Similarity — Exercise 15(B)

In the given figure, DE is parallel to BC. If AD : BD = 3 : 5 then DE : BC is :

1. 3 : 8

2. 3 : 5

3. 5 : 3

4. 8 : 3

Answer

From figure,

In △ DAE and △ BAC,

⇒ ∠DAE = ∠BAC (Common angle)

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

∴ △ DAE ~ △ BAC (By A.A. postulate)

Given,

AD : BD = 3 : 5

Let AD = 3x and BD = 5x.

From figure,

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

We know that,

Corresponding sides of similar triangle are in proportion.

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

Hence, Option 1 is the correct option.

If AD = AE and BD = CE then :

1. △ ADE ~ △ ACB

2. △ ABC ~ △ ACB

3. △ ABD ~ △ ABC

4. △ ADE ~ △ ABC

Answer

Given,

⇒ AD = AE = x (let)

⇒ BD = CE = y (let)

From figure,

⇒ AB = AD + BD = x + y

⇒ AC = AE + EC = x + y

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC

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

∴ △ ADE ~ △ ABC (By S.A.S. postulate)

Hence, Option 4 is the correct option.

In the given figure,

AB = 10 cm, CD = 8 cm = OB, then OD is equal to :

1. 10 cm

2. 3.2 cm

3. 6.4 cm

4. 8 cm

Answer

From figure,

In △ AOB and △ COD,

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

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

∴ △ AOB ~ △ COD (By A.A. postulate)

We know that,

Corresponding sides of similar triangle are in proportion.

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

Hence, Option 3 is the correct option.

In the given figure, AE : EC = 2 : 3 and BC = 20 cm then DE is equal to :

1. 10 cm

2. 12 cm

3. 8 cm

4. 16 cm

Answer

From figure,

In △ ADE and △ ABC,

⇒ ∠DAE = ∠BAC (Common angle)

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

∴ △ ADE ~ △ ABC (By A.A. postulate)

Given,

AE : EC = 2 : 3

Let AE = 2x and EC = 3x.

From figure,

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

We know that,

Corresponding sides of similar triangle are in proportion.

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

Hence, Option 3 is the correct option.

Two congruent triangles are :

1. not equal in area

2. similar

3. not similar

4. not similar but congruent

Answer

We know that,

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

Hence, Option 2 is the correct option.

In the following figure, point D divides AB in the ratio 3 : 5. Find:

(i) $\dfrac{\text{AE}}{\text{EC}}$

(ii) $\dfrac{\text{AD}}{\text{AB}}$

(iii) $\dfrac{\text{AE}}{\text{AC}}$

Also if,

(iv) DE = 2.4 cm, find the length of BC.

(v) BC = 4.8 cm, find the length of DE.

Answer

(i) Given,

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

By basic proportionality theorem we have :

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

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

Hence, AE : EC = 3 : 5.

(ii) Given,

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

Let AD = 3x and DB = 5x.

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

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

Hence, AD : AB = 3 : 8.

(iii) Given,

$\Rightarrow \dfrac{AE}{EC} = \dfrac{3}{5} \\[1em] \Rightarrow \dfrac{EC}{AE} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{EC}{AE} + 1 = \dfrac{5}{3} + 1 \\[1em] \Rightarrow \dfrac{EC + AE}{AE} = \dfrac{5 + 3}{3} \\[1em] \Rightarrow \dfrac{AC}{AE} = \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{AE}{AC} = \dfrac{3}{8}.$

Hence, AE : AC = 3 : 8.

(iv) In ∆ADE and ∆ABC,

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

∠A = ∠A [Common angles]

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

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

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

Hence, BC = 6.4 cm.

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

So, we have

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

Hence, DE = 1.8 cm.

In the given figure, PQ || AB; CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find :

(i) $\dfrac{CP}{PA}$

(ii) PQ

(iii) If AP = x, then the value of AC in terms of x.

Answer

(i) Given PQ || AB,

By basic proportionality theorem :

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

Hence, ratio = 4 : 3.

(ii) In ∆CPQ and ∆CAB,

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

∠PCQ = ∠ACB [Common angle]

∴ ∆CPQ ~ ∆CAB [By AA].

From figure,

CB = CQ + QB = 4.8 + 3.6 = 8.4

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

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

Hence, PQ = 3.6 cm

(iii) As, ∆CPQ ~ ∆CAB.

