CBSE NCERT Solutions for Class 10 Mathematics Chapter 6 ... · Class- X-CBSE-Mathematics Triangles...
Transcript of CBSE NCERT Solutions for Class 10 Mathematics Chapter 6 ... · Class- X-CBSE-Mathematics Triangles...
Class- X-CBSE-Mathematics Triangles
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CBSE NCERT Solutions for Class 10 Mathematics Chapter 6
Back of Chapter Questions
EXERCISE 6.1
1. Fill in the blanks using the correct word given in 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 re similar, if (a) their
corresponding angles are _________ and (b) their corresponding sides are
_________. (equal, proportional)
Solution:
(i) similar, since size of circles may be different, but shape will be always
same.
(ii) similar, since size of squares may be different, but shape will be always
same.
(iii) All equilateral triangles are similar because of their same shape.
(iv) Two polygons of same number of sides are similar, if (a) their
corresponding angles are equal and (b) their corresponding sides are
proportional.
2. Give two different examples of pair of
(i) Similar figures.
(ii) Non-similar figures.
Solution:
(i) Two equilateral triangles with sides 1cmand 2cm.
Two squares with sides 1cmand 2cm.
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(ii) Trapezium and Square
Triangle and Parallelogram
3. State whether the following quadrilaterals are similar or not:
Solution:
Corresponding sides of two quadrilaterals are proportional i. e. 1: 2 but their
corresponding angles are not equal. Hence, quadrilateral PQRS and ABCD are not
similar.
EXERCISE 6.2
1. In Fig. (i) and (ii), DE ∥ BC. Find EC in (i) and AD in (ii).
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(i)
(ii)
Solution:
(i)
Let EC = 𝑥cm
Since DE ∥ BC
Hence, using basic proportionality theorem,
AD
DB=
AE
EC
⇒1.5
3=
1
𝑥
⇒ 𝑥 =3 × 1
1.5
⇒ 𝑥 = 2
Hence, EC = 2cm
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(ii)
Let AD = 𝑥
Since DE ∥ BC,
Hence, using basic proportionality theorem,
AD
DB=
AE
EC
⇒𝑥
7.2=
1.8
5.4
⇒ 𝑥 =1.8 × 7.2
5.4
⇒ 𝑥 = 2.4
Hence, AD = 2.4cm
2. Eand 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 = 3.9cm, EQ = 3cm, PF = 3.6cmandFR = 2.4cm
(ii) PE = 4cm, QE = 4.5cm, PF = 8cmandRF = 9cm
(ii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cmandPF = 0.36cm
Solution:
(i)
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Given, PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4
Now,
PE
EQ=
3.9
3= 1.3
PF
FR=
3.6
2.4= 1.5
Since, PE
EQ≠
PF
FR
Hence, EF is not parallel to QR.
(ii)
Given, PE = 4, QE = 4.5, PF = 8, RF = 9
PE
EQ=
4
4.5=
8
9
PF
FR=
8
9
Since PE
EQ=
PF
FR
Hence,EF ∥ QR (using basic proportionality theorem)
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(iii)
Given, PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36
∴ 𝐸𝑄 = 𝑃𝑄 − 𝑃𝐸 = 1.28 − 0.18 = 1.1 and 𝐹𝑅 = 𝑃𝑅 − 𝑃𝐹 = 2.56 −0.36 = 2.2
PE
EQ=
0.18
1.1=
18
110=
9
55
PF
FR=
0.36
2.2=
9
55
Since PE
EQ=
PF
FR
Hence, EF ∥ QR (using basic proportionality theorem)
3. In Fig. if LM ∥ CB and LN ∥ CD, prove that
AM
AB=
AN
AD
Solution:
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In the given figure
Since LM ∥ CB.
Hence, using basic proportionality theorem,
AM
MB=
AL
LC … (i)
Again, since LN ∥ CD
Hence, using basic proportionality theorem,
AN
ND=
AL
LC … (ii)
From (i) and (ii)
AM
MB=
AN
ND
⇒ MB
AM=
ND
AN
⇒ MB
AM+ 1 =
ND
AN+
⇒MB + AM
AM=
ND + AN
AN
⇒AB
AM=
AD
AN
⇒AM
AB=
AN
AD
4. In Figure, DE ∥ AC andDF ∥ AE. Prove that
BF
FE=
BE
EC
Solution:
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In ΔABC
Since DE ∥ AC
Hence, BD
DA=
BE
EC … (i) (using basic proportionality theorem)
In ∆BAE,
Since DF ∥ AE
Hence, BD
DA=
BF
FE … (ii) (using basic proportionality theorem)
From (i) and(ii)
BE
EC=
BF
FE
5. In Figure, DE ∥ OQ andDF ∥ OR. Show thatEF ∥ QR.
Solution:
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In ∆POQ
Since DE ∥ OQ
PE
EQ=
PD
DO … (i)[Using basic proportionality theorem]
In ∆POR
Since DF ∥ OR
PF
FR=
PD
DO … (ii)[Using basic proportionality theorem]
From (i)and (ii)
PE
EQ=
PF
FR
Using converse of basic proportionality theorem
EF ∥ QR
6. In Figure A, B andC are points onOP, OQ andOR respectively such thatAB ∥ PQ
andAC ∥ PR. Show that BC ∥ QR.
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Solution:
In ∆POQ
Since AB ∥ PQ,
Hence, OA
AP=
OB
BQ … (i)[Using basic proportionality theorem]
In ∆POR
Since AC ∥ PR
Hence, OA
AP=
OC
CR … (ii)[Using basic proportionality theorem]
From (i)and (ii)
OB
BQ=
OC
CR
Hence, BC ∥ QR
(Using converse of basic proportionality theorem)
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7. Prove that a line drawn through the mid-point of one side of a triangle parallel to
another side bisects the third side. (Recall that you have proved it in ClassIX).
Solution:
Let in the given figure PQ is a line segment drawn through mid-point P of line AB
such that PQ ∥ BC
Hence, AP = PB
Now, using basic proportionality theorem
AQ
QC=
AP
PB
⇒AQ
QC=
AP
AP
⇒AQ
QC= 1
⇒ AQ = QC
Hence, Q is the mid-point of AC.
