NCERT Solutions for Maths Class 10 Chapter 6

6.1

Question 1:

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)

(i) Similar

(ii) Similar

(iii) Equilateral

(iv) (a) Equal  (b) Proportional

Question 2: Give two different examples of pair of

(i) Similar figures

(ii) Non-similar figures
Answer:

(i) Two equilateral triangles with sides \(1\) cm and \(2\) cm

Two squares with sides 1 cm and 2 cm

(ii) Trapezium and square 

Triangle and parallelogram

Question 3:  State whether the following quadrilaterals are similar or not:

Answer: Quadrilateral \(PQRS\) and \(ABCD\) are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

Exercise 6.2

Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

(ii) 

Answer: 

(i) 

Let \(EC=x\) cm

It is given that \(DE||BC\)

By using basic proportionality theorem, we obtain

\(\frac{AD}{BD}=\frac{AE}{EC}\)

\(\frac{1.5}{3}=\frac{1}{x}\)

\(x=\frac{3\times1}{1.5}\)

\(x=2\)

\(∴EC=2\) cm

(ii) 

Let \(AD=x\) cm

It is given that \(DE||BC\)

By using basic proportionality theorem, we obtain

\(\frac{AD}{BD}=\frac{AE}{EC}\)

\(\frac{x}{7.2}=\frac{1.8}{5.4}\)

\(x=\frac{1.8\times7.2}{5.4}\)

\(x=2.4\)

\(∴AD=2.4\) cm

Question 2: \(E\) and \(F\) are points on the sides \(PQ\) and \(PR\) respectively of a \(ΔPQR\). 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 \(PF = 0.63 cm\)

Answer:

(i)

Given that, \(PE=3.9 cm\), \(EQ=3 cm\), \(PF=3.6 cm\), \(FR=2.4 cm\)

\(\frac{PE}{EQ}=\frac{3.9}{3}=1.3\)

\(\frac{PF}{FR}=\frac{3.6}{2.4}=1.5\)

Hence, \(\frac{PE}{EQ}\ne\frac{PF}{FR}\)

Therefore, \(EF\) is not to \(QR\).

(ii) 

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

\(\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\)

\(=\frac{PF}{FR}=\frac{8}{9}\)

Hence, \(\frac{PE}{EQ}=\frac{PF}{FR}\)

Therefore, \(EF\) is parallel to \(QR\).

(iii)

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

\(\frac{PE}{PQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}\)

\(\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\)

Hence, \(\frac{PE}{PQ}=\frac{PF}{PR}\)

Therefore, \(EF\) is parallel to \(QR\).

Question 3: In the following figure, if \(LM || CB\) and \(LN || CD\), prove that

Answer: 

In the given figure, \(LM || CB\)

By using basic proportionality theorem, we obtain

\(\frac{AM}{AB}=\frac{AL}{AC}\)    ..(i)

Similarly, \(LN||CD\)

\(∴\frac{AN}{AD}=\frac{AL}{AC}\)    ..(ii)

From (i) and (ii), we obtain

\(\frac{AM}{AB}=\frac{AN}{AD}\)

Question 4: In the following figure, \(DE || AC\) and \(DF || AE\). Prove that \(\frac{BF}{FE}=\frac{BE}{EC}\)

Answer: 

In ΔABC, DE || AC

\(∴\frac{BD}{DA}=\frac{BE}{EC}\) (Basic Proportionality Theorem) ..(i) 

In \(∆BAE,\space DF∥AE\)

\(∴\frac{BD}{DA}=\frac{BF}{FE}\)  (Basic Proportionality Theorem) …(ii)

From (i) and (ii), we obtain

\(\frac{BE}{EC}=\frac{BF}{FE}\)

Question 5: In the following figure, \(DE || OQ\) and \(DF || OR\), show that \(EF || QR\). 

Answer: 

In   \(POQ,\space DE || OQ\)

\(∴\frac{PE}{EQ}=\frac{PD}{DO}\) (Basic proportionality theorem). …(i)

In \(∆POR,\space DF∥OR\)

\(∴\frac{PF}{FR}=\frac{PD}{DO}\) (Basic proportional theorem) …(ii)

From (i) and (ii), we obtain

\(\frac{PE}{EQ}=\frac{PF}{FR}\)

\(∴EF∥QR\) (Converse of basic proportional theorem)

Question 6: In the following figure, \(A,\space B\) and \(C\) are points on \(OP,\space OQ\) and \(OR\) respectively such that \(AB||PQ\) and \(AC||PR\). Show that \(BC||QR\).

Answer: 

In  \(POQ,\space AB || PQ\)

\(∴\frac{OA}{AP}=\frac{OB}{BQ}\)    (Basic proportionality theorem) …(i)

In \(∆POR,\space AC∥PR\)

\(∴\frac{OA}{AP}=\frac{OC}{CR}\)  (By basic proportionality theorem)..(ii)

From (ii) and (ii), we obtain

\(\frac{OB}{BQ}=\frac{OC}{CR}\)

\(∴BC∥WR\)    (By the converse of basic proportionality theorem)

Question 7: 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).

Answer:

Consider the given figure in which \(PQ\) is a line segment drawn through the mid-point \(P\) of line \(AB\), such that  \(PQ∥BC\).

By using basic proportionality theorem, we obtain

\(\frac{AQ}{QC}=\frac{AP}{PB}\)

\(\frac{AQ}{QC}=\frac{1}{1}\)     (\(P\) is the mid-point of \(AB. ∴AP=PB\)

\(⇒AQ=QC\)

Or, \(Q\) is the mid-point of \(AC\).

Question 8: Using Converse of basic proportionality theorem, 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).

Answer: 

Consider the given figure in which \(PQ\) is a line segment joining the mid-points \(P\) and \(Q\) of line \(AB\) and \(AC\) respectively.

i.e., \(AP = PB\) and \(AQ = QC\)

It can be observed that

\(\frac{AP}{PB}=\frac{1}{1}\)

And \(\frac{AQ}{QC}=\frac{1}{1}\)

\(∴\frac{AP}{PB}=\frac{AQ}{QC}\)

Hence, by using basic proportionality theorem, we obtain PQ∥BC

Question 9: \(ABCD\) is a trapezium in which \(AB || DC\) and its diagonals intersect each other at the point \(O\). Show that  \(\frac{AO}{BO}=\frac{CO}{DO}\)

Draw a line \(EF\) through point \(O\), such that \(EF∥CD\)

In \(ΔADC,\space EO∥CD\)

By using basic proportionality theorem, we obtain

\(\frac{AE}{ED}=\frac{AO}{OC}\)    ….(i)

In \(ΔABD,\space OE∥ AB\)

So, by using basic proportionality theorem, we obtain

\(\frac{ED}{AE}=\frac{OD}{BO}\)

\(⇒\frac{AE}{ED}=\frac{BO}{OD}\) …(ii)

From equations (i) and (ii), we obtain

\(\frac{AO}{OC}=\frac{BO}{OD}\)

\(⇒\frac{AO}{BO}=\frac{OC}{OC}\)

Question 10: The diagonals of a quadrilateral \(ABCD\) intersect each other at the point \(O\) such that

\(\frac{AO}{BO}=\frac{CO}{DO}\)   Show that \(ABCD\) is a trapezium.