We have,

$\dfrac{CP}{AC} = \dfrac{CQ}{CB} \\[1em] \dfrac{CP}{AC} = \dfrac{4.8}{8.4} \\[1em] \dfrac{CP}{AC} = \dfrac{4}{7}.$

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

Given, AP = x

or, 3 parts = x

⇒ 1 part = $\dfrac{x}{3}$

⇒ 7 parts = $\dfrac{7x}{3}$.

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

A line PQ is drawn parallel to the side BC of ΔABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.

Answer

Let AP = x cm.

From figure,

PB = AB - AP = 9 - x cm.

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

Given PQ || AB,

By basic proportionality theorem :

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

Hence, AP = 6.3 cm.

In ΔABC, D and E are the points on sides AB and AC respectively.

Find whether DE || BC, if

(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm.

(ii) AB = 6.3 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.

Answer

(i) From figure,

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

In ∆ADE and ∆ABC,

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

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

(ii) From figure,

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

In ∆ADE and ∆ABC,

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

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

In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If 'x' be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of 'x'.

Answer

Given,

ΔABC ~ ΔADE

Given,

AE : EC = 4 : 7

Let AE = 4y and EC = 7y.

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

From figure,

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

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

As ΔABC ~ ΔADE, we have :

∠ABC = ∠ADE and ∠ACB = ∠AED

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

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

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

AP = x

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

In ∆ADP and ∆ABQ,

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

∠APD = ∠AQB [Both = 90°]

∴ ∆ADP ~ ∆ABQ [By AA]

$\Rightarrow \dfrac{AD}{AB} = \dfrac{AP}{AQ} \\[1em] \Rightarrow \dfrac{4}{11} = \dfrac{x}{AQ} \\[1em] \Rightarrow AQ = \dfrac{11}{4}x.$

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

A line segment DE is drawn parallel to base BC of ∆ABC which cuts AB at point D and AC at point E. If AB = 5BD and EC = 3.2 cm, find the length of AE.

Answer

Given,

⇒ AB = 5BD

⇒ AD + BD = 5BD

⇒ AD = 5BD - BD

⇒ AD = 4BD

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

Given DE || BC,

by basic proportionality theorem :

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

Hence, AE = 12.8 cm

In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.

Answer

In ∆BEF,

DC || FE

So, by basic proportionality theorem,

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

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

Let BD = ax and DF = 3a.

From figure,

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

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

In ∆AFB and CDF,

∠AFB = ∠CFD [Common angles]

∠ABF = ∠CDF [Corresponding angles are equal]

∴ ∆AFB ~ ∆CDF [By AA]

Since, corresponding sides of similar triangles are proportional.

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

In ∆BCD and ∆BEF ,

∠FBE = ∠DBC [Common angles]

∠CDB = ∠EFB [Corresponding angles are equal]

∴ ∆BEF ~ ∆BCD [By AA]

Since, corresponding sides of similar triangles are proportional.

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

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

Let BD = ay and BF = 4.5a

From figure,

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

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

From (1) and (5) we get,

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

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

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

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

Substituting value of x in (4) we get,

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

Hence, x = 5 cm and y = 2.8125 cm

In the figure, given below, PQR is a right-angled triangle right-angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.

Answer

Given XY || QR,

By basic proportionality theorem :

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

From figure,

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

Since, PQR is a right-angled triangle.

By pythagoras theorem we get,

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

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

In the following figure, M is the mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that : PE = 2PD.

Answer

In ∆BME and ∆DMC,

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

∠MCD = ∠MBE [Alternate angles are equal]

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

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

∴ BE = DC [By C.P.C.T]

Since, opposite sides of parallelogram are equal.

∴ AB = DC

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

In ∆DCP and ∆EPA,

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

∠CDP = ∠AEP [Alternate angles are equal]

∴ ∆DCP ~ ∆EAP [By AA]

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

$\Rightarrow \dfrac{DC}{EA} = \dfrac{PD}{PE} \\[1em] \Rightarrow \dfrac{EA}{DC} = \dfrac{PE}{PD} \\[1em] \Rightarrow \dfrac{PE}{PD} = \dfrac{AB + BE}{DC} \\[1em] \Rightarrow \dfrac{PE}{PD} = \dfrac{2DC}{DC} \\[1em] \Rightarrow \dfrac{PE}{PD} = 2 \\[1em] \Rightarrow PE = 2PD.$

Hence proved that PE = 2PD.

The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

Answer

From figure,

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

In △DEC and △EAF

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

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

∴ △DEC ~ △EAF [By AA]

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

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

Since, ABCD is a ||gm.

AB = CD and BC = AD.

From figure,

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

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

= 12 + 10 + 12 + 10

= 44 cm.

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