8. 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).
Solution:
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Let in the given figure PQ is a line segment joining mid-points PandQ of line
ABand AC respectively.
Hence, AP = PBandAQ = QC
Now, since AP
PB=
AP
AP= 1and
AQ
QC=
AQ
AQ= 1
Hence, AP
PB=
AQ
QC
Now, using converse of basic proportionality theorem PQ ∥ BC
9. ABCD is a trapezium in whichAB ∥ DC and its diagonals intersect each other at the
point O. Show that
AO
BO=
CO
DO
Solution:
Let a line segment EF is drawn through point O such that EF ∥ CD
In ∆ADC
EO ∥ CD
Hence, using basic proportionality theorem
AE
ED=
AO
OC … (1)
Similarly in ∆BDC
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FO ∥ CD
Hence, using basic proportionality theorem
BF
FC=
BO
OD … (2)
Now consider trapezium ABCD
As FE ∥ CD
So, AE
ED=
BF
FC … (3)
Now from equation (1), (2), (3)
AO
OC=
BO
OD
⇒AO
BO=
OC
OD
10. The diagonals of a quadrilateralABCD intersect each other at the pointO such that AO
BO=
CO
DO. Show that ABCD is a trapezium.
Solution:
Draw a line segment OE ∥ AB
In ∆ABC
Since OE ∥ AB
Hence,AO
OC=
BE
EC
But by the given relation, we have:
AO
BO=
CO
DO
⇒AO
OC=
OB
OD
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Hence, OB
OD=
BE
EC
So, using converse of basic proportionality theorem, EO ∥ DC
Therefore AB ∥ OE ∥ DC
⇒ AB ∥ CD
Therefore, ABCD is a trapezium.
EXERCISE 6.3
1. State which pairs of triangles in Fig. 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:
(i)
(ii)
(iii)
(iv)
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(v)
(vi)
Solution:
(i) ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Hence by AAA rule ∆ABC~∆PQR
(ii) AB
QR=
2
4=
1
2
BC
RP=
2.5
5=
1
2
CA
PQ=
3
6=
1
2
Since, AB
QR=
BC
RP=
CA
PQ
Hence, by SSS rule
∆ABC~∆QRP
(iii) Triangles are not similar as the corresponding sides are not proportional.
(iv) MN
PQ=
2.5
5=
1
2
ML
QR=
5
10=
1
2
∠M = ∠Q = 70o
Hence, by SAS rule
∆MNL~∆PQR
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(v) Triangles are not similar as the corresponding sides are not proportional.
(vi) In ∆DEF
∠D + ∠E + ∠F = 180o(Sum of angles of a triangle is 180o)
⇒ 70o + 80o + ∠F = 180o
⇒ ∠F = 30o
Similarly in ∆PQR
∠P + ∠Q + ∠R = 180o (Sum of angles of a triangle is 180o)
⇒ ∠P + 80o + 30o = 180o
⇒ ∠P = 70o
Now, since
∠D = ∠P = 70o
∠E = ∠Q = 80o
∠F = ∠R = 30o
Hence, by AAA rule
∆DEF~∆PQR
2. In Figure, ∆ODC~∆OBA, ∠BOC = 125oand ∠CDO = 70o. Find ∠DOC, ∠DCO
and∠OAB.
Solution:
Since DOB is a straight line
Hence, ∠DOC + ∠COB = 180o
⇒ ∠DOC = 180o − 125o = 55o
In∆DOC,
∠DCO + ∠CDO + ∠DOC = 180o
⇒ ∠DCO + 70o + 55o = 180o
⇒ ∠DCO = 55o
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Since, ∆ODC~∆OBA,
Hence, ∠OCD = ∠OAB [Corresponding angles equal in similar triangles]
Hence, ∠OAB = 55o
3. DiagonalsAC andBD of a trapeziumABCD withAB ∥ DC intersect each other at the
pointO. Using a similarity criterion for two triangles, show that OA
OC=
OB
OD
Solution:
In ∆DOCand∆BOA
AB||CD
Hence,∠CDO = ∠ABO[Atlernate interior angles]
∠DCO = ∠BAO[Atlernate interior angles]
∠DOC = ∠BOA [Vertically opposite angles]
Hence,∆DOC~∆BOA [AAArule]
⇒DO
BO=
OC
OA [Corresponding sides are proportional]
⇒OA
OC=
OB
OD
4. In Figure, QR
QS=
QT
PR and∠1 = ∠2. Show that ∆PQS~∆TQR.
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Solution:
In ∆PQR
∠PQR = ∠PRQ
Hence, PQ = PR … (i)
Given
QR
QS=
QT
PR
Using (i)
QR
QS=
QT
PQ… (ii)
Also, ∠RQT = ∠PQS = ∠1
Hence, by SAS rule
∆PQS~∆TQR
5. S andT are points on sidesPR andQR of ∆PQRsuch that∠P = ∠RTS. Show that
∆RPQ~∆RTS.
Solution:
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In ∆RPQand∆RTS
∠QPR = ∠RTS [Given]
∠R = ∠R [Common angle]
∠RQP = ∠RST [Remaining angle]
Hence, ∆RPQ~∆RTS [byAAA rule]
6. In Fig., if∆ABE ≅ ∆ACD, show that∆ADE~∆ABC.
Solution:
Since ∆ABE ≅ ∆ACD
Therefore AB = AC … (1)
AE = AD
⇒ AD = AE … (2)
Now, in ∆ADE and ∆ABC,
Dividing equation (2)by (1)
AD
AB=
AE
AC
∠A = ∠A [Common angle]
Hence,∆ADE~∆ABC [by SAS rule]
7. In Fig., altitudesAD andCEof∆ABC intersect each other at the pointP. Show that:
(i) ∆AEP~∆CDP
(ii) ∆ABD~∆CBE
(iii) ∆AEP~∆ADB
(iii) ∆PDC~∆BEC
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Solution:
(i)
In ∆AEP and ∆CDP
∠CDP = ∠AEP = 90o
∠CPD = ∠APE(Vertically opposite angles)
∠PCD = ∠PAE(Remaining angle)
Hence, by AAArule,
∆AEP~∆CDP
(ii)
In ∆ABDand∆CBE
∠ADB = ∠CEB = 90o
∠ABD = ∠CBE (Common angle)
∠DAB = ∠ECB(Remaining angle)
Hence, by AAArule,
∆ABD~∆CBE
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(iii)
In ∆AEPand∆ADB
∠AEP = ∠ADB = 90o
∠PAE = ∠DAB (Common angle)
∠APE = ∠ABD (Remaining angle)
Hence, by AAA rule,
∆AEP~∆ADB
(iv)
In ∆PDCand∆BEC
∠PDC = ∠BEC = 90o
∠PCD = ∠BCE (Common angle)
∠CPD = ∠CBE (Remaining angle)
Hence, by AAArule,
∆PDC~∆BEC
8. Eis a point on the sideAD produced of a parallelogram ABCD andBE intersectsCD
atF. Show that ∆ABE~∆CFB.
