
01. Real Numbers
9
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9


02. Polynomials
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03. Linear Equation
9
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9


04. Quadratic Equation
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


05. Arithmetic Progressions
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06. Some Applications of Trigonometry
7
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11

Lecture7.12

Lecture7.13

Lecture7.14

Lecture7.15

Lecture7.16

Lecture7.17


08. Triangles
15
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15


09. Circles
8
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8


10. Areas Related to Circles
10
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10


11. Introduction to Trigonometry
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12. Surface Areas and Volumes
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
12
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12


14. Probability
9
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Triangles Exercise 8.1 – 8.6
Q.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 are similar, if
(a) their corresponding angles are ___ and
(b)their corresponding sides are ___ (equal , proportional)
Sol.
(i) similar
(ii) similar
(iii) equilateral
(iv) equal, proportional
Q.2 Give two different examples of pair of
(i) similar figures
(ii) nonsimilar figures.
Sol.
(i) Two different examples of pair of similar figures are :
(a) any two rectangles (b) any two squares
(ii) Two different examples of pair of nonsimilar figures are :
(a) a scelene and an equilateral triangle.
(b) an equilateral triangle and a right angle triangle
Q.3 State whether the following quadrilaterals are similar or not :
Sol. On looking at the given figures of the quadrilaterals, we can say that they are not similar.
Q.1 In figure, (i) and (ii), DE  BC. Find EC in (i) and AD in (ii).
Sol.
(i) In fig. (i),
since DE  BC,
ADDB=AEEC⇒153=1EC
⇒ EC=315=3×1015= 2 cm.
(ii) In fig. (ii),
since DE  BC,
ADDB=AEEC⇒AD7.2=1.85.4
⇒ AD=1854×7210= 2.4 cm.
Q.2 E and F are points on the sides PQ and PB respectively of a Δ PQR. For each of the following cases, state whether EF  QR :
(i) PE = 3.9 cm, EQ = 4 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 = 0.36 cm
Sol.
(i) We have,
PE = 3.9 cm, EQ = 4 cm,
PF = 3.6 cm and FR = 2.4 cm
Now, PEEQ=3.94=0.97 cm
and, PFFR=3.62.4+32=1.2 cm
⇒ PEEQ=PFFR
⇒ EF does not divide the sides PQ and PR of ΔPQR in the same ratio.
Therefore, EF is not parallel to QR.
(ii) We have, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, PEEQ=44.5=4045=89
and, PFFR=89
⇒ PEEQ=PFEQ
Thus, EF divides sides PQ and PR of ΔPQR in the same ratio.
Therefore, by the converse of BasicProportionality Theorem we have EF  QR.
(iii) We have, PQ = 1.28 cm, PR = 2.56 cm
PF = 0.18 cm and, PF = 0.36 cm
Therefore EQ = PQ – PE = (1.28 – 0.18) cm.
= 1.10 cm
and, ER = PR – PF = (2.56 – 0.36)
= 2.20 cm
Now, PEEQ=0.181.10=18110=955
and, PFFR=0.362.20=36220=955
⇒ PEEQ=PFFR
Thus, EF divides sides PQ and PR of ΔPQR in the same ratio.
Therefore, by the converse of Basic Proportionality Theorem, we have EF  QR
Q.3 In figure, if LM  CB and LN  CD, prove that AMAB=ANAD.
Sol.
In ΔABC , we have
LM  CB [Given]
Therefore by Proportionality Theorem, we have
AMAB=ALAC
In ΔACD, we have
LN  CD [Given]
Therefore By Basic Proportionality Theorem, we have
ALAC=ANAD ………….. (2)
From (1) and (2), we obtain that
AMAB=ANAD
Q.4 In figure, DE  AC and DF  AE. Prove that BFFE=BEEC
Sol.
In ΔBCA, we have
DE  AC [Given]
Therefore By Basic Proportionality Theorem, we have
BEEC=BDDA ………… (1)
In Δ BEA, we have
DF  AE [Given]
Therefore by Basic Proportionality Theorem, we have
BFFE=BDDA …………. (2)
From (1) and (2), we obtain that
BFFE=BEEC
Q.5 In figure, DE  OQ and DF  OR. Show that EF  QR. Sol.
In ΔPQO, we have
DE  OQ [Given]
Therefore By Basic Proportionality Theorem, we have
PEEQ=PDDO ………… (1)
In ΔPOR, we have
DF  OR [Given]
Therefore By Basic Proportionality Theorem, we have
PDDO=PFFR ………….. (2)
From (1) and (2), we obtain that
PEEQ=PFFR
⇒EFQR [By the converse of BPT]
Q.6 In figure, A, B and C are points on OP, OQ and OR respectively such that AB  PQ and AC  PR. Show that BC  QR.
Sol.
Given : O is any point within ΔPQR, AB  PQ and AC  PR
To prove : BC  QR
Construction : Join BC
Proof : In ΔOPQ, we have
AB  PQ [Given]
Therefore By Basic Proportionality Theorem, we have
OAAP=OBBQ ……… (1)
In Δ OPR, we have
AC  PR [Given]
Therefore by Basic Proportionality Theorem, we have
OAAP=OCCR …………. (2)
From (1) and (2), we obtain that
OBBQ=OCCR
Thus, in ΔOQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by
the converse of Basic Proportionality Theorem, we have,
BC  QR
Q.7 Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Sol.
