NCERT-CLASS-10-TRIANGLE-SOLUTION-EXPLANATION

CHAPTER 6: TRIANGLES SOLUTION AND EXPLANATION

Introduction:

The first thing is to know about the similarity. It is not like the congruency of triangles, which you have read in your 10th class. Here, things are a little different. Suppose that we want to draw a mountain in the notebook; can we draw it of a similar size as we see in reality? The answer is absolutely no! We draw it by taking some ratio on the notebook so it looks similar to the actual mountain.

There are several examples like it. Another example is when we see a builder plans to make an apartment, and he starts with a raw model on the board, which is prepared with some ratio.

Similarly, in triangles, we look for those figures that are in some proportion, not exactly of the same size. 

Exercise 6.1 Questions

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 Answer 1: (i) All circles are similar.

For congruency, their radii must be the same.

(ii) All sides are similar.

As their sides may be proportional but not the same.

(iii) All the equilateral triangles are similar, as their interior angles are the same. (60 degrees)

(a) Their corresponding angles are equal, and 

(b) Their corresponding sides are proportional.

Question 2

(i) Similar figures (two different-sized squares and rectangles)

(ii) A circle and a square are not similar.

Question 3

In the figure, we can see

PS/AD = 1.5/3 = 1/2

PQ/AB = 1.5/3 = 1/2

QR/BC = 1.5/3 = 1/2

SR/DC = 1.5/3 = 1/2

The ratios are equal, but the corresponding angles are not equal. Hence, the figures are not similar.

Exercise 6.2

Basic Proportionality Theorem 

To prove this theorem, we need to understand the required construction.

Area of triangle = 1/2 * (base) * altitude

We can find the area of an acute-angled triangle easily, but in an obtuse-angled triangle, we need a trick.

In this figure, we have drawn two altitudes, AD on BC and CE on AB

So area (1/2)*BC*AD = 1/2 *AB*CE (altitudes are taken for each side separately).

It is an obtuse-angled triangle. To find the area, we draw a perpendicular from the vertex and extend the base. Then use the base and the altitude to find the area as written above.

Proof of Theorem 6.1

Construction: In the given figure, DM is drawn perpendicular to AC, and EN is perpendicular to AB. Join BE and DC. 

Area of triangle ADE, taking AD as base, EN as perpendicular

Area = 1/2 * (AD * EN)

Area of triangle BDE (obtuse-angled triangle)

= 1/2 (BD*EN)

(Area of triangle ADE)/(Area of triangle BDE) = (1/2)*AD*EN/(1/2)*BD*EN = AD/BD ...[1]

Again, the area of triangle ADE is taking AE as the base = (1/2)*AE*DM

Area of an obtuse-angled triangle CDE

= (1/2)*CE*DM

(Area of ADE)/(Area of CDE) = (1/2)*AE*DM/(1/2)*CE*DM = AE/CE...[2]

DE is parallel to the base BC

The area of two triangles between the two parallels and on the same base is equal

Area triangle BDE = Area of triangle DEC

Therefore, from [1] and [2] 

AD/BD = AE/EC

Taking reciprocal and adding 1 

BD/AD + 1 = EC/AE + 1

AB/AD = AC/AE

AD/AB = AE/AC    (additional)

Exercise 6.2 solution

Question 1:

(i) DE is parallel to BC, using B.P.T.

