Similarity of Triangles — Exercise 16(A)

In the given figure, XY || BC. Given that AX = 3 cm, XB = 1.5 cm and BC = 6 cm.

(i) Calculate $\dfrac{AY}{YC}$.

(ii) Calculate XY.

Answer

(i) By basic proportionality theorem,

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

Since, XY || BC

$\therefore \dfrac{AY}{YC} = \dfrac{AX}{XB} \\[1em] = \dfrac{3}{1.5} \\[1em] = \dfrac{2}{1}.$

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

(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.

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

Hence, XY = 4 cm.

In the given figure, DE || BC.

(i) If AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm, find EC.

(ii) If $\dfrac{AD}{DB} = \dfrac{3}{5}$ and AC = 5.6 cm, find AE.

(iii) If AD = x cm, DB = (x − 2) cm, AE = (x + 2) cm and EC = (x − 1) cm, find the value of x.

Answer

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,

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

Hence, EC = 3.6 cm.

(ii) Given,

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

AC = 5.6 cm

Since, DE || BC by basic proportionality theorem,

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

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,

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

Hence, x = 4 cm.

D and E are points on the sides AB and AC respectively of ΔABC. For each of the following cases, state whether DE ∥ BC :

(i) AD = 5.7 cm, BD = 9.5 cm, AE = 3.6 cm and EC = 6 cm.

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 9.6 cm and EC = 2.4 cm.

(iii) AB = 11.7 cm, BD = 5.2 cm, AE = 4.4 cm and AC = 9.9 cm.

(iv) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.

Answer

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,

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

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,

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

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,

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

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,

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

We conclude that DE is parallel to BC

Hence, DE is parallel to BC.

In the given figure, it is given that ∠ABD = ∠CDB = ∠PQB = 90°. If AB = x units, CD = y units and PQ = z units, prove that $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$.

Answer

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.

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

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.

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

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

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

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

In ΔABC, AD is the bisector of ∠A. If BC = 10 cm, BD = 6 cm and AC = 6 cm, find AB.

Answer

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

In ΔBCE,

Since AD ∥ CE, by Basic Proportionality Theorem:

$\dfrac{BD}{DC} = \dfrac{AB}{AE}$ ...(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):

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

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,

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

Hence, length of AB is 9 cm.

In the given figure, AC ∥ DE ∥ BF. If AC = 24 cm, EG = 8 cm, GB = 16 cm, BF = 30 cm.

(i) Prove that ΔGED ∼ ΔGBF.

(ii) Find DE.

(iii) Find DB : AB.

Answer

(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.

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

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.

$\Rightarrow \dfrac{DB}{AB} = \dfrac{DE}{AC} \\[1em] \Rightarrow \dfrac{DB}{AB} = \dfrac{15}{24} \\[1em] \Rightarrow \dfrac{DB}{AB} = \dfrac{5}{8}.$

Hence, DB : AB = 5 : 8.

In the adjoining figure, ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that :

(i) DP : PL = DC : BL.

(ii) DL : DP = AL : DC.

Answer

(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.

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

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.

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

DL : DP = AL : DC

Hence, proved DL : DP = AL : DC.

In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F.

Prove that DF × FE = FB × FA.

Answer

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.

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

DF × FE = FB × FA

Hence, proved that DF × FE = FB × FA.

In the adjoining figure (not drawn to scale), PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5 cm.

(i) Show that ΔPQR ∼ ΔPST.

(ii) Calculate ST, if QR = 5.8 cm.

Answer

(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,

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

∠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.

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

Hence, ST = 2.9 cm.

In the adjoining figure, ABCD is a parallelogram in which AB = 16 cm, BC = 10 cm and L is a point on AC such that CL : LA = 2 : 3. If BL produced meets CD at M and AD produced at N, prove that :

(i) ΔCLB ∼ ΔALN.

(ii) ΔCLM ∼ ΔALB.

Answer

(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 the given figure, AB ∥ PQ and AC ∥ PR. Prove that BC ∥ QR.

Answer

In ΔOPQ,

AB ∥ PQ [Given]

By Basic Proportionality Theorem,

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

In ΔOPR,

AC ∥ PR [Given]

By Basic Proportionality Theorem,

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

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

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

In ΔOQR,

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

By Converse of Basic Proportionality Theorem,

BC ∥ QR

Hence, proved that BC ∥ QR.

In the given figure, medians AD and BE of ΔABC meet at G and DF ∥ BE. Prove that :

(i) EF = FC.

(ii) AG : GD = 2 : 1.

Answer

(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.

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

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,

$\dfrac{AE}{EF} = \dfrac{AG}{GD}$ .....(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,

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

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

In the given figure, DE ∥ BC and BD = DC.

(i) Prove that DE bisects ∠ADC.

(ii) If AD = 4.5 cm, AE = 3.9 cm and DC = 7.5 cm, find CE.

(iii) Find the ratio AD : DB.

Answer

(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,

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

Hence, CE = 6.5 cm.

(iii) By basic proportionality theorem,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{AD}{DB} = \dfrac{3.9}{6.5} \\[1em] \Rightarrow \dfrac{AD}{DB} = \dfrac{3}{5}.$

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

In the given figure, BA ∥ DC. Show that ΔOAB ∼ ΔODC. If AB = 4 cm, CD = 3 cm, OC = 5.7 cm and OD = 3.6 cm, find OA and OB.

Answer

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.

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

Considering,

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

Considering,

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

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

In the given figure, ∠ABC = 90° and BD ⟂ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

Answer

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.

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

Hence, BC = 8.1 cm.