Answer:

Let us consider the following figure for the given question.

Draw a line \(OE || AB\)

In \(ΔABD,\space OE || AB\)

By using basic proportionality theorem, we obtain

\(\frac{AE}{ED}=\frac{BO}{OD}\)  …(i)

However, it is given that

\(\frac{AO}{OC}=\frac{OB}{OD}\)  …(ii)

From equations (i) and (ii), we obtain

\(\frac{AE}{ED}=\frac{AP}{OC}\)

\(⇒ EO || DC\)   [By the converse of basic proportionality theorem]

\(⇒ AB || OE || DC\)

\(⇒ AB || CD\)

∴ ABCD is a trapezium.

Exercise 6.3

Question 1: State which pairs of triangles in the following 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:

(i) 

(ii) 

(iii)

(iv)

(v)

(vi)

Answer: 

(i) \(∠A = ∠P = 60^0\)
\(∠B = ∠Q = 80^0\)
\(∠C = ∠R = 40^0\)

The  \(\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}  PQR\)     [By AAA similarity criterion]

(ii) \(∴∆ABC-∆QRP\) [By SSS similarity criterion]

(iii) The given triangles are not similar as the corresponding sides are not proportional.

(iv) The given triangles are not similar as the corresponding sides are not proportional.

(v) The given triangles are not similar as the corresponding sides are not

(vi) In ΔDEF,

\(∠D +∠E +∠F = 180^0\)

(Sum of the measures of the angles of a triangle is (\(180^0\)). \(70^0 + 80^0 +∠F = 180^0 ∠F = 30^0\)

Similarly, in ΔPQR,

\(∠P +∠Q +∠R = 180^0\)

(Sum of the measures of the angles of a triangle is (\(180^0\)).  \(∠P + 80^0 +30^0 = 180^0∠P = 70^0\)

In ΔDEF and ΔPQR,

\(∠D = ∠P\) (Each\( 70^0\))
\(∠E = ∠Q\) (Each \(80^0\))
\(∠F = ∠R\) (Each \(30^0\))

∴ ΔDEF ∼ ΔPQR [By AAA similarity criterion]

Question 2: In the following figure, \(ΔODC∼ΔOBA, ∠BOC=125^0\) and \(∠CDO=70^0\).

Find \(∠DOC, ∠DCO\) and \(∠OAB\).

Answer:

\(DOB\) is a straight line.

\(∴ ∠DOC + ∠COB = 180^0\)
\(⇒ ∠DOC = 180° – 125^0\)
\(= 55^0\)

In \(ΔDOC, ∠DCO + ∠CDO + ∠DOC = 180^0\)

(Sum of the measures of the angles of a triangle is \(180^0\)).

\(⇒ ∠DCO + 70^0+ 55^0 = 180^0\)

\(⇒ ∠DCO = 55^0\)

It is given that \(ΔODC ∼ ΔOBA\)

\(∴ ∠OAB = ∠ OCD\) [Corresponding angles are equal in similar triangles.]

\(⇒ ∠OAB = 55^0\)

Question 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  \(\frac{AO}{OC}=\frac{OB}{OD}\)

Answer: 

In \(ΔDOC\) and \(ΔBOA\)

\(∠CDO =  ∠ABO\) [Alternate interior angles as AB  || CD]

\(∠DCO =  ∠BAO\) [Alternate interior angles as AB  || CD]

\(∠DOC = ∠BOA\) [Vertically opposite angles]

\(∴ ΔDOC ∼ ΔBOA\) [AAA similarity criterion]

\(∴\frac{DO}{BO}=\frac{OC}{OA}\) [Corresponding sides are proportional]

\(⇒\frac{OA}{OC}=\frac{OB}{OD}\)

Question 4: In the following figure,  \(\frac{QR}{QS}=\frac{QT}{PR}\)  and \(∠1=∠2\) Show that \(∆PQS~∆TQR\)

Answer:

In ΔPQR, ∠PQR = ∠PRQ

\(∴ PQ = PR\)  ..(i)

Given,

\(\frac{QR}{QS}=\frac{QT}{PR}\)

Using (i), we obtain

\(\frac{QR}{QS}=\frac{QT}{QP}\)       …(ii)

In ∆PQS and ∆TQR,

\(frac{QR}{QS}=\frac{QT}{QP}\)  [Using (ii)]

\(∠Q=∠Q\)

\(∴∆PQS~∆TQR\) [SAS similarity criterion

Question 5: S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.

Answer:

In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ∼ ΔRTS (By AA similarity criterion)

Question 6:

In the following figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.

Answer: 

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1)
And, AD = AE [By CPCT]  ..(i)
In ΔADE and ΔABC,

\(\frac{AD}{AB}=\frac{AE}{AC}\)  [Dividing equation (ii) and (i)

∠A = ∠A [Common angle]

∴ ΔADE ∼ ΔABC [By SAS similarity criterion]

Question 7: In the following figure, altitudes AD and CE of ΔABC intersect each other at the point  P. Show that: 

(i) \(ΔAEP ∼ ΔCDP\)
(ii) \(ΔABD ∼ ΔCBE\)
(iii) \(ΔAEP ∼ ΔADB\)

(v) \(ΔPDC ∼ ΔBEC\)

Answer: 

(i)

In ΔAEP and ΔCDP,

\(∠AEP = ∠CDP\) (Each \(90^0\))

\(∠APE = ∠CPD\) (Vertically opposite angles)

Hence, by using AA similarity criterion,

\(ΔAEP ∼ ΔCDP \)

(ii) 

In ΔABD and ΔCBE,

\(∠ADB = ∠CEB\) (Each \(90^0\))
\(∠ABD = ∠CBE\) (Common)

Hence, by using \(AA\) similarity criterion,

\(ΔABD ∼ ΔCBE\)

(iii) 

In ΔAEP and ΔADB,

\(∠AEP = ∠ADB\) (Each \(90^0\))
\(∠PAE = ∠DAB\) (Common)

Hence, by using \(AA\) similarity criterion,

\(ΔAEP ∼ ΔADB\)

(iv) 

In ΔPDC and ΔBEC,

\(∠PDC = ∠BEC\) (Each \(90^0\))

\(∠PCD = ∠BCE\) (Common angle)

Hence, by using \(AA\) similarity criterion,

ΔPDC ∼ ΔBEC

Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB

Answer:

In ΔABE and ΔCFB,

\(∠A = ∠C\) (Opposite angles of a parallelogram)

\(∠AEB = ∠CBF\) (Alternate interior angles as \(AE || BC\))

\(∴ ΔABE ∼ ΔCFB\) (By AA similarity criterion)

Question 9: In the following figure, \(ABC\) and \(AMP\) are two right triangles, right angled at \(B\) and \(M\) respectively, prove that:

(i) \(ΔABC ∼ ΔAMP\)

(ii) \(\frac{CA}{PA}=\frac{BC}{MP}\)