AD/BD = AE/CE

1.5/3 = 1/EC

1.5EC = 3

EC = 3/1.5

= 2 cm

(ii) AD/BD = AE/EC

AD/7.2 = 1.8/5.4

AD/7.2 = 1/3

AD = 7.2/3 = 2.4 cm

Question 2

Given that

(i) PE = 9 cm, EQ = 3 cm, PF = 6 cm, FR = 2.4 cm

PE/EQ = 3.9/3 = 1.3/1

PF/FR = 3.6/2.4 = 3/2

Since these ratios are not equal, EF is not parallel to QR

(ii) PE = 4 cm, QE = 4.5 cm, PF =8cm, RF = 9cm

PE/QE = 4/4.5 = 8/9

PF/RF = 8/9

PE/QE = PF/RF = 8/9

Hence, EF is parallel to QR

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18, PF = 0.36cm

PE/PQ = 0.18/1.28 = 18/128 = 9/64

PF/PR = 0.36/2.56 = 36/256 = 9/64

PE/PQ = PF/PR 

Hence, EF is parallel to QR

Questions 3 & 4

Solution Q3

 Given that LM is parallel to Cb and LN is parallel to CD 

To prove: AM/AB = AN/AD

Proof

 In the triangles ABC

LM  is parallel to BC

Using B.P.T. or Theorem 6.1

We can write 

AM/AB = AL/AC (extended version of B.P.T. as we have done above by adding 1 after reversing both sides)          [1]

and in triangle ACD, LN is parallel to CD, using B.P.T. (extended)

AN/ND = AL/AC         [2]

From [1] and [2], we can see that the RHS are equal, so the LHS will be equal

Hence, AM/AB = AN/AD proved

Solution 4:

Given DE || AC and DF|| AE

To prove: BF/FE = BE/EC

Proof: In the triangle ABC, DE||AC.

using B.P.T.

BD/AD = BE/EC          [1]

and in triangle ABE, DF||AE, using B.P.T., we get

BD/AD = BF/FE                [2]

from [1] and [2]

we get

BF/FC = BE/EC

Solution Q5

Given: DE||OQ, DF||OR

To prove:  EF||QR

 Proof

 In Triangle POQ

DE||OQ, using B.P.T.

PE/EQ = PD/DO       [1]

In triangle POR

AC|| PR, Using B.P.T.

PD/DO = PF/FR  [2]

From [1] and [2]

PE/EQ = PF/FR 

By converse of the Basic Proportionality theorem

EF || QR

Proved

Solution Q6

Given: AB|| PQ, AC ||PR

To prove: BC BC|| QR

Proof:

In triangle OPQ, AB || || PQ, using B.P.T.

OA/AP = OB/BQ   [1]

In triangle OPR

AC|| PR

So, using B.P.T. 

OA/AP = OC/CR  [2]

from [1] and [2]

OB/BQ = OC/CR

By the converse of B.P.T.

we get BC || QR

Solution Q7:

Given: D is the midpoint of AB, and DE||AC.

To Prove: DE bisects AC, or E is the midpoint of AC.

Proof:

Since D is the midpoint of AB, then AD = BD

or AD/BD = 1      [1]

Since DE||BC, using BPT

AD/BD = AE/EC   [2]

from [1] and [2]

AE/EC = AD/BD = 1

AE = EC 

E is the midpoint of AC.

Solution Q8 

Consider the above figure Q7

Given that: D and E are the midpoints of the sides AB and AC

To prove: DE||BC

Proof:

In the triangle ABC

D is the midpoint of AB; 

therefore, AD = BD

 or AD/BD = 1

and E is the midpoint of the AC. 

therefore, 

AE = EC

AE/EC = 1

Hence, 

AD/BD = AE/EC = 1

By the converse of the BPT 

DE || BC

Solution Q9:

Given that: AB||DC

To prove: AO/BO = CO/DO

Proof:

In the trapezium AB||DC 

AC can be considered as transversal. 

angle DCO = angle OAC (alternate interior)

and angle DOC = AOB (vertically opposite)

So triangle DOC ~ triangle BOA (AA criteria similarity)

Hence,

DC/AB = DO/OB = OC/OA

=> DO/OB = OC/OA

or AO/OB = OC/OD (Cross multiplication method)

Solution Q10:

Given: AO/BO = CO/DO

To prove: ABCD is a trapezium

Proof:  (Consider figure 9

 In triangle AOB and Triangle COD

Since, AO/BO = CO/DO

and angle AOB =  angle COD (Vertically opposite)

Hence, Triangle AOB ~ Triangle COD.

