Similarity of Triangles

(i) By basic proportionality theorem,

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

Since, XY || BC

$$\begin{aligned}\therefore \frac{AY}{YC} = \frac{AX}{XB} \\[1em] = \frac{3}{1.5} \\[1em] = \frac{2}{1}.\end{aligned}$$

Hence, $\frac{AY}{YC} = \frac{2}{1}$.

(ii) In ΔAXY and ΔABC,

∠AXY = ∠ABC [Corresponding angles are equal]

∠XAY = ∠BAC [Common ]

∴ ΔAXY ∼ ΔABC.

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

$$\begin{aligned}\Rightarrow \frac{AX}{AB} = \frac{XY}{BC} \\[1em] \Rightarrow \frac{AX}{AX + XB} = \frac{XY}{BC} \\[1em] \Rightarrow \frac{3}{3 + 1.5} = \frac{XY}{6} \\[1em] \Rightarrow \frac{3}{4.5} = \frac{XY}{6} \\[1em] \Rightarrow XY = \frac{3}{4.5} \times 6 \\[1em] \Rightarrow XY = 4 \text{ cm.}\end{aligned}$$

Hence, XY = 4 cm.

By basic proportionality theorem,

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

(i) Given,

AD = 3.6 cm

AB = 9 cm

AE = 2.4 cm

Since, DE || BC by basic proportionality theorem,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{AD}{AB - AD} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{3.6}{9 - 3.6} = \frac{2.4}{EC}\\[1em] \Rightarrow \frac{3.6}{5.4} = \frac{2.4}{EC}\\[1em] \Rightarrow EC = \frac{5.4 \times 2.4}{3.6}\\[1em] \Rightarrow EC = 3.6 \text{ cm.}\end{aligned}$$

Hence, EC = 3.6 cm.

(ii) Given,

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

AC = 5.6 cm

Since, DE || BC by basic proportionality theorem,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{AE}{AC - AE} \\[1em] \Rightarrow \frac{3}{5} = \frac{AE}{5.6 - AE}\\[1em] \Rightarrow 3 \times (5.6 - AE) = 5AE \\[1em] \Rightarrow 16.8 - 3AE = 5AE \\[1em] \Rightarrow 5AE + 3AE = 16.8 \\[1em] \Rightarrow 8AE = 16.8 \\[1em] \Rightarrow AE = \frac{16.8}{8} \\[1em] \Rightarrow AE = 2.1 \text{ cm.}\end{aligned}$$

Hence, AE = 2.1 cm.

(iii) Given,

AD = x cm

DB = (x − 2) cm

AE = (x + 2) cm

EC = (x − 1) cm

Since, DE || BC by basic proportionality theorem,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{x}{(x - 2)} = \frac{(x + 2)}{(x - 1)} \\[1em] \Rightarrow x(x - 1) = (x + 2) (x - 2) \\[1em] \Rightarrow x^2 - x = (x^2 - (2)^2) \\[1em] \Rightarrow x^2 - x = x^2 - 4 \\[1em] \Rightarrow x^2 - x - x^2 + 4 = 0 \\[1em] \Rightarrow - x + 4 = 0 \\[1em] \Rightarrow x = 4 \text{ cm.}\end{aligned}$$

Hence, x = 4 cm.

By basic proportionality theorem,

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

(i) Given,

AD = 5.7 cm

BD = 9.5 cm

AE = 3.6 cm

EC = 6 cm.

Check for proportionality,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{5.7}{9.5} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{3}{5} = 0.6 \\[1em] \Rightarrow \frac{AE}{EC} = \frac{3.6}{6.0} \\[1em] \Rightarrow \frac{AE}{EC} = \frac{3}{5} = 0.6 \\[1em] \therefore \frac{AD}{DB} = \frac{AE}{EC}.\end{aligned}$$

We conclude that DE is parallel to BC

Hence, DE is parallel to BC.

(ii) Given,

AB = 5.6 cm

AD = 1.4 cm

AC = 9.6 cm

EC = 2.4 cm.

From figure,

DB = AB - AD = 5.6 - 1.4 = 4.2 cm

AE = AC - EC = 9.6 - 2.4 = 7.2 cm

Check for proportionality,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{1.4}{4.2} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{1}{3} \\[1em] \Rightarrow \frac{AE}{EC} = \frac{7.2}{2.4} \\[1em] \Rightarrow \frac{AE}{EC} = 3 \\[1em] \therefore \frac{AD}{DB} \ne \frac{AE}{EC}.\end{aligned}$$

We conclude that DE is not parallel to BC

Hence, DE is not parallel to BC.

(iii) Given,

AB = 11.7 cm

BD = 5.2 cm

AE = 4.4 cm

AC = 9.9 cm.

From figure,

AD = AB - BD = 11.7 - 5.2 = 6.5 cm

EC = AC - AE = 9.9 - 4.4 = 5.5 cm

Check for proportionality,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{6.5}{5.2} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{5}{4} = 1.25 \\[1em] \Rightarrow \frac{AE}{EC} = \frac{4.4}{5.5} \\[1em] \Rightarrow \frac{AE}{EC} = \frac{4}{5} = 0.8 \\[1em] \therefore \frac{AD}{DB} \ne \frac{AE}{EC}.\end{aligned}$$

We conclude that DE is not parallel to BC

Hence, DE is not parallel to BC.

(iv) Given,

AB = 10.8 cm

BD = 4.5 cm

AC = 4.8 cm

AE = 2.8 cm.

AD = AB - BD = 10.8 - 4.5 = 6.3 cm

EC = AC - AE = 4.8 - 2.8 = 2 cm

Check for proportionality,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{6.3}{4.5} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{7}{5} = 1.4 \\[1em] \Rightarrow \frac{AE}{EC} = \frac{2.8}{2.0} \\[1em] \Rightarrow \frac{AE}{EC} = \frac{7}{5} = 1.4 \\[1em] \therefore \frac{AD}{DB} = \frac{AE}{EC}.\end{aligned}$$

We conclude that DE is parallel to BC

Hence, DE is parallel to BC.

In ΔPQD and ΔABD,

∠ABD = ∠PQD = 90° [From figure]

∠ADB = ∠PDQ [Common angles]

∴ ΔPQD ∼ ΔABD (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{PQ}{AB} = \frac{QD}{BD} \\[1em] \Rightarrow \frac{z}{x} = \frac{QD}{BD} \text{ .....(1)}\end{aligned}$$

In ΔPQB and ΔCDB,

∠CDB = ∠PQB = 90° [Given]

∠CBD = ∠PBQ [Common angles]

∴ ΔPQB ∼ ΔCDB (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{PQ}{CD} = \frac{BQ}{BD} \\[1em] \Rightarrow \frac{z}{y} = \frac{BQ}{BD} \text{ ....(2)}\end{aligned}$$

Add equations (1) and (2) we get,

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

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

Construction: Draw a line through C parallel to AD, meeting BA produced at E.

In ΔBCE,

Since AD ∥ CE, by Basic Proportionality Theorem:

$\frac{BD}{DC} = \frac{AB}{AE}$ ...(i)

Also, since AD ∥ CE:

∠BAD = ∠AEC [Corresponding angles are equal]

∠DAC = ∠ACE [Alternate interior angles are equal]

Since AD is bisector of ∠A, ∠BAD = ∠DAC.

Therefore, ∠AEC = ∠ACE.

In ΔACE, sides opposite to equal angles are equal:

AE = AC ...(ii)

Substitute (ii) into (i):

$\frac{BD}{DC} = \frac{AB}{AC}$

Given,

AC = 6 cm

BC = 10 cm

BD = 6 cm

DC = BC - BD [from figure]

DC = 10 - 6 = 4 cm

Let length of AB be x,

$$\begin{aligned}\Rightarrow \frac{BD}{DC} = \frac{AB}{AC} \\[1em] \Rightarrow \frac{6}{4} = \frac{x}{6} \\[1em] \Rightarrow \frac{3}{2} = \frac{x}{6} \\[1em] \Rightarrow \frac{3}{2} \times 6 = x \\[1em] \Rightarrow x = 9.\end{aligned}$$

Hence, length of AB is 9 cm.

(i) In ΔGED and ΔGBF,

∠DGE = ∠BGF [Vertically opposite angles are equal]

∠GED = ∠GBF [Alternate angles are equal]

∴ ΔGED ∼ ΔGBF (By A.A. axiom)

Hence, proved ΔGED ∼ ΔGBF.

(ii) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{DE}{BF} = \frac{EG}{GB} \\[1em] \Rightarrow \frac{DE}{30} = \frac{8}{16} \\[1em] \Rightarrow \frac{DE}{30} = \frac{1}{2} \\[1em] \Rightarrow DE = \frac{1}{2} \times 30 \\[1em] \Rightarrow DE = 15 \text{ cm}.\end{aligned}$$

Hence, DE = 15 cm.

(iii) In ΔDBE and ΔABC,

∠EBD = ∠CBA [Common angles]

∠BDE = ∠BAC [Corresponding angles are equal, Since AC ∥ DE]

ΔDBE ∼ ΔABC (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{DB}{AB} = \frac{DE}{AC} \\[1em] \Rightarrow \frac{DB}{AB} = \frac{15}{24} \\[1em] \Rightarrow \frac{DB}{AB} = \frac{5}{8}.\end{aligned}$$

Hence, DB : AB = 5 : 8.

(i) Given,

ABCD is a parallelogram.

∴ AB || DC and AD || BC

In ΔDPC and ΔLPB,

∠PDC = ∠PLB [Alternate angles are equal]

∠DPC = ∠LPB [Vertically opposite angles are equal]

∴ ΔDPC ∼ ΔLPB (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$\frac{DP}{LP} = \frac{DC}{LB}$

DP : PL = DC : BL

Hence, proved DP : PL = DC : BL.

(ii) In ΔALD and ΔCPD,

∠ALD = ∠PDC [Alternate interior angles, AB || DC]

∠DAL = ∠PCD [Opposite angles of a parallelogram are equal]

∴ ΔLAD ∼ ΔDCP (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$\frac{DL}{DP} = \frac{AL}{DC}$

DL : DP = AL : DC

Hence, proved DL : DP = AL : DC.

Since, ABCD is a || gm

∴ AD || BC

In ΔADF and ΔEBF,

∠ADF = ∠EBF [Alternate angles are equal]

∠AFD = ∠EFB [Vertically opposite angles are equal]

∴ ΔADF ∼ ΔEBF (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$\frac{DF}{FB} = \frac{FA}{FE}$

DF × FE = FB × FA

Hence, proved that DF × FE = FB × FA.

(i) Given,

PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5 cm.

PR = PS + SR = 4 + 2 = 6 cm

PQ = PT + TQ = 3 + 5 = 8 cm

In ΔPQR and ΔPST,

$$\begin{aligned}\Rightarrow \frac{PQ}{PS} = \frac{8}{4} = 2 \\[1em] \Rightarrow \frac{PR}{PT} = \frac{6}{3} = 2 \\[1em] \therefore \frac{PQ}{PS} = \frac{PR}{PT}\end{aligned}$$

∠QPR = ∠SPT [Common angle]

∴ ΔPQR ∼ ΔPST [By S.A.S. axiom]

Hence, ΔPQR ∼ ΔPST.

(ii) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{QR}{ST} = \frac{PQ}{PS} \\[1em] \Rightarrow \frac{5.8}{ST} = \frac{8}{4} \\[1em] \Rightarrow ST = \frac{5.8}{2} \\[1em] \Rightarrow ST = 2.9 \text{ cm.}\end{aligned}$$

Hence, ST = 2.9 cm.

(i) In ΔCLB and ΔALN,

∠CBL = ∠ANL [Alternate angles are equal]

∠CLB = ∠ALN [Vertically opposite angles are equal]

∴ ΔCLB ∼ ΔALN (By A.A. axiom)

Hence, proved that ΔCLB ∼ ΔALN.