Answer: 

In ΔABC and ΔAMP,

\(∠ABC = ∠AMP\) (Each \(90^0\))
\(∠A = ∠A\) (Common)

\(∴ ΔABC ∼ ΔAMP\) (By AA similarity criterion)

\(⇒\frac{CA}{PA}=\frac{BC}{MP}\)  (Corresponding sides similar triangles are proportional)

Question 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) \(\frac{CD}{GH}=\frac{AC}{FG}\)

(ii) \(ΔDCB ∼ ΔHGE\)

(iii) \(ΔDCA ∼ ΔHGF\)

Answer: 

It is given that \(ΔABC ∼ ΔFEG\)

\(∴ ∠A = ∠F, ∠B = ∠E\) and \(∠ACB = ∠FGE ∠ACB = ∠FGE \)

\(∴ ∠ACD = ∠FGH\) (Angle bisector)

And, \(∠DCB = ∠HGE\) (Angle bisector)

In ΔACD and ΔFGH,

\(∠A=∠F\)            (Proved above)

\(∠ACD=∠FGH \) (Proved above)

\(∴∆ACD~∆FGH\)  (By AA similarity criterion)

\(⇒\frac{CD}{GH}=\frac{AC}{FG}\)

In ΔDCB and ΔHGE,

\(∠DCB  = ∠HGE \) (Proved above)

\(∠B  = ∠E\)  (Proved above)

∴   ΔDCB ∼ ΔHGE (By AA similarity criterion) In ΔDCA and ΔHGF,

∠ACD  = ∠FGH  (Proved above)

∠A  = ∠F  (Proved above)

∴ ΔDCA  ∼ ΔHGF  (By AA similarity criterion)

Question 11: 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 ΔABD ∼ ΔECF

Answer:

Iṭ is given that \(ABC\) is an isosceles triangle.

\(∴ AB  = AC\)\(⇒∠ABD  = ∠ECF\)

In ΔABD and ΔECF,

\(∠ADB = ∠EFC\) (Each \(90^0\))

\(∠BAD  = ∠CEF \) (Proved above)

\(∴ ΔABD  ∼ ΔECF\)  (By using AA similarity criterion)

Question 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 the given figure). Show that ΔABC ∼ ΔPQR

Answer: 

Median divides the opposite side.

\(∴BD=\frac{BC}{2}\) and \(QM=\frac{QR}{2}\)

Given that,

\(\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}\)

\(⇒\frac{AB}{PQ}=\frac{\frac{1}{2}  QR}{\frac{1}{2} QR}=\frac{AD}{PM}\)

\(⇒\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}\)

In \(\Delta ABD\) and \(\Delta PQM\)

\(\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}\)

\(∴Δ ABC∼\Delta PQM\) (by SSS similarity criterion)

\(⇒\angle ABD∼\angle PQM\) (Corresponding angles of similar triangles)

In \(ΔABC\) and \(ΔPQR\)

\(\angle ABC\) and \(\angle PQM\) (Proved above)

\(∴Δ ABC∼ΔPQR\) (By SAS similarity criterion)

\(\frac{AB}{PQ}=\frac{BC}{QR}\)

Question 13: D is a point on the side BC of a triangle ABC such that  ∠ADC  =  ∠BAC. Show that

Answer:

In ΔADC and ΔBAC,

∠ADC  = ∠BAC  (Given)

∠ACD  = ∠BCA  (Common angle)

∴ ΔADC  ∼ ΔBAC  (By AA similarity criterion)

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

\(∴\frac{CA}{CB}=\frac{CD}{CA}\)

\(⇒CA^2=CB×CD\)

Question 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

Answer:

Given that,

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

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML.

Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC  = BE and AB  = EC  (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ=LR

It was given that

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

\(⇒\frac{AB}{PQ}=\frac{BE}{QL}=\frac{1AD}{2PM}\_

\(⇒\frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{Pl}\)

\(∴∆ABE~∆PQL\) (By SS similarity criterion)

\(∴ ΔABE ∼ ΔPQL\) (By SSS similarity criterion)

\(∴ ∠BAE = ∠QPL \)  … (1)

Similarly, it can be proved that \(ΔAEC ∼ ΔPLR\) and \(∠CAE = ∠RPL\) … (2)

Adding equation (1) and (2), we obtain

\(∠BAE  + ∠CAE  = ∠QPL  + ∠RPL\)

\(⇒∠CAB = ∠RPQ \)… (3)

In ΔABC and

ΔPQR,

\(\frac{AB}{PQ}=\frac{AC}{PR}\)  (Given)

\(∠CAB  = ∠RPQ\)  [Using equation  (3)]

\(∴ ΔABC  ∼ ΔPQR\)  (By SAS similarity criterion)

Question 15: A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer:

Let \(AB\) and \(CD\) be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, \(∠DCF  =  ∠BAE\)

And, \(∠DFC  =  ∠BEA\)

\(∠CDF = ∠ABE \) (Tower and pole are vertical to the ground)

\(∴ ΔABE  ∼ ΔCDF \) (AAA similarity criterion)

\(⇒\frac{AB}{CD}=\frac{BE}{DF}\)
\{(⇒\frac{AB}{6 m}=\frac{28}{4}\)

\(⇒AB=42 m\)

Therefore, the height of the tower will be 42 metres.

Question 16:

(i) If \(AD\) and \(PM\) are medians of triangles \(ABC\) and \(PQR\),, respectively where ∆ABC~∆PQR prove that

Answer: 

It is given that \(ΔABC ∼ ΔPQR\)

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

\(∴\frac{AB}{PQ} =\frac{AC}{PR} =\frac{BC}{QR}\)    … (1)

Also, \(∠A  =  ∠P,  ∠B  =  ∠Q,  ∠C  =  ∠R\)  …  (2)

Since AD and PM are medians, they will divide their opposite sides.

\(∴ BD=\frac{BC}{2}\)  and \(QM=\frac{QR}{2}\)… (3)

From equations (1) and (3), we obtain

\(\frac{AB}{PQ}=\frac{BD}{QM}\)  ..(4)

In ΔABD and ΔPQM,

\(∠B  = ∠Q\)        [Using equation  (2)]

\(\frac{AB}{PQ}=\frac{BD}{QM}\)  [Using equation (4)]

∴ ΔABD  ∼ ΔPQM  (By SAS similarity criterion)

\(⇒\frac{AB}{PO}=\frac{BD}{OM}=\frac{AD}{PM}\)

Exercise 6.4

Question 1: Let ∆ABC and DEF  their areas be, respectively, 64 cm²  and 121 cm².

If EF = 15.4 cm, find BC.
Answer:

In is given that ∆ABC~∆DEF

\(∴ \frac{ar(ΔABC)}{ar(ΔABC)}=(\frac{AB}{DE})^2=(\frac{BC}{EF})^2=(\frac{AC}{DF})^2\)

Given that,

\(EF=15.4\) cm

\(ar(ΔABC)=64\) cm²

\(ar(ΔDEF)=121\) cm²

\(∴ \frac{ar(ABC)}{ar(DEF)}=(\frac{BC}{EF})^2\)

\(⇒(\frac{64cm^2}{121 cm^2})=\frac{BC^2}{(15.4cm)^2}\)

\(⇒\frac{BC}{15.4}=(\frac{8}{11})cm\)

\(⇒BC=(\frac{8\times15.4}{11})cm\)

\(⇒(8\times1.4)cm\)

\(⇒11.2 cm\)

Question 2: 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.