=> Angle OAB = Angle OCD

which are alternate angles formed when AB||DC is intersected by a transversal AC

Hence, AB|| DC

So, quadrilateral ABCD is a trapezium.

Exercise 6.3

Solution 1(i) 

In triangle ABC and PQR

Angle A  = Angle P 

Angle B = Angle Q

Angle C = Angle R 

Triangle ABC ~ Triangle PQR (AA)

Solution 1(ii) 

In Triangle ABC and PQR

AB/QR = 2/4 =1/2

BC/RP = 2.5/5 = 1/2

AC/QP = 3/6 = 1/2

hence, AB/QR = BC/RP = AC/QP = 1/2.

triangle ABC ~ triangle QRP (SSS)

Solution (iii)

In triangles LMP and DEF

LP/DF = 3/6 = 1/2

MP/DE = 2/4 = 1/2

LM/EF = 2.7/5 

since 

LP/DF = MP/DE not equal to LM/EF

Hence, triangles are not similar.

Solution (iv)

In triangles LMN and RQP

MN/QP = 2.5/5 = 1/2

ML/QR = 5/10 = 1/2

Angle M = Q = 70

Hence, triangle LMN ~ RQP (SAS)

Solution (v)

Triangle ABC and Triangle DEF are not similar; in triangle ABC, the side AC is not given for comparison with the corresponding side of triangle DEF.

Solution (vi)

 ▲DEF 

∠F = 180 - (70 + 80) = 30 degrees

In ▲PQR

∠P = 180 - (80 + 30) = 70

∠D = ∠P = 70 degrees

∠E = ∠Q = 80 degrees

Hence, ▲DEF ~ ▲PQR (AA)

Solution Q2

In the figure

DOB is a straight line

therefore, ∠DOC = 180 - (125) = 55 degrees [Linear pair]

DC॥AB and DB is transversal

∠OBA = 70 (alternate interior)

∠DOA = 125 degrees [V.O.A.]

∠DOA = ∠D + ∠C [exterior = sum of opposite two interior angles]

∠C = 125-70 = 55 degrees

and again, ∠A = ∠C = 55 degrees (Alternate angles, DC॥AB, AC transversal)

Solution 3

 Given: ⬜ABCD is a trapezium in which AB⏸DC and diagonals AC and BD intersect each other at O

To prove: OA/OC = OB/OD

Proof: AB⏸DC, and AC and BD are transversals.

In ▲ AOB and ▲ COD 

∠OAB = ∠OCD (alternate)

∠AOB = DOC (alternate)

Therefore, ▲ AOB ~ ▲ OCD (AA)

Hence,

OB/OD = OA/OC proved.

Solution 4

Given: QR/QS = QT/PR, Angle 1 = Angle 2

To Prove: ▲PQS ~ ▲TQR

Proof:

In ▲PQR 

∠1 = ∠2

Hence, PQ = PR

We can write, 

QR/QS = QT/PR

QR/QS = QT/PQ

So, in ▲PQS & ▲TQR

∠1 = ∠2

and we have, 

QR/QS = QT/PQ

=>  ▲PQS ~ ▲TQR (SAS)

Solution 5:

Given: S and T are the points on the sides PR and QR of such that ∠P = ∠RTS.

To prove: ▲RPQ ~ ▲RTS

Proof: In ▲RPQ and ▲RTS

∟RPQ = ∟ RTS (Given)

∟PRQ = ∟ SRT (common)

Therefore, ▲RPQ ~ ▲RTS (AA)

Solution 6:

To prove:

Given: ▲ABE ≅ ▲ACD

To prove: ▲ADE~ ▲ACD

Proof: Since

▲ABE ≅ ▲ACD

Therefore, AB = AC

=> ∟ABC = ∟ACD [CPCT]

∟A = ∟A [COMMON]

AE = AD [CPCT]

∟ABE = ∟ACD [CPCT]

In ▲ADE~ ▲ACD

∟DAE = ∟CAD   [common]

AE/AD = 1 [Since AE = AD]

AB/AC = 1  [Since AB = AC]

AE/AD = AB/AC = 1 [from above]

AE/AB = AD/AC

Hence, by the SAS criteria

▲ADE~ ▲ACD

Solution 7:

Given: AD and CE intersect each other at P, ∟D = 90, Angle E = 90.