(ii) In ΔCLM and ΔALB,

∠LMC = ∠LAB [Alternate angles, since CM ∥ AB, transversal LM]

∠CLM = ∠ALB [Vertically opposite angles are equal]

∴ ΔCLM ∼ ΔALB (By A.A. axiom)

Hence, proved that ΔCLM ∼ ΔALB.

In ΔOPQ,

AB ∥ PQ [Given]

By Basic Proportionality Theorem,

$\frac{OA}{AP} = \frac{OB}{BQ}$ .... (i)

In ΔOPR,

AC ∥ PR [Given]

By Basic Proportionality Theorem,

$\frac{OA}{AP} = \frac{OC}{CR}$ .... (ii)

From (i) and (ii), we get:

$\frac{OB}{BQ} = \frac{OC}{CR}$

In ΔOQR,

Since $\frac{OB}{BQ} = \frac{OC}{CR}$

By Converse of Basic Proportionality Theorem,

BC ∥ QR

Hence, proved that BC ∥ QR.

(i) In ∆CFD and ∆CEB,

∠CDF = ∠CBE [Corresponding angles are equal]

∠FCD = ∠ECB [Common]

∴ ∆CFD ∼ ∆CEB (By A.A. axiom)

Since, corresponding sides of similar triangles are proportional.

$$\begin{aligned}\Rightarrow \frac{CF}{CE} = \frac{CD}{CB} \\[1em] \Rightarrow \frac{CF}{CE} = \frac{1}{2} \text{ [∵ D is the mid-point of BC]} \Rightarrow CE = 2CF.\end{aligned}$$

From figure,

⇒ CE = CF + FE

⇒ 2CF = CF + FE

⇒ CF = FE.

Hence, proved that EF = FC.

(ii) In ∆AFD,

GE ∥ DF

By basic proportionality theorem we have,

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

Now, AE = EC [∵ BE is media,n so E is the mid-point of AC]

As, AE = EC = 2EF [As, EF = FC].

Substituting value of AE in (1) we get,

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

Hence, proved that AG : GD = 2 : 1.

(i) Given, DE ∥ BC.

⇒ ∠ADE = ∠DBC [Corresponding angles are equal] ... (1)

⇒ ∠EDC = ∠DCB [Alternate interior angles are equal] ... (2)

Given, BD = DC.

⇒ ∠DBC = ∠DCB [Angles opposite to equal sides are equal] ... (3)

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

∠ADE = ∠EDC

Thus, DE bisects ∠ADC.

Hence, DE bisects ∠ADC.

(ii) Given,

AD = 4.5 cm, AE = 3.9 cm, DC = 7.5 cm

Given,

BD = DC = 7.5 cm

By basic proportionality theorem we have,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{4.5}{7.5} = \frac{3.9}{EC} \\[1em] \Rightarrow 4.5 \times EC = 3.9 \times 7.5 \\[1em] \Rightarrow 4.5 \times EC = 29.25 \\[1em] \Rightarrow EC = \frac{29.25}{4.5} \\[1em] \Rightarrow EC = 6.5 \text{ cm.}\end{aligned}$$

Hence, CE = 6.5 cm.

(iii) By basic proportionality theorem,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{3.9}{6.5} \\[1em] \Rightarrow \frac{AD}{DB} = \frac{3}{5}.\end{aligned}$$

Hence, the ratio AD : DB is 3 : 5.

Considering ΔAOB and ΔDOC,

∠AOB = ∠COD [Vertically opposite angles are equal]

∠A = ∠D [Alternate angles are equal]

∴ ΔAOB ∼ ΔDOC (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \frac{OA}{OD} = \frac{OB}{OC} = \frac{AB}{DC}$

Considering,

$$\begin{aligned}\Rightarrow \frac{AB}{DC} = \frac{OA}{OD} \\[1em] \Rightarrow \frac{4}{3} = \frac{OA}{3.6} \\[1em] \Rightarrow OA = \frac{4 \times 3.6}{3} \\[1em] \Rightarrow OA = \frac{14.4}{3} \\[1em] \Rightarrow OA = 4.8 \text{ cm}.\end{aligned}$$

Considering,

$$\begin{aligned}\Rightarrow \frac{AB}{DC} = \frac{OB}{OC} \\[1em] \Rightarrow \frac{4}{3} = \frac{OB}{5.7} \\[1em] \Rightarrow OB = \frac{4 \times 5.7}{3} \\[1em] \Rightarrow OB = \frac{22.8}{3} \\[1em] \Rightarrow OB = 7.6 \text{ cm}.\end{aligned}$$

Hence, OA = 4.8 cm and OB = 7.6 cm.

We have, ∠ABC = 90° and BD ⟂ AC

In ΔABC and ΔBDC,

∠ABC = ∠BDC [Each 90°]

∠ACB = ∠BCD [Common]

∴ ΔABC ∼ ΔBDC (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{AB}{BD} = \frac{BC}{DC} \\[1em] \Rightarrow \frac{5.7}{3.8} = \frac{BC}{5.4} \\[1em] \Rightarrow BC = \frac{5.7 \times 5.4}{3.8}\\[1em] \Rightarrow BC = \frac{30.78}{3.8}\\[1em] \Rightarrow BC = 8.1 \text{ cm}.\end{aligned}$$

Hence, BC = 8.1 cm.

(i) Considering ΔADE and ΔABC,

∠BAC = ∠DAE [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

Hence, proved that ΔADE ∼ ΔABC.

(ii) Given,

$$\begin{aligned}\Rightarrow AD = \frac{1}{2} BD \\[1em] \Rightarrow AD = \frac{1}{2}(AB - AD) \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \frac{AD}{AB} = \frac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.\end{aligned}$$

Since triangles ADE and ABC are similar so, ratio of their corresponding sides will be equal.

$$\begin{aligned}\therefore \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \therefore \frac{1}{3} = \frac{DE}{4.5} \\[1em] \therefore DE = \frac{4.5}{3} \\[1em] \therefore DE = 1.5 \text{ cm}.\end{aligned}$$

Hence, the length of DE = 1.5 cm.

(iii) We know that,

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

$$\begin{aligned}\Rightarrow \frac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \frac{DE^2}{BC^2} \\[1em] = \Big(\frac{DE}{BC}\Big)^2 \\[1em] = \Big(\frac{1}{3}\Big)^2 \\[1em] = \frac{1}{9}.\end{aligned}$$

Hence, $\frac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \frac{1}{9}$.

(iv) From figure,

Area of trap.(BCED) = area(Δ ABC) - area(Δ ADE)

$$\begin{aligned}\Rightarrow \frac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \frac{9}{1} \\[1em] \Rightarrow \text{ar(ΔABC)} = 9[\text{ar(ΔADE)}]\\[1em] \Rightarrow \text{ar(trap. BCED)} = 9[\text{ar(ΔADE)}] - \text{ar(ΔADE)} \\[1em] \Rightarrow \text{ar(trap. BCED)} = 8[\text{ar(ΔADE)}] \\[1em] \Rightarrow \frac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \frac{1}{8}.\end{aligned}$$

Hence, $\frac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \frac{1}{8}$.

(i) Given,

ΔABC ∼ ΔPQR

ar(ΔABC) = 49 cm²

ar(ΔPQR) = 25 cm²

AB = 5.6 cm

We know that,

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

$$\begin{aligned}\therefore \frac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\frac{AB}{PQ}\Big)^2 \\[1em] \Rightarrow \frac{49}{25} = \Big(\frac{5.6}{PQ}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{49}{25}} = \frac{5.6}{PQ} \\[1em] \Rightarrow \frac{7}{5} = \frac{5.6}{PQ} \\[1em] \Rightarrow PQ = \frac{5 \times 5.6}{7}\\[1em] \Rightarrow PQ = \frac{28}{7}\\[1em] \Rightarrow PQ = 4 \text{ cm.}\end{aligned}$$

Hence, PQ = 4 cm.

(ii) Given,

ΔABC ∼ ΔPQR

ar(ΔABC) = 28 cm²

ar(ΔPQR) = 63 cm²

PR = 8.4 cm

We know that,

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

$$\begin{aligned}\therefore \frac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\frac{AC}{PR}\Big)^2 \\[1em] \Rightarrow \frac{28}{63} = \Big(\frac{AC}{8.4}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{4}{9}} = \frac{AC}{8.4} \\[1em] \Rightarrow \frac{2}{3} = \frac{AC}{8.4} \\[1em] \Rightarrow AC = \frac{2 \times 8.4}{3}\\[1em] \Rightarrow AC = \frac{16.8}{3}\\[1em] \Rightarrow AC = 5.6 \text{ cm.}\end{aligned}$$

Hence, AC = 5.6 cm.

(iii) Given,

ΔABC ∼ ΔPQR

BC = 4 cm

QR = 5 cm

ar(ΔABC) = 32 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.

$$\begin{aligned}\therefore \frac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\frac{BC}{QR}\Big)^2 \\[1em] \Rightarrow \frac{32}{\text{ar(ΔPQR)}} = \Big(\frac{4}{5}\Big)^2 \\[1em] \Rightarrow \frac{32}{\text{ar(ΔPQR)}} = \frac{16}{25} \\[1em] \Rightarrow \text{ar(ΔPQR)} = \frac{32 \times 25}{16} \\[1em] \Rightarrow \text{ar(ΔPQR)} = 25 \times 2 \\[1em] \Rightarrow \text{ar(ΔPQR)} = 50 \text{ cm}^2.\end{aligned}$$

Hence, ar(ΔPQR) = 50 cm².

Let the length of altitude of other triangle be x cm

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$$\begin{aligned}\therefore \frac{\text{Area of first Δ}}{\text{Area of second Δ}} = \Big(\frac{\text{Altitude of first Δ}}{\text{Altitude of second Δ}}\Big)^2 \\[1em] \Rightarrow \frac{48}{75} = \Big(\frac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \frac{16}{25} = \Big(\frac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{16}{25}} = \frac{3.6}{x} \\[1em] \Rightarrow \frac{4}{5} = \frac{3.6}{x} \\[1em] \Rightarrow x = \frac{18}{4} \\[1em] \Rightarrow x = 4.5 \text{ cm.}\end{aligned}$$

Hence, the length of altitude of other triangle = 4.5 cm.

Given,

AB ⟂ BC and DE ⟂ BC

Thus AB ∥ DE.

∠ABC = ∠DEC [Given]

∠ACB = ∠DCE [Common angle in both triangles]

∴ ΔABC ∼ ΔDEC (By A.A. axiom)

From figure,

DC = AC - AD = 24 - AD

We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{DE}{AB} = \frac{DC}{AC} \\[1em] \Rightarrow \frac{3}{9} = \frac{24 - AD}{24} \\[1em] \Rightarrow 24 - AD = \frac{3 \times 24}{9} \\[1em] \Rightarrow 24 - AD = \frac{72}{9} \\[1em] \Rightarrow 24 - AD = 8 \\[1em] \Rightarrow AD = 24 - 8 \\[1em] \Rightarrow AD = 16 \text{ cm.}\end{aligned}$$

Hence, AD = 16 cm.

Considering ΔADE and ΔABC,

∠A = ∠A [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

We know that,

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

Let the area of ΔABC be x cm².

$$\begin{aligned}\therefore \frac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \Big(\frac{{DE}}{{BC}}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \frac{{DE}^2}{{BC}^2} \\[1em] \Rightarrow \frac{20}{x} = \frac{4^2}{6^2} \\[1em] \Rightarrow \frac{20}{x} = \frac{16}{36} \\[1em] \Rightarrow x = \frac{20 \times 36}{16} \\[1em] \Rightarrow x = \frac{720}{16} \\[1em] \Rightarrow x = 45 \text{ cm}^2\end{aligned}$$

Hence, area of ΔABC = 45 cm².