Given : A ΔABC, in which D is the midpoint of side AB and the line DE is drawn parallel to BC, meeting AC in E. To Prove : AE = EC
Proof : Since DE  BC
Therefore by Basic Proportionality Theorem, we have
ADDB=AEEC ………… (1)
But, AD = DB [Because D is the midpoint of AB]
⇒ ADDB = 1
From (1), AEEC = 1 ⇒ AE = EC
Hence, E is the midpoint of the third side AC.
Q.8 Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is, parallel to the third side. (Recall that you have done it in Class IX).
Sol.
Given : A ΔABC, in which D and E are the midpoints of sides AB and AC respectively.
To prove : DE  BC
Proof : Since, D and E are the midpoints of AB and AC respectively.
Therefore AD = DB and AE = EC
Since, AD = DB
Therefore ADDB = 1 and AE = EC
Therefore AEEC = 1
Therefore ADDB=AEEC=1
i.e. ADDB=AEEC
Thus, in ΔABC, D and E are points dividing the sides AB and AC in the same ratio, Therefore, by the converse
of Basic Proportionality Theorem, we have,
DE  BC.
Q.9 ABCD is a trapezium in which AB  DC and its diagonals intersect each other at the point O. Show that AOBO=CODO
Sol.
Given : A trapezium ABCD, in which AB  DC and its diagonals AC and BD intersect each other at O.
To prove : AOBO=CODO
Construction : Through O, draw OE  AB i.e. OE  DC
Proof : In ΔADC, we have OE  DC [Construction]
Therefore by Basic Proportionality Theorem, we have
AEED=AOCO …………….. (1)
Again, in ΔABD, we have
OE  AB [Construction]
Therefore by Basic Proportionality Theorem, we have
EDAE=DOBO⇒AEED=BODO …………… (2)
From (1) and (2), we obtain that
AOCO=BODO⇒AOBO=CODO
Q.10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO. Show that ABCD is a trapezium.
Sol.
Given : A quadrilateral ABCD in which its diagonals AC and BD intersect each other at O such that
AOBO=CODO i.e. AOCO=BODO
To prove : Quadrilateral ABCD is a trapezium.
Construction : Through O draw OE  AB meeting AD in E.
Proof : In ΔADB, we have
OE  AB [Construction]
Therefore by Basic Proportionality Theorem, we have
DEEA=ODBO
⇒EADE=BODO
⇒EADE=BODO=AOCO[ThereforeAOCO=BODO(given)]
⇒EADE=AOCO
Thus, in ΔADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore, by
the converse of Basic Proportionality Theorem, we have
EO  DC
But, EO  AB [Construction]
Hence, AB  DC
Therefore Quadrilateral ABCD is a trapezium.
Q.1 State which pairs of triangle 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 :Sol.
(i) In Δs ABC and PQR, we observe that
∠A=∠P=60∘, ∠B=∠Q=80∘ and ∠C=∠R=40∘
Therefore by AAA criterion of similarity,
ΔABC∼ΔPQR
(ii) In Δs ABC and PQR, we observe that
ABQR=BCRP=CAPQ=12
Therefore by SSS criterion of similarity, ΔABC∼ΔQRP
(iii) In Δs LMP and DEF, we observe that the ratio of the sides of these triangles are not equal.
So, the two triangles are not similar.
(iv) In Δs MNL and QPR, we observe that
∠M=∠Q=70∘
But, MNPQ≠MLQR
Therefore These two triangles are not similar as they do not satisfy SAS criterion of similarity.
(v) In Δs ABC and FDE, we have
∠A=∠F=80∘
But, ABDF≠ACEF [Because AC is not given]
So, by SAS criterion of similarity, these two triangless are not similar.
(vi) In Δs DEF and PQR
∠D=∠P=70∘ [Because ∠P=180∘−80∘−30∘=70∘]
∠E=∠Q=80∘
So, by AAA criterion of similarity,
ΔDEF∼ΔPQR
Q.2 In figures, ΔODC∼ΔOBA, ∠BOC=125∘ and ∠CDO=70∘. Find ∠DOC, ∠DCO and ∠OAB.
Sol.
Since BD is a line and OC is a ray on it,
Therefore ∠DOC+∠BOC=180∘
⇒ ∠DOC+125∘=180∘
⇒ ∠DOC=180∘−125∘=55∘
In ΔCDO, we have
∠CDO+∠DOC+∠DCO=180∘
⇒ 70∘+55∘+∠DCO=180∘
⇒ ∠DCO=180∘−125∘=55∘
It is given that ΔODC∼ΔOBA
∠OBA=∠ODC,
∠OAB=∠OCD
⇒ ∠OBA=70∘ and ∠OAB=55∘
Hence, ∠DOC=55∘, ∠DCO=55∘
and ∠OAB=55∘
Q.3 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 OAOC=OBOD.
Sol.
Given : ABCD is a trapezium in which AB  DC.
To prove : OAOC=OBOD
Proof : In Δs OAB and OCD, we have
∠AOB=∠COD [Vert. opp. ∠s]
∠OAB=∠OCD [Alternate ∠s]
and ∠OBA=∠ODC [Alternate ∠s]
Therefore by AAA criterion of similarity,
ΔOAB∼ΔODC
Hence, OAOC=OBOD
[Because in case of two similar triangles, the ratio of their corresponding sides are equal]
Q.4 In figure, QRQS=QTPR and ∠1=∠2. Show that ΔPQS∼ΔTQR
Sol.