Solution:
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∆ABEand∆CFB
∠A = ∠C (Opposite angles of parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE ∥ BC)
∠ABE = ∠CFB (Alternate interior angles as AB ∥ DC)
Hence, by AAArule,
∆ABE~∆CFB
9. In Figure,ABC andAMP are two right triangles, right angled atB andMrespectively.
Prove that:
(i) ∆ABC~∆AMP
(ii) CA
PA=
BC
MP
Solution:
(i) In ∆ABCand∆AMP
∠ABC = ∠AMP = 90o
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∠A = ∠A(Common angle)
∠ACB = ∠APM(Remaining angle)
Hence, by AAArule,
∆ABC~∆AMP
(ii) Since, ∆ABC~∆AMP
Hence, CA
PA=
BC
MP (Corresponding sides are proportional)
10. CD andGH are respectively the bisectors of∠ACB and∠EGF such thatD andH lie
on sidesAB andFE of∆ABC and∆EFG respectively. If∆ABC~∆FEG, show that:
(i) CD
GH=
AC
FG
(ii) ∆DCB~∆HGE
(iii) ∆DCA~∆HGF
Solution:
Since ∆ABC~∆FEG
Hence, ∠A = ∠F
∠B = ∠E
∠ACB = ∠FGE
⇒∠ACB
2=
∠FGE
2
⇒ ∠ACD = ∠FGH(Angle bisector)
And ∠DCB = ∠HGE (Angle bisector)
In∆ACDand∆FGH
∠A = ∠F
∠ACD = ∠FGH (Angle bisector)
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∠ADC = ∠FHG (Remaining angle)
Hence, by AAA rule,
∆ACD~∆FGH
So, CD
GH=
AC
FG (Corresponding sides are proportional)
In∆DCBand∆HGE
∠B = ∠E
∠DCB = ∠HGE (Angle bisector)
∠BDC = ∠EHG (Remaining angle)
Hence, by AAA rule,
∆DCB~∆HGE
11. In Fig.,E is a point on sideCB produced of an isosceles triangleABC with AB =
AC. IfAD ⊥ BC andEF ⊥ AC, prove that∆ABD~∆ECF.
Solution:
In∆ABD and∆ECF,
Since, AB = AC (isosceles triangles)
So, ∠ABD = ∠ECF
∠ADB = ∠EFC = 90o
∠BAD = ∠CEF (Remaining angle)
Hence, by AAA rule,
∆ABD~∆ECF
12. SidesAB andBC and medianAD of a triangleABC are respectively proportional to
sidesPQ andQR and medianPM of ∆PQR(see Fig.). Show that ∆ABC~∆PQR.
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Solution:
Median divides opposite side.
So, BD =BC
2 andQM =
QR
2
Given that
AB
PQ=
BC
QR=
AD
PM
So, AB
PQ=
BD
QM=
AD
PM
Hence, by SSSrule
∆ABD~∆PQM
Hence, ∠ABD = ∠PQM (Corresponding angles of similar triangles)
Hence, ∠ABC = ∠PQR
And AB
PQ=
BC
QR
Hence, by SASrule
∆ABC~∆PQR
13. Dis a point on the sideBC of a triangleABC such that∠ADC = ∠BAC. Show that
CA2 = CB. CD.
Solution:
In ∆ACDand∆BAC
Given that ∠ADC = ∠BAC
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∠ACD = ∠BCA (Common angle)
∠CAD = ∠CBA (Remaining angle)
Hence, by AAA rule,
∆ADC~∆BAC
So, corresponding sides of similar triangles will be proportional to each other
CA
CB=
CD
CA
Hence CA2 = CB × CD
14. SidesAB andAC and medianAD of a triangleABC are respectively proportional to
sides PQand PR and medianPM of another trianglePQR. Show that∆ABC~∆PQR.
Solution:
Given that
AB
PQ=
AC
PR=
AD
PM
Let us extend ADandPM up to point EandL respectively such that AD = DE and
PM = ML. Now join Bto E, CtoE,QtoLandRto L.
We know that medians divide opposite sides.
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So, BD = DCand QM = MR
Also, AD = DE (by construction)
And PM = ML(By construction)
So, in quadrilateral ABEC, diagonals AEandBC bisects each other at point D.
Also,in quadrilateral PQLR, diagonals PLandQR bisects each other at point M.
So, quadrilaterals ABCD and PQLR are a parallelogram.
AC = BE and AB = EC (Since it is a parallelogram, opposite sides will be
equal)
Also PR = QL and PQ = LR (Since it is a parallelogram, opposite sides will be
equal)
In ∆ABEand∆PQL,
AB
PQ=
BE
QL=
AE
PL (
AC
PR=
BE
QL𝑎𝑛𝑑
AD
PM=
2AD
2PM=
AE
PL)
Hence, by SSS rule,
∆ABE~∆PQL
Similarly, ∆AEC~∆PLR
Hence, ∠BAE = ∠QPL and ∠EAC = ∠LPR
Hence, ∠BAC = ∠QPR
Now, in ∆ABCand∆PQR,
AB
PQ=
AC
PR and ∠BAC = ∠QPR
Hence, by SAS rule,
∆ABC~∆PQR
15. A vertical pole of length6m casts a shadow4m long on the ground and at the
same time a tower casts a shadow28m long. Find the height of the tower.