(i) Given,

AB = 6 cm

AL = 2 cm

LB = AB - AL = 6 - 2 = 4 cm

AC = AM + MC

AM = AC - MC

AM = 9 - MC

In ΔAML and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{AL}{LB} = \frac{AM}{MC} \\[1em] \Rightarrow \frac{2}{4} = \frac{9 - MC}{MC} \\[1em] \Rightarrow 2MC = 4(9 - MC) \\[1em] \Rightarrow 2MC = 36 - 4MC \\[1em] \Rightarrow 2MC + 4MC = 36 \\[1em] \Rightarrow 6MC = 36 \\[1em] \Rightarrow MC = \frac{36}{6} \\[1em] \Rightarrow MC = 6 \text{ cm.}\end{aligned}$$

Hence, CM = 6 cm.

(ii) Given

In ΔALM and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

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

$$\begin{aligned}\therefore \frac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\frac{AL}{AB}\Big)^2 \\[1em] \Rightarrow \frac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\frac{2}{6}\Big)^2 \\[1em] \Rightarrow \frac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \frac{4}{36} \\[1em] \Rightarrow \frac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \frac{1}{9}.\end{aligned}$$

Let ar(ΔALM) = x, then ar(ΔABC) = 9x.

From figure,

ar(trap. LBCM) = ar(ΔABC) - ar(ΔALM)

= 9x - x

= 8x.

$\Rightarrow \frac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \frac{x}{8x} = \frac{1}{8}$.

Hence, $\frac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \frac{1}{8}$.

(i) Given,

∠ABC = ∠AED = 90° [Given]

∠BAC = ∠DAE [Common angle]

∴ ΔABC ∼ ΔAED (By A.A. axiom)

Hence, proved that ΔABC ∼ ΔAED.

(ii) ΔABC is right-angled triangle, applying pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ BC² = AC² - AB²

⇒ BC² = (15)² - (12)²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 cm

Area of ΔABC = $\frac{1}{2}$ × Base × height

= $\frac{1}{2}$ × 12 × 9

= 54 cm²

Since, ΔABC ∼ ΔAED,

We know that,

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

$$\begin{aligned}\therefore \frac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \frac{{ED}^2}{{BC}^2} \\[1em] \Rightarrow \frac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \frac{3^2}{9^2} \\[1em] \Rightarrow \frac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \frac{9}{81} \\[1em] \Rightarrow \frac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \frac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \text{ar(ΔABC)} \times \frac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \frac{54}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = 6 \text{ cm}^2\end{aligned}$$

Hence, proved that ar(ΔAED) = 6 cm².

(iii) From figure,

ar(quad. BCED) = ar(ΔABC) - ar(ΔAED)

ar(quad. BCED) = 54 - 6

ar(quad. BCED) = 48 cm².

$\Rightarrow \frac{\text{ar(quad. BCED)}}{\text{ar.(ΔABC)}} = \frac{48}{54} = \frac{8}{9}$.

Hence, proved that ar(quad BCED) : ar(ΔABC) = 8 : 9.

(i) Considering ΔPQR and ΔPST.

∠P = ∠P [Common angles]

∠PQR = ∠PST [Both are equal to 90°]

∴ ΔPQR ∼ ΔPST (By A.A. axiom)

Hence, proved that ΔPQR ∼ ΔPST.

(ii) We know that,

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

$$\begin{aligned}\Rightarrow \frac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \frac{{PQ}^2}{{PS}^2} \\[1em] \Rightarrow \frac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \frac{5^2}{2^2} \\[1em] \Rightarrow \frac{\text{Area of ΔPQR}}{\text{Area of ΔPQR} - \text{Area of SRQT}} = \frac{25}{4} \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 (\text{ Area of ΔPQR} - \text{Area of SRQT}) \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 \text{ Area of ΔPQR} - 25\text{Area of SRQT} \\[1em] \Rightarrow 25\text{Area of SRQT} = 25 \text{ Area of ΔPQR} - 4\text{ Area of ΔPQR}\\[1em] \Rightarrow 25\text{Area of SRQT} = 21 \text{ Area of ΔPQR} \\[1em] \Rightarrow \frac{\text{Area of ΔPQR}}{\text{Area of SRQT}} = \frac{25}{21}.\end{aligned}$$

Hence, area of ΔPQR : Area of quadrilateral SRQT = 25 : 21.

(i) In ΔPQL and ΔRPM

∠LPQ = ∠MRP [Given]

∠LQP = ∠RPM [Given]

∴ ΔPQL ∼ ΔRPM (y A.A. axiom)

Hence, proved that ΔPQL ∼ ΔRPM.

(ii) Since, ΔPQL ∼ ΔRPM 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 ΔPQL and ΔRQP

∠LPQ = ∠QRP [Given]

∠Q = ∠Q [Common]

∴ ΔPQL ∼ ΔRQP (By A.A. axiom)

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.

(i) Since, BD and CE are medians.

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

By mid-point theorem,

ED ∥ BC and ED = $\frac{1}{2}$ BC [By mid-point theorem]

$\Rightarrow \frac{ED}{BC} = \frac{1}{2}$

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

In triangle EGD and BGC,

∠EGD = ∠BGC [Vertically opposite angles are equal]

∠DEG = ∠GCB [Alternate angles are equal]

∴ ΔEGD ∼ ΔCGB by (By A.A. axiom)

Hence, proved that ΔEGD ∼ ΔCGB.

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

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

Hence, proved that BG = 2GD.

(i) In ∆APD and ∆BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Both equal to 90°)

Hence, proved that ∆APD ∼ ∆BPC.

(ii) In ∆ADB,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = $\sqrt{64}$ = 8 cm.

⇒ PB = BD - PD = 8 - 4.5 = 3.5 cm

Hence, BD = 8 cm and PB = 3.5 cm.

(iii) In ∆APD,

By pythagoras theorem,

⇒ AP² = AD² + DP²

⇒ AP² = 6² + (4.5)²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = $\sqrt{56.25}$ = 7.5 cm

Hence, length of AP = 7.5 cm.

(iv) We know that,

Ratio of area of similar triangles is equal to the square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APD}}{\text{Area of △BPC}} = \frac{AD^2}{BC^2} \\[1em] = \frac{6^2}{(2.4)^2} \\[1em] = \frac{6 \times 6}{2.4 \times 2.4} \\[1em] = \frac{1 \times 1}{0.4 \times 0.4} \\[1em] = \frac{10 \times 10}{4 \times 4} \\[1em] = \frac{100}{16} \\[1em] = \frac{25}{4} \\[1em] = 25 : 4.\end{aligned}$$

Hence, area ∆APD : area ∆BPC = 25 : 4.

In ΔRLQ and ΔPLN,

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

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

∴ ΔRLQ ∼ ΔPLN (By A.A. axiom)

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 ΔRLM and ΔPLQ,

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

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

∴ ΔRLM ∼ ΔPLQ (By A.A. axiom)

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

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

Hence, PN = 15 cm and RM = 10 cm.

Since ΔABC ∼ ΔPQR

So, their respective sides will be in proportion

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

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

In ΔABM and ΔPQN,

∠ABM = ∠PQN [Since, ABC and PQR are similar]

∠AMB = ∠PNQ = 90° [Given ]

∴ ΔΑΒΜ ∼ ΔPQN by AA similarity

$\frac{AM}{PN} = \frac{AB}{PQ}$ .....(1)

Since, AX and PY are medians so they will divide their opposite sides.

BX = $\frac{BC}{2}$ and QY = $\frac{QR}{2}$

Therefore, we have:

$\frac{AB}{PQ} = \frac{BX}{QY}$

∠ABC = ∠PQR

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal.

Hence, ∆ABX ∼ ∆PQY (by SAS similarity rule).So,

$\frac{AM}{PQ} = \frac{AX}{PY}$ .....(2)

From (1) and (2),

$\frac{AM}{PN} = \frac{AX}{PY}$

Hence, proved that $\frac{AM}{PN} = \frac{AX}{PY}$

Area of ΔADE = Area of ΔABC + Area of trapezium BCED = 25 + 24 = 49 cm².

Given,

BC ∥ DE.

In ΔABC and ΔADE,

∠ABC = ∠ADE [Corresponding angles are equal]

∠ACB = ∠AED [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE by (By A.A. axiom)

We know that,

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

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

Hence, BC = 10 cm.

It is given that DE ∥ BC.

∠ADE = ∠ABC [Corresponding angles are equal]

∠AED = ∠ACB [Corresponding angles are equal]

∴ ΔADE ∼ ΔАВС (By A.A. axiom)

We know that,

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

$\frac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \frac{(BC)^2}{(DE)^2}$

Subtracting 1 from both sides, we get:

$$\begin{aligned}\Rightarrow \frac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} - 1 = \frac{(5)^2}{(3)^2} - 1 \\[1em] \Rightarrow \frac{\text{ar(ΔABC)} - \text{ar(ΔADE)}}{\text{ar(ΔADE)}} = \frac{25}{9} - 1 \\[1em] \Rightarrow \frac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \frac{25 - 9}{9} \\[1em] \Rightarrow \frac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \frac{16}{9} \\[1em] \Rightarrow \frac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \frac{9}{16}.\end{aligned}$$

Hence, ar(ΔADE) : ar(trap. BCED) = 9 : 16.

We know that,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of the third side.

Given,

In Δ ABC,

D is mid-point of side AB and E is the mid-point of the side AC.

∴ DE ∥ BC and DE = $\frac{1}{2}$ BC

D is mid-point of side AB.

∴ AB = 2AD

Let us consider ΔADE and ΔABC

∠DAE = ∠BAC [Common angle]

∠ADE = ∠ABC [Corresponding angle are equal]

∴ ΔADE ∼ ΔABC by AA similarity.

$$\begin{aligned}\Rightarrow \frac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \frac{AD^2}{AB^2} \\[1em] \Rightarrow \frac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \frac{AD^2}{(2AD)^2} \\[1em] \Rightarrow \frac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \frac{1}{4}.\end{aligned}$$

Hence, ar(ΔADE) : ar(ΔABC) = 1 : 4.

(i) Considering ΔPMN and ΔPQR,

∠P = ∠P [Common angles]

∠PMN = ∠PQR [Corresponding angles are equal]

∴ ΔPMN ∼ ΔPQR by AA similarity.

Given,

$$\begin{aligned}\frac{PM}{MQ} = \frac{2}{3} \\[1em] \Rightarrow \frac{PM}{PQ - PM} = \frac{2}{3} \\[1em] \Rightarrow 3PM = 2(PQ - PM) \\[1em] \Rightarrow 3PM = 2PQ - 2PM \\[1em] \Rightarrow 3PM + 2PM = 2PQ \\[1em] \Rightarrow 5PM = 2PQ \\[1em] \Rightarrow \frac{PM}{PQ} = \frac{2}{5}.\end{aligned}$$

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\frac{MN}{QR} = \frac{PM}{PQ} = \frac{2}{5}$.

Hence, $\frac{MN}{QR} = \frac{2}{5}$

(ii) Considering ΔOMN and ΔORQ,

∠MON = ∠QOR (Vertically opposite angles are equal)

∠OMN = ∠ORQ (Alternate angles are equal)

Hence, by AA similarity ΔOMN ∼ ΔORQ.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \frac{MN^2}{QR^2} \\[1em] \Rightarrow \frac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \frac{2^2}{5^2} \\[1em] \Rightarrow \frac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \frac{4}{25} \\[1em]\end{aligned}$$

Hence, the ratio of the Area of ΔOMN : Area of ΔORQ = 4 : 25.