We have, QRQS=QTPR
⇒ QTQR=PRQS ………….. (1)
Also, ∠1=∠2 [Given]
⇒ PR = PQ ………….. (2) [Because sides opp. to equal ∠s are equal]
From (1) and (2), we get
QTQR=PQQS⇒PQQT=QSQR ……….. (3)
Therefore in Δs PQS and TQR, we have
PQQT=QSQR and ∠PQS=∠TQR=∠Q
Therefore by SAS criterion of similarity,
ΔPQS∼ΔTQR.
Q.5 S and T are points on sides PR and QR of ΔPQR such that ∠P=∠RTS. Show that ΔRPQ∼ΔRTS.
Sol.
In Δs RPQ and RTS, we have
∠RPQ=∠RTS [Given]
∠PRQ=∠TRS [Common]
Therefore by AA criterion of similarity,
ΔRPQ∼ΔRTS
Q.6 In figure, if ΔABE≅ΔACD, show that ΔADE∼ΔABC.
Sol.
It is given that
ΔABE≅ΔACD
Therefore AB = AC [Because corresponding parts of congruent triangles are equal]
and, AE = AD
⇒ ABAD=ACAE
⇒ ABAC=ADAE ………. (1)
Therefore In Δs ADE and ABC, we have
ABAC=ADAE [Because of (1)]
and, ∠BAC=∠DAE [Common]
Thus, by SAS criterion of similarity,
ΔADE∼ΔABC
Q.7 In figure, altitudes AD and CE of ΔABC intersect each other at the point P, Show that :
(i) ΔAEP∼ΔCDP
(ii) ΔABD∼ΔCBE
(iii) ΔAEP∼ΔADB
(iv) ΔPDC∼ΔBEC
Sol.
(i) In Δs AEP and CDP, we have
∠AEP=∠CDP=90∘ [Because CE⊥AB and AD⊥BC]
∠APE=∠CPD [Vert. opp. ∠s]
Therefore by AA – criterion of similarity, we have
ΔAEP∼ΔCDP
(ii) In Δs ABD and CBE, we have
∠ABD=∠CBE [Common angle]
∠ADB=∠CEB=90∘
Therefore by AAcriterion of similarity, we have
ΔABD∼ΔCBE
(iii) In Δs AEP and ADB, we have
∠AEP=∠ADB=90∘ [Becauese AD⊥BC and CE⊥AB]
∠PAE=∠DAB [Common angle]
Therefore by AAcriterion of similarity, we have
ΔAEP∼ΔADB
(iv) In Δs PDC and BEC, we have
∠PDC=∠BEC=90∘ [Because AD⊥BC and CE⊥AB]
∠PCD=∠ECB [Common angle]
Therefore AAcriterion of similarity, we have
ΔPDC∼ΔBEC
Q.8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB.
Sol.
In Δs ABE and CFB, we have
∠AEB=∠CBF [Alt. ∠s]
∠A=∠C [Opp. ∠s of a gm]
Therefore by AAcriterion of similarity, we have
ΔADE∼ΔCFB
Q.9 In figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
(i) ΔABC∼ΔAMP
(ii) CAPA=BCMP
Sol.
(i) In Δs ABC and AMP, we have
∠ABC=∠AMP=90∘ [Given]
and, ∠BAC=∠MAP [Common ∠s]
Therefore by AA, criterion of similarity, we have
ΔABC∼ΔAMP
(ii) We have, ΔABC∼ΔAMP [As proved above]
⇒ CAPA=BCMP
Q.10 CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC∼ΔFEG, show that :
(i) CDGH=ACFG
(ii) ΔDCB∼ΔHGE
(iii) ΔDCA∼ΔHGF
Sol.
We have, ΔABC∼ΔFEG
⇒ ∠A=∠F ………….. (1)
and, ∠C=∠G
⇒ 12∠C=12∠G
⇒ ∠1=∠3 and ∠2=∠4 ………….. (2)
[Because CD and GH are bisectors of ∠C and ∠G respectively.
Therefore In Δs DCA and HGF, we have
∠A=∠F [From (1)]
∠2=∠4 [From (2)]
Therefore by AAcriterion of similarity, we have
ΔDCA∼ΔHGF
which proves the (iii) part.
We have, ΔDCA∼ΔHGF [As proved above]d
⇒ ACFG=CDGH
⇒ CDGH=ACFG
which proves the (i) part.
In Δs DCB and HGE, we have
∠1=∠3 [From (2)]
∠B=∠E [Because ΔABC∼ΔFEG]
Therefore by AAcriterion of similarity, we have
ΔDCB∼ΔHGE
which proves, the (ii) part.
Q.11 In 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 ΔABD∼ΔECF.
Sol.
Here, ΔABC is isosceles with AB = AC
∠B=∠C
In Δs ABD and ECF, we have
∠ABD=∠ECF [Because ∠B=∠C]
∠ADB=∠EFC=90∘ [Because AD⊥BC and EF⊥AC]
Therefore by AAcriterion of similarity,
ΔABD∼ΔECF.
Q.12 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show that ΔABC∼ΔPQR.
Sol.
Given : AD is the median of Δ ABC and PM is the median of Δ PQR such that
ABPQ=BCQR=ADPM
To prove : ΔABC∼ΔPQR
Proof : BD=12BC [Given]
and, QM=12QR [Given]
Also, ABPQ=BCQR=ADPM [Given]
⇒ ABPQ=2BD2QM=ADPM
⇒ ABPQ=BCQR=ADPM
By SSScriterion of similarity, we have
ΔABD∼ΔPQM
⇒ ∠B=∠Q [Similar Δs have corresponding ∠s]
and, ABPQ=BCQR [Given]
Therefore by SAScriterion of similarity, we have
ΔABC∼ΔPQR
Q.13 D is a point on the side BC of a triangle ABC such that ∠ADC=∠BAC. Show that CA2=CB.CD.
