NCERT Solutions for Class 10 Math Chapter 6 - Triangles

Answer:

(i) Similar

(ii) Similar

(iii) Equilateral

(iv) (a) Equal

(b) Proportional

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

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.

Answer:

(i)

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

(ii)

Let AD = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

Answer:

(i)

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

(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Answer:

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

Answer:

In ΔABC, DE || AC

Answer:

In Δ POQ, DE || OQ

Answer:

In Δ POQ, AB || PQ

Answer:

Consider the given figure in which l is a line  drawn through the mid-point P of line segment AB meeting AC at Q, such that .

Or, Q is the mid-point of AC.

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

Hence, by using basic proportionality theorem, we obtain

Answer:

Draw a line EF through point O, such that

In ΔADC,

By using basic proportionality theorem, we obtain

In ΔABD,

So, by using basic proportionality theorem, we obtain

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

Answer:

Let us consider the following figure for the given question.

Draw a line OE || AB

In ΔABD, OE || AB

By using basic proportionality theorem, we obtain

However, it is given that

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

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

Answer:

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]

(ii)

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

(iv) In ∆ MNL and ∆ QPR, we observe that,

     $$\begin{gathered} \frac{\mathrm{MN}}{\mathrm{QP}} = \frac{\mathrm{ML}}{\mathrm{QR}} = \frac{1}{2} \\ \angle \mathrm{M} = \angle \mathrm{Q} = 70° \\ \therefore ∆\mathrm{MNL} ~ ∆\mathrm{QPR} \left[\mathrm{By} \mathrm{SAS} \mathrm{similarity} \mathrm{criterion}\right] \end{gathered}$$
 

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

(vi) In ΔDEF,

∠D +∠E +∠F = 180º

(Sum of the measures of the angles of a triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ΔPQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ΔDEF and ΔPQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ΔDEF ∼ ΔPQR [By AAA similarity criterion]

Answer:

DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° − 125°

= 55°

In ΔDOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ∼ ΔOBA.

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

⇒ ∠OAB = 55°

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]

Answer:

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,

Answer:

In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ∼ ΔRTS (By AA similarity criterion)

Answer:

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1)

And, AD = AE [By CPCT] (2)

In ΔADE and ΔABC,

[Dividing equation (2) by (1)]

∠A = ∠A [Common angle]

∴ ΔADE ∼ ΔABC [By SAS similarity criterion]

Answer:

(i)

In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔCDP

(ii)

In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

ΔABD ∼ ΔCBE

(iii)

In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔADB

(iv)

In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

ΔPDC ∼ ΔBEC

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)

Answer:

In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC ∼ ΔAMP (By AA similarity criterion)

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)

∴ ΔACD ∼ ΔFGH (By AA similarity criterion)

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)

Answer:

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠ABD = ∠ECF (Proved above)

∴ ΔABD ∼ ΔECF (By using AA similarity criterion)

Answer:

Median divides the opposite side.

∴

Given that,

In ΔABD and ΔPQM,

(Proved above)

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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.

Answer:

Given that,

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

∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠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,

(Given)

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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)

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

Answer:

It is given that ΔABC ∼ ΔPQR

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

∴ … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

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

∴ … (3)

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

… (4)

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

[Using equation (4)]

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

⇒

Answer:

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)

Answer:

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

We know that area of a triangle =

.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

Answer:

Let us assume two similar triangles as ΔABC ∼ ΔPQR.

Answer:

D and E are the mid-points of ΔABC.

Answer:

Let us assume two similar triangles as ΔABC ∼ ΔPQR. Let AD and PS be the medians of these triangles.

ΔABC ∼ ΔPQR

…(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,

∴ BD = DC =

And, QS = SR =

Equation (1) becomes

… (3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And, [Using equation (3)]

∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

Therefore, it can be said that

… (4)

From equations (1) and (4), we may find that

And hence,

Answer:

Let ABCD be a square of side a.

Therefore, its diagonal

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 60º 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.

Answer:

We know that equilateral triangles have all its angles as 60º 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

Hence, the correct answer is (C).

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 triangles =

Hence, the correct answer is (D).