(i) In ΔPQR and ΔSPR,

⇒ ∠PSR = ∠QPR [Given ]

⇒ ∠PRQ = ∠PRS [Common angle]

∴ ΔPQR ∼ ΔSPR by AA similarity.

Hence, proved that ΔPQR ∼ ΔSPR.

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

$$\begin{aligned}\Rightarrow \frac{QR}{PR} = \frac{PR}{SR} \\[1em] \Rightarrow \frac{QR}{6} = \frac{6}{3} \\[1em] \Rightarrow \frac{QR}{6} = \frac{36}{3} = 12 \text{ cm.}\end{aligned}$$

Also,

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

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.

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

Hence, $\frac{\text{Area of ΔPQR}}{\text{Area of ΔPQR}} = \frac{4}{1}$

(a) In ΔABC and ΔAED,

⇒ ∠ABC = ∠AED [Both = 90°]

⇒ ∠BAC = ∠DAE [Common angles]

∴ ΔABC ~ ΔAED (By AA similarity postulate)

Hence, proved that ΔABC ~ ΔAED.

(b) Given,

AB = 20 cm, AC = 25 cm, DE = 3 cm

In ΔABC,

By pythagoras theorem,

⇒ AB² + BC² = AC²

⇒ (20)² + BC² = (25)²

⇒ 400 + BC² = 625

⇒ BC² = 625 - 400

⇒ BC² = 225

⇒ BC = $\sqrt{225}$

⇒ BC = 15 cm.

We know that,

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

$\frac{AB}{AE} = \frac{BC}{DE} = \frac{AC}{AD}$

Solving,

$$\begin{aligned}\Rightarrow \frac{AB}{AE} = \frac{BC}{DE} \\[1em] \Rightarrow \frac{20}{AE} = \frac{15}{3} \\[1em] \Rightarrow \frac{20 \times 3}{15} = AE \\[1em] \Rightarrow AE = \frac{20}{5} \\[1em] \Rightarrow AE = 4\text{ cm}.\end{aligned}$$

Substituting values in $\frac{BC}{DE} = \frac{AC}{AD}$ we get :

$$\begin{aligned}\Rightarrow \frac{BC}{DE} = \frac{AC}{AD} \\[1em] \Rightarrow \frac{15}{3} = \frac{25}{AD} \\[1em] \Rightarrow 5 = \frac{25}{AD} \\[1em] \Rightarrow AD = \frac{25}{5} \\[1em] \Rightarrow AD = 5\text{ cm}.\end{aligned}$$

Hence, BC = 15 cm, AE = 4 cm, AD = 5 cm.

(c) Given,

Area on ground = 576 m²

By formula,

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

Area of ΔABC = $\frac{1}{2}$ × AB × BC

= $\frac{1}{2} \times 20 \times 15$

= 150 cm².

Area of ΔAED = $\frac{1}{2}$ × AE × DE

$= \frac{1}{2} \times 4 \times 3$

$= \frac{1}{2} \times 12$

= 6 cm².

From figure,

⇒ Area of Quadrilateral (BCED) = Area of ΔABC - Area of ΔAED

= 150 - 6 = 144 cm².

⇒ Actual ground area = 576 m²

= 576 × 10000 cm² = 5760000 cm²

Let scale factor be k.

By formula,

k² = $\frac{\text{Area of BCED}}{\text{Actual area of BCED on ground}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow k^2 = \frac{144}{5760000} \\[1em] \Rightarrow k^2 = \frac{1}{40000} \\[1em] \Rightarrow k = \sqrt{\frac{1}{40000}}\\[1em] \Rightarrow k = \frac{1}{200}\\[1em]\end{aligned}$$

Hence, scale factor equals 1 : 200.

Given,

ΔABC ∼ ΔDEF.

Thus, corresponding angles are equal.

∠B = ∠E = 83°

We know that, sum of all angles in triangle is equal to 180°.

∠A + ∠B + ∠C = 180°

47° + 83° + ∠C = 180°

130° + ∠C = 180°

∠C = 180° - 130°

∠C = 50°.

Hence, option 1 is the correct option.

Given,

ΔABC ∼ ΔQRP.

$\therefore \frac{AB}{QR} = \frac{BC}{RP}$

Hence, option 4 is the correct option.

Given,

The two triangles are ΔABC and ΔPQR.

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

By SSS similarity criterion, if the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and the triangles are similar.

To find the correspondence of vertices:

The side AB corresponds to QR. The vertex opposite to AB is C, and opposite to QR is P.

C = P

The side BC corresponds to PR. The vertex opposite to BC is A, and opposite to PR is Q.

A = Q

The side CA corresponds to PQ. The vertex opposite to CA is B, and opposite to PQ is R.

B = R

∴ΔPQR ∼ ΔCAB

Hence, option 1 is the correct option.

$\frac{AB}{DE} = \frac{24}{12} = 2$

$\frac{AC}{DF} = \frac{18}{9} = 2$

Thus,

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

∠BAC = ∠EDF [Given ]

∴ ΔABC ∼ ΔEDF (By S.A.S. axiom)

Hence, option 2 is the correct option.

Given,

⇒ ∠C = ∠F

Since, the sides containing the angle may or may not be proportional,

i.e., we don't know if $\frac{AC}{DF}$ and $\frac{BC}{EF}$ are equal or not equal.

∴ Similarity of given triangles cannot be determined.

Hence, Option 4 is the correct option.

Given,

$\frac{AB}{FD} = \frac{BC}{DE} = \frac{CA}{EF}$

∴ ΔABC ∼ ΔDEF BY SSS theorem.

Side AB corresponds to side FD. Therefore, vertex A corresponds to vertex F and vertex B corresponds to vertex D.

Side BC corresponds to side DE. Therefore, vertex B corresponding to D. It also implies vertex C corresponds to vertex E. 

Since the triangles are similar, the corresponding angles are also similar,

∠A = ∠F and ∠B = ∠D

Hence, option 3 is the correct option.

Given,

ΔPOQ and ΔROT

∠POQ = ∠TOR [vertically opposite angles are equal]

∠OPQ = ∠ORT [Alternate interior angles are equal]

∴ ΔPOQ ∼ ΔROT (By A.A. axiom)

We know that,

The ratio of corresponding sides in similar triangles are proportional.

$\frac{PQ}{RT} = \frac{OP}{OR} = \frac{OQ}{OT}$

Hence, option 2 is the correct option.

Checking option 3,

Given,

In ΔABC, DE is parallel to BC if and only if it divides the sides AB and AC in the same ratio

From figure,

DB = AB - AD = 18 - 8 = 10 cm

$\frac{AD}{DB} = \frac{8}{10} = \frac{4}{5}$

$\frac{AE}{EC} = \frac{12}{15} = \frac{4}{5}$

$\therefore \frac{AD}{DB} = \frac{AE}{EC}$

AB = 18 cm, AD = 8 cm, AE = 12 cm, EC = 15 cm DE is parallel to BC holds true.

Hence, option 3 is the correct option.

By basic proportionality theorem,

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the same ratio.

Since DE ∥ BC in ΔABC,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{3}{5} = \frac{AE}{AC - AE} \\[1em] \Rightarrow \frac{3}{5} = \frac{AE}{5.6 - AE} \\[1em] \Rightarrow 3 \times (5.6 - AE) = 5AE \\[1em] \Rightarrow 16.8 - 3AE = 5AE \\[1em] \Rightarrow 5AE + 3AE = 16.8 \\[1em] \Rightarrow 8AE = 16.8 \\[1em] \Rightarrow AE = \frac{16.8}{8} \\[1em] \Rightarrow AE = 2.1 \text{ cm.}\end{aligned}$$

Hence, option 1 is the correct option.

By basic proportionality theorem,

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the same ratio.

Since DE ∥ BC in ΔABC,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{(7x - 4)}{(3x + 4)} = \frac{(5x - 2)}{3x} \\[1em] \Rightarrow 3x \times (7x - 4) = (5x - 2) \times (3x + 4) \\[1em] \Rightarrow 21x^2 - 12x = 15x^2 + 20x -6x -8 \\[1em] \Rightarrow 21x^2 - 12x - 15x^2 - 20x + 6x + 8 = 0\\[1em] \Rightarrow 21x^2 - 15x^2 - 12x - 20x + 6x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 26x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 24x - 2x + 8 = 0 \\[1em] \Rightarrow 6x(x - 4) - 2(x - 4) = 0 \\[1em] \Rightarrow (6x - 2)(x - 4) = 0 \\[1em] \Rightarrow (6x - 2) = 0 \text{ or } (x - 4) = 0 \text{[Using Zero-product rule]}\\[1em] \Rightarrow 6x= 2 \text{ or } x = 4 \\[1em] \Rightarrow x= \frac{2}{6} \text{ or } x = 4 \\[1em] \Rightarrow x= \frac{1}{3} \text{ or } x = 4 \\[1em]\end{aligned}$$

x cannot be $\frac{1}{3}$ as it yields negative AD and AE values. Therefore x = 4.

Hence, option 3 is the correct option.

Given,

∠A = 30°

∠C = 50°

In triangle ABC,

∠A + ∠B + ∠C = 180°

∠B = 180° - (∠A + ∠C)

∠B = 180° - (30° + 50°)

∠B = 180° - 80°

∠B = 100°.

As the triangles are similar, the corresponding angles will be equal.

So, ∠F = ∠B = 100°

Since the triangles are similar, the ratio of corresponding sides is proportional,

$$\begin{aligned}\Rightarrow \frac{AB}{DF} = \frac{AC}{DE} = \frac{BC}{FE} \\[1em] \text{Let's consider} \\[1em] \Rightarrow \frac{AB}{DF} = \frac{AC}{DE} \\[1em] \Rightarrow \frac{5}{7.5} = \frac{8}{DE} \\[1em] \Rightarrow DE = \frac{8 \times 7.5}{5} \\[1em] \Rightarrow DE = \frac{60}{5} \\[1em] \Rightarrow DE = 12 \text{ cm.}\end{aligned}$$

DE = 12 cm, ∠F = 100°

Hence, option 2 is the correct option.

Given, in ΔDEF and ΔPQR:

∠D = ∠Q

∠E = ∠R

∴ ΔDEF ∼ ΔQRP by AA similarity.

The ratio of corresponding sides is:

$\frac{DE}{QR} = \frac{DF}{QP}$

$\frac{DE}{PQ} = \frac{EF}{RP}$ do not hold true.

Hence, option 1 is the correct option.

From figure,

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

In ΔABC and ΔADE,

∠A = ∠A [Common angle]

∠ABC = ∠ADE [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE (By A.A. axiom).

In similar triangles, ratios of corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AE}{AC} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{2}{5} = \frac{DE}{10} \\[1em] \Rightarrow DE = \frac{2 \times 10}{5} \\[1em] \Rightarrow DE = \frac{20}{5} \\[1em] \Rightarrow DE = 4 \text{ cm.}\end{aligned}$$

Hence, option 1 is the correct option.

Given,

ΔDEF ∼ ΔABC

Perimeter of ΔABC = AB + BC + CA = 3 + 2 + 2.5 = 7.5 cm

Since the two triangles are similar, we can write:

$$\begin{aligned}\Rightarrow \frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}} = \frac{BC}{EF} \\[1em] \Rightarrow \frac{7.5}{\text{Perimeter of ΔDEF}} = \frac{2}{4} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \frac{7.5 \times 4}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \frac{30}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = 15 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

Given,

∆ABC ∼ ∆PQR

∴ ∠A = ∠P

⇒ $\frac{∠A}{2} = \frac{∠P}{2}$

⇒ ∠BAD = ∠QPS

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]

In ∆ABD ∼ ∆PQS,

⇒ ∠BAD = ∠QPS

⇒ ∠B = ∠Q

∴ ∆ABD ∼ ∆PQS (By A.A. axiom)

Hence, Option 2 is the correct option.

Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB ∥ CD.

By basic proportionality theorm,

$$\begin{aligned}\Rightarrow \frac{PD}{DB} = \frac{PC}{CA} \\[1em] \Rightarrow \frac{PD}{DB} = \frac{6}{4} \\[1em] \Rightarrow \frac{PD}{DB} = \frac{3}{2} \\[1em] \Rightarrow PD : DB = 3 : 2.\end{aligned}$$

Hence, option 3 is the correct option.

Since, DE // BC,

By basic proportionality theorem,

$$\begin{aligned}\Rightarrow \frac{AD}{DB} = \frac{AE}{EC} \\[1em] \Rightarrow \frac{3}{1} = \frac{3.3}{EC} \\[1em] \Rightarrow EC = \frac{3.3}{3} \\[1em] \Rightarrow EC = 1.1 \text{ cm.}\end{aligned}$$

From figure,

AC = AE + EC

AC = 3.3 + 1.1 = 4.4 cm.

Hence, option 3 is the correct option.

From figure,

AB = AE + EB = 8 + 7 = 15 cm

AC = AD + DC = 10 + 2 = 12 cm

In ΔABC and ΔADE

∠BAC = ∠DAE [Common angle]

∠ABC = ∠ADE [Given]

∴ ΔABC ∼ ΔADE (By A.A. axiom).

We know that,

In similar triangles the ratio of corresponding sides are equal.

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \\[1em] \text{Let's Consider} \\[1em] \Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{10}{15} = \frac{DE}{9} \\[1em] \Rightarrow DE = \frac{10 \times 9}{15} \\[1em] \Rightarrow DE = \frac{90}{15} \\[1em] \Rightarrow DE = 6 \text{ cm.}\end{aligned}$$

Hence, option 1 is the correct option.

Given,

ΔDEF ∼ ΔABC

Perimeter of ΔABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm.

Since the triangles are similar,

$$\begin{aligned}\therefore \frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}} = \frac{BC}{EF} \\[1em] \Rightarrow \frac{22.5}{\text{Perimeter of ΔDEF}} = \frac{6}{8} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \frac{22.5 \times 8}{6} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \frac{180}{6} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = 30 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Construction: Draw a line through C parallel to AD, meeting BA produced at E.

Since AD ∥ EC and BE is the transversal:

∠BAD = ∠AEC [Corresponding angles are equal]

Since AD ∥ EC and AC is the transversal:

∠DAC = ∠ACE [Alternate interior angles]

Given AD is the bisector of ∠A:

∠BAD = ∠DAC

∴ ∠AEC = ∠ACE

In ΔACE, sides opposite to equal angles are equal.

∴ AE = AC = 14 cm

Let DC be x,

Now, in ΔBCE, we have AD ∥ EC. By Basic Proportionality Theorem,

$$\begin{aligned}\Rightarrow \frac{BD}{DC} = \frac{AB}{AE} \\[1em] \Rightarrow \frac{BD}{DC} = \frac{10}{14} \\[1em] \Rightarrow \frac{BD}{DC} = \frac{10}{14} \\[1em] \Rightarrow \frac{BD}{DC} = \frac{5}{7} \\[1em]\end{aligned}$$

Let CD = x.

Then BD = BC - CD = 6 - x.

Substituting these values into the ratio:

$$\begin{aligned}\Rightarrow \frac{6 - x}{x} = \frac{5}{7} \\[1em] \Rightarrow 7(6 - x) = 5x \\[1em] \Rightarrow 42 - 7x = 5x \\[1em] \Rightarrow 42 = 5x + 7x \\[1em] \Rightarrow 12x = 42 \\[1em] \Rightarrow x = \frac{42}{12} \\[1em] \Rightarrow x = 3.5 \text{ cm.}\end{aligned}$$

Thus, CD = 3.5 cm.

Hence, option 1 is the correct option.

Considering ΔAPB and ΔCPD,

$\frac{AP}{PD} = \frac{6}{5} \text{ and } \frac{BP}{CP} = \frac{3}{2.5} = \frac{6}{5}$

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

∴ ΔAPB ∼ ΔDPC (By SAS similarity)

Hence, ∠PAB = ∠PDC = 30°

⇒ ∠PBA = 180° - (∠PAB + ∠APB)

⇒ ∠PBA = 180° - (30° + 50°)

⇒ ∠PBA = 180° - 80°

⇒ ∠PBA = 100°.

Hence, option 4 is the correct option.

Given,

Consider ΔAOB and ΔCOD

∠AOB = ∠COD [Vertically opposite angles are equal]

∠OBA = ∠ODC [Alternate angles are equal]

Therefore, by AA rule of similarity ΔAOB ~ ΔCOD.

Hence, option 4 is the correct option.

Given,

The ratio of corresponding sides be 1 : 3.

Heights are linear measures corresponding to sides, so the ratio of the heights will also be 1 : 3.

Hence, option 1 is the correct option.

Given,

ΔABC and ΔDEF

∠A = ∠E = 40° [Equal angles]

AB : ED = AC : EF [Corresponding sides are equal]

∴ ΔABC ∼ ΔEDF by SAS similarity. Therefore,

⇒ ∠A = ∠E = 40°

⇒ ∠B = ∠D = x

⇒ ∠C = ∠F = 65°

The sum of all the angles in a triangle is 180°:

⇒ ∠A + ∠B + ∠C = 180°

⇒ 40° + x + 65° = 180°

⇒ x = 180° - 40° - 65°

⇒ x = 75°

⇒ ∠B = ∠D = 75°.

Hence, option 3 is the correct option.

Given,

OA = (2x + 4) cm

OB = (9x − 21) cm

OC = (2x − 1) cm

OD = 3 cm

In ΔOAB and ΔOCD

∠COD = ∠AOB [Vertically Opposite Angles are equal]

∠OAB = ∠OCD [alternate interior angles are equal]

∴ ΔOAB ∼ ΔOCD by AA similarity. Then ratios of corresponding sides are equal :

$$\begin{aligned}\Rightarrow \frac{OA}{OC} = \frac{OB}{OD} \\[1em] \Rightarrow \frac{(2x + 4)}{(2x - 1)} = \frac{(9x - 21)}{3} \\[1em] \Rightarrow 3 \times (2x + 4) = (2x - 1) \times (9x - 21) \\[1em] \Rightarrow 6x + 12 = 18x^2 - 42x -9x +21 \\[1em] \Rightarrow 18x^2 - 42x -9x +21 - 6x - 12 = 0\\[1em] \Rightarrow 18x^2 - 42x - 9x - 6x +21 - 12 = 0\\[1em] \Rightarrow 18x^2 - 57x + 9 = 0\\[1em] \Rightarrow 18x^2 - 54x - 3x + 9 = 0\\[1em] \Rightarrow 18x(x - 3) - 3(x - 3) = 0\\[1em] \Rightarrow (18x - 3)(x - 3) = 0\\[1em] \Rightarrow (18x - 3) = 0 \text{ or }(x - 3) = 0 \\[1em] \Rightarrow 18x = 3 \text{ or } x = 3 \\[1em] \Rightarrow x = \frac{3}{18} \text{ or } x = 3 \\[1em] \Rightarrow x = \frac{1}{6} \text{ or } x = 3.\end{aligned}$$

x cannot be $\frac{1}{6}$ as it yields negative OB and OC values. Therefore x = 3.

Hence, option 2 is the correct option.

Given,

Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.

Suppose EA and CA are the shadows of DE and BC respectively.

Now, In ΔABC and ΔADE,

∠ACB = ∠AED = 90°

∠A = ∠A [Common]

∴ ΔABC ∼ ΔADE by AA similarity.Then ratios of corresponding sides are equal:

$$\begin{aligned}\Rightarrow \frac{CA}{AE} = \frac{BC}{DE} \\[1em] \Rightarrow \frac{CA}{2} = \frac{12.5}{5} \\[1em] \Rightarrow CA = \frac{12.5 \times 2}{5} \\[1em] \Rightarrow CA = \frac{25}{5} \\[1em] \Rightarrow CA = 5 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Area of a right angled triangle = $\frac{1}{2} \times \text{ base } \times \text{ height }$

Area of triangle ABC = $\frac{1}{2} \times AB \times AC$

Area of triangle ABC = $\frac{1}{2} \times bc$ .......(1)

Also,

Area of triangle ABC = $\frac{1}{2} \times BC \times AD$

Area of triangle ABC = $\frac{1}{2} \times ap$ .....(2)

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

$\frac{1}{2}$ bc = $\frac{1}{2}$ ap

bc = ap

a = $\frac{bc}{p}$

Applying Pythogoras theorem to right-angled triangle ABC:

$$\begin{aligned}\Rightarrow BC^2 = AC^2 + AB^2 \\[1em] \Rightarrow a^2 = b^2 + c^2 \\[1em] \Rightarrow \Big(\frac{bc}{p}\Big)^2 = b^2 + c^2 \\[1em] \Rightarrow \frac{b^2c^2}{p^2} = b^2 + c^2 \\[1em] \Rightarrow \frac{1}{p^2} = \frac{b^2 + c^2}{b^2c^2} \\[1em] \Rightarrow \frac{1}{p^2} = \frac{b^2}{b^2c^2} + \frac{c^2}{b^2c^2}\\[1em] \Rightarrow \frac{1}{p^2} = \frac{1}{c^2} + \frac{1}{b^2} .\end{aligned}$$

Hence, option 4 is the correct option.

Given,

In ΔABC and ΔDBA,

∠ADC = ∠ADB = 90°

∠B = ∠B [Common]

ΔABC ∼ ΔDBA [By AA similarity]

From the similarity ΔABC ∼ ΔDBA, the ratio of corresponding sides is equal:

$$\begin{aligned}\therefore \frac{AB}{DB} = \frac{BC}{BA} = \frac{AC}{DA} \\[1em] \text{ Considering first two terms } \\[1em] \Rightarrow \frac{AB}{DB} = \frac{BC}{BA} \\[1em] \Rightarrow AB^2 = BC \times BD .\end{aligned}$$

Hence, option 2 is the correct option.

Given,

In ΔABC and ΔAXY

∠BAC = ∠XAY [Common angle]

∠AXY = ∠ABC [Corresponding angles are equal]

∴ ΔABC ∼ ΔAXY (By A.A. axiom)

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

$\frac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \Big(\frac{AX}{AB}\Big)^2$ .....(1)

X and Y divides triangle ABC into two equal parts. Therefore,

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

$\frac{\text{Area of ΔAXY}}{\text{Area of ΔABC}} = \frac{1}{2}$ .....(2)

Equating Eqn(1) and Eqn(2) :

$$\begin{aligned}\Rightarrow \frac{1}{2} = \Big(\frac{AX}{AB}\Big)^2 \\[1em] \Rightarrow \frac{AX}{AB} = \frac{1}{\sqrt{2}}.\end{aligned}$$

Hence, option 1 is the correct option.

In ΔABD,

∠B + ∠BAD = 90°

∠BAD = 90° - ∠B ........(1)

In ΔABC,

Since, BC is the hypotenuse, thus angle A = 90°

From figure,

∠DAC = ∠BAC - ∠BAD

∠DAC = 90° - ∠BAD

Substituting value of ∠BAD from equation (1) in above equation, we get :

∠DAC = 90° - (90° - ∠B) = ∠B.

In ΔABD and ΔCAD,

∠DAC = ∠B (Proved above)

∠ADB = ∠ADC (Both equal to 90°)

∴ ΔABD ∼ ΔCAD by AA similarity.

Thus, (I) is true.

or we can say that ΔADB ∼ ΔCDA.

Thus, (II) is true.

