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Here, AC and PR represents the bigger and smaller poles, and BC and QR represents their shadows respectively. Now, ∆ACB – ∆PRQ [ ∵ Vertical poles and their shadows form similar figures] 

∴ CB/RQ =AC/PR  [Corresponding sides of similar tringles]

∴ x/6 = 8/4

∴ x = (8 x 6)/4

∴ x = 12m

∴ The shadow of the bigger pole will be 12 metres long at that time.

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

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

∴ Required ratio = 3 : 5

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

∴ Required ratio = 3 : 5

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

∴ Required ratio = (3)² : (5)² = 9 : 25

Proof: 

In ∆ABC, 

Point D is on angle bisector of ∠A. [Given] 

∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle] 

A(∆ABD)/A(∆ACD) = (AB x DM) (AC x DN) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights] 

∴ A(∆ABD)/A(∆ACD) = AB/AC (ii) [From (i)] 

Also, ∆ABD and ∆ACD have equal height. 

∴ A(∆ABD)/A(∆ACD) = BD/CD (iii) [Triangles having equal height ]

∴ AB/AC = BD/DC  [From (ii) and (iii)]

Answer is False

Answer is False

Answer is True

Answer is (A) 25 : 16

∆ABC ~ ∆DEF

(ar.∆ABC)/(ar.∆DEF) = AB²/DE²

= (10/8)2

= 25/16

ar. (∆ABC) : ar.(∆DEF) = 25 : 16

Given : In two triangles ABC and ∆PQR D is mid-point of BC and M is mid-point of QR.

and \(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) = \(\frac { AD }{ PM }\) ……(i)

To prove : ∆ABC ~ ∆PQR

Construction : Produce AD up to E such that AD = DE. Join BE and CE and produce PM upto N such that PM = MN. Join QN and NR.

Proof : Diagonals AE and BC of quadrilateral ABEC bisect each other at point D.

∴ quadrilateral ABEC is a parallelogram.

∴ BE = AC …(ii)

Similarly PQNR is a parallelogram

∴ QN = PR …..(iii)

Dividing equation (ii) by (iii)

\(\frac { BE }{ QN }\) = \(\frac { AC }{ PR }\) …..(iv)

Now, \(\frac { AD }{ PM }\) = \(\frac { 2AD }{ 2PM }\) = \(\frac { AD }{ PM }\) = \(\frac { AE }{ PN }\) …(v)

In equation (i), (iv) and (v),

\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)

Thus, in ∆ABE and ∆PQN

\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)

∴ ∆ABE ~ ∆PQN (By SSS)

∴ ∠BAE = ∠QPN …(vi)

Similarly

∆AEC ~ ∆PNR

∠EAC = ∠NPR …(vii)

Adding equation (vi) and (vii)

∠BAE + ∠EAC = ∠QPN + ∠NPR

⇒ ∠BAC = ∠QPR

Now, in ∆ABC and ∠PQR

\(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) [From equation (i)]

∠A = ∠P

∴ By SAS similarity criterion

∆ABC ~ ∆PQR

(i) Given : Two similar triangles ABC and PQR CD ⊥ AB and RS ⊥ PQ.

To prove : ∆ADC ~ ∆PSR

Proof : \(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\)

⇒ \(\frac { CA }{ RP }\) = \(\frac { 2AD }{ 2PS }\) (Since D and S. are medians of AB and PQ)

Now, in ∆ACD and ∆PRS,

\(\frac { CA }{ RP }\) = \(\frac { AD }{ PS }\)​​​​​

and ∠A = ∠P (∵ ∆ABC ~ ∆PQR)

∠ADC = ∠PSR = 90°

Thus, by AA similarity Criterion.

∆ADC ~ ∆PSR.

(ii) Given :

∆ADC ~ ∆PSR.

To prove : \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

Proof : Given that ∆ABC and ∆PQR are similar. 

So ∠A = ∠P ….(i) (∵ Corresponding angle S of similar triangles are same.)

\(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\) = \(\frac { CB }{ RQ }\) ….(ii)

No, In ∆CAD and ∆RPS,

∠A = ∠P (given)

∠CDA = ∠RSP = 90°

By A-A similarity criterion

∆CDA ~ ∆RSP

\(\frac { CA }{ RP }\) = \(\frac { CD }{ RS }\)

From equation (ii) and (iii)

\(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

This statement is false because for similarity of triangles two sides and angle included them should be equal.

∆ABC ~ ∆FDE (Given)

∴ ∠A = ∠F, ∠B = ∠D and ∠C = ∠E

Side AB ↔ FD, Side BC ↔ DE, side AC ↔ FE

∴ AB/FD = BC/DE = AC/FE

Thus AB/DE = BC/EF = CA/FD is not possible.

If a line parallel to a side of a triangle intersects the other sides in two distinct points, then the line divides those sides in proportion.

