{\Rightarrow} \frac{AC}{DF} = \frac{2}{3} \\[1em] \Rightarrow DF = \frac{3}{2}AC \\[1em] = \frac{12}{2} \\[1em] = 6 \text{ cm}. \\[1em]\end{aligned}$$

Hence, DE = 4.5 cm and DF = 6 cm.

Let △ABC and △PQR be two isosceles triangles with AB = AC and PQ = PR.

Then,

$\frac{AB}{AC} = \frac{PQ}{PR} = 1$.

or,

$\frac{AB}{PQ} = \frac{AC}{QR}$

Also,

∠A = ∠P (Given)

∴ △ABC ~ △PQR (By SAS)

We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore\frac{\text{Area of ∆ABC}}{\text{Area of ∆PQR}} = \Big(\frac{AB}{PQ}\Big)^2 \\[1em] \Rightarrow \frac{16}{25} = \Big(\frac{AB}{PQ}\Big)^2 \\[1em] \Rightarrow \frac{AB}{PQ} = \sqrt{\frac{16}{25}} = \frac{4}{5}.\end{aligned}$$

We know that,

The ratio between sides of similar triangle is equal to ratio of their altitudes.

Hence, ratio between altitudes = 4 : 5.

(i) In △APO and △ABC,

∠APO = ∠ABC [Corresponding angles are equal]

∠PAO = ∠BAC [Common angle]

∴ △APO ~ △ABC [By AA].

Given,

$\frac{AP}{PB} = \frac{2}{3}$

Let AP = 2x and PB = 3x.

From figure,

AB = AP + PB = 2x + 3x = 5x.

We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore\frac{\text{Area of ∆APO}}{\text{Area of ∆ABC}} = \Big(\frac{AP}{AB}\Big)^2 \\[1em] = \Big(\frac{2x}{5x}\Big)^2 \\[1em] = \Big(\frac{2}{5}\Big)^2 \\[1em] = \frac{4}{25}.\end{aligned}$$

Hence, area △APO : area △ABC = 4 : 25.

(ii) Since, PO || BC.

In △ABC,

By basic proportionality theorem,

$\frac{AO}{OC} = \frac{AP}{PB} = \frac{2}{3}$.

In △APO and △COQ,

∠AOP = ∠COQ [Vertical opposite angles are equal]

∠APO = ∠CQO [Alternate angles are equal]

∴ △APO ~ △COQ [By AA].

We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore\frac{\text{Area of ∆APO}}{\text{Area of ∆COQ}} = \Big(\frac{AO}{OC}\Big)^2 \\[1em] = \Big(\frac{2}{3}\Big)^2 \\[1em] = \frac{4}{9}.\end{aligned}$$

Hence, area △APO : area △COQ = 4 : 9.

(i) In △ABC and △EBD,

⇒ ∠EDB = ∠ACB [Given]

⇒ ∠DBE = ∠ABC [Common]

∴ △ABC ~ △EBD

From figure,

⇒ BC = BE + EC = 6 + 4 = 10 cm.

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

$$\begin{aligned}\Rightarrow \frac{AB}{BE} = \frac{BC}{BD} \\[1em] \Rightarrow AB = \frac{BC \times BE}{BD} \\[1em] \Rightarrow AB = \frac{10 \times 6}{5} \\[1em] \Rightarrow AB = 12 \text{ cm}.\end{aligned}$$

Hence, length of AB = 12 cm.

(ii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of ∆ABC}}{\text{Area of ∆BED}} = \Big(\frac{AB}{BE}\Big)^2 \\[1em] \frac{\text{Area of ∆ABC}}{9} = \Big(\frac{12}{6}\Big)^2 \\[1em] \text{Area of ∆ABC} = \Big(\frac{12}{6}\Big)^2 \times 9 \\[1em] \text{Area of ∆ABC} = 36 \text{ cm}^2.\end{aligned}$$

Hence, area of ∆ABC = 36 cm².