In ΔADB and ΔCAB,

∠ADB = ∠CAB = 90°

∠ABD = ∠CBA [Common angle]

∴ ΔADB ∼ ΔCAB by AA similarity.

Thus, (III) is true.

Hence, option 4 is the correct option.

Given,

Let, A₁ and A₂ be the areas of two triangle. s₁ and s₂ be the length of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$$\begin{aligned}\Rightarrow \frac{A_1}{A_2} = \Big(\frac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \frac{49}{64} = \Big(\frac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \frac{s_1}{s_2} = \sqrt{\frac{49}{64}} \\[1em] \Rightarrow \frac{s_1}{s_2} = \frac{7}{8}.\end{aligned}$$

Ratio between sides = 7 : 8.

Hence, option 1 is the correct option.

Given,

Let, A₁ and A₂ be the areas of two triangle. h₁ and h₂ be the height of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding heights.

$$\begin{aligned}\Rightarrow \frac{A_1}{A_2} = \Big(\frac{h_1}{h_2}\Big)^2 \\[1em] \Rightarrow \frac{12}{48} = \Big(\frac{2.1}{h_2}\Big)^2 \\[1em] \Rightarrow \frac{1}{4} = \Big(\frac{2.1}{h_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{1}{4}} = \frac{2.1}{h_2} \\[1em] \Rightarrow \frac{1}{2} = \frac{2.1}{h_2} \\[1em] \Rightarrow h_2 = 2.1 \times 2 \\[1em] \Rightarrow h_2 = 4.2 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Let, A₁ and A₂ be the areas of two triangle. s₁ and s₂ be the length of their corresponding sides.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$$\begin{aligned}\Rightarrow \frac{A_1}{A_2} = \Big(\frac{s_1}{s_2}\Big)^2 \\[1em] \Rightarrow \frac{81}{144} = \Big(\frac{27}{s_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{81}{144}} = \frac{27}{s_2} \\[1em] \Rightarrow \frac{9}{12} = \frac{27}{s_2} \\[1em] \Rightarrow s_2 = \frac{27 \times 12}{9} \\[1em] \Rightarrow s_2 = \frac{324}{9} \\[1em] \Rightarrow s_2 = 36 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

Given,

ΔABC and ΔDEF are similar to each other.

Since the triangles are similar, the ratios of Areas of triangles is equal to squares of corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of triangle ABC}}{\text{Area of triangle DEF}} = \Big(\frac{(\sqrt{2} + 1)}{\sqrt{3}}\Big)^2 \\[1em] = \frac{(\sqrt{2} + 1)^2}{(\sqrt{3})^2} \\[1em] = \frac{(\sqrt{2})^2 + (1)^2 + 2 \times \sqrt{2} \times 1}{3} \\[1em] = \frac{2 + 1 + 2\sqrt{2}}{3} \\[1em] = \frac{3 + 2\sqrt{2}}{3}.\end{aligned}$$

Area of triangle ABC : Area of triangle DEF = $(3 + 2\sqrt{2}) : 3$.

Hence, option 1 is the correct option.

Given,

AB = 18 cm

BC = 15 cm

Since the triangles are similar, the ratios of Areas of triangles is equal to the ratio of the squares of corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\frac{BC}{PR}\Big)^2 \\[1em] \Rightarrow \frac{9}{4} = \Big(\frac{15}{PR}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{9}{4}} = \frac{15}{PR} \\[1em] \Rightarrow \frac{3}{2} = \frac{15}{PR} \\[1em] \Rightarrow PR = \frac{15 \times 2}{3} \\[1em] \Rightarrow PR = \frac{30}{3} \\[1em] \Rightarrow PR = 10 \text{ cm.}\end{aligned}$$

Hence, option 3 is the correct option.

Given,

∠CAB = 90°

AC = 75 cm = 0.75 m

AB = 1 m

BC = 1.25 m

In ΔCAB and ΔADB,

∠CAB = ∠ADB = 90° [AD ⟂ BC]

∠CBA = ∠ABD [Common angle]

∴ ΔCAB ∼ ΔADB by AA similarity. Then ratios of corresponding sides are equal:

$$\begin{aligned}\Rightarrow \frac{AC}{AD} = \frac{BC}{AB} \\[1em] \Rightarrow \frac{AC}{BC} = \frac{AD}{AB} \\[1em] \Rightarrow \frac{0.75}{1.25} = \frac{AD}{1} \\[1em] \Rightarrow AD = \frac{0.75}{1.25} \\[1em] \Rightarrow AD = 0.6 \text{ m.}\end{aligned}$$

AD = 0.6 m = 60 cm.

Hence, option 2 is the correct option.

Let AB = 3.3 m be the lamp post and QP = 1.1 m be the position of boy after 4 seconds.

AP is the distance moved in 4s at 0.8m/s = 4(0.8) = 3.2 m

PM is the length of the shadow of boy.

In triangle AMB and PMQ,

∠MAB = ∠MPQ = 90°

∠AMB = ∠PMQ [Common angle]

∴ ΔAMB ∼ ΔPMQ [By AA similarity].

Then ratios of corresponding sides of similar triangles are equal.

$$\begin{aligned}\Rightarrow \frac{AM}{PM} = \frac{AB}{PQ} \\[1em] \Rightarrow \frac{AP + PM}{PM} = \frac{AB}{PQ} \\[1em] \Rightarrow \frac{3.2 + x}{x} = \frac{3.3}{1.1} \\[1em] \Rightarrow 3.2 + x = 3x \\[1em] \Rightarrow 3.2 = 2x \\[1em] \Rightarrow x = \frac{3.2}{2} \\[1em] \Rightarrow x = 1.6 \text{ m}.\end{aligned}$$

Hence, option 2 is the correct option.

In △ APQ and △ ABC,

⇒ ∠PAQ = ∠BAC (Common angle)

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

∴ △ APQ ~ △ ABC (By A.A. axiom)

From figure,

Let AP = x km.

We know that,

Corresponding sides of similar triangle are proportional.

$$\begin{aligned}\Rightarrow \frac{AP}{AB} = \frac{PQ}{BC} \\[1em] \Rightarrow \frac{x}{AP + PB} = \frac{20}{50} \\[1em] \Rightarrow \frac{x}{x + 18} = \frac{2}{5} \\[1em] \Rightarrow 5x = 2(x + 18) \\[1em] \Rightarrow 5x = 2x + 36 \\[1em] \Rightarrow 5x - 2x = 36 \\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \frac{36}{3} \\[1em] \Rightarrow x = 12.\end{aligned}$$

AB = AP + BP = 12 + 18 = 30 km.

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{AB}{90} = \frac{30}{90} = \frac{1}{3} \text{ hr} = \frac{1}{3} \times 60$ = 20 minutes.

Hence, Option 4 is the correct option.

Given,

Triangle P is similar to triangle Q

Perimeter of Triangle P = 3 + 4 + 5 = 12 units

Since the triangles are similar, the corresponding sides have a scale factor of k . One of the length of side of triangle Q is 60 units. The Corresponding scale factors will be:

Case 1 :

k = $\frac{60}{3}$ = 20

Case 2 :

k = $\frac{60}{4}$ = 15

Case 3 :

k = $\frac{60}{5}$ = 12

Among the three cases, smallest sclae factor = 12.

Smallest possible perimeter of Triangle (Q) = 12 × 12 = 144 units.

Hence, option 2 is the correct option.

Solving option 2,

Since the triangles are similar,

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

$$\begin{aligned}\Rightarrow \frac{Area_1}{Area_2} = \Big(\frac{\text{Perimeter}_1}{\text{Perimeter}_2}\Big)^2 \\[1em] \Rightarrow \frac{25}{64} = \Big(\frac{\text{Perimeter}_1}{\text{Perimeter}_2}\Big)^2 \\[1em] \Rightarrow \sqrt{\frac{25}{64}} = \frac{\text{Perimeter}_1}{\text{Perimeter}_2} \\[1em] \Rightarrow \frac{\text{Perimeter}_1}{\text{Perimeter}_2} =\frac{5}{8} \\[1em]\end{aligned}$$

Hence proved.

Hence, option 2 is the correct option.

Given,

Since D, E, F are midpoints of BC, AC and AB respectively.

According to the Mid-point Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and is half its length.

Thus, by mid-point theorem,

DE ∥ AB

DE = $\frac{1}{2}$AB

EF ∥ BC

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

FD ∥ AC

DF = $\frac{1}{2}$AC

Ratios can be written as,

$$\begin{aligned}\Rightarrow \frac{DF}{AC} = \frac{EF}{BC} = \frac{DE}{AB} \\[1em] \Rightarrow \frac{\frac{1}{2}AC}{AC} = \frac{\frac{1}{2}BC}{BC} = \frac{\frac{1}{2}AB}{AB} \\[1em] \Rightarrow \frac{1}{2} = \frac{1}{2} = \frac{1}{2}\end{aligned}$$

∴ ΔDFE ∼ ΔACB is true, but ΔDFE ≁ ΔABC.

Hence, option 4 is the correct option.

We compare the ratios of the corresponding sides of the first triangle with sides 6, 9, 12 and the second triangle with sides 8, 12, 16.

k = $\frac{6}{8} = \frac{9}{12} = \frac{12}{16} = \frac{3}{4}$.

Therefore, the two triangles are similar with scale factor of k = $\frac{3}{4}$. The ratio of Areas of two triangle is equal to square of it's scale factor.

$$\begin{aligned}\Rightarrow \frac{\text{Area}_1}{\text{Area}_2} = k^2 \\[1em] \Rightarrow \frac{A}{\text{Area}_2} = \Big(\frac{3}{4}\Big)^2 \\[1em] \Rightarrow \frac{A}{\text{Area}_2} = \frac{9}{16} \\[1em] \Rightarrow \text{Area}_2 = \frac{16}{9}A.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

△ PQR ~ △ TSR.

From figure,

⇒ ∠PRQ = ∠SRT (Common angle)

The order of vertices of two similar triangles are written in such a way that the corresponding vertices occupy the same position.

∴ ∠PQR = ∠TSR

∴ △ PQR ~ △ TSR (By A.A. axiom)

Hence, Option 3 is the correct option.

Directions (Q. 44 to 47): These questions are based on the following information :

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. It is given that AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm and PQ = 2 cm.

From figure,

AB = AP + PB = 1 + 3 = 4 cm

AC = AQ + QC = 1.5 + 4.5 = 6 cm

In ΔAPQ and ΔABC,

$\Rightarrow \frac{AP}{AB} = \frac{1}{4}$

$\Rightarrow \frac{AQ}{AC} = \frac{1.5}{6} = \frac{1}{4}$

∠PAQ = ∠BAC [Common angles]

ΔAPQ ∼ ΔABC [By SAS Similarity]

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

$$\begin{aligned}\Rightarrow \frac{PQ}{BC} = \frac{AQ}{AC} \\[1em] \Rightarrow \frac{PQ}{BC} = \frac{1}{4} \\[1em] \Rightarrow \frac{2}{BC} = \frac{1}{4} \\[1em] \Rightarrow BC = 2 \times 4 \\[1em] \Rightarrow BC = 8 \text{ cm}.\end{aligned}$$

Perimeter = AB + AC + BC

= 4 + 6 + 8

= 18 cm.

Hence, option 3 is the correct option.

We know that,

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

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

Hence, option 4 is the correct option.

In ΔAPQ and ΔABC,

$\frac{AP}{AB} = \frac{AQ}{AC} = \frac{1}{4}$

∠PAQ = ∠BAC [Common angles]

∴ ΔAPQ ∼ ΔABC by SAS Similarity.

Hence, option 1 is the correct option.

Two pairs of corresponding sides are proportional ($\frac{AP}{AB} = \frac{AQ}{AC}$) and icluded angle is also equal (angle A).