Answer:

Since \(AB || CD\)

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB  = ∠COD  (Vertically opposite angles)

∠OAB  = ∠OCD  (Alternate interior angles)

∠OBA  = ∠ODC  (Alternate interior angles)

∴ ΔAOB  ∼ ΔCOD  (By AAA similarity criterion)

\(∴ \frac{ar(\Delta AOB)}{ar(\Delta COD)}=(\frac{B}{CD})^2\)

Since \(AB=2CD\)

\(∴ \frac{ar(\Delta AOB)}{ar(\Delta COD)}=(\frac{2CD}{CD})^2\)

\(=\frac{4}{1}\)

\(=4:1\)

Question 3: In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac{area (∆ABC)}{area(∆DBC)}=\frac{AO}{DO}\)

Answer:

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

We know that area of a triangle \(=\frac{1}{2}\times Base\times Height\)
\(∴\frac{ar(∆ABC}{ar(∆DBC)}=\frac{\frac{1}{2}BC\times AP}{\frac{1}{2}BC\times DM}=\frac{AP}{DM}\)
In \(\Delta APO\) and \(\Delta DMO\),
\(\angle APB=\angle DMO\) (Each \(=90^0\)
\(\angle AOP=\angle DOM\) (Vertically opposite angles)
\(∴\angle APB∼\angle DMO\) (By AA similarity criterion)
\(∴\frac{AP}{DM}=\frac{AO}{DO}\)
\(∴\frac{ar(\Delta ABC)}{ar(\Delta DBC)}=\frac{AO}{DO}\)

Question 4: If the areas of two similar triangles are equal, prove that they are congruent.
Answer: Let us assume two similar triangles as \(\Delta ABC∼\Delta PQR\).
\(\frac{ar(\Delta ABC)}{ar(\Delta PQR}=(\frac{AB}{PQ})^2=(\frac{BC}{QR})^2=(\frac{AC}{PR})^2\) …(ii)
Given this value in equation (i), we obtain
\(1=(\frac{AB}{PQ})^2=(\frac{BC}{QR})^2=(\frac{AC}{PR})^2\)
\(⇒AB=PQ,\space BC=QR\) and \(AC=PR\)
\(∴∆ABC≅∆PQR\) (By SSS congruence criterion)

Question 5: D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Answer:

D and E are the mid-points of \(\Delta ABC\)
\(∴DE\parallel AC\) and \(DE=\frac{1}{2}AC\)
In \(\Delta BED\) and \(\Delta BCA\)
\(\angle BED=\angle BCA\) (Corresponding angles)
\(\angle BDE=\angle BAC\) (Corresponding angles)
\(\angle EBD=\angle CBA\) (Common angles)
\(∴\Delta BED∼\Delta BCA\) (AAA similarity criterion)
\(\frac{ar(\angle BED)}{ar(\angle BCA)}=(\frac{DE}{AC})^2\)
\(⇒\frac{ar(\angle BED)}{ar(\angle BCA)}=\frac{1}{4}\)
\(⇒ar(\angle BED)=\frac{1}{4}ar(\angle BCA)\)
Similarly, \(ar(∆CFE=\frac{1}{4} ar(CBA)\) and

\(ar(∆ADF)=\frac{1}{4} ar (∆ABC)\)
Also,

\(ar(∆DEF)=ar(∆ABC)-[ar(∆BED)\)

\(+ar(∆CFE)+ar(∆ADF)]\)
\(⇒ar(∆DEF)=ar(∆ABC)-\frac{3}{4} ar(∆ABC)\)

\(=\frac{1}{4} ar(∆ABC)\)
\(⇒\frac{ar(∆DEF)}{ar(∆ABC) }=\frac{1}{4}\)

Question 6: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. 

Answer:

Let us assume two similar triangles as

\(\Delta BAC∼\Delta PQR\)
\(∴\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\) …(i)
\(\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R\) … (ii)
Since AD and PS are medians,
\(BD=DC=\frac{BC}{2}\)
And, \(QS=SR=\frac{QR}{2}\)
Equation (i) becomes
\(\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}\) …. (iii)
In \(\Delta ABC\) and \(\Delta PQS\),
\(\angle B=\angle Q\) [Using equation (ii)]
And, \(\frac{AB}{PQ}=\frac{BS}{WS}\) [Using equation (iii)]
\(∴\Delta ABD∼\Delta PQS\) (SAS similarity criterion)
Therefore, it can be said that
\(\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}\) ..(iv)
\(\frac{ar(ar\Delta ANC)}{ar(\Delta ABC)}=(\frac{AB}{PQ})^2=(\frac{BC}{QR})^2=(\frac{AC}{PR})^3\)
From equations (i) and (iv), we may find that
\(\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}\)
And hence,
\(\frac{ar(\Delta ABC)}{ar(\Delta PQR)}=(\frac{AD}{PS})^2\)

Question 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.

Answer:

Let ABC be a square of side \(a\).
Therefore, its diagonal \(=\sqrt2a\)
Two desired equilateral triangles are formed as ΔABE and ΔDBF. Side of an equilateral triangle, ΔABE, described on one of its sides \(= a\)
Side of an equilateral triangle, ΔDBF, described on one of its diagonals.
We know that equilateral triangles have all its angles as \(60^0\) and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
\(\frac{Area\space of \space \Delta ABE}{Area\space of\space\Delta DBF}=(\frac{a}{\sqrt2a})^2=\frac{1}{2}\)
Question 8: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is

(i)  2:1

(ii)  2:1

(iii)  4:1

(iv)  1:4

Answer: 

We know that equilateral triangles have all its angles as \(60^0\) and all its sides of the  same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio
between the sides of these triangles.

Let side of ΔABC = x

Therefore, side of \(\Delta BDE=\frac{x}{2}\)

\(∴\frac{ar(\Delta ABC)}{ar(\Delta BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}\)

Question 9: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(i) 2:3

(ii) 4:9

(iii) 81:16

(iv) 16:81

Answer: If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangle \(=(\frac{4}{9})^2=\frac{16}{81}\)

Exercise 6.5

Question 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

 Answer: 

(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides, we will obtain 49, 576, and 625. 49 + 576 = 625

Or, \(7^2+24^2=25^2\)

The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides, we will obtain 9, 64, and 36.

However, \(9 + 36 ≠ 64\)

Or, \(3^+6^2≠ 8^2\)

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem. Hence, it is not a right triangle.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However, \(2500 + 6400 ≠ 10000\)

Or, \(50^2 + 80^2 ≠ 100^2\)

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem. Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25. Clearly, 144 +25=169

Or,

The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 13 cm. 