Sol.
In Δs ABC and DAC, we have
∠ADC=∠BAC and ∠C=∠C
Therefore by AAcriterion of similarity, we have
ΔABC∼ΔDAC
⇒ ABDA=BCAC=ACDC
⇒ CBCA=CACD
⇒ CA2=CB×CD
Q.14 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 ΔABC∼ΔPQR.
Sol. Same as question 12.
Q.15 A vertical pole of 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.
Sol.
Let AB be the vertical pole and AC be the shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE = x metres.
We have, AB = 6 m, AC = 4 cm and DF = 28 m
In Δs ABC and DEF, we have
∠A=∠D=90∘,
and ∠C=∠F
[Because each is the angular elevation of the sun]
Therefore by AAcriterion of similarity,
ΔABC∼ΔDEF
⇒ABDE=ACDF
⇒ 6x=428
⇒ 6x=17
⇒ x=6×7=42
Hence, the height of the tower is 42 metres.
Q.16 If AD and PM are medians of triangles ABC and PQR respectively, where ΔABC∼ΔPQR, prove that ABPQ=ADPM
Sol.
Given : AD and PM are the medians of Δs ABC and PQR respectively, where
ΔABC∼ΔPQR
To prove : ABPQ=ADPM
Proof : In Δs ABD and PQM, we have
∠B=∠Q [Because ΔABC∼ΔPQR]
ABPQ=12BC12QR [Since AD and PM are the median of BC and QR respectively]
⇒ ABPQ=BDQM
Therefore by SAS – criterion of similarity
ΔABD∼ΔPQM
⇒ ABPQ=BDQM=ADPM
⇒ ABPQ=ADPM
Q.1 Let ΔABC∼ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, Find BC.
Sol.
We have, Area(ΔABC)Area(ΔDEF)=BC2EF2
⇒ 64121=BC2(15.4)2
⇒ 811=BC15.4
⇒ BC=(811×15.4) cm = 11.2 cm
Q.2 Diagonals of a trapezium ABCD with AB  DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Sol.
In Δs AOB and COD, we have
∠AOB=∠COD [Vert. opp. ∠s]
and, ∠OAB=∠OCD [Alernate ∠s]
Therefore by AAcriterion of similarity, we have
ΔAOB∼ΔCOD
⇒ Area(ΔAOB)Area(ΔCOD)=AB2DC2
⇒ Area(ΔAOB)Area(ΔCOD)=(2DC)2DC2=41
Hence, area (ΔAOB) : area (ΔCOD) = 4 : 1
Q.3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar(ΔABC)ar(ΔDBC)=AODO
Sol.
Given : Two Δs ABC and DBC which stand on the same base but on the opposite sides of BC.
To prove : Area(ΔABC)Area(ΔDBC)=AODO
Construction : Draw AE⊥BC and DF⊥BC
Proof : In Δs AOE and DOF, we have
∠AEO=∠DFO=90∘
∠AOE=∠DOF [Vertically opp. ∠s]
Therefore by AAcriterion of similarity, we have
ΔAOE∼ΔDOF
⇒ AEDF=AOOD ……….. (1)
Now, Area(ΔABC)Area(ΔDBC)=12×BC×AE12×BC×DF
⇒ AEDF=AOOD [Using (1)]
Therefore Area(ΔABC)Area(ΔDBC)=AOOD
Q.4 If the areas of two similar triangles are equal, prove that they are congruent.
Sol.
Given : Two Δs ABC and DEF such that
ΔABC∼ΔDEF
and, Area (ΔABC) = Area (ΔDEF)
To prove : ΔABC≅ΔDEF
Proof : ΔABC∼ΔDEF
⇒ ∠A=∠D, ∠B=∠E, ∠C=∠F
and, ABDE=BCEF=ACDF
To establish ΔABC≅ΔDEF, it is sufficient to prove that
AB = DE, BC = EF and AC = DF
Now, Area (ΔABC) = Area (ΔDEF)
⇒ Area(ΔABC)Area(ΔDEF)=1
⇒ AB2DE2=BC2EF2=AC2DF2=1
⇒ ABDE=BCEF=ACDF=1
⇒ AB = DE, BC = EF, AC = DF
Hence, ΔABC≅ΔDEF.
Q.5 D, E and F are respectively the midpoints of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.
Sol.
Since, D and E are the midpoints of the sides AB and BC of ΔABC respectively.
Therefore DE  AC
⇒ DE  FC …………. (1)
Since, D and F are the midpoints of the sides AB and AC of ΔABC respectively.
Therefore DF  BC ⇒ DF  EC ………… (2)
From (1) and (2), we can say that DECF is a parallelogram.
Similarly, ADEF is a parallelogram
Now, in Δs DEF and ABC, we have
∠DEF=∠A [Opp. ∠s of gm ADEF]
and, ∠EDF=∠C [Opp. ∠s of gm DECF]
Therefore by AAcriterion of similarity, we have
ΔDEF∼ΔABC
⇒ Area(ΔDEF)Area(ΔABC)=DE2AC2
=(12AC)2AC2=14 [Because DE ======= ]
Hence, Area (ΔDEF) : Area (ΔABC) = 1 : 4.
Q.6 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Sol.
Given : ΔABC∼ΔPQR, AD and PM are the medians of Δs ABC and PQR respectively.