This is the S.A.S. (Side-Angle-Side) Similarity Axiom.

Hence, option 4 is the correct option.

Directions (Q. 48 to 51): Using the given diagram answer the following questions.

In ΔPQR, AB ∥ QR, QP ∥ CB and AR intersects CB at O.

In ΔARQ and ΔORC,

∠BRC = ∠ARQ [Common angle]

∠AQR = ∠BCR [Corresponding angles, AB ∥ QR, BC is the transversal]

ΔARQ ∼ ΔORC [By AA Similarity]

Hence, option 1 is the correct option.

In ΔPQR and ΔBCR,

∠PRQ = ∠BRC [Common angle]

∠PQR = ∠BCR [Corresponding angles, CB ∥ QP, BC is the transversal]

ΔPQR ∼ ΔBCR [By AA or AAA similarity]

Hence, option 2 is the correct option.

RQ = RC + CQ = 4 + 6 = 10 cm.

RB = 3 cm

RC = 4 cm

From 46 que we have ΔPQR ∼ ΔBCR,

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

$$\begin{aligned}\Rightarrow \frac{RP}{RB} = \frac{RQ}{RC} \\[1em] \Rightarrow \frac{RP}{3} = \frac{10}{4} \\[1em] \Rightarrow RP = \frac{5}{2} \times 3 \\[1em] \Rightarrow RP = 7.5 \text{ cm.}\end{aligned}$$

Hence, option 3 is the correct option.

Using ΔPQR ∼ ΔBCR,

$$\begin{aligned}\Rightarrow \frac{PQ}{BC} = \frac{RQ}{RC} \\[1em] = \frac{10}{2} \\[1em] = \frac{5}{2} = 5:2\end{aligned}$$

Hence, option 4 is the correct option.

Directions (Q. 52 to 55): Answer these questions on the basis of the following information:

Through the mid-point M of the side CD of a ∥gm ABCD, the line BM is drawn, intersecting AC on L and AD produced in E.

In ΔBMC and ΔEMD,

∠EDM = ∠MCB [Alternate angles are equal]

∠EMD = ∠BMC [Vertically opposite angles are equal]

CM = DM (As M is the mid-point of CD)

ΔBMC ≅ ΔEMD [By A.A. axiom]

Congruent triangles are similar, have equal areas and perimeters.

∴ All statements are correct

Hence, option 4 is the correct option.

In ΔAEL and ΔCBL,

∠EAL = ∠BCL[Alternate angles are equal]

∠ALE = ∠CLB [Vertical opposite angles are equal]

ΔAEL ∼ ΔCBL [By A.A. axiom]

Hence, option 1 is the correct option.

The similarity ΔAEL ∼ ΔCBL was established by using two pairs of corresponding angles.

Hence, option 1 is the correct option.

AD = BC [ABCD is a parallelogram]

BC = DE [ΔBMC ≅ ΔEMD]

From figure,

AE = AD + DE

AE = BC + BC

AE = 2BC

Using ΔAEL ∼ ΔCBL,

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

$$\begin{aligned}\Rightarrow \frac{EL}{BL} = \frac{AE}{CB} \\[1em] \Rightarrow \frac{EL}{BL} = \frac{2BC}{CB} \\[1em] \Rightarrow \frac{EL}{BL} = \frac{2}{1}\\[1em] \Rightarrow EL : BL = 2 : 1.\end{aligned}$$

Hence, option 2 is the correct option.

Directions (Q. 56 to 59): Study the following diagram carefully and answer the given questions:

In ΔABC, D and E are points on AB and AC respectively such that AD = a, DB = 3a, AE = b and EC = 3b. DQ ∥ EA and EP ∥ DA are drawn. QP is joined.

In ΔADE and ΔABC,

∠A is common.

$\frac{AD}{AB} = \frac{a}{a + 3a} = \frac{1}{4}$

$\frac{AE}{AC} = \frac{b}{b + 3b} = \frac{1}{4}$

By SAS Similarity of Triangles, ΔADE ∼ ΔABC.

In ΔADE and ΔDAQ,

∠DQA = ∠ADE [Corresponding angle]

∠QDA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔDAQ by AA similarity

In ΔADE and ΔEPA,

∠EPA = ∠ADE [Corresponding angle]

∠PEA = ∠DAE [Alternate Interior Angle]

∴ ΔADE ∼ ΔEPA by AA similarity

Hence, option 1 is the correct option.

Since ΔADE ∼ ΔABC, corresponding sides are proportional.

$\frac{DE}{BC} = \frac{AD}{AB}$

Given AD = a and DB = 3a, so AB = 4a.

$$\begin{aligned}\Rightarrow \frac{2}{BC} = \frac{a}{4a} \\[1em] \Rightarrow \frac{2}{BC} = \frac{1}{4}\\[1em] \Rightarrow BC = 8 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\Rightarrow \frac{\text{ Perimeter(ΔADE)}}{\text{ Perimeter(ΔABC)}} = \frac{AD}{AB} \\[1em] = \frac{a}{4a} \\[1em] = \frac{1}{4} = 1:4.\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}\Rightarrow \frac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = \Big(\frac{AB}{AD}\Big)^2 \\[1em] \Rightarrow \frac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\frac{4a}{a}\Big)^2 \\[1em] \Rightarrow \frac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}}= \Big(\frac{4}{1}\Big)^2 \\[1em] \Rightarrow \frac{\text{ Area(ΔABC)}}{\text{ Area(ΔADE)}} = 16 \\[1em] \Rightarrow \text{ Area(ΔABC)} = 16 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \text{ Area quad. DBCE} = \text{ Area(ΔABC)} - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 16 (\text{ Area(ΔADE)}) - \text{ Area(ΔADE)} \\[1em] \Rightarrow \text{ Area quad. DBCE} = 15 (\text{ Area(ΔADE)}) \\[1em] \Rightarrow \frac{\text{ Area(ΔADE)}}{\text{ Area quad. DBCE}} = \frac{1}{15} \\[1em]\end{aligned}$$

Hence, option 3 is the correct option.

Directions (Q. 60 to 63): Study the given information and answer the questions that follow:

In the given figure ABCD is a trapezium in which DC is parallel to AB. AB = 16 cm and DC = 8 cm, OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x − 1) cm.

Given,

In ΔAOB and ΔCOD,

∠AOB = ∠COD [Vertically opposite angle are equal]

∠OAB = ∠OCD [Alternate interior angle are equal]

∴ ΔAOB ∼ ΔCOD (By A.A. axiom)

Hence, option 2 is the correct option.

Given,

ΔAOB ∼ ΔCOD.Since the triangles are similar, the ratios of the corresponding sides are equal,

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

Hence, option 1 is the correct option.

Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

Ratio = $\frac{CD}{AB} = \frac{8}{16} = \frac{1}{2}$ = 1 : 2.

Hence, option 2 is the correct option.

Given,

ΔAOB ∼ ΔCOD. Since the triangles are similar, the ratios the corresponding sides are equal,

$\frac{CD}{AD} = \frac{OC}{OA} = \frac{OD}{OB} = \frac{1}{2}$

Solving,

$\frac{OC}{OA} = \frac{1}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{(x - 1)}{11} = \frac{1}{2} \\[1em] \Rightarrow 2 \times (x - 1) = 11 \\[1em] \Rightarrow 2x - 2 = 11 \\[1em] \Rightarrow 2x = 11 + 2 \\[1em] \Rightarrow 2x = 13 \\[1em] \Rightarrow x = \frac{13}{2} \\[1em] \Rightarrow x = 6.5\end{aligned}$$

Solving,

$\frac{OD}{OB} = \frac{1}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{5}{(y + 3)} = \frac{1}{2} \\[1em] \Rightarrow 2 \times 5 = (y + 3) \\[1em] \Rightarrow 10 = y + 3 \\[1em] \Rightarrow y = 10 - 3 \\[1em] \Rightarrow y = 7.\end{aligned}$$

x = 6.5, y = 7

Hence, option 3 is the correct option.

Given, ΔABC ∼ ΔEFG

So, the corresponding sides are in proportion.

$$\begin{aligned}\Rightarrow \frac{AB}{EF} = \frac{BC}{FG}\\[1em] \Rightarrow \frac{15}{75} = \frac{3}{FG}\\[1em] \Rightarrow \frac{1}{5} = \frac{3}{FG}\\[1em] \Rightarrow FG = 3 \times 5\\[1em] \Rightarrow FG = 15\text{ cm}\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In ∆APQ and ∆ABC,

⇒ ∠APQ = ∠ABC [Both equal to 90°]

⇒ ∠BAC = ∠PAQ [Common angles]

∴ ∆APQ ∼ ∆ABC (By A.A. axiom)

We know that,

The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆APQ}}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{18}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC}= \frac{(4 + 3)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{49}{9} \times 18 = 98 \text{ cm}^2.\end{aligned}$$

From figure,

Area QPBC = Area of ∆ABC − Area of ∆APQ

= 98 - 18 = 80 cm².

Hence, Option 3 is the correct option.

In ΔABC and ΔPQR,

⇒ ∠BAC = ∠QPR [Given]

⇒ ∠ABC = ∠PQR [Given]

∴ ΔABC ~ ΔPQR (By A.A. axiom)

So, Assertion (A) is true and reason (R) is false.

Hence, option 3 is the correct option.

In ΔEDB and ΔACB,

∠EDB = ∠ACB (Given)

∠EBD = ∠ABC (Common angle)

∴ ΔEDB ∼ ΔACB (By A.A. axiom)

From figure, BC = BE + EC = 6 + 4 = 10 cm.

Since corresponding sides of similar triangles are proportional,

$$\begin{aligned}\Rightarrow \frac{DB}{CB} = \frac{EB}{AB} \\[1em] \Rightarrow \frac{5}{10} = \frac{6}{AB} \\[1em] \Rightarrow AB = 6 \times 2 = 12 \text{ cm}.\end{aligned}$$

So, Assertion (A) is true.

If two triangles have two pairs of corresponding angles equal, then the triangles are similar by the AA axiom, and this is exactly the criterion used above to find AB.

So, Reason (R) is true and is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Given, DE ∥ BC.

In ΔADE and ΔABC,

∠ADE = ∠ABC (Corresponding angles are equal)

∠DAE = ∠BAC (Common angle)

∴ ΔADE ∼ ΔABC (By A.A. axiom)

From figure, AD = 3 units, DB = 4 units, so AB = AD + DB = 3 + 4 = 7 units.

Since corresponding sides of similar triangles are proportional,

$$\begin{aligned}\Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \\[1em] \Rightarrow \frac{3}{7} = \frac{x}{14} \\[1em] \Rightarrow x = \frac{3 \times 14}{7} = 6.\end{aligned}$$

So, Assertion (A) is true.

Similar triangles only require corresponding angles to be equal and corresponding sides to be proportional; they need not be congruent. So similar triangles are not always congruent.

So, Reason (R) is false.

Hence, option 3 is the correct option.

In ΔADB and ΔBDC,

∠ADB = ∠BDC = 90°

∠DBA = ∠DCB (Angles complementary to ∠DBC)

∴ ΔADB ∼ ΔBDC (By A.A. axiom)

Since corresponding sides of similar triangles are proportional,

$$\begin{aligned}\Rightarrow \frac{AD}{BD} = \frac{BD}{DC} \\[1em] \Rightarrow \frac{4}{6} = \frac{6}{DC} \\[1em] \Rightarrow DC = \frac{36}{4} = 9 \text{ cm}.\end{aligned}$$

From figure, AC = AD + DC = 4 + 9 = 13 cm and height BD = 6 cm.

$\text{Area of ΔABC} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 13 \times 6 = 39 \text{ cm}^2.$

The Assertion states the area is 40 cm², which is incorrect (it is 39 cm²).