Question 2: \(PQR\) is a triangle right angled at \(P\) and \(M\) is a point on \(QR\) such that \(PM ⊥ QR\). Show that \(PM^2 = QM × MR.\)

Answer: 

Let \(\angle MPR=x\)

In \(\Delta MPR\)

\(\angle MRP=180^0-90^0-x\)

\(\angle MRP=90^0-x\)

Similarly, in \(\Delta MPQ\)

\(\angle MPQ=90^0-\angle MPR\)

\(=90^0-x\)

\(\angle MQP=180^0-90^0-(90^0-x)\)

\(\angle MQP=x\)

In \(\Delta QMP\) and \(\Delta PMR\),

\(\angle MPQ=\angle MRP\)

\(\angle PMQ=\angle MPR\)

\(\angle MQP=\angle MPR\)

\(∴\Delta QMP∼\Delta PMR\)               (By AAA similarty criterion)

\(⇒\frac{QM}{PM}=\frac{MP}{MR}\)

\(⇒PM^2=QM\times MR\)

Question 3: In the following figure, \(ABD\) is a triangle right angled at \(A\) and \(AC  ⊥ BD\). Show that 

(i) \(AB^2=BC\times BD\)

(ii) \(AC^2=BC\times DC\)

(iii) \(AD^2=BD\times CD\)

Answer: 

(i) In \(\Delta ADB\) and \(\Delta CAB\)

\(\angle DAB=\angle CAB\)               (Each \(90^0\))

\(∴\Delta ADB∼\Delta CAB\)                (AA Similarity criterion)

\(\angle ABD=\angle CBA\)               (Common angle)

\(⇒\frac{AB}{CB}=\frac{BD}{AB}\)

\(⇒AB^2=CB\times BD\)

(ii) Let \(\angle CAB=x\)

In \(\Delta CBA\)

\(\angle CBA=180^0-90^0-x\)

\(\angle CBA=90^0-x\)

Similarly, in \(\Delta CAD\),

\(\angle CAD=90^0-\angle CAB\)

\(=90^0-x\)

\(\angle CDA=180^0-90^0-(90^0-x)\)

\(\angle CDA=x\)

In \(\Delta CBA\) and \(\Delta CAD\)

\(\angle CDA=\angle CAD\)

\(\angle CBA=\angle CDA\)

\(\angle ACB=\angle DCA\)       (Each \(90^0\))

\(∴\Delta CBA∼\Delta CAB\)       (By AAA rule)

\(⇒\frac{AC}{DC}=\frac{BC}{AC}\)

\(⇒AC^2=DC\times BC\)

(iii) In \(\Delta DCA\) and \(\Delta DAB\),

\(\angle DCa=\angle DBA\)    (Each \(90^0\))

\(\angle CDA=\angle ADB\)    (Common angle)

\(∴\Delta DCA∼\Delta DAB\)

\(⇒\frac{DC}{DA}=\frac{DA}{DB}\)

\(⇒AD^2=BD\times CD\)

Question 4: \(ABC\) is an isoceles triangle right angled at \(C\). prove that \(AB^2=2AC^2\).

Answer: 

Given that \(\Delta ABC\) is isosceles triangle.

\(∴ AC=CB\)

Applying Pythagoras theorem in \(\Delta ABC\) (i.e., right angles at point C), we obtain

\(AC^2+CB^2=AB^2\)

\(⇒AC^2+AC^2=aB^2\)        (AC=CB)

\(⇒2AC^2=AB^2\)

Question 5: \(ABC\) is an isosceles triangle with \(AC = BC\). If \(AB^2 = 2 AC^2\), prove that ABC is a right triangle.

Answer: 

Given that,

\(AB^2=2AC^2\)

\(⇒AB^2=AC^2+AC^2\)

\(⇒AB^2=AC^2+BC^2\)      (As \(AC=BC\))

The triangles is satisfying the Pythagoras theorem.

Therefore, the given triangle is a right-angled triangle.

Question 6:  \(ABC\) is an equilateral triangle of side \(2a\). Find each of its altitudes.

Answer:

Let \(AD\) be the altitide in the given equilateral triangle, \(\Delta ABC\).

We know that altitude bisects opposite side.

\(∴BD=DC=a\)

In \(\Delta ADB\)

\(\angle ADB=90^0\)

Applying Pythagoras theorem, we obtain

\(AD^2+BD^2=AB^2\)
\(⇒AD^2+a^2=(2a)^2\)

\(⇒AD^2+a^2=4a^2\)

\(⇒AD^2=3a^2\)

\(⇒AD=a\sqrt3\)

In an equilateral triangle, all the altitudes are equal in length.

Therefore, the length of each altitude will be \(\sqrt3a\)

Question 7: Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer:

In \(ΔAOB,\space ΔBOC,\space ΔCOD,\space ΔAOD\)

Applying Pythagoras theorem, we obtain

\(AB^2=AO^2+OB^2\)               …(1)

\(BC^2=BO^2+OC^2\)                …(2)

\(CD^2=CO^2+OD^2\)               …(3)

\(AD^2=AO^2+OD^2\)               …(4)

Adding all these equations, we obtain

\(AB^2+BC^2+CD^2 AD^2\)

\(=2(AO^2+OB^2+OC^2+OD^2)\)

\(=[(\frac{AC}{2})^2+(\frac{BD}{2})^2+(\frac{AC}{2})^2+(\frac{BD}{2})^2]\)   (Diagonals bisect each other)

\(=2[\frac{(AC)^2}{2}+\frac{(BD)^2}{2}]\)

\(=(AC)^2+(BD)^2\)

Question 8:  In the following figure, \(O\) is a point in the interior of a triangle \(ABC,\space OD ⊥ BC,\space OE ⊥ AC\) and \(OF ⊥ AB\). Show that

(i) \(OA^2+OB^2+OC^2-OD^2-OE^2-OF^2\)

\(=AF^2+BD^2+CE^2\)

(ii) \(AF^2+BD^2+CE^2=AE^2+CD^2+BF^2\)

Answer: 

Join \(OA\space OB\) and \(OC\).

(i) Applying Pythagoras theorem in \(\Delta AOF\), we obtain

\(OA^2=OF^2+AF^2\)

Similarly, in \(ΔBOD\),

\(OB^2=OD^2\)

Similarly, in \(ΔCOE\),

\(OC^2=OE^2+EC^2\)

Adding these equations,

\(OA^2+OB^2+OC^2\)

\(=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2\)

\(OA^2+OB^2+OC^2-OD^2-OE^2 0OF^2\)

\(=AF^2+BD^2+EC^2\)

(ii) From the above result,

\(AF^2+BD^2+EC^2=(OA^2-OE^2 )+(OC^2-OD^2 )+(OB^2-OF^2)\)

\(∴AF^2+BD^2+EC^2=AE^2+CD^2+BF^2\)

Question 9: A ladder \(10\space m\) long reaches a window \(8\space m\) above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer: 

Let \(OA\) be the wall and \(AB\) be the ladder.