To prove : Area(ΔABC)Area(ΔPQR)=AD2PM2
Proof : Since ΔABC∼ΔPQR
Area(ΔABC)Area(ΔPQR)=(AB)2(PQ)2 ……………. (1)
But, sideABsidePQ=medianADmedianPM ………….. (2)
From (1) and (2), we have
Area(ΔABC)Area(ΔPQR)=AD2PM2
Q.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 the diagonals.
Sol.
Given : A square ABCD. Equilateral Δs BCE and have been drawn on side and the diagonals AC respectively.
To prove : Area (ΔBCE) 12(AreaΔACF)
Proof : ΔBCE∼ΔACF
[Being equilateral so similar by AAA criterion of similarity]
⇒ Area(ΔBCE)Area(ΔACF)=BC2AC2
⇒ Area(ΔBCE)Area(ΔACF)=BC2(2√BC)2
[Because Diagonal = 2–√ side ⇒ AC = 2–√ BC]
⇒ Area(ΔBCE)Area(ΔACF)=12
Tick the correct answer and justify :
Q.8 ABC and BDE are two equilateral triangles such that D is the midpoint of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 : 1 (b) 1 : 2
(c) 4 : 1 (d) 1 : 4
Sol.
Since ΔABC and ΔBDE are equilateral, they are equiangular and hence
ΔABC∼ΔBDE
⇒ Area(ΔABC)Area(ΔBDE)=BC2BD2
⇒ Area(ΔABC)Area(ΔBDE)=(2BD)2BD2
[Because D is the midpoint of BC therefore BC = 2BD]
⇒ Area(ΔABC)Area(ΔBDE)=41
Therefore (C) is the correct answer.
Q.9 Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
(A) 2 : 3 (B) 4 : 9
(C) 81 : 16 (D) 16 : 81
Sol.
Since the ratio of the areas of two similar triangles equal to the ratio of the squares of any two corresponding sides. Therefore,
ratio of areas = (4)2:(9)2=16:81
Therefore (D) is the correct answer
Q.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) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Sol.
(i) Let a = 7 cm, b = 24 cm and c = 25 cm
Here the larger side is c = 25 cm
We have, a2+b2=72+242=49+576=625=c2
So, the triangle with the given sides is a right triangle. Its hypotenuse = 25 cm.
(ii) Let a = 3 cm, b = 8 cm and c = 6 cm
Here the larger side is b = 8 cm
We have, a2+c2=32+62=9+36=45≠b2
So, the triangle with the given sides is not a right triangle.
(iii) Let a = 50 cm, b = 80 cm and c = 100 cm
Here the larger side is c = 100 cm
We have, a2+b2=502+802=2500+6400 = 8900 ≠ c2
So, the triangle with the given sides is not a right triangle.
(iv) Let a = 13 cm, b = 12 cm and c = 5 cm
Here the larger side is a = 13 cm
We have, b2+c2=122+52=144+25=169=a2
So, the triangle with the given sides is a right triangle. Its hypotenuse = 13 cm.
Q.2 PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that PM2=QM.MR.
Sol.
Given : PQR is a triangle right angled at P and PM⊥QR.
To prove : PM2=QM.MR
Proof : Since PM⊥QR
Therefore ΔPQM∼ΔPRM
⇒ PMQM=MRPM
⇒ PM2=QM.MR
Q.3 In figure, ABD is a triangle right angled at A and AC⊥BD. SHow that
(i) AB2=BC.BD
(ii) AC2=BC.DC
(iii) AD2=BD.CD
Sol.
Given : ABD is a triangle right angled at A and AC⊥BD.
To prove :
(i) AB2=BC.BD
(ii) AC2=BC.DC
(iii) AD2=BD.CD
Proof :
(i) Since AC⊥BD
Therefore ΔABC∼ΔADC and each triangle is similar to ΔABD
Because ΔABC∼ΔABD
Therefore ABBD=BCAB ⇒ AB2=BC.BD
(ii) Since, ΔABC∼ΔADC
Therefore ACBC=DCAC ⇒ AC2=BC.DC
(iii) Since, ΔACD∼ΔABD
ADCD=BDAD ⇒ AD2=BD.CD
Q.4 ABC is an isosceles triangle right angled at C. Prove that AB2=2AC2.
Sol.
Since, ABC is an isosceles right triangle, right angled at C.
Therefore AB2=AC2+BC2
⇒ AB2=AC2+AC2 [Because BC = AC, given]
⇒ AB2=2AC2
Q.5 ABC is an isosceles triangle with AC = BC. If AB2=2AC2, prove that ABC is a right triangle.
Sol.
Since, ABC is an isosceles triangle with AC = BC and AB2=2AC2
⇒ AB2=AC2+AC2
⇒ AB2=AC2+BC2 [Because AC = BC, given]
Therefore ΔABC is right angled at AC.
Q.6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Sol.
Let ABC be an equilateral triangle of side 2a units.
Draw AD⊥BC. Then, D is the midpoint of BC.
⇒ BD=12BC=12×2a=a
Since, ABD is a right triangle, right angled at D.
Therefore AB2=AD2+BD2
⇒ (2a)2=AD2+(a)2
⇒ AD2=4a2−a2=3a2
Therefore each of its altitude = 3–√a.
Q.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.
Sol.
Let the diagonals AC and BD of rhombus ABCD interesect each other at O. Since the diagonals of a rhombus bisect each other at right angles.
Therefore ∠AOB=∠BOC=∠COD = ∠DOA=90∘
and, AO = CO, BO = OD
Since, AOB is a right triangle, right angled at O.