Solution:
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Let AB be a tower and CD be a pole
Shadow of ABisBE
Shadow of CDisDF
The sun ray will fall on tower and pole at same angle.
So, ∠DCF = ∠BAE and ∠DFC = ∠BEA
∠CDF = ∠ABE = 90o (Tower and pole are vertical to ground)
Hence, by AAA rule,
∆ABE~∆CDF
Therefore AB
CD=
BE
DF
⇒AB
6=
28
4
⇒ AB = 42
Hence, height of tower = 42meters
16. IfAD andPM are medians of trianglesABC andPQR, respectively
where∆ABC~∆PQR, prove that AB
PQ=
AD
PM
Solution:
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Since ∆ABC~∆PQR
So, their respective sides will be in proportion
Or, AB
PQ=
AC
PR=
BC
QR … (1)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)
Since, ADandPM are medians. So, they will divide their opposite sides equally.
Hence, BD =BC
2 and QM =
QR
2… (3)
From equation (1)and(3)
AB
PQ=
BD
QM
∠B = ∠Q (From equation2)
Hence, by SASrule
∆ABD~∆PQM
Hence, AB
PQ=
AD
PM (Corresponding sides are proportional)
EXERCISE 6.4
1. Let ∆ABC~∆DEF and their areas be, respectively,64cm2 and121cm2. IfEF =15.4cm, findBC.
Solution:
Since, ∆ABC~∆DEF
Hence, area(∆ABC)
area(∆DEF)= (
AB
DE)
2
= (BC
EF)
2
= (AC
DF)
2
Since EF = 15.4,area(∆ABC) = 64; area(∆DEF) = 121
Hence, 64
121=
BC2
15.42
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⇒BC
15.4=
8
11
⇒ BC =8×15.4
11= 8 × 1.4 = 11.2cm.
2. Diagonals of a trapeziumABCD withAB ∥ DC intersect each other at the pointO.
IfAB = 2CD, find the ratio of the areas of trianglesAOB andCOD.
Solution:
Since AB ∥ CD
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite angles)
Hence, by AAA rule,
∆AOB~∆COD
Hence, area(∆AOB)
area(∆COD)= (
AB
CD)
2
Since AB = 2CD
area(∆AOB)
area(∆COD)=
4
1= 4: 1
3. In Figure,ABC andDBC are two triangles on the same baseBC. IfAD intersectsBC
atO, show that
ar(ABC)
ar(DBC)=
AO
DO.
Solution:
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We know that area of a triangle =1
2× Base × height
Since ∆ABCand∆DBC are on same base,
Hence, ratio of their areas will be same as ratio of their heights.
Let us draw two perpendiculars APandDM on BC.
In ∆APO and ∆DMO
∠APO = ∠DMO = 90o ∠AOP = ∠DOM (Vertically opposite angles)
∠OAP = ∠ODM (Remaining angle)
Hence, by AAArule
∆APO ~ ∆DMO
Hence, AP
DM=
AO
DO
Hence, area(∆ABC)
area(∆DBC)=
AP
DM=
AO
DO
4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Let us assume that ∆ABC ~ ∆PQR
Now, area(∆ABC)
area(∆PQR)= (
AB
PQ)
2
= (BC
QR)
2
= (AC
PR)
2
Since area(∆ABC) =area(∆PQR)
Hence, AB = PQ
BC = QR
AC = PR
Since, corresponding sides of two similar triangles are of same length
Hence, ∆ABC ≅ ∆PQR (by SSSrule)
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5. D, E andF are respectively the mid-points of sidesAB, BC andCA of∆ABC. Find the
ratio of the areas of∆DEF and∆ABC.
Solution:
Since D and E are mid-points of AB and BC of ∆ABC
Hence, DE ∥ AC andDE =1
2AC (by mid-point theorem)
Similarly, EF =1
2AB and DF =
1
2BC
Now in ∆ABC and ∆EFD
AB
EF=
BC
FD=
CA
DE= 2
Hence, by SSSrule
∆ABC~∆EFD
Hence, area(∆ABC)
area(∆DEF)= (
AC
DE)
2
= 4
⇒area(∆DEF)
area(∆ABC)=
1
4= 1: 4
6. Prove that the ratio of the areas of two similar triangles is equal to the square of
the ratio of their corresponding medians.
Solution:
Class- X-CBSE-Mathematics Triangles
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Let us assume that ∆ABC~∆PQR. Let ADand PSbe the medians of these triangles.
So, AB
PQ=
BC
QR=
AC
PR… (1)
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
Since, AD and PS are medians
So, BD = DC =BC
2 and QS = SR =
QR
2
So, equation (1) becomes
AB
PQ=
BD
QS=
AC
PR
Now in ∆ABDand ∆PQS
∠B = ∠Q and, AB
PQ=
BD
QS
Hence, ∆ABD~∆PQS
Hence, AB
PQ=
BD
QS=
AD
PS… (2)
Since, area(∆ABC)
area(∆PQR)= (
AB
PQ)
2
⇒area(∆ABC)
area(∆PQR)= (
AD
PS)
2
[from equation (2)]
7. Prove that the area of an equilateral triangle described on one side of a square is
equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Let ABCD be a square of side 𝑎. Therefore it's diagonal = √2𝑎
Let ∆ABEand∆DBF are two equilateral triangles.
Hence, AB = AE = BE = 𝑎 and DB = DF = BF = √2𝑎
We know that all angles of equilateral triangles are 60o.
Class- X-CBSE-Mathematics Triangles
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Hence, all equilateral triangles are similar to each other.
Hence, ratio of areas of these triangles will be equal to the square of the ratio
between sides of these triangles.
areaof∆ABE
areaof∆DBF= (
𝑎
√2𝑎)
2
=1
2
Hence, areaof∆ABE =1
2(areaof∆DBF)
8. ABC andBDE are two equilateral triangles such thatD is the mid-point ofBC. Ratio
of the areas of trianglesABC andBDE is
(A) 2: 1
(B) 1: 2
(C) 4: 1
(D) 1: 4
Solution: (C)
Since all angle of equilateral triangles are 60o. Hence, all equilateral triangles are
similar to each other. So ratio between areas of these triangles will be equal to the
square of the ratio between sides of these triangles.