Therefore, by Pythagoras theorem,

\(AB+OA^2+BO^2\)

\((10\space m)^2=(8\space m)^2+OB^2\)

\(100\space m^2=64\space m^2+OB^2\)

\(OB^2=36\space m^2\)

\(OB=6\space m\)

Therefore, the distance of the foot of the ladder from the base of the wall is \(6\space m\).

Question 10: A guy wire attached to a vertical pole of height \(18\space m\) is \(24\space m\) 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? 

Answer: 

Let \(OB\) be the pole and \(AB\) be the wire.

By Pythagoras theorem,

\(AB^2=OB^2+OA^2\)

\((24\space m)^2=(18\space m)^2+OA^2\)

\(OA^2=(567-324)\space m^2=252\space m^2\)

\(OA=\sqrt{252}\space  m=\sqrt{6×6×7}\space m=6\sqrt7\space  m\)

Therefore, the distance from the base is \(6\sqrt7\space  m\).

Question 11: An aeroplane leaves an airport and flies due north at a speed of \(1,000\space km\) per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of \(1,200\space km\) per hour. How far apart will be the two planes after \(1\frac{1}{2}\) hours?

Answer: 

Distance travelled by the plane flying towards \(1\frac{1}{2}\) hrs. north in

\(=1000\times1\frac{1}{2}=1500\) km.

Similarly, distance travelled by the plane flying towards west \(1\frac{1}{2}\) hrs. in

\(=1200\times1\frac{1}{2}=1800\) km.

Let these distance be represented by \(OA\) and \(OB\) respectively.

Applying Pythagoras theorem,

Distance between these planes after \(1\frac{1}{2}\) hrs.

\(AB=\sqrt{OA^2+OB^2}\)

\(=[\sqrt{(1500)^2+(1800)^2}]\) km

\(=\sqrt{2250000+3240000}\) km.

Therefore, the distance between these planes will be \(300\sqrt{61}\) km after \(1\frac{1}{2}\) hrs.

Question 12: Two poles of heights \(6\space m\) and \(11\space m\) stand on a plane ground. If the distance between the feet of the poles is \(12\space m\), find the distance between their tops.

Answer:

Let \(CD\) and \(AB\) be the poles of height \(11\space m\) and \(6\space m\).

Therefore, \(CP = 11 – 6 = 5\space m\)

From the figure, it can be observed that \(AP = 12\space m\)

Applying Pythagoras theorem for \(ΔAPC\), we obtain

\(AP^2+PC^2+AC^2\)

\((12\space m)^2+(5\space m)^2=AC^2\)

\(AC^2=(144+25)\space m^2=169\space m^2\)

\(AC=13\space m\)

Therefore, the distance between their tops is \(13\space m\).

Question 13: \(D\) and \(E\) are points on the sides \(CA\) and \(CB\) respectively of a triangle \(ABC\) right angled at \(C\). Prove that \(AE^2 + BD^2 = AB^2 + DE^2\)

Answer: 

Applying Pythagoras theorem in \(ΔACE\), we obtain

\(AC^2+CE^2=AE^2\)                                        …(1)

Applying Pythagoras theorem in \(∆BCD\), we obtain

\(BC^2+CD^2=BD^2\)                                       …(2)

Using equation (1) and equation (2), we obtain

\(AC^2+CE^2+BC^2+CD^2=AE^2+BD^2\)      …(3)

Applying Pythagoras theorem in \(∆CDE\), we obtain

\(DE^2=CD^2+CE^2\)

Applying Pythagoras theorem in \(∆ABC\), we obtain

\(AB^2=AC^2+CB^2\)

Putting the values in equation (3), we obtain

\(DE^2=AB^2=AE^2+BD^2\)

Question 14: The perpendicular from \(A\) on side \(BC\) of a \(ΔABC\) intersect \(BC\) at \(D\) such that \(DB = 3 CD\). Prove that \(2 AB^2 = 2 AC^2 + BC^2\)

Answer: Applying Pythagoras theorem for \(ΔACD\),

we obtain

\(AC^2=AD^2+DC^2\)

\(AD^2=AC^2-DC^2\)                …(1)

Applying Pythagoras theorem in \(ΔABD\), we obtain

\(AB^2=AD^2+DB^2\)

\(AD^2=AB^2-DB^2\)                …(2)

From equation (1) and equation (2), we obtain

\(AC^2-DC^2=AB^2-DB^2\)      …(3)

It is given that \(3DC=DB\)

\(∴DC=\frac{BC}{4}\)  and \(DB=\frac{3BC}{4}\)

Putting these values in equation (3), we obtain

\(AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2\)

\(AC^2-\frac{BC^2}{16}=AB^2-\frac{9BC^2}{16}\)

\(16AC^2-BC^2=16AB^2-9BC^2\)

\(16AB^2-16AC^2=8BC^2\)

\(2AB^2=2AC^2+BC^2\)

Question 15: In an equilateral triangle \(ABC,\space D\) is a point on side \(BC\) such that \(BD = BC\). Prove that \(9 AD^2 = 7 AB^2\)

Answer: 

Let the side of the equilateral triangle be a, and \(AE\) be the altitude of \(ΔABC\).

\(∴BE=EC=\frac{BC}{2}=\frac{a}{2}\)

And, \(AE=\frac{a\sqrt3}{2}\)

Given that, \(BD=\frac{1}{2}BC\)

\(∴BD=\frac{a}{3}\)

\(DE=BE-BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}\)

Applying Pythagoras theorem in \(ΔADE\), we obtain \(AD^2 = AE^2 + DE^2\)

\(AD^2=(\frac{a\sqrt3}{2})^2+(\frac{a}{6})^2\)

\(=(\frac{3a^2}{4})+(\frac{a^2}{36})\)

\(=\frac{28a^2}{36}\)

\(=\frac{7}{9}AB^2\)

\(⇒9AD^2=7AB^2\)

Question 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.

 Answer:

Let the side of the equilateral triangle be a, and \(AE\) be the altitude of \(ΔABC\).

\(∴ BE = EC \frac{BC}{2}=\frac{a}{2}\)

Applying Pythagoras theorem in \(ΔABE\),

we obtain \(AB^2 = AE^2 + BE^2\)

\(a^2=AE^2+(\frac{a}{2})^2\)

\(AE^2=a^2-\frac{a^2}{4}\)

\(AE^2=\frac{3a^2}{4}\)

\(4AE^2 = 3a^2\)

\(⇒4 ×\) (Square of altitude) \(= 3 ×\) (Square of one side)

Question: Tick the correct answer and justify: In \(ΔABC,\space AB =6\sqrt3\space  cm\), \(AC=12\space cm\) and \(BC=6\space cm\).

The angle \(B\) is:

(i) \(120^0\)

(ii) \(60^0\)

(iii) \(90^0\)

(iv) \(45^0\)

Answer: 

Given that,

\(AB =6\sqrt3\space  cm\), \(AC = 12\sqrt cm\), and \(BC = 6\space cm\)

It can be observed that

\(AB^2 = 108\)

\(AC^2 = 144\)

And, \(BC^2 = 36\)

\(AB^2 +BC^2 = AC^2\)

The given triangle, \(ΔABC\), is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at \(B\).