Therefore AB2=OA2+OB2
⇒ AB2=(12AC)2+(12BD)2 [Because OA = OC and OB = OD]
⇒ 4AB2=AC2+BD2 ……………. (1)
Siumilarly, we have
4BC2=AC2+BD2 …………… (2)
4CD2=AC2+BD2 …………… (3)
and 4AD2=AC2+BD2 …………….(4)
Adding all these results, we get
4(AB2+BC2+CD2+AD)2=4(AC2+BD2)
⇒ AB2+BC2+CD2+DA2=AC2+BD2
Q.8 In figure, O is a point in the interior of a triangle ABC, 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
Sol.
Join AO, BO and CO
(i) In right Δs OFA, ODB and OEC, we have
OA2=AF2+OF2
OB2=BD2+OD2
and, OC2=CE2+OE2
Adding all these, we get
OA2+OB2+OC2=AF2+BD2+CE2+OF2+OD2+OE2
⇒ OA2+OB2+OC2−OD2−OE2−OF2
= AF2+BD2+CE2
(ii) In right Δs ODB and ODC, we have
OB2=OD2+BD2
and, OC2=OD2+CD2
⇒ OB2−OC2=BD2−CD2 …………… (1)
Suimilarly, we have
OC2−OA2=CE2−AE2 ……………. (2)
and, OA2−OB2=AF2−BF2 …………….. (3)
Adding equations (1), (2) and (3), we get
(OB2−OC2)+(OC2−OA2)+(OA2−OB2)
= (BD2−CD2)+(CE2−AE2)+(AF2−BF2)
⇒ (BD2+CE2+AF2)−(AE2+CD2+BF2) = 0
⇒ AF2+BD2+CE2
= AE2+BF2+CD2.
Q.9 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Sol.
Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a right triangle, right angled at C.
Therefore AB2=AC2+BC2
⇒ 102=AC2+82
⇒ AC2=100−64
⇒ AC2=36
⇒ AC=6
Hence, the foot of the ladder is at a distance 6 m from the base of the wall.
Q.10 A guy wire attached to a vertical pole of height 18 m is 24 m 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 ?
Sol.
Let AB (= 24 m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.
Because AB2=AC2+BC2
⇒ 242=AC2+182
⇒ AC2=576−324
⇒ AC2=252
⇒ AC=252−−−√=67–√
Hence, the stake may be placed at distance of 67–√ m from the base of the pole.
Q.11 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1½ hours ?
Sol.
Let the first aeroplane starts from O and goes upto A towards north where OA=(1000×32) km = 1500 km.
Let the second aeroplane starts from O at the same time and goes upto B towards west where OB=(1200×32) km = 1800 km.
According to the problem the required distane = BA.
In right angled ΔABC, by Pythagoras theorem, we have,
AB2=OA2+OB2
= (1500)2+(1800)2
= 2250000+3240000
= 5490000=9×61×100×100
⇒ AB=3×10061−−√=30061−−√ km.
Q.12 Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Sol.
Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m.
Draw CE⊥AB and join AC
Therefore CE = DB = 12 m,
AE = AB – BE = AB – CD = (11–6) m = 5 m.
In right angled ΔACE, by Pythagoras theorem, we have
AC2=CE2+AE2
= (12)2+(5)2
= 144+25=169
⇒ AC=169−−−√=13
Hence, the distance between the tops of the two poles is 13 m.
Q.13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2+BD2=AB2+DE2.
Sol.
In right angled Δs ACE and DCB, we have
AE2=AC2+CE2
and, BD2=DC2+BC2
⇒ AE2+BD2=(AC2+CE2)+(DC2+BC2)
⇒ AE2+BD2=(AC2+BC2)+(DC2+CE2)
⇒ AE2+BD2=AB2+DE2
[By Pythagoras theorem, AC2+BC2=AB2 and DC2+CE2=DE2]
Q.14 The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB2=2AC2+BC2.
Sol.
We have, DB = 3CD
Now, BC = DB + CD
⇒ BC = 3CD + CD [Because BD = 3CD]
⇒ BC = 4CD
Therefore CD=14BC and,
DB=3CD=34BC …………. (1)
Since, ΔABD is a right triangle, right angled at D. Therefore, by Pythagoras theorem, we have
AB2=AD2+DB2 …………… (1)
Similarly, from ΔACD, we have
AC2=AD2+CD2 ……….. (2)
(1) – (2) gives,
AB2−AC2=DB2−CD2
⇒ AB2−AC2=(34BC)2−(14BC)2 [Using (1)]
⇒ AB2−AC2=(916−116)BC2
⇒ AB2−AC2=12BC2
⇒ 2AB2−2AC2=BC2
⇒ 2AB2=2AC2+BC2
Which is the required result.
Q.15 In an equilateral triangle ABC, D is a point on side BC such that BD=13BC. Prove that 9AD2=7AB2
Sol.
Let ABC be an equilateral triangle and let D be a point on BC such that
BD=13BC
Draw AE⊥BC. Join AD.
In Δs AEB and AEC, we have
AB = AC [Because ΔABC is equilateral]
∠AEB=∠AEC [Because each = 90°]
and, AE = AE
Therefore by SAScriterion of similarity, we have
ΔAEB∼ΔAEC
⇒ BE = EC
Thus, we have
BD=13BC,DC=23BC
and, BE=EC=12BC …………….. (1)
Since, ∠C=60∘
Therefore ΔADC is an acute triangle.
Therefore AD2=AC2+DC2−2DC×EC
= AC2+(23BC)2−2×23BC×12BC [Using (1)]
= AC2+49BC2−23BC2
= AB2+49AB2−23AB2 [Because AB = BC = AC]
= (9+4−6)AB29=79AB2
⇒ 9AD2=7AB2, which is the required result.
