NCERT Solutions for Class 10 Maths Chapter 6 Triangles

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Written by Team Trustudies
Updated at 2021-02-21


NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.1

Q1 ) Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

(i) Similar
(ii) Similar
(iii)Equilateral
(iv)(a) Equal
(b) Proportional

Q2 ) Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

(i)Two Equilateral Triangle and Two Rectangle




(ii) 1)Triangle and rhombus


2)Rectangle and circle

Q3 ) State whether the following quadrilaterals are similar or not :
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.2

Q1 ) In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

(i) Given, in Triangle ABC, DE II BC
ADDB=AEEC
[Using Basic proportionality theorem]
?1.53=1EC
?EC=31.5
?EC = 3×1015 = 2 cm
Hence, EC = 2 cm.


(ii) Given, in Triangle ABC, DE is parallel to BC
= ADDB=AEEC
[Using Basic proportionality theorem]
?AD7.2=1.85.4
?AD=1.8×7.25.4
=(1810)×(7210)×(1054)
=2410

AD = 2.4 cm

Hence, AD = 2.4 cm.

Q2 ) E and F are points on the sides PQ and PR respectively of a ?PQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, in ?PQR, E and F are two points on side PQ and PR respectively.

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,
PEEQ=3.93=3930=1310=1.3
And
PFFR=3.62.4=3624=32=1.5
? PEEQ?PFFR

Hence, EF is not parallel to QR.


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

Therefore, by using Basic proportionality theorem, we get,

PEQE=44.5=4045=89
And,
PFRF=89

So, we get here,

PEQE=PFRF

Hence, EF is parallel to QR.


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

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And,
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
? PEEQ=0.181.10=18110=955………….(i)
And,
PEFR=0.362.20=36220=955…………(ii)

So, we get here,
PEEQ=PFFR

Hence, EF is parallel to QR.

Q3 ) In the figure, if LM || CB and LN || CD, prove that AMAB=ANAD.
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

AMAB=ALAC…………………….. (i)

Similarly, given, LN || CD and using basic proportionality theorem,

ANAD=ALAC……………………………(ii)

From equation (i) and (ii), we get,
AMAB=ANAD

Hence, proved.

Q4 ) In the figure, DE||AC and DF || AE. Prove that BFFE=BEEC.
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

In ?ABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

BDDA=BEEC ………………………………………………(i)

In ?ABC, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

BDDA=BFFE ………………………………………………(ii)

From equation (i) and (ii), we get
BEEC=BFFE

Hence, proved.

Q5 ) In the figure, DE||OQ and DF||OR, show that EF||QR.
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given,
In ?PQO, DE || OQ

So by using Basic Proportionality Theorem,
PDDO=PEEQ ………………. (i)

Again given, in ?PQO, DE || OQ ,

So by using Basic Proportionality Theorem,
PDDO=PFFR ………………… (ii)

From equation (i) and (ii), we get,
PEEQ=PFFR

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ?PQR.

Q6 ) In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given here,
In ?OPQ, AB || PQ

By using Basic Proportionality Theorem,
OAAP=OBBQ…………….(i)

Also given,
In ?OPR, AC || PR

By using Basic Proportionality Theorem
OAAP=OCCR……………(ii)

From equation (i) and (ii), we get,
OBBQ=OCCR

Therefore, by converse of Basic Proportionality Theorem,

In ?OQR, BC || QR.

Q7 ) Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.(Recall that you have proved it in Class IX).



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, in ?ABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.
AD=DB
=>ADDB = 1 …………………………. (i)

In ?ABC, DE || BC,

By using Basic Proportionality Theorem,

Therefore, ADDB=AEEC

From equation (i), we can write,
=> 1 = AEEC
=> AE = EC

Hence, proved, E is the midpoint of AC.

Q8 ) Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, in ?ABC, D and E are the mid points of AB and AC respectively, such that,

AD=BD and AE=EC.

We have to prove that: DE || BC.

Since, D is the midpoint of AB
AD=DB
=>ADBD = 1……………………………….. (i)

Also given, E is the mid-point of AC.
AE=EC
=> AEEC = 1

From equation (i) and (ii), we get,
ADBD=AEEC

By converse of Basic Proportionality Theorem,
DE || BC

Hence, proved.

Q9 ) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO=CODO.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AOBO=CODO

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In ?ADC, we have OE || DC

Therefore, By using Basic Proportionality Theorem
AEED=AOCO ……………..(i)

Now, In ?ABD, OE || AB

Therefore, By using Basic Proportionality Theorem
DEEA=DOBO…………….(ii)

From equation (i) and (ii), we get,
AOCO=BODO
=>AOBO=CODO

Hence, proved.