Proof:  (i) In ▲AEP and ▲CDP

∟AEP = ∟CDP (each 90)

∟APE = ∟CPD (V.O.A)

therefore, 

▲AEP ~▲CDP

(ii) In ▲ABD and ▲CBE

∟ABD = ∟CBE (common)

∟ADB = ∟BEC (90 degrees)

Hence,  ▲ABD ~ ▲CBE

(iii)  ▲AEP and ▲ADB

∟AEP = ∟ADB (each 90 degrees)

∟PAE = ∟DAB (common)

Therefore, ▲AEP ~ ▲ADB (AA criteria)

(iv) In ▲PDC and ▲BEC

∟PDC = ∟BEC (each 90degrees)

∟PCD =∟BCE (common)

Therefore,  ▲PDC ~ ▲BEC (AA criteria) 

Solution 8

Given: 🀱 ABCD is  a parallelogram, AB ⏸CD, AD produced to E, BE intersects DC at F

To prove: 🛆ABE ~ 🛆CFB

Proof:   In 🛆 ABE and 🛆CFB

BC || AD, BE transversal

∟CBF = ∟AEB (AD produced to E, AD will remain || BC) [alternate]

∟EAB = ∟BCF ( opposite angles of parallelogram ABCD)

Hence, 🛆ABE ~ 🛆CFB (AA criteria)

Solution 9: 

Given: 🛆ABC and  🛆AMP are right-angled triangles 

To prove: 🛆ABC ~ 🛆AMP

and CA/PA = BC/MP

Proof:  In 🛆ABC and  🛆AMP

∠ABC = ∠AMP (given)

∠BAC = ∠PAM (each 90)

Therefore, 🛆ABC ~ 🛆AMP (AA criteria)

Hence, 

AC/AP = BC/MP (ratio of corresponding sides )

Solution 10:

Given: CD and GH are the bisectors of the ∠ACB and ∠FGE

and 🛆ABC~ 🛆FEG

To prove: (i) CD/GH = AC/FG

(ii) 🛆DCB ~🛆HGE

(iii) 🛆DCA~🛆HGF

Proof: 

This question's proof can be started with taking triangles for similarity

🛆ABC~ 🛆FEG          [a]

∠ABC = ∠FEG (as triangles are similar)   [1]

ㄥACB = ∠FGE       [2]

Since CD is the bisector of the ∠ACB and GH is the bisector of the ㄥFGE

therefore,

1/2 ㄥACB = 1/2∠FGE              [3]

Now, in 🛆DCB and 🛆HGE

∠DBC = ∠HEG  [from 1]

∠DCB = ∠HEG  [from 2]

hence,  🛆DCB ~ 🛆HGE (AA)

Now in 🛆DCA and 🛆HGF

∠DAC = ∠HFG   [from a]

∠ACD = ∠FGH  [3]

Hence,

🛆DCA~🛆HGF (AA criteria)          [4]

So we can write from the proved parts

DC/GG = AC/FG [4] 

proof completed 

Solution 11: 

Given: AB =AC, E is a point on CB produced, AD  is perpendicular to BC, and EF is perpendicular to AC

To prove: ▲ABD ~ ▲ECF

Proof:  Given that in ▲ABC 

AB = AC

so, ㄥABC = ㄥACB (isoceles) 

In ▲ABD  and  ▲ECF

ㄥABD = ㄥECF     (as ㄥABC =ㄥACB)

ㄥADB = EFC [each 90 degrees]

Hence, ▲ABD ~ ▲ECF [ AA criteria]

Solution 12:

Given: AB/ PQ = BC/QR = AD/PM 

To prove: ▲ABC  ~ ▲PQR

Proof: 

Given that AB/ PQ = AC/PR = AD/PM 

AD and PM are the medians, so D and M are the midpoints 

BC =2BD , QR =2QM

AB/ PQ = BC/QR = AD/PM 

re writing

AB/ PQ = 2BD/2QM = AD/PM 

AB/ PQ = BD/QM = AD/PM 

Hence, .