\(∴ ∠B  =  90^0\)

Exercise 6.6

Question 1: In the given figure, \(PS\) is the bisector of \(∠QPR\) of \(ΔPQR\). Prove that

Answer: 

Let us draw a line segment \(RT\) parallel to \(SP\) which intersects extended line segment \(QP\) at point \(T\).

Given that, \(PS\) is the angle bisector of \(∠QPR\).

\(∠QPS  = ∠SPR\)                          …(1)

By construction,

\(∠SPR  = ∠PRT \) (As PS  || TR)    …(2)
\(∠QPS  = ∠QTR\)  (As PS  || TR)  …(3)

Using these equations, we obtain

\(∠PRT  = ∠QTR\)

\(∴ PT  = PR\)

By construction,

\(PS || TR\)

By using basic proportionality theorem for \(ΔQTR\),

\(\frac{QS}{SR}=\frac{QP}{PT}\)

\(⇒\frac{QS}{SR}=\frac{PQ}{PR}\)             (PT=TR)

Question 2: In the given figure, \(D\) is a point on hypotenuse \(AC\) of \(ΔABC,\space DM ⊥ BC\) and \(DN ⊥ AB\), Prove that:

(i) \(DM^2 = DN.MC\)

(ii) \(DN^2 = DM.AN\)

Answer:

(i) Let us join \(DB\).

We have, \(DN || CB\), \(DM || AB\), and \(∠B = 90^0\)

\(∴DMBN\) is a rectangle.

\(∴DN = MB\) and \(DM  = NB\)

The condition to be proved is the case when \(D\) is the foot of the perpendicular drawn from \(B\) to \(AC\).

\(∴∠CDB = 90^0\)

\(⇒ ∠2 + ∠3 = 90^0\)                   … (1)

In \(ΔCDM\)

\(∠1  + ∠2  + ∠DMC  =  180^0\)

\(⇒ ∠1 + ∠2 = 90^0\)                     … (2)

In \(ΔDMB\)

\(∠3  + ∠DMB  + ∠4  =  180^0\)

\(⇒∠3  + ∠4  =  90^0\)                      …  (3)

From equation (1) and (2), we obtain

\(∠1  = ∠3\)

From equation (1) and (3), we obtain

\(∠2  = ∠4\)

\(∠1 = ∠3\)     (Proved above)

\(∠2  = ∠4\)  (Proved above)

\(∴ΔDCM  ∼ ΔBDM\)  (AA similarity criterion)

\(⇒\frac{BM}{DM}=\frac{DM}{MC}\)

\(⇒\frac{DN}{DM}=\frac{DM}{MC}\)     (BM=DN)

\(⇒DM^2=DN×MC\)

(ii) In right triangle \(DBN\)

\(∠5  + ∠7  =  90^0\)             …  (4)

In right triangle \(DAN\)

\(∠6  + ∠8  =  90^0\)             …  (5)

\(∴  ∠ADB = 90^0\)

\(⇒∠5  + ∠6  =  90^0\)          …  (6)

From equation (4) and (6), we obtain

\(∠6  = ∠7\)

From equation (5) and (6), we obtain

\(∠8  = ∠5\)

\(∠6 = ∠7\)       (Proved above)

\(∠8  = ∠5\)       (Proved above)

\(∴ΔDNA  ∼ ΔBND\)  (AA similarity criterion)

\(⇒\frac{AN}{DN}=\frac{DN}{NB}\)

\(⇒ DN^2 = AN × NB\)

\(⇒DN^2  = AN  × DM\)     (As \(NB  = DM\))

Question 3: In the given figure, \(ABC\) is a triangle in which  \(∠ABC>  90^0\) and \(AD  ⊥ CB\) produced.

Prove that \(AC^2 = AB^2 + BC^2 + 2BC.BD.\)

Answer: 

Applying Pythagoras theorem in \(ΔADB\), we obtain \(AB^2 = AD^2 + DB^2\) … (1)

Applying Pythagoras theorem in \(ΔACD\), we obtain \(AC^2 = AD^2 + DC^2\)

\(AC^2 = AD^2 + (DB + BC)^2\)

\(AC^2 = AD^2 + DB^2 + BC^2 + 2DB × BC\)

\(AC^2 = AB^2 + BC^2 + 2DB × BC\)  [Using equation (1)]

Question 4: In the given figure, \(ABC\) is a triangle in which \(∠ABC < 90^0\) and \(AD ⊥ BC\).

Prove that \(AC^2 = AB^2 + BC^2 − 2BC.BD\)

Answwer: Applying Pythagoras theorem in \(ΔADB\),

we obtain,

\(AD^2 + DB^2 = AB^2\)

\(⇒AD^2 = AB^2 − DB^2\)   … (1)

Applying Pythagoras theorem in \(ΔADC\), we obtain \(AD^2 + DC^2 = AC^2\)

\(AB^2 − BD^2 + DC^2 = AC^2\)    [Using equation (1)]

\(AB^2 − BD^2 + (BC − BD)^2 = AC^2\)

\(AC^2 = AB^2 − BD^2 + BC^2 + BD^2 −2BC × BD\) \(= AB^2 + BC^2 − 2BC × BD\)

Question 5: In the given figure, \(AD\) is a median of a triangle \(ABC\) and \(AM  ⊥ BC\). Prove that:

(i) \(AC^2=AD^2+\frac{BC}{DM}+(\frac{BC}{2})^2\)

(ii) \(AB^2=AD^2-BC.DM+(\frac{BC}{2})^2\)

(iii) \(AC^2+AB^2=2AD^2+\frac{1}{2}BC^2\)

Answer: 

(i) Applying Pythagoras theorem in \(\Delta AMD\), we obtain

\(AM^2+MD^2=AD^2\)  …..(1)

Applying Pythagoras theorem in \(\Delta AMC\), we obtain

\(AM^2+MC^2=AC^2\)

\(AM^2+(MD+DC)^2=AC^2\)

\((AM^2+MD^2)+DC^2+2\space MD.DC=AC^2\)

\(AD^2+DC^2+2MD.DC\space MD.DC=AC^2\)   [Using equation (1)]

Using the result, \(DC=\frac{BC}{2}\), we obtain

\(AD^2+(\frac{BC}{2})^2+2MD.(\frac{BC}{2})=AC^2\)

\(AD^2+(\frac{BC}{2})^2+MD\times BC=AC^2\)

(ii) Applying Pythagoras theorem in \(ΔABM\), we obtain

\(AB^2=AM^2+MB^2\)

\(=(AD^2-DM^2)+MB^2\)

\(=(AD^2-DM^2)+(BD-MD)^2\)

\(=AD^2-DM^2+BD^2+MD^2-2BD\times MD\)

\(=AD^2+BD^2-2BD\times MD\)

\(AD^2+(\frac{BC}{2})^2-2(\frac{BC}{2}\times MD\)

\(=AD^2+(\frac{BC}{2})^2-BC\times MD\)

(iii) Applying Pythagoras theorem in \(ΔABM\), we obtain

\(AM^2 + MB^2 = AB^2\)… (1)

Applying Pythagoras theorem in \(ΔAMC\), we obtain

\(AM^2 + MC^2 = AC^2\) … (2)

Adding equations (1) and (2), we obtain

\(2AM^2 + MB^2 + MC^2 = AB^2 + AC^2\)

\(2AM^2 + (BD − DM)^2 + (MD + DC)^2 = AB^2 + AC^2\)

\(2AM^2+BD^2 + DM^2 − 2BD.DM + MD^2 + DC^2 + 2MD.DC\)

\(= AB^2 +AC^2\space 2AM^2 + 2MD^2 + BD^2 + DC^2 + 2MD (− BD + DC)\)

\(= AB^2 + AC^2\)

\(2(AM^2+MD^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2MD(-\frac{BC}{2}+\frac{BC}{2})\)

\(=AB^2+AC^2\)

\(2AD^2+\frac{Bc^2}{2}=AB^2+AC^2\)

Question 6: Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer:

Answer: Let \(ABCD\) be a parallelogram.