The ratio is 1 : 1 : 2 .

The ratio is 1 : √3 : 2.

Answer is (C) SSA

Proof

∠APQ = ∠ABC = 60° [Given] 

∴ ∠APQ ≅ ∠ABC

∴ side PQ || side BC (i) [Corresponding angles test] 

In ∆ABC,

sidePQ || sideBC [From (i)]

∴ AP/PB = AQ/QC [Basic proportionality theorem]

Proof: 

In ∆PMR,

QM = QR = a ---- [Given]

\(\therefore\) Q is midpoint of seg MR.

\(\therefore\) seg PQ is the median

\(\therefore\) PM² + PR² = 2PQ² + 2QM² ---- [By Apollonius theorem]

\(\therefore\) PM² + a² = 2a² + 2a² ---- [Substituting the given values]

\(\therefore\) PM² + a² = 4a² 

\(\therefore\) PM² = 4a² - a²

\(\therefore\) PM² = 3a²

\(\therefore\) PM = √3 a ---- [Taking square root on both sides]

Similarly, we can prove

PN = √3 a

\(\therefore\) PM = PN = √3 a

line AB || line CD || line FE [Given] 

∴ BD/DF = AC/CE  [Property of three parallel lines and their transversals] 

∴ 8/4 = 12/X 

∴ X = (12 x 4)/8

 ∴ X = 6 units 

Now, AE AC + CE [A – C – E] 

= 12 + x 

= 12 + 6 

= 18 units 

∴ x = 6 units and AE = 18 units

In ∆MNP, seg NQ bisects ∠N. [Given] 

∴ PN/MN = QP/MQ [Property of angle bisector of a triangle] 

∴ 40/25 = QP/14

 ∴ QP = (40 x 14)/25 

∴ QP = 22.4 units

side AB || side PQ || side DC [Given] 

∴ AP/PD = BQ/QC  [Property of three parallel lines and their transversals] 

∴ 15/12 = BQ/14 

∴ BQ = (15 x 14)/12 

∴ BQ = 17.5 units

side AB || side DC [Given] 

and seg BD is their transversal. 

∴ ∠DBA ≅ ∠BDC [Alternate angles] 

∴ ∠OBA ≅ ∠ODC (i) [D – O – B] 

In ∆OBA and ∆ODC 

∠OBA ≅ ∠ODC [From (i)] 

∠BOA ≅ ∠DOC [Vertically opposite angles] 

∴ ∆OBA ~ ∆ODC [AA test of similarity] 

∴ OB/OD = AB/DC  = [Corresponding sides of similar triangles] 

∴  15/OD = 20/6

 ∴ OD = x = (15 x 6)/20

∴ OD = 4.5 units 

Answer is (A) 4 cm

∴ ∠P = ∠R

∴ QR = PQ

⇒ PQ = QR = 4 cm

PN + NR = PR [P – N – R] 

∴ PN + 8 = 20 

∴ PN = 20 – 8 = 12 

Also, PM + MQ = PQ [P – M – Q] 

∴ 15 + MQ = 25

∴ MQ = 25 - 15 = 10

PN/NR = 12/8

∴ PN/NR = 3/2   ....(i)

PM/MQ = 15/10

∴ PM/MQ = 3/2   ...(ii)

In ∆PQR,

PN/NR = PM/MQ    ...[Form (i) and (ii)

∴ line NM || side RQ [Converse of basic proportionality theorem]

Correct answer is

(B) ∆PQR – ∆CAB

∆ABC ~ ∆XYZ by ABC ↔ YZX.

∆ABC and ∆ADB have same base AB.

∴ A(∆ABC)/A(∆ADB) = BC/AD    .....[Tringles having equal base]

= 4/8

∴ A(∆ABC)/A(∆ADB) = 1/2

Let the base, height and area of the first triangle be b₁ , h₁ , and A₁ respectively. Let the base, height and area of the second triangle be b₂ , h₂ and A₂ respectively.

A₁/A₂ = (b₁ x h₁)/(b₂ x h₂) ...[Ratio of areas of two tringles is equal to the product of their bases and corresponding ]

A₁/A₂ = (9 x 5)/(10 x 6)

= 45/60

A₁/A₂ = 3/4

∴ The ratio of areas of the tringles is 3:4.

Given : In ∆ABC, point L,M, N respectively lie OA, OB and OC such that LM || AB and MN || BC.

To prove : LN || AC

Proof : In ∆ABO

LM || AB (given)

OL/LA = OM/MB…..(ii) 

(By Basic prop. theorem)

Again, In ∆BCO

MN || BC

OM/MB = ON/NC ….(ii) (By Basc prop. theorem)

From equation (i) and (ii),

⇒ OL/LA = OM/MB = ON/NC

OL/LA = ON/NC (By Converse of B. P. theorem)

⇒ LN || AC