Let side of ∆ABC = 𝑎
Hence, side of ∆BDE =𝑎
2
Hence, area(∆ABC)
area(∆BDE)= (
𝑎𝑎
2
)2
=4
1= 4: 1
9. Sides of two similar triangles are in the ratio4: 9. Areas of these triangles are in
the ratio
(A) 2: 3
(B) 4: 9
(C) 81: 16
Class- X-CBSE-Mathematics Triangles
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(D) 16: 81
Solution: (D)
If, two triangles are similar to each other, ratio between areas of these triangles
will be equal to the square of the ratio between sides of these triangles.
Given that sides are in the ratio4: 9.
Hence, ratio between areas of these triangles = (4
9)
2
=16
81= 16: 81
EXERCISE 6.5
1. Sides of triangles are given below. Determine which of them are right triangles. In
case of a right triangle, write the length of its hypotenuse.
(i) 7cm, 24cm, 25cm
(ii) 3cm, 8cm, 6cm
(iii) 50cm, 80cm, 100cm
(iv) 13cm, 12cm, 5cm
Solution:
(i) Given that sides are 7cm, 24cm and 25cm.
Squaring the lengths of these sides we get 49,576 and 625.
Clearly, 49 + 576 = 625or 72 + 242 = 252.
Since, given triangle satisfies Pythagoras theorem. So, it is a right triangle.
As we know that the longest side in a right triangle is the hypotenuse.
Hence, length of hypotenuse = 25cm.
(ii) Given that sides are 3cm, 8cm and 6cm.
Squaring the lengths of these sides we may get 9,64 and 36.
Clearly, sum of squares of lengths of two sides is not equal to square of
length of third side. Hence, given triangle is not satisfying Pythagoras
theorem. So, it is not a right triangle
(iii) Given that sides are 50cm, 80cmand 100cm.
Squaring the lengths of these sides we may get 2500,6400and 10000.
Clearly, sum of squares of lengths of two sides is not equal to square of
length of third side. Therefore, given triangle is not satisfying Pythagoras
theorem. So, it is not a right triangle.
(iv) Given that sides are 13cm, 12cm and 5cm.
Squaring the lengths of these sides we may get 169,144 and 25.
Class- X-CBSE-Mathematics Triangles
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Clearly, 144 + 25 = 169 or 122 + 52 = 132.
Since, given triangle is satisfying Pythagoras theorem. So, it is a right
triangle. As we know that the longest side in a right triangle is the
hypotenuse.
Hence, length of hypotenuse = 13cm.
2. PQR is a triangle right angled atP andM is apoint onQR such thatPM ⊥ QR. Show
that PM2 = QM ∙ MR.
Solution:
Let ∠MPR = 𝑥
In ∆MPR
∠MRP = 180o − 90o − 𝑥
⇒ ∠MRP = 90o − 𝑥
Similarly in ∆MPQ
∠MPQ = 90o − ∠MPR = 90o − 𝑥
∠MQP = 180o − 90o − (90o − 𝑥)
⇒ ∠MQP = 𝑥
Now in ∆MPQ and ∆MRP, we may observe that
∠MPQ = ∠MRP
∠PMQ = ∠RMP
∠MQP = ∠MPR
Hence, by AAA rule,
∆MPQ~∆MRP
Hence, QM
PM=
MP
MR
⇒ PM2 = QM ∙ MR
3. In Fig., ABD is a triangle right angled atA andAC ⊥ BD. Show that
Class- X-CBSE-Mathematics Triangles
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(i) AB2 = BC ∙ BD
(ii) AC2 = BC ∙ DC
(iii) AD2 = BD ∙ CD
Solution:
(i) In ∆ABCand ∆ABD
∠CBA = ∠DBA (common angles)
∠BCA = ∠BAD = 90°
∠BAC = ∠BDA (remaining angle)
Therefore, ∆ABC~∆ABD (by AAA)
∴ AB
BD=
BC
AB
⇒ AB2 = BC ∙ BD
(ii) Let ∠CAB = 𝑥
In ∆CBA
∠CBA = 180o − 90o − 𝑥
∠CBA = 90o − 𝑥
Similarly in ∆CAD
∠CAD = 90o − ∠CAB = 90o − 𝑥
∠CDA = 180o − 90o − (90o − 𝑥)
∠CDA = 𝑥
Now in ∆CBAand ∆CAD,we may observe that
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA = 90o
Therefore ∆CBA~∆CAD(by AAA rule)
Therefore, AC
DC=
BC
AC
Class- X-CBSE-Mathematics Triangles
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⇒ AC2 = DC × BC
(iii) In∆DCA&∆DAB
∠DCA = ∠DAB = 90°
∠CDA = ∠ADB (Common angle)
∠DAC = ∠DBA(remaining angle)
∆DCA~∆DAB (byAAAproperty)
Therefore, DC
DA=
DA
DB
⇒ AD2 = BD × CD
4. ABC is an isosceles triangle right angled atC. Prove that AB2 = 2AC2.
Solution:
Given that ∆ABC is an isosceles triangle.
Therefore, AC = CB
Applying Pythagoras theorem in ∆ABC (i.e. right angled at point C)
AC2 + CB2 = AB2
2AC2 = AB2 (as AC = CB)
5. ABC is an isosceles triangle withAC = BC. IfAB2 = 2AC2, prove thatABC is a
right triangle.
Solution:
Given that
AB2 = 2AC2
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⇒ AB2 = AC2 + AC2
⇒ AB2 = AC2 + BC2 (as AC = BC)
Since triangle is satisfying the Pythagoras theorem
Therefore, given triangle is a right angled triangle.
6. ABC is an equilateral triangle of side2𝑎. Find each of its altitudes.
Solution:
Let AD be the altitude in given equilateral ∆ABC.
We know that altitude bisects the opposite side.
So, BD = DC = a
in ∆ADB
∠ADB = 90o
Now applying Pythagoras theorem
AD2 + BD2 = AB2
⇒ AD2 + a2 = (2a)2
⇒ AD2 + a2 = 4a2
⇒ AD2 = 3a2
⇒ AD = a√3
Since in an equilateral triangle, all the altitudes are equal in length.