Let us draw perpendicular \(DE\) on extended side \(AB\) and \(AF\) on side \(DC\).

Applying Pythagoras theorem in \(ΔDEA\), we obtain

\(DE^2 + EA^2 = DA^2\)                                … (i)

Applying Pythagoras theorem in \(ΔDEB\),

we obtain \(DE^2 + EB^2 = DB^2\)

\(DE^2 + (EA + AB)^2 = DB^2\)

\((DE^2 + EA^2) + AB^2 + 2EA × AB = DB^2\)

\(DA^2 + AB^2 + 2EA × AB = DB^2\)             … (ii)

Applying Pythagoras theorem in \(ΔADF\), we obtain

\( AD^2 = AF^2 + FD^2\)

Applying Pythagoras theorem in \(ΔAFC\), we obtain

\(AC^2 = AF^2 + FC^2\)

\(= AF^2 + (DC − FD)^2\)

\(= AF^2 + DC^2 + FD^2 − 2DC × FD\)

\(= (AF^2 + FD^2) + DC^2 − 2DC × FD\)

\(AC^2 = AD^2 + DC^2 − 2DC × FD\)              … (iii)

Since \(ABCD\) is a parallelogram,

\(AB = CD\)                                               … (iv)

And, \(BC = AD\)                                      … (v)

In \(ΔDEA\) and \(ΔADF\),

\(∠DEA  = ∠AFD\)  (Both  \(90^2\))

\(∠EAD  = ∠ADF  (EA  || DF)\)

\(AD = AD\)      (Common)

\(∴ ΔEAD ΔFDA \)     (AAS congruence criterion)

\(⇒EA  = DF\)                                            …  (vi)

Adding equations (i) and (iii), we obtain

\(DA^2 + AB^2 + 2EA × AB + AD^2 + DC^2 − 2DC × FD\)

\(= DB^2 + AC^2\)

\(DA^2 + AB^2 + AD^2 + DC^2 + 2EA × AB − 2DC × FD\)

\(= DB^2 + AC^2\)

\(BC^2 + AB^2 + AD^2 + DC^2 + 2EA × AB − 2AB × EA\)

\(= DB^2 + AC^2\)

[Using equations (iv) and (vi)]

\(AB^2 + BC^2 + CD^2 + DA^2\)

\( = AC^2 + BD^2\)

Question 7: In the given figure, two chords \(AB\) and \(CD\) intersect each other at the point \(P\). prove that:

(i) \(ΔAPC ∼ ΔDPB\)

(ii) \(BP = CP.DP\)

Answer: 

Let us join \(CB\).

(i) In \(ΔAPC\) and \(ΔDPB\),

\(∠APC  = ∠DPB\)  (Vertically opposite angles)

\(∠CAP  = ∠BDP\)  (Angles in the same segment for chord \(CB\))

\(ΔAPC  ∼ ΔDPB\)  (By AA similarity criterion)

(ii) We have already proved that

\(ΔAPC  ∼ ΔDPB\)

We know that the corresponding sides of similar triangles are proportional.

\(∴\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}\)

\(∴AP.PB=PC.DP\)

\(⇒\frac{AP}{DP}=\frac{PC}{PB}\)

Question 8: In the given 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) \(PA.PB = PC.PD\)

(i) In \(ΔPAC\) and \(ΔPDB\),

\(∠P  = ∠P\)  (Common)

\(∠PAC = ∠PDB\) (Exterior angle of a cyclic quadrilateral is \(∠PCA = ∠PBD\) equal to the opposite interior angle)

\(∴ ΔPAC  ∼ ΔPDB \)

(ii) We know that the corresponding sides of similar triangles are proportional.

\(∴\frac{PA}{PD}=\frac{AC}{DB}=\frac{PC}{PB}\)

\(∴ PA.PB = PC.PD\)

\(⇒\frac{PA}{PD}=\frac{PC}{PB}\)

Question 9: In the given figure, \(D\) is a point on side \(BC\) of \(ΔABC\) such that m . Prove that \(AD\) is the bisector of \(∠BAC\).

Answer: Let us extend \(BA\) to \(P\) such that \(AP = AC\). Join \(PC\).

It is given that,

\(\frac{BD}{CD}=\frac{AB}{AC}\)

\(⇒\frac{BD}{CD}=\frac{AP}{AC}\)

By using the converse of basic proportionality theorem, we obtain \(AD || PC\)

\(⇒ ∠BAD = ∠APC\)   (Corresponding angles)                           … (1)

And, \(∠DAC = ∠ACP\) (Alternate interior angles)                   … (2)

By Construction, we have

\(AP=AC\)

\(⇒\angle APC=\angle ACP\)      …(3)

On comparing equations (1), (2), and (3), we obtain

\(∠BAD = ∠APC\)

\(⇒ AD\) is the bisector of the angle \(BAC\)

Question 10:

Nazima is fly fishing in 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 taut, ho much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer: 

Let \(AB\) be the height of the tip of the fishing rod from the water surface.

Let \(BC\) be the horizontal distance of the fly from the tip of the fishing rod. Then, \(AC\) is the length of
the string. \(AC\) can be found by applying Pythagoras theorem in \(ΔABC\).

\(AC^2 = AB^2 + BC^2\)

\(AB^2 = (1.8 m)^2 + (2.4 m)^2\)

\(AB^2 = (3.24 + 5.76) m^2\)

\(AB^2 = 9.00 m^2\)

\(⇒AB\sqrt9m=3m\)

Thus, the length of the string out is \(3\space m\).

She pulls the string at the rate of \(5\space cm\) per second.

Therefore, string pulled in \(12\) seconds \(= 12 × 5 = 60\space cm = 0.6\space m\)

Let the fly be at point (D\) after \(12\) seconds.

Length of string out after \(12\)seconds is \(AD\).

\(AD = AC − \) String pulled by Nazima in \(12\) seconds

\(=(3.00-0.6\space m\)

\(=2.4(1.8\space m)^2+BD^2=(2.4\space m)^2\)

\(=(1.587+1.2)\space m\)

\(=2.787\space m\)

\(=2.79\space m\)