So, Assertion (A) is false.

Areas of two similar triangles are indeed proportional to the squares of their corresponding sides, which is a true statement.

So, Reason (R) is true.

Hence, option 4 is the correct option.

In ΔABC and ΔDAC,

∠ABC = ∠DAC (Given)

∠BCA = ∠DCA (Common angle)

∴ ΔABC ∼ ΔDAC (By A.A. axiom)

Since corresponding sides of similar triangles are proportional,

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

So, Assertion (A) is true.

Both the SAS criterion and the SSS criterion are valid tests for the similarity of two triangles, so the Reason is a true statement. However, the similarity above was established using the AA axiom, so the Reason is not the correct explanation of the Assertion.

So, Reason (R) is true but is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

In ΔABC and ΔPQR,

⇒ ∠BAC = ∠QPR [Given]

⇒ ∠ABC = ∠PQR [Given]

∴ ΔABC ~ ΔPQR (By A.A. axiom)

So, Assertion (A) is true and reason (R) is false.

Hence, option 3 is the correct option.

(a) In △ ABD and △ CBA,

⇒ ∠ABD = ∠CBA (Common angle)

⇒ ∠BAD = ∠ACB (Given)

∴ △ ABD ~ △ CBA (By A.A. axiom)

Hence, proved that △ ABD ~ △ CBA.

(b) We know that,

Corresponding sides of similar triangle are proportional.

$$\begin{aligned}\therefore \frac{AB}{BC} = \frac{BD}{BA} \\[1em] \Rightarrow \frac{8}{x + 4} = \frac{4}{8} \\[1em] \Rightarrow 4(x + 4) = 8 \times 8 \\[1em] \Rightarrow 4x + 16 = 64 \\[1em] \Rightarrow 4x = 64 - 16 \\[1em] \Rightarrow 4x = 48 \\[1em] \Rightarrow x = \frac{48}{4} = \text{ 12 cm}.\end{aligned}$$

Hence, x = 12 cm.

Given,

ABCD is a rectangle where side BC is twice side AB.

⇒ BC = 2AB

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (2AB)²

⇒ AC² = AB² + 4AB²

⇒ AC² = 5AB²

⇒ AC = $\sqrt{5}$ AB.

We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$$\begin{aligned}\therefore \frac{\text{area of △BAP}}{\text{area of △ACQ}} = \frac{BA^2}{AC^2} \\[1em] \Rightarrow \frac{\text{area of △BAP}}{\text{area of △ACQ}} = \frac{BA^2}{(\sqrt{5}BA)^2} \\[1em] \Rightarrow \frac{\text{area of △BAP}}{\text{area of △ACQ}} = \frac{BA^2}{5BA^2} \\[1em] \Rightarrow \frac{\text{area of △BAP}}{\text{area of △ACQ}} = \frac{1}{5} \\[1em]\end{aligned}$$

Area of △BAP : Area of △ACQ = 1 : 5.

Hence, Area of △BAP : Area of △ACQ = 1 : 5.

Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ ∆ ABC and ∆ AB'C' are similar triangles.

(a) We know that,

Ratio of corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{AB}{AB'} = \frac{BC}{B'C'} \\[1em] \Rightarrow \frac{AB}{AB'} = \frac{3}{5}\end{aligned}$$

Let AB = 3x and AB' = 5x

From figure,

⇒ AB' = AB + BB'

⇒ 5x = 3x + BB'

⇒ BB' = 5x - 3x = 2x.

⇒ AB : BB' = 3x : 2x = 3 : 2.

Hence, AB : BB' = 3 : 2.

(b) As,

$$\begin{aligned}\Rightarrow \frac{AB}{BB'} = \frac{3}{2} \\[1em] \Rightarrow \frac{AB}{4} = \frac{3}{2} \\[1em] \Rightarrow AB = \frac{3}{2} \times 4 \\[1em] \Rightarrow AB = 6 \text{ cm}.\end{aligned}$$

Hence, AB = 6 cm.

(c) Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ BC || B'C'

In ∆ ABC and ∆ AB'C',

⇒ ∠BAC = ∠B'AC' (Common angle)

⇒ ∠ABC = ∠AB'C' (Corresponding angle are equal)

∴ ∆ ABC ~ ∆ AB'C' (By A.A. axiom).

Hence, proved that ∆ ABC ~ ∆ AB'C'.

(d) We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\frac{AB}{AB'}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\frac{3}{5}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \frac{9}{25}.\end{aligned}$$

Let area of ∆ ABC = 9a and area of ∆ AB'C' = 25a.

Area of quadrilateral BB'C'C = Area of ∆ AB'C' - Area of ∆ ABC = 25a - 9a = 16a.

∴ Area of ∆ ABC : Area of quadrilateral BB'C'C = 9a : 16a = 9 : 16.

Hence, area of ∆ ABC : area of quadrilateral BB'C'C = 9 : 16.

(a) In Δ AFE,

⇒ ∠AFE = ∠ACB = 105° (Corresponding angle are equal)

⇒ ∠EAF = $\frac{∠BAD}{2} = \frac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

By angle sum property of triangle,

⇒ ∠AFE + ∠EAF + ∠AEF = 180°

⇒ 105° + 35° + ∠AEF = 180°

⇒ 140° + ∠AEF = 180°

⇒ ∠AEF = 180° - 140° = 40°.

∠AGF = ∠ADC = 40° (Corresponding angles are equal)

In Δ AGF,

⇒ ∠GAF = $\frac{∠BAD}{2} = \frac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

In Δ AEF and Δ AGF,

⇒ ∠EAF = ∠GAF = 35° (AC being the bisector)

⇒ ∠AEF = ∠AGF = 40° (Proved above)

∴ Δ AEF ~ Δ AGF (By A.A. axiom)

Hence, proved that Δ AEF ~ Δ AGF.

(b) In Δ ACD,

⇒ ∠CAD = 35°

⇒ ∠ADC = 40°

By angle sum property of triangle,

⇒ ∠CAD + ∠ADC + ∠DCA = 180°

⇒ 35° + 40° + ∠DCA = 180°

⇒ ∠DCA + 75° = 180°

⇒ ∠DCA = 180° - 75° = 105°.

In Δ ACD and Δ ACB,

⇒ ∠ACB = ∠ACD = 105° (Proved above)

⇒ ∠BAC = ∠DAC = 35° (Proved above)

∴ Δ ACD ~ Δ ACB (By A.A. axiom)

(i) Given,

⇒ AE : EB = 2 : 3

Let AE = 2x and EB = 3x.

∴ AB = AE + EB = 2x + 3x = 5x.

We know that,

Ratio of corresponding sides are proportional.

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

Hence, AG : AD = 2 : 5.

(ii) In △ ABC and △ ADC,

⇒ ∠BAC = ∠DAC (Both equal to 35°)

⇒ ∠ACB = ∠ACD (Both equal to 105°)

⇒ ∠ABC = ∠ADC (Both equal to 40°)

∴ △ ABC and △ ADC are congruent.

We know that,

Area of congruent triangles are equal.

Let area of △ ABC and area of △ ADC = x.

∴ Area of △ ABC : Area of △ ADC = x : x = 1 : 1.

Hence, area of △ ABC : area of △ ADC = 1 : 1.

(iii) From figure,

Area of quadrilateral ABCD = Area of △ ABC + Area of △ ADC = x + x = 2x.

∴ Area of quadrilateral ABCD : Area of △ ACB = 2x : x = 2 : 1.

Hence, area of quadrilateral ABCD : area of △ ACB = 2 : 1.

(a) In △ ACD and △ FCE,

⇒ ∠ACD = ∠FCE (Common angles)

⇒ ∠ADC = ∠FEC (Corresponding angles are equal)

∴ △ ACD ~ △ FCE (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$$\begin{aligned}\therefore \frac{FC}{AC} = \frac{EC}{CD} \\[1em] \Rightarrow \frac{FC}{AC} = \frac{EC}{EC + ED} \\[1em] \Rightarrow \frac{FC}{AC} = \frac{3}{3 + 18} \\[1em] \Rightarrow \frac{FC}{AC} = \frac{3}{21} \\[1em] \Rightarrow \frac{FC}{AC} = \frac{1}{7}.\end{aligned}$$

Let FC = x and AC = 7x.

From figure,

AF = AC - FC = 7x - x = 6x.

$\therefore \frac{AF}{FC} = \frac{6x}{x} = \frac{6}{1}$ = 6 : 1.

Hence, AF : FC = 6 : 1.

(b) Since, △ ACD ~ △ FCE

$$\begin{aligned}\therefore \frac{EF}{AD} = \frac{EC}{CD} \\[1em] \Rightarrow \frac{EF}{35} = \frac{3}{21} \\[1em] \Rightarrow EF = \frac{3}{21} \times 35 \\[1em] \Rightarrow EF = \frac{105}{21} = 5 \text{ cm}.\end{aligned}$$

Hence, EF = 5 cm.

(c) We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$$\begin{aligned}\therefore \frac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\frac{DC}{EC}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\frac{21}{3}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \frac{441}{9} \\[1em] \Rightarrow \frac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \frac{49}{1}.\end{aligned}$$

Let area of △ ADC = 49a and area of △ FCE = a.

Area of trapezium ADEF = Area of △ ADC - Area of △ FCE = 49a - a = 48a.

$\therefore \frac{\text{Area of trapezium ADEF}}{\text{Area of △ FCE}} = \frac{48a}{a} = \frac{48}{1}.$

Hence, area of trapezium ADEF : area of △ EFC = 48 : 1.

(d) In △ AGF and △ ABC,

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

⇒ ∠GAF = ∠BAC (Common angles)

∴ △ AGF ~ △ ABC (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \frac{AC}{AF} = \frac{BC}{GF}$

From part (a),

⇒ FC = x and AC = 7x

⇒ AF = AC - FC = 7x - x = 6x.

$\Rightarrow \frac{BC}{GF} = \frac{7x}{6x} = \frac{7}{6}.$

Hence, BC : GF = 7 : 6.

At the given time, all objects and their shadows form similar right triangles, so the ratio of height to shadow is the same for every object.

For Amit's house : $\frac{\text{height}}{\text{shadow}} = \frac{20}{10} = 2$.

So, height of tower = 2 × (its shadow) = 2 × 50 = 100 m.

Hence, option (c) is the correct option.

When Amit's house (20 m) casts a shadow of 12 m, the common ratio of height to shadow at that time is

$\frac{\text{height}}{\text{shadow}} = \frac{20}{12} = \frac{5}{3}.$

For the tower (height 100 m), shadow = $\frac{\text{height}}{\text{ratio}} = \frac{100}{\frac{5}{3}} = 100 \times \frac{3}{5} = 60$ m.

Hence, option (b) is the correct option.

At the time when Amit's house casts a shadow of 10 m, the ratio of height to shadow is 2 (from Question 1), and Sumit's house casts a shadow of 20 m.

So, height of Sumit's house = 2 × 20 = 40 m.

Hence, option (c) is the correct option.

When the tower (height 100 m) casts a shadow of 40 m, the common ratio of height to shadow at that time is

$\frac{\text{height}}{\text{shadow}} = \frac{100}{40} = \frac{5}{2}.$

For Sumit's house (height 40 m), shadow = $\frac{40}{\frac{5}{2}} = 40 \times \frac{2}{5} = 16$ m.

Hence, option (b) is the correct option.

When the tower (height 100 m) casts a shadow of 30 m, the common ratio of height to shadow at that time is

$\frac{\text{height}}{\text{shadow}} = \frac{100}{30} = \frac{10}{3}$

For Amit's house (height 20 m), shadow = $\frac{20}{\frac{10}{3}} = 20 \times \frac{3}{10} = 6$ m.

Hence, option (c) is the correct option.