Q.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.
Sol.
Let ABC be an equilateral triangle and let AD⊥BC.
In ΔADB and ΔADC, we have
AB = AC [Given]
∠B=∠C [Each = 60°]
and, ∠ADB=∠ADC [Each = 90°]
By RHS criterion of congruence, we have
ΔADB≅ΔADC
⇒ BD=DC
⇒ BD=DC=12BC
Since ΔADB is a right triangle, right angled at D, by Pythagoras theorem, we have
AB2=AD2+BD2
⇒ AB2=AD2+(12BC)2
⇒ AB2=AD2+14BC2
⇒ AB2=AD2+AB24 [Because BC = AB]
⇒ 34AB2=AD2
⇒ 3AB2=4AD2
Hence, the result.
Q.17 Tick the correct answer and justify : In ΔABC, AB=63–√ cm, AC = 12 cm and BC = 6 cm. The angles A and B are respectively :
(a) 90° and 30°
(b) 90° and 60°
(c) 30° and 90°
(d) 60° and 90°
Sol.
In ΔABC, we have
AB = 63–√ cm,
AC = 12 cm
and BC = 6 cm …………… (1)
Now, AB2+BC2=(63–√)2+(6)2
= 36 × 3 + 36 = 108 + 36
= 144=(AC)2
Thus, ΔABC is a right triangle, right angled at B.
Therefore ∠B=90∘
Let D be the midpoint of AC. We know that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.
AD = BD = CD
⇒ CD = BD = 6 cm [Because CD=12AC]
Also BC = 6 cm
Therefore in ΔBDC, we have
BD = CD = BC
⇒ ΔBDC is equilateral ⇒ ∠ACB=60∘
Therefore ∠A=180∘−(∠B+∠C)
= 180° – (90° + 60°) = 30°
Thus, ∠A=30∘ and ∠B=90∘
Therefore (C) is the correct choice.
(*These exercises are not for examination point of view)
Q.1 In Figure, PS is the bisector of ∠QPR of ΔPQR. Prove that QSSR=PQPR
Sol.
Given PQR is a triangle and PS is the internal bisector of ∠QPR meeting QR at S.
Therefore ∠QPS=∠SPR
To prove : QSSR=PQPR
Construction : Draw RT  SP to cut QP produced at T.
Proof : Since PS  TR and PR cuts them, hence, we have
∠SPR=∠PRT ………… (1) [Alternate ∠s]
and, ∠QPS=∠PTR …………… (2) [Corresponding ∠s]
But, ∠QPS=∠SPR [Given]
Therefore ∠PRT=∠PTR [From (1) and (2)]
⇒ PT = PR ………. (3) [Because sides opp. to equal ∠s are equal]
Now, in ΔQRT, we have
RT  SP [By construction]
Therefore QSSR=PQPT [By Basic Proportionality Theorem]
⇒ QSSR=PQPR [From (3)]
Q.2 In figure, D is a point on hypotenuse AC of ΔABC, BD⊥AC, DM⊥BC and DN⊥AB. Prove that
(i) DM2=DN.MC
(ii) DN2=DM.AN
Sol.
We have, AB⊥BC and DM⊥BC
⇒ AB  DM
Similarly, we have
BC⊥AB and DN⊥AB
⇒ CB  DN
Hence, quadrilateral BMDN is a rectangle.
Therefore BM = ND
(i) In ΔBMD, we have
∠1+∠BMD+∠2=180∘
⇒ ∠1+90∘+∠2=180∘⇒∠1+∠2=90∘
Similarly, in ΔDMC, we have
∠3+∠4=90∘
Since BD⊥AC. Therefore,
∠2+∠3=90∘
Now, ∠1+∠2=90∘ and ∠2+∠3=90∘
⇒ ∠1+∠2=∠2+∠3
⇒ ∠1=∠3
Also, ∠3+∠4=90∘ and ∠2+∠3=90∘
⇒ ∠3+∠4=∠2+∠3⇒∠2=∠4
Thus, in Δs BMD and DMC, we have
∠1=∠3 and ∠2=∠4
Therefore By AAcriterion of similarity, we have
ΔBMD∼ΔDMC
⇒ BMDM=MDMC
⇒ DNDM=DMMC [Because BM = ND]
⇒ DM2=DN×MC
(ii) Proceeding as in (i), we can prove that
ΔBND∼ΔDNA
⇒ BNDN=NDNA
⇒ DMDN=DNAN [Because BN = DM]
⇒ DN2=DM×AN
Q.3 In figure, ABC is a triangle in which ∠ABC>90∘ and AD⊥CB produced. Prove that AC2=AB2+BC2+2BC.BD
Sol.
Given : ABC is a triangle in which ∠ABC>90∘ and AD⊥CB produced.
To Prove : AC2=AB2+BC2+2BC.BD
Proof : Since ΔADB is a right triangle, rightangled at D, therefore, by Pythagoras theorem, we have
AB2=AD2+DB2 ……….. (1)
Again, ΔADC is a right triangle, rightangled at D. Therefore, by Pythagoras theorem, we have
AC2=AD2+DC2
⇒ AC2=AD2+(DB+BC)2
⇒ AC2=AD2+DB2+BC2+2DB.BC
⇒ AC2=(AD2+DB2)+BC2+2BC.BD
⇒ AC2=AB2+BC2+2BC.BD [Using (1)]
which proves the required result.
Q.4 In figures, ABC is a triangle in which ∠ABC<90∘ and AD⊥BC. Prove that AC2=AB2+BC2−2BC.BD
Sol.