So, length of each altitude will be √3a
7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of
the squares of its diagonals.
Solution:
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In ∆AOB, ∆BOC, ∆COD, ∆AOD
Applying Pythagoras theorem
AB2 = AO2 + OB2
BC2 = BO2 + OC2
CD2 = CO2 + OD2
AD2 = AO2 + OD2
Adding all these equations,
AB2 + BC2 + CD2 + AD2 = 2(AO2 + OB2 + OC2 + OD2)
= 2 ((AC
2)
2
+ (BD
2)
2
+ (AC
2)
2
+ (BD
2)
2
) (diagonals bisect each other.)
= 2 ((AC)2
2+
(BD)2
2)
= (AC)2 + (BD)2
8. In Fig.,O is a point in the interior of a triangleABC, OD ⊥ BC, OE ⊥ AC and OF ⊥AB. Show that
(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
(i) In ∆AOF
Applying Pythagoras theorem
OA2 = OF2 + AF2
Similarly in ∆BOD
OB2 = OD2 + BD2
similarly in ∆COE
OC2 = OE2 + EC2
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Adding these equations
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
⇒ OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + EC2
(ii) As from above result
AF2 + BD2 + EC2 = (OA2 − OE2) + (OC2 − OD2) + (OB2 − OF2)
Therefore, AF2 + BD2 + EC2 = AE2 + CD2 + BF2
9. A ladder10m long reaches a window8m above the ground. Find the distance of
the foot of the ladder from base of the wall.
Solution:
Let OAbe the wall and ABbe the ladder
Therefore by Pythagoras theorem,
AB2 = OA2 + BO2
⇒ 102 = 82 + OB2
⇒ 100 = 64 + OB2
⇒ OB2 = 36
⇒ OB = 6
Therefore, distance of foot of ladder from of the wall = 6m
10. A guy wire attached to a vertical pole of height18m is24m long and has a stake
attached to the other end. How far from the base of the pole should the stake be
driven so that the wire will be taut?
Solution:
Class- X-CBSE-Mathematics Triangles
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Let OB be the pole and AB be the wire.
Therefore, by Pythagoras theorem,
AB2 = OB2 + OA2
⇒ 242 = 182 + OA2
⇒ OA2 = 576 − 324
⇒ 𝑂𝐴 = √252 = √6 × 6 × 7 = 6√7
Therefore distance from base = 6√7m
11. An aeroplane leaves an airport and flies due north at a speed of 1000km per hour.
At the same time, another aeroplane leaves the same airport and flies due west at a
speed of1200 km per hour. How far apart will be the two planes after11
2 hours?
Solution:
Distance traveled by the plane flying towards north in 11
2hrs
= 1,000 × 11
2= 1,500km
Distance traveled by the plane flying towards west in 11
2hrs
= 1,200 × 11
2= 1,800km
Let these distances are represented by OA and OB respectively.
Now applying Pythagoras theorem
Distance between these planes after 11
2hrs, AB = √OA2 + OB2
= √(1,500)2 + (1,800)2 = √2250000 + 3240000
= √5490000 = √9 × 610000 = 300√61
So, distance between these planes will be 300√61km. after 11
2hrs.
Class- X-CBSE-Mathematics Triangles
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12. Two poles of heights6m and11m stand on a plane ground. If the distance between
the feet of the poles is12m, find the distance between their tops.
Solution:
Let CD and AB be the poles of height 11𝑚and 6m.
Therefore, CP = 11 − 6 = 5m
From the figure we may observe that AP = 12m
In ∆APC, by applying Pythagoras theorem
AP2 + PC2 = AC2
⇒ 122 + 52 = AC2
⇒ AC2 = 144 + 25 = 169
⇒ AC = 13 Therefore, distance between their tops = 13m.
13. D andE are points on the sidesCA andCB respectively of a triangleABC right
angled atC. Prove thatAE2 + BD2 = AB2 + DE2.
Solution:
In∆ACE,
AC2 + CE2 = AE2 … (i)
In ∆BCD,
BC2 + CD2 = BD2 … (ii)
Adding (i) and (ii)
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 … (iii)
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⇒ CD2 + CE2 + AC2 + BC2 = AE2 + BD2
In ∆CDE
DE2 = CD2 + CE2
In ∆ABC
AB2 = AC2 + CB2
Putting the values in equation (iii)
DE2 + AB2 = AE2 + BD2
14. The perpendicular fromA on side BCof a∆ABC intersectsBC atD such thatDB =3CD(see Fig.). Prove that 2AB2 = 2AC2 + BC2.
Solution:
Given that 3DC = DB
DC =BC
4[DB: DC = 3: 1] … (1)
and
DB =3BC
4… (2)
In ∆ACD
AC2 = AD2 + DC2
AD2 = AC2 − DC2 … (3)
In ∆ABD
AB2 = AD2 + DB2
AD2 = AB2 − DB2 … (4)
From equation (3) and (4)
AC2 − DC2 = AB2 − DB2
Since, given that 3DC = DB
Class- X-CBSE-Mathematics Triangles
Page - 45
AC2 − (BC
4)
2
= AB2 − (3BC
4)
2
(from(1)and (2))
⇒ AC2 −BC2
16= AB2 −
9BC2
16
⇒ 16AC2 − BC2 = 16AB2 − 9BC2
⇒ 16AB2 − 16AC2 = 8BC2
⇒ 2AB2 = 2AC2 + BC2
15. In an equilateral triangleABC, Dis a point on sideBC such thatBD =1
3BC. Prove
that 9AD2 = 7AB2.
Solution:
Let side of equilateral triangle be a, and AEbe the altitude of∆ABC
So, BE = EC =BC
2=
a
2
And, AE =a√3
2
Given that BD =1
3BC =
a
3
So, DE = BE − BD =a
2−
a
3=
a
6
Now, in ∆ADEby applying Pythagoras theorem
AD2 = AE2 + DE2
⇒ AD2 = (a√3
2)
2
+ (a
6)
2
= (3a2
4) + (
a2
36) =
28a2
36
or, 9AD2 = 7AB2
16. In an equilateral triangle, prove that three times the square of one side is equal to
four times the square of one of its altitudes.