(i) Let ∠CAD = x.

So, ∠DAB = 90° - x.

In △ABD,

⇒ ∠DAB + ∠ADB + ∠ABD = 180° [By angle sum property]

⇒ 90° - x + 90° + ∠ABD = 180°

⇒ 180° - x + ∠ABD = 180°

⇒ ∠ABD = x + 180° - 180°

⇒ ∠ABD = x.

In △ADB and △CDA,

⇒ ∠CAD = ∠ABD (Both = x)

⇒ ∠CDA = ∠ADB (Both = 90°)

∴ △ADB ~ △CDA [By AA]

Hence, proved that △ADB ~ △CDA.

(ii) Since, △ADB ~ △CDA and corresponding sides of similar triangles are proportional to each other.

$\frac{BD}{AD} = \frac{AD}{CD}$

⇒ AD² = BD × CD

⇒ AD² = 18 × 8

⇒ AD² = 144

⇒ AD = $\sqrt{144}$ = 12 cm.

Hence, AD = 12 cm.

(iii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore\frac{\text{Area of ∆ADB}}{\text{Area of ∆CDA}} = \Big(\frac{AD}{CD}\Big)^2 \\[1em] = \Big(\frac{12}{8}\Big)^2 \\[1em] = \Big(\frac{3}{2}\Big)^2 \\[1em] = \frac{9}{4}.\end{aligned}$$

Hence, ratio of the area of △ADB to area of △CDA = 9 : 4.

(i) In △ADE and △ACB,

⇒ ∠AED = ∠ABC [Both = 90°]

⇒ ∠EAD = ∠CAB [Common angle]

∴ △ADE ~ △ACB [By AA].

Hence, proved that △ADE ~ △ACB.

(ii) In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ 13² = AB² + 5²

⇒ 169 = AB² + 25

⇒ AB² = 144

⇒ AB = $\sqrt{144}$ = 12 cm.

Since, △ADE ~ △ACB and corresponding sides of similar triangles are proportional to each other.

$$\begin{aligned}\therefore \frac{AE}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{DE}{5} = \frac{4}{12} \\[1em] \Rightarrow DE = \frac{20}{12} = \frac{5}{3} = 1\frac{2}{3} \text{ cm}. \\[1em] \Rightarrow \frac{AD}{AC} = \frac{AE}{AB} \\[1em] \Rightarrow \frac{AD}{13} = \frac{4}{12} \\[1em] \Rightarrow AD = \frac{52}{12} \\[1em] \Rightarrow AD = \frac{13}{3} = 4\frac{1}{3} \text{ cm}.\end{aligned}$$

Hence, DE = $1\frac{2}{3} \text{ cm and } AD = 4\frac{1}{3}$ cm.

(iii) From figure,

Area of △ADE = $\frac{1}{2} \times AE \times DE$

$$\begin{aligned}= \frac{1}{2} \times 4 \times \frac{5}{3} \\[1em] = \frac{10}{3} \text{ cm}^2 \\[1em] \text{Area of △ABC} = \frac{1}{2} \times BC \times AB \\[1em] = \frac{1}{2} \times 5 \times 12 \\[1em] = 30 \text{ cm}^2.\end{aligned}$$

Area of quadrilateral BCED = Area of △ABC - Area of △ADE

$$\begin{aligned}= 30 - \frac{10}{3} \\[1em] = \frac{90 - 10}{3} \\[1em] = \frac{80}{3} \text{ cm}^2. \\[1em] \frac{\text{Area of △ADE}}{\text{Area of quadrilateral BCED}} = \frac{\frac{10}{3}}{\frac{80}{3}} \\[1em] = \frac{10}{80} = \frac{1}{8}.\end{aligned}$$

Hence, area of △ADE : area of quadrilateral BCED = 1 : 8.

(i) In △ABG, DE || AB.

So, by basic proportionality theorem we get,

$\frac{DG}{AD} = \frac{GE}{BE}$ .....(1)

In △BCG, EF || BC.