Q10 ) The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO. Show that ABCD is a trapezium.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
AOBO=CODO.

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In ?DAB, EO || AB

Therefore, By using Basic Proportionality Theorem
DEEA=DOOB ……………………(i)

Also, given,
AOBO=CODO
=> AOCO=BODO
=> COAO=DOBO
=> DOOB=COAO …………………………..(ii)

From equation (i) and (ii), we get
DEEA=COAO

Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
=> AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.3

Q1 ) State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Figure



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

i) Given, in ? ABC and ? PQR,

?A = ?P = 60°
?B = ?Q = 80°
?C = ?R = 40°

Therefore by AAA similarity criterion,

? ABC is similar to ? PQR


(ii) Given, in ? ABC and ? PQR,
ABQR=BCRP=CAPQ
By SSS similarity criterion,

? ABC is similar to ? QRP


(iii) Given, in ? LMP and ? DEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MPDE=24=12
PLDF=36=12
LMEF=2.75=2750
Here , MPDE=PLDF ? LMEF

Therefore, ? LMP and ? DEF are not similar.


(iv) In ? MNL and ? QPR, it is given,
MNQP=MLQR=12
?M = ?Q = 70°
Therefore, by SAS similarity criterion

? MNL is similar to ? QPR


(v) In ? ABC and ? DEF, given that,
AB = 2.5, BC = 3, ?A = 80°, EF = 6, DF = 5, ?F = 80°
Here , ABDF=2.55=12
And, BCEF=36=12

?B ? ?F

Hence, ? ABC and ? DEF are not similar.


(vi) In ? DEF, by sum of angles of triangles, we know that,
?D + ?E + ?F = 180°
70° + 80° + ?F = 180°
?F = 180° – 70° – 80°
?F = 30°

Similarly, In ? PQR,
?P + ?Q + ?R = 180 (Sum of angles of ? )
?P + 80° + 30° = 180°
?P = 180° – 80° -30°
?P = 70°

Now, comparing both the triangles, ? DEF and ? PQR, we have
?D = ?P = 70°
?E = ?Q = 80°
?F = ?R = 30°

Therefore, by AAA similarity criterion,

Hence, ? DEF is similar to ? PQR

Q2 ) In the figure, ? ODC ? ? OBA, ? BOC = 125° and ? CDO = 70°. Find ? DOC, ? DCO and ? OAB.
Figure



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

As we can see from the figure, DOB is a straight line.

Therefore, ? DOC + ? COB = 180°
? DOC = 180° – 125° (Given, ? BOC = 125°)
= 55°

In ? DOC, Sum of the measures of the angles of a triangle is 180º

Therefore, ? DCO + ? CDO + ? DOC = 180°
? DCO + 70º + 55º = 180°(Given, ? CDO = 70°)
? DCO = 55°

It is given that, ? ODC ? OBA,
Therefore, ? ODC ~ OBA.

Hence, Corresponding angles are equal in similar triangles

? OAB = ? OCD
? OAB = 55°

Q3 ) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

In ?DOC and ?BOA,

AB || CD, thus alternate interior angles will be equal,
?CDO = ?ABO

Similarly,

?DCO = ?BAO

Also, for the two triangles ?DOC and ?BOA, vertically opposite angles will be equal;
?DOC = ?BOA
Hence, by AAA similarity criterion,
?DOC ~ ?BOA

Thus, the corresponding sides are proportional.
DO/BO = OC/OA
OA/OC = OB/OD

Hence, proved.

4. In the fig.6.36, QR/QS = QT/PR and ? 1 = ? 2. Show that ? PQS ~ ? TQR.
fig. 6.36



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

In ? PQR,
? PQR = ? PRQ
PQ = PR ………………………(i)
Given,
QR/QS = QT/PR
Using equation (i), we get
QR/QS = QT/QP……………….(ii)
In ? PQS and ? TQR, by equation (ii),
QR/QS = QT/QP
? Q = ? Q
? PQS ~ ? TQR [By SAS similarity criterion]

Q5 ) S and T are point on sides PR and QR of ? PQR such that ? P = ? RTS. Show that ? RPQ ~ ? RTS.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, S and T are point on sides PR and QR of ? PQR

And ? P = ? RTS.

In ? RPQ and ? RTS,
? RTS = ? QPS (Given)
? R = ? R (Common angle)
? RPQ ~ ? RTS (AA similarity criterion)

Q6 ) In the figure, if ? ABE ? ? ACD, show that ? ADE ~ ? ABC.
figure



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ? ABE ? ? ACD.

AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)

In ? ADE and ? ABC, dividing eq.(ii) by eq(i),
AD/AB = AE/AC
? A = ? A [Common angle]

? ADE ~ ? ABC [SAS similarity criterion]

Q7 ) In the figure, altitudes AD and CE of ? ABC intersect each other at the point P. Show that:
(i) ? AEP ~ ? CDP
(ii) ? ABD ~ ? CBE
(iii) ? AEP ~ ? ADB
(iv) ? PDC ~ ? BEC.
Quadrilaterals



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, altitudes AD and CE of ? ABC intersect each other at the point P.
(i) In ? AEP and ? CDP,
? AEP = ? CDP (90° each)
? APE = ? CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
? AEP ~ ? CDP

(ii) In ? ABD and ? CBE,
? ADB = ? CEB ( 90° each)
? ABD = ? CBE (Common Angles)
Hence, by AA similarity criterion,
? ABD ~ ? CBE

(iii) In ? AEP and ? ADB,
? AEP = ? ADB (90° each)
? PAE = ? DAB (Common Angles)
Hence, by AA similarity criterion,
? AEP ~ ? ADB

(iv) In ? PDC and ? BEC,
? PDC = ? BEC (90° each)
? PCD = ? BCE (Common angles)
Hence, by AA similarity criterion,
? PDC ~ ? BEC
Hence, by AA similarity criterion,
? PDC ~ ? BEC

Q8 ) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ? ABE ~ ? CFB.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

fig. 6.36

In ? ABE and ? CFB,

? A = ? C
(Opposite angles of a parallelogram)
? AEB = ? CBF
(Alternate interior angles as AE || BC)

? ABE ~ ? CFB (AA similarity criterion)

Q9 ) In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
figure
(i) ? ABC ~ ? AMP
(ii) CA/PA = BC/MP



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ABC and AMP are two right triangles, right angled at B and M respectively.


(i) In ? ABC and ? AMP, we have,
? CAB = ? MAP (common angles)
? ABC = ? AMP = 90° (each 90°)
? ABC ~ ? AMP (AA similarity criterion)


(ii) As, ? ABC ~ ? AMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP

Q10 ) CD and GH are respectively the bisectors of ? ACB and ? EGF such that D and H lie on sides AB and FE of ? ABC and ? EFG respectively. If ? ABC ~ ? FEG, Show that: (i) CD/GH = AC/FG (ii) ? DCB ~ ? HGE (iii) ? DCA ~ ? HGF



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, CD and GH are respectively the bisectors of ? ACB and ? EGF such that D and H lie on sides AB and FE of ? ABC and ? EFG respectively.


(i) From the given condition,

? ABC ~ ? FEG.
? A = ? F, ? B = ? E, and ? ACB = ? FGE

Since, ? ACB = ? FGE
? ACD = ? FGH (Angle bisector)
And, ? DCB = ? HGE (Angle bisector)

In ? ACD and ? FGH,
? A = ? F
? ACD = ? FGH
? ACD ~ ? FGH (AA similarity criterion)
CD/GH = AC/FG


(ii) In ? DCB and ? HGE,
? DCB = ? HGE (Already proved)
? B = ? E (Already proved)
? DCB ~ ? HGE (AA similarity criterion)


(iii) In ? DCA and ? HGF,
? ACD = ? FGH (Already proved)
? A = ? F (Already proved)
? DCA ~ ? HGF (AA similarity criterion)

Q11 ) In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ? BC and EF ? AC, prove that ? ABD ~ ? ECF.
fig. 6.36



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ABC is an isosceles triangle.

AB = AC
? ABD = ? ECF

In ? ABD and ? ECF,
? ADB = ? EFC (Each 90°)
? BAD = ? CEF (Already proved)
? ABD ~ ? ECF (using AA similarity criterion)

Q12 ) Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ? PQR (see Fig 6.41). Show that ? ABC ~ ? PQR.
fig. 6.36



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ? ABC and ? PQR, AB, BC and median AD of ? ABC are proportional to sides PQ, QR and median PM of ? PQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ? ABC ~ ? PQR

As we know here,
AB/PQ = BC/QR = AD/PM
ABPQ=12BC12QR=ADPM
AB/PQ = BC/QR = AD/PM
(D is the midpoint of BC. M is the midpoint of QR)
? ABD ~ ? PQM [SSS similarity criterion]

? ABD = ? PQM
[Corresponding angles of two similar triangles are equal]
? ABC = ? PQR

In ? ABC and ? PQR
AB/PQ = BC/QR ………………………….(i)
? ABC = ? PQR ……………………………(ii)

From equation (i) and (ii), we get,
? ABC ~ ? PQR [SAS similarity criterion]

Q13 ) D is a point on the side BC of a triangle ABC such that ? ADC = ? BAC. Show that CA2=CB.CD



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, D is a point on the side BC of a triangle ABC such that ? ADC = ? BAC.

fig. 6.36

In ? ADC and ? BAC,
? ADC = ? BAC (Already given)
? ACD = ? BCA (Common angles)
? ADC ~ ? BAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
CA/CB = CD/CA
CA2=CB.CD
Hence, proved.