Solution:
Class- X-CBSE-Mathematics Triangles
Page - 46
Let side of equilateral triangle be 𝑎. And AE be the altitude of ∆ABC
So, BE = EC =BC
2=
𝑎
2
And, AE =𝑎√3
2
Now in ∆ABEby applying Pythagoras theorem
AB2 = AE2 + BE2
⇒ 𝑎2 = AE2 + (𝑎
2)
2
⇒ AE2 = 𝑎2 −𝑎2
4
⇒ AE2 =3𝑎2
4
⇒ 4AE2 = 3𝑎2
or, 4AE2 = 3 × square of one side
17. Tick the correct answer and justify: In∆ABC, AB = 6√3cm, AC = 12cm andBC =6cm. The angleB is:
(A) 120o
(B) 60o
(C) 90o
(D) 45o
Solution: (C)
Class- X-CBSE-Mathematics Triangles
Page - 47
Given that AB = 6√3cm, AC = 12cm and BC = 6cm
We may observe that
AB2 = 108AC2 = 144
And BC2 = 36
AB2 + BC2 = AC2
Thus the given ∆ABC is satisfying Pythagoras theorem Therefore triangle is a
right angled triangle right angled at B
Therefore, ∠B = 90o.
EXERCISE 6.6 (Optional)*
1. In Fig.,PS is the bisector of∠QPR of ∆PQR. Prove that
QS
SR=
PQ
PR
Solution:
Let us draw a line segment RT parallel to SP which intersects extended line
segment QP at point T.
Given that PS is angle bisector of ∠QPR
∠QPS = ∠SPR (1)
∠SPR = ∠PRT (As PS || TR, alternate interior angles) (2)
∠QPS = ∠QTR (As PS || TR, corresponding angles) (3)
Class- X-CBSE-Mathematics Triangles
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Using these equations we may find
∠PRT = ∠QTR from (2) and (3)
So, PT = PR (Since ΔPTR is isosceles triangle)
Now in ΔQPS andΔQTR
∠QSP = ∠QRT (As PS || TR)
∠QPS = ∠QTR (As PS || TR)
∠Q is common
ΔQPS~ΔQTR (by AAA property)
So, QR
QS=
QT
QP
⇒ QR
QS− 1 =
QT
QP− 1
⇒𝑆𝑅
𝑄𝑆=
PT
QP
⇒QS
SR=
QP
PT
⇒QS
SR=
PQ
PR
2. In Fig.,D is a point on hypotenuseAC of ∆ABC, such thatBD ⊥ AC, DM ⊥ BC and
DN ⊥ AB. Prove that:
(i) DM2 = DN ∙ MC
(ii) DN2 = DM ∙ AN
Solution:
(i) Let us join DB.
Class- X-CBSE-Mathematics Triangles
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DN||CB
DM||AB
So, DN = MB
DM = NB
Then ∠CDB = ∠ADB = 90°
∠2 + ∠3 = 90o … (1)
In ΔCDM
∠1 + ∠2 + ∠DMC = 180°
∠1 + ∠2 = 90° … (2)
In ΔDMB
∠3 + ∠DMB + ∠4 = 180°
∠3 + ∠4 = 90° … (3)
From equation (1) and (2)
∠1 = ∠3
From equation (1) and (3)
∠2 = ∠4
ΔBDM~ΔDCM
BM
DM=
DM
MC
⇒DN
DM=
DM
MC
⇒ DM2 = DN × MC
(ii) Similarly in ΔDBN
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∠4 + ∠3 = 90o …(4)
In ∆DAN
∠5 + ∠6 = 90o…(5)
In ΔDAB
∠4 + ∠5 = 90o …(6)
From equation (4) and (6)
∠3 = ∠5
From equation (5) and (6)
∠4 = ∠6
ΔDNA~ΔBND
AN
DN=
DN
NB
⇒ DN2 = AN × NB
⇒ DN2 = AN × DM
(as NB = DM)
3. In Fig., ABC is a triangle in which ∠ABC > 90o and AD ⊥ CB produced. Prove
that
AC2 = AB2 + BC2 + 2BC ∙ BD.
Solution:
In ΔADB, applying Pythagoras theorem
AB2 = AD2 + DB2 … (1)
In ΔACD, applying Pythagoras theorem
AC2 = AD2 + DC2
⇒ AC2 = AD2 + (DB + BC)2
⇒ AC2 = AD2 + DB2 + BC2 + 2DB × BC
Now using equation (1)
Class- X-CBSE-Mathematics Triangles
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AC2 = AB2 + BC2 + 2BC. BD
4. In Fig., ABC is a triangle in which ∠ABC < 90o and AD ⊥ BC. Prove that AC2 =AB2 + BC2 − 2BC. BD.
Solution:
In ΔADB, applying Pythagoras theorem
AD2 + DB2 = AB2
⇒ AD2 = AB2 − DB2 … (1)
In ΔADC applying Pythagoras theorem
D2 + DC2 = AC2 (2)
Now using equation (1)
AB2 − BD2 + DC2 = AC2
⇒ AB2 − BD2 + (BC − BD)2 = AC2
⇒ AC2 = AB2 − BD2 + BC2 + BD2 − 2BC. BD
= AB2 + BC2 − 2BC. BD
5. In Fig., ADis a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) AC2 = AD2 + BC ∙ DM + (BC
2)
2
(ii) AB2 = AD2 − BC ∙ DM + (BC
2)
2
(iv) AC2 + AB2 = 2AD2 +1
2BC2
Solution:
(i) In ΔAMD
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AM2 + MD2 = AD2 … (1)
In ΔAMC
AM2 + MC2 = AC2 … (2)
⇒ AM2 + (MD + DC)2 = AC2
⇒ (AM2 + MD2) + DC2 + 2MD. DC = AC2
Using equation (1) we may get
AD2 + DC2 + 2MD. DC = AC2
Now using the result, DC =BC
2
AD2 + (BC
2)
2
+ 2MD. (BC
2) = AC2
⇒ AD2 + (BC
2)
2
+ MD × BC = AC2
(ii) In ΔABM, applying Pythagoras theorem
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD)2
= AD2 − DM2 + BD2 + MD2 − 2BD. MD
= AD2 + BD2 − 2BD. MD
= AD2 + (BC
2)
2
− 2 (BC
2) × MD
= AD2 + (BC
2)
2
− BC × MD
(iii) In ΔAMB
AM2 + MB2 = AB2 … (1)
In ΔAMC
AM2 + MC2 = AC2 … (2)
Adding equation (1) and (2)
2AM2 + MB2 + MC2 = AB2 + AC2
⇒ 2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2
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⇒ 2AM2 + BD2 + DM2 − 2BD. DM + MD2 + DC2 + 2MD. DC= AB2 + AC2
⇒ 2AM2 + 2MD2 + BD2 + DC2 + 2MD(−BD + DC) = AB2 + AC2
⇒ 2(AM2 + MD2) + (BC
2)
2
+ (BC
2)
2
+ 2MD (−BC
2+
BC
2) = AB2 + AC2
⇒ 2AD2 +BC2
2= AB2 + AC2
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the
sum of the squares of its sides.