So, by basic proportionality theorem we get,

$\frac{GE}{BE} = \frac{FG}{CF}$ .....(2)

From (1) and (2) we get,

$$\begin{aligned}\Rightarrow \frac{DG}{AD} = \frac{FG}{CF} \\[1em] \Rightarrow \frac{AD}{DG} = \frac{CF}{FG}.\end{aligned}$$

Hence, proved that $\frac{AD}{DG} = \frac{CF}{FG}.$

(ii) In △DFG and △ACG,

⇒ $\frac{AD}{DG} = \frac{CF}{FG}$ [Proved above]

⇒ ∠DGF = ∠AGC [Common angle]

∴ △DFG ~ △ACG [By SAS]

Hence, proved that △DFG ~ △ACG.

(i) In △PQR and △SPR,

⇒ ∠PSR = ∠QPR [Given]

⇒ ∠PRQ = ∠PRS [Common angle]

∴ △PQR ~ △SPR [By AA]

Hence, proved that △PQR ~ △SPR.

(ii) Since, △PQR ~ △SPR and corresponding sides of similar triangle are proportional to each other.

$$\begin{aligned}\Rightarrow \frac{QR}{PR} = \frac{PR}{SR} \\[1em] \Rightarrow \frac{QR}{6} = \frac{6}{3} \\[1em] \Rightarrow QR = \frac{36}{3} = 12 \text{ cm}. \\[1em] \text{ Also}, \\[1em] \Rightarrow \frac{PQ}{SP} = \frac{PR}{SR} \\[1em] \Rightarrow \frac{8}{SP} = \frac{6}{3} \\[1em] \Rightarrow SP = \frac{24}{6} = 4 \text{ cm}.\end{aligned}$$

Hence, QR = 12 cm and PS = 4 cm.

(iii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$$\begin{aligned}\therefore\frac{\text{Area of ∆PQR}}{\text{Area of ∆SPR}} = \Big(\frac{PQ}{SP}\Big)^2 \\[1em] = \Big(\frac{8}{4}\Big)^2 \\[1em] = (2)^2 \\[1em] = 4.\end{aligned}$$

Hence, $\frac{\text{area of △PQR}}{\text{area of △SPR}}$ = 4 : 1.

Since the lamp post, tower, and Rohan are all vertical to the ground and sunlight makes the same angle, the triangles formed are similar.

(i) Thus,

△ABC ∼ △DEF

We know that,

In similar triangles, corresponding sides are proportional (in the same ratio).

$$\begin{aligned}\Rightarrow \frac{AB}{BC} = \frac{DE}{EF} \\[1em] \Rightarrow \frac{5}{4} = \frac{DE}{28} \\[1em] \Rightarrow DE = \frac{5}{4} \times 28 = 5 \times 7 = 35 \text{ m}.\end{aligned}$$

Hence, the height of the tower = 35 m.

(ii) Thus,

△ABC ∼ △PQR

In similar triangles, corresponding sides are proportional (in the same ratio).

$$\begin{aligned}\Rightarrow \frac{AB}{BC} = \frac{PQ}{QR} \\[1em] \Rightarrow \frac{5}{4} = \frac{PQ}{1} \\[1em] \Rightarrow PQ = \frac{5}{4} \times 1 = \frac{5}{4} = 1.25 \text{ m}.\end{aligned}$$

Hence, Rohan's height = 1.25 m.

(iii) When tower casts a shadow of 35 m,

Due to similarity,

$$\begin{aligned}\Rightarrow \frac{\text{Height of tower}}{\text{Length of shadow of tower}} = \frac{\text{Height of lamp}}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow \frac{35}{35} = \frac{5}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow 1 = \frac{5}{\text{Length of shadow of lamp}} \\[1em] \Rightarrow \text{Length of shadow of lamp} = 5 \text{ m}.\end{aligned}$$

Hence, length of shadow of lamp = 5 m.