▲ABD ~ ▲PQM (SSS criteria)

=> ㄥABD = ㄥPQM

Now in ▲ABC and ▲PQR

AB/PQ = BC/QR (Given)

and ㄥB = ㄥQ [ ㄥABD = ㄥPQM]

Hence, ▲ABC  ~ ▲PQR [SAS]

Solution 13: 

Given: In ▲ABC, D is a point on BC such that 

∟BAC =∟ADC

To prove: CA*CA = CB*CD

Proof:

In ▲ABC and ▲ADC

∠BAC = ∠ADC

∠BCA = ∠ACD

∆BCA ~∆CDA (AA criteria)

BC/AC = AB/AD = AC/DC

Considering BC/AC = AC/DC

AC*AC = BC*DC                        proved

Solution 14

Given: AB/PQ = AC/PR = AE/PM

To Prove: ▲ABC ~▲PQR

Construction: Produce AE to D such that AE DE, and join CD

 Produce PM to N such that PM = MN, join NR

Proof: In ▲ABE and ▲DEC

BE = EC   [ AE is median]

AE =ED [ Constructed]

∟BEA =∟DEC (Vertically opposite angles)

Hence, from above

▲ABE ≌ ▲DEC (SAS criteria of Congruency)

=> DC = AB (CPCT)

∟BAE = ∟EDC (CPCT)

Similarly, in ▲PQM and ▲RMN

QM = RM [ PM median]

ㄥPMQ =  ㄥNMR

PM = MN [Constructed]

Therefore, from above 

▲PQM  ≌  ▲RMN [SAS criteria of Congruency]

ㄥQPM = ㄥMNR

PQ = NR

Now, from the given conditions, 

AB/PQ = AE/PM = AC/PR

CD/PQ = 2AE/2PM = AC/PR

CD/PQ = AD/PN = AC/PR

Hence, ▲ADC ~▲PNR

=> ㄥDAC = ㄥNPR

=>ㄥCDA = ㄥRNP

=> ㄥCAD = ㄥRPN               [1]

Similarly, we can prove by constructing another triangle

∟BAE = ∟QPM  [2]

As we have proved above

∟BAE = ∟EDC 

ㄥCAD = ㄥRPN                   [2]

Adding [1] and [2]

∟CAD +∟BAE = ∟RPN +∟QPM

we  get,

∟A = ∟P

So, from given AB/PQ = AC/PR and ∟A =∟P

▲ABC~ ▲PQR (SAS criteria of similarity)

Solution 15: 

The sun can be considered at the same altitude  for both the Ploe and Tower

hence,

∟C = ∟R (Sun's angle )

∟B = ∟Q (each 90 )

Hence, ▲ABC ~▲PQR

AB/BC = PQ/QR

6/4 = 28/QR

QR = 28*4/6 = 56/3 = 18.6m 

Solution16: 

Given: AD and PM are the medians of the triangles ▲ABC and ▲PQR

To prove: AB/PQ = AD/PM

Proof: Since ▲ABC ~ ▲PQR (Given)

AB/PQ = BC/QR = AC/PR

and ∟B = ∟Q ( since both triangles are similar)

so, AB/PQ = 2BD/ 2QM ( BC= 2BD, QR = 2 QM)

AB/PQ = BD/DR  

So, 🛆ABD ~ 🛆PQM (SAS) 

=> AB/PQ = BD/QM = AD/PM

Proved