Q14 ) Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ? ABC ~ ? PQR.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given: Two triangles ? ABC and ? PQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM
We have to prove, ? ABC ~ ? PQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

figure

In ? ABD and ? CDE, we have
AD = DE [By Construction.]
BD = DC [Since, AP is the median]
and, ? ADB = ? CDE
[Vertically opposite angles]

? ABD ? ? CDE [SAS criterion of congruence]
AB = CE [By CPCT] …………………………..(i)

Also, in ? PQM and ? MNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ? PMQ = ? NMR
[Vertically opposite angles]
? PQM = ? MNR [SAS criterion of congruence]
PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM

From equation (i) and (ii),
CE/RN = AC/PR = AD/PM
CE/RN = AC/PR = 2AD/2PM
CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
? ACE ~ ? PRN [SSS similarity criterion]

Therefore, ? 2 = ? 4
Similarly, ? 1 = ? 3
? 1 + ? 2 = ? 3 + ? 4
? A = ? P …………………………………………….(iii)

Now, In ? ABC and ? PQR, we have
AB/PQ = AC/PR (Already given)

From equation (iii),
? A = ? P
? ABC ~ ? PQR [ SAS similarity criterion]

Q15 ) A vertical pole of a length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m

Graph 1

In ? ABC and ? DEF,
? C = ? E (angular elevation of sum)
? B = ? F = 90°
? ABC ~ ? DEF (AA similarity criterion)

AB/DF = BC/EF
(If two triangles are similar corresponding sides are proportional)
6/h = 4/28
h =( 6×28)/4
h = 6 × 7
h = 42 m

Hence, the height of the tower is 42 m.

If AD and PM are medians of triangles ABC and PQR, respectively where ? ABC ~ ? PQR prove that AB/PQ = AD/PM.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ? ABC ~ ? PQR

figure

We know that the corresponding sides of similar triangles are in proportion.

AB/PQ = AC/PR = BC/QR……………………………(i)
Also, ? A = ? P, ? B = ? Q, ? C = ? R ………….…..(ii)

Since AD and PM are medians, they will divide their opposite sides.
BD = BC2 and QM = QR2 ……………..………….(iii)

From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)

In ? ABD and ? PQM,

From equation (ii), we have
? B = ? Q

From equation (iv), we have,
AB/PQ = BD/QM
? ABD ~ ? PQM (SAS similarity criterion)
AB/PQ = BD/QM = AD/PM

NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.4

Q1 ) Let ? ABC ~ ? DEF and their areas be, respectively, 64 cm2and 121 cm2. If EF = 15.4 cm, find BC.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ? ABC ~ ? DEF,
Area of ? ABC = 64 cm2
Area of ? DEF = 121 cm2
EF = 15.4 cm

Areaof?ABCAreaof?DEF=AB2DE2

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
? AC2DF2=BC2EF2
? 64121=BC2EF2
? (811)2=(BC15.4)2
? 811=BC15.4
? BC = 8×15.411
? BC = 8 × 1.4
? BC = 11.2 cm

Q2 ) Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

figure

In ? AOB and ? COD, we have
? 1 = ? 2 (Alternate angles)
? 3 = ? 4 (Alternate angles)
? 5 = ? 6 (Vertically opposite angle)
? AOB ~ ? COD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

Areaof(?AOB)Areaof(?COD)=AB2CD2
= (2CD)2CD2 [ AB = CD]
Areaof(?AOB)Areaof(?COD)
= 4CD2CD=41

Hence, the required ratio of the area of ? AOB and ? COD = 4:1

Q3 ) In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (? ABC)/area (? DBC) = AO/DO.
fig. 6.36



NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Answer :

Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.

We have to prove: Area (? ABC)/Area (? DBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

figure

We know that area of a triangle = 12 × Base × Height
ar(?ABC)ar(?DEF=12BC(AP)12BC(DM)=APDM

In ? APO and ? DMO,
? APO = ? DMO (Each 90°)
? AOP = ? DOM
(Vertically opposite angles)
? APO ~ ? DMO (AA similarity criterion)
APDM=AODO
Area(?ABC)Area(?DBC)=AODO .

Q4 ) If the areas of two similar triangles are equal, prove that they are congruent.



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