Solution:
Let ABCD be a parallelogram
Let us draw perpendicular DE on extended side BA and AF on side DC.
In ΔDEA
DE2 + EA2 = DA2 … (i)
In ΔDEB
DE2 + EB2 = DB2
⇒ DE2 + (EA + AB)2 = DB2
⇒ (DE2 + EA2) + AB2 + 2EA. AB = DB2
⇒ DA2 + AB2 + 2EA. AB = DB2 … (ii)
In ΔADF
AD2 = AF2 + FD2
In ΔAFC
Class- X-CBSE-Mathematics Triangles
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AC2 = AF2 + FC2
= AF2 + (DC − FD)2
= AF2 + DC2 + FD2 − 2DC ∙ FD
= (AF2 + FD2) + DC2 − 2DC. FD
⇒ AC2 = AD2 + DC2 − 2DC ∙ FD … (iii)
Since ABCD is a parallelogram
AB = CD (iii)
And BC = AD (iv)
In ΔDEA and ΔADF
∠DEA = ∠AFD
∠EAD = ∠FDA(EA ∥ DF)
∠EDA = ∠FAD(AF ∥ ED)
AD is common in both triangles.
Since respective angles are same and respective sides are same
ΔDEA ≅ ΔAFD
SOEA = DF
Adding equation (ii) and (iii)
⇒ DA2 + AB2 + 2EA. AB + AD2 + DC2 − 2DC. FD = DB2 + AC2
⇒ DA2 + AB2 + AD2 + DC2 + 2EA. AB − 2DC. FD = DB2 + AC2
⇒ BC2 + AB2 + AD2 + DC2 + 2EA. AB − 2AB. EA = DB2 + AC2
⇒ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
7. In Fig., two chordsAB andCD intersect each other at the point P. Prove that:
(i) ∆APC~∆DPB
(ii) AP ∙ PB = CP ∙ DP
Solution:
Class- X-CBSE-Mathematics Triangles
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Let us join CB
(i) In ΔAPC and ΔDPB
∠APC = ∠DPB {Vertically opposite angles}
∠CAP = ∠BDP {Angles in same segment for chord CB}
ΔAPC~ΔDPB {ByAA similarly criterion}
(ii) We know that corresponding sides of similar triangles are proportional
∴AP
DP=
PC
PB=
CA
BD
⇒AP
DP=
PC
PB
∴ AP. PB = PC ∙ DP
8. In Fig., two chordsABandCD of a circle intersect each other at the pointP (when
produced) outside the circle. Prove that
(i) ∆PAC~∆PDB (ii) PA ∙ PB = PC ∙ PD
Solution:
(i) In ΔPAC and ΔPDB
∠P = ∠P(Common)
∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is equal to
opposite interior angle)
Class- X-CBSE-Mathematics Triangles
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∠PCA = ∠PBD (remaining angles)
ΔPAC ∼ ΔPDB (by AAA)
(ii) We know that corresponding sides of similar triangles are proportional.
PA
PD=
AC
DB=
PC
PB
⇒PA
PD=
PC
PB
∴ PA. PB = PC. PD
9. In Fig.,D is a point on sideBC of ∆ABC such that BD
CD=
AB
AC. Prove thatAD is the
bisector of ∠BAC.
Solution:
In ΔDBA and ΔDCA
BD
CD=
AB
AC (Given)
AD = AD(Common)
So, ΔDBA ∼ ΔDCA(By SSS)
Now, corresponding angles of similar triangle will be equal.
∠BAD = ∠CAD
AD is angle bisector of ∠BAC
10. Nazima is fly fishing in a stream. The tip of her fishing rod is1.8m above the
surface of the water and the fly at the end of the string rests on the water3.6m
away and2.4mfrom a point directly under the tip of the rod. Assuming that her
string (from the tip of her rod to the fly) is taut, how much string does she have
out (see Fig.)? If she pulls in the string at the rate of5cm per second, what will be
the horizontal distance of the fly from her after12 seconds?
Class- X-CBSE-Mathematics Triangles
Page - 57
A
Solution:
Let AB be the height of tip of fishing rod from water surface. Let BC Be the
horizontal distance of fly from the tip of fishing rod.
Then, AC is the length of string.
AC Can be found by applying Pythagoras theorem in ΔABC
AC2 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2
AC2 = 3.24 + 5.76
AC2 = 9.00
Thus, length of string out is3m.
Now, she pulls string at rate of 5cm per second.
So, String Pulled in 12Second = 12 × 5 = 60cm = 0.6m
Class- X-CBSE-Mathematics Triangles
Page - 58
Let after 12second, fly be at point D.
Length of string out after12second is AD
AD = AC − String pulled by Nazima in 12second
= 3.00 − 0.6
= 2.4
In ΔADB
AB2 + BD2 = AD2
⇒ (1.8)2 + BD2 = (2.4)2
⇒ BD2 = 5.76 − 3.24 = 2.52
⇒ BD = 1.587
Horizontal distance of fly = BD + 1.2
= 1.587 + 1.2
= 2.787
= 2.79m