Given : ABC is a triangle in which ∠ABC<90∘ and AD⊥BC.
To prove : AC2=AB2+BC2−2BC.BD
Proof : Since ΔADB is a right triangle, rightangled at D, therefore, by Pythagoras theorem, we have,
AB2=AD2+BD2 ………….. (1)
Again, ΔADC is a right triangle, rightangled at D, therefore, by Pythagoras theorem, we have,
AC2=AD2+DC2
⇒ AC2=AD2+(BC−BD)2
⇒ AC2=AD2+(BC2+BD2−2BC.BD)
⇒ AC2=(AD2+BD2)+BC2−2BC.BD
⇒ AC2=AB2+BC2−2BC.BD [Using (1)]
which proves the required result.
Q.5 In figure, AD is a median of a triangle ABC and AM⊥BC. Prove that
(i) AC2=AD2+BC.DM+(BC2)2
(ii) AB2=AD2−BC.DM+(BC2)2
(iii) AC2=AB2=2AD2+12BC2
Sol.
Since ∠AMD=90∘, therefore, ∠ADM<90∘ and ∠ADC>90∘.
Thus, ∠ADC is acute and ∠ADC is obtuse.
(i) In ΔADC, ∠ADC is an obtuse angle.
Therefore AC2=AD2+DC2+2DC.DM
⇒ AC2=AD2+(BC2)2+2.BC2.DM
⇒ AC2=AD2+(BC2)2+BC.DM
⇒ AC2=AD2+BC.DM+(BC2)2 ………….. (1)
(ii) In ΔABD, ∠ADM is an acute angle.
Therefore AB2=AD2+BD2−2BD.DM
⇒ AB2=AD2+(BC2)2−2.BC2.DM
⇒ AB2=AD2−BC.DM+(BC2)2 ………… (2)
(iii) From (1) and (2), we get
AB2+AC2=2AD2+12BC2
Q.6 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Sol.
We know that if AD is a median of ΔABC, then AB2+AC2=2AD2+12BC2
[See above question part (iii)]
Sine the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of Δs
ABC and ADC respectively.
Therefore AB2+BC2=2BO2+12AC2 ……….. (1)
and, AD2+CD2=2DO2+12AC2 ………. (2)
Adding (1) and (2), we get
AB2+BC2+CD2+AD2=2(BO2+DO2)+AC2
⇒ AB2+BC2+CD2+AD2=2(14BD2+14BD2)+AC2 [Because DO=12BD]
⇒ AB2+BC2+CD2+AD2=AC2+BD2
Q.7 In figure, two chords AB and CD intersect each other at the point P. Prove that
(i) ΔAPC∼ΔDPB
(ii) AP.PB = CP.DP
Sol.
(i) In Δs APC and DPB, we have
∠APC=∠DPB [Vert. opp. ∠s]
∠CAP=∠BDP [Angles in the same segment of a circle are equal]
Therefore by AAcriterion of similarity, we have
ΔAPC∼ΔDPB
(ii) Since ΔAPC∼ΔDPB
Therefore APDP=CPPB
⇒ AP×PB=CP×DP
Q.8 In figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle, Prove that
(i) ΔPAC∼ΔPDB
(ii) PA.PB = PC.PD
Sol.
(i) In Δs PAC and PDB, we have
∠APC=∠BPD [Common]
∠PAC=∠PDB [Because ∠BAC=180∘−∠PAC and ∠PDB=∠CDB=180∘−∠BAC=180∘−(180∘−∠PAC)=∠PAC]
Therefore by AAcriterion of similarity, we have
ΔPAC∼ΔDPB
(ii) Since ΔPAC∼ΔDPB
PAPD=PCPB
⇒ PA.PB = PC.PD
Q.9 In figure, D is a point on side BC of ΔABC such that BDCD=ABAC. Prove that AD is the bisector of ∠BAC.
Sol.
Given : ABC is a triangle and D is a point on BC such that
BDCD=ABAC
To prove : AD is the internal bisector of ∠BAC.
Construction : Produce BA to E such that AE = AC. Join CE.
Proof : In ΔAEC, since AE = AC, hence
∠AEC=∠ACE ……….. (1)
[Because Angles opp. to equal sides of a Δ are equal]
Now, BDCD=ABAC [Given]
⇒ BDCD=ABAE [Because AE = AC, construction]
Therefore by converse of Basic Proportionality Theorem, we have
DA  CE
Now, since CA is a transeversal, we have
∠BAD=∠AEC ……… (2) [Corresponding ∠s]
and, ∠DAC=∠ACE …… (3) [Alternate angles]
Also, ∠AEC=∠ACE [From (1)]
Hence, ∠BAD=∠DAC [From (2) and (3)]
Thus, AD bisects, ∠BAC internally.
Q.10 Nazima is fly fishing i a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taur, how much string does she have out (see figure) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ?
Sol.
In fact, we want to find AC.
By Pythagoras theorem, we have
AC2=(2.4)2+(1.8)2
⇒ AC2=5.76+3.24=9.00
⇒ AC = 3 m
Therefore length of string she have out = 3 m.
Length of the string pulled at the rate of 5 cm/sec in 12 seconds.
= (5 × 12) cm
= 60 cm = 0.60 m
Therefore remaining string left out = (3 – 0.6) m = 2.4 m
In 2nd case let us find PB
PB2=PC2−BC2
= (2.4)2−(1.8)2
= 5.76−3.24=2.52
⇒ PB=2.52−−−−√=1.59 (nearly)
Hence, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m
= 2.79 m (nearly)
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