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,

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, 3² + 6² ≠ 8²

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, 50² + 80² ≠ 100²

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.

​

Answer:

Answer:

(i) In ,

∴ (AA similarity criterion)

(ii)

(iii)

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

Answer:

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

Answer:

Given that,

Answer:

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = a

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

Therefore, the length of each altitude will be.

Answer:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

Answer:

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

(ii) From the above result,

Answer:

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Answer:

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

Therefore, the distance from the base is m.

Answer:

Distance travelled by the plane flying towards north in

Similarly, distance travelled by the plane flying towards west in

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after, AB =

Therefore, the distance between these planes will be km after.

Answer:

Let CD and AB be the poles of height 11 m and 6 m.

Therefore, CP = 11 − 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtain

Therefore, the distance between their tops is 13 m.

Answer:

Applying Pythagoras theorem in ΔACE, we obtain

Answer:

Applying Pythagoras theorem for ΔACD, we obtain

Applying Pythagoras theorem in ΔABD, we obtain

Answer:

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

∴ BE = EC = =

And, AE =

Given that, BD = BC

∴ BD =

DE = BE − BD =

Applying Pythagoras theorem in ΔADE, we obtain

AD² = AE² + DE²

⇒ 9 AD² = 7 AB²

Answer:

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

∴ BE = EC = =

Applying Pythagoras theorem in ΔABE, we obtain

AB² = AE² + BE²

4AE² = 3a²

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

Answer:

Given that, AB =cm, AC = 12 cm, and BC = 6 cm

It can be observed that

AB² = 108

AC² = 144

And, BC² = 36

AB² +BC² = AC²

The given triangle, ΔABC, is satisfying Pythagoras theorem.

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

∴ ∠B = 90°

Hence, the correct answer is (C).

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,

$\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{QP}}{\mathrm{PT}}$
$\Rightarrow \frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}} \left(\because \mathrm{PT}=\mathrm{PR}\right)$

Answer:

(i)Let us join DB.

We have, DN || CB, DM || AB, and ∠B = 90°

∴ 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°

⇒ ∠2 + ∠3 = 90° … (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), we obtain

∠1 = ∠3

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

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

⇒ DM² = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

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

∠6 = ∠7

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

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ∼ ΔBND (AA similarity criterion)

⇒ DN² = AN × NB

⇒ DN² = AN × DM (As NB = DM)

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AB² = AD² + DB² … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC² = AD² + DC²

AC² = AD² + (DB + BC)²

AC² = AD² + DB² + BC² + 2DB × BC

AC² = AB² + BC² + 2DB × BC [Using equation (1)]

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

⇒ AD² = AB² − DB² … (1)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB² − BD² + DC² = AC² [Using equation (1)]

AB² − BD² + (BC − BD)² = AC²

AC² = AB² − BD² + BC² + BD² −2BC × BD

= AB² + BC² − 2BC × BD

Answer:

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, , we obtain

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

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

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC) = AB² + AC²

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

DE² + EA² = DA² … (i)

Applying Pythagoras theorem in ΔDEB, we obtain

DE² + EB² = DB²

DE² + (EA + AB)² = DB²

(DE² + EA²) + AB² + 2EA × AB = DB²

DA² + AB² + 2EA × AB = DB² … (ii)

Applying Pythagoras theorem in ΔADF, we obtain

AD² = AF² + FD²

Applying Pythagoras theorem in ΔAFC, we obtain

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 … (iv)

And, BC = AD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEAD ΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

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

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

[Using equations (iv) and (vi)]

AB² + BC² + CD² + DA² = AC² + BD²

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.

∴ AP. PB = PC. DP

Answer:

(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.

∴ PA.PB = PC.PD

Answer:

Let us extend BA to P such that AP = AC. Join PC.

It is given that,

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

⇒ ∠APC = ∠ACP … (3)

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

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.

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.

AC² = AB² + BC²

AB² = (1.8 m)² + (2.4 m)²

AB² = (3.24 + 5.76) m²

AB² = 9.00 m²

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

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

= 2.4 m

In ΔADB,

AB² + BD² = AD^2­

(1.8 m)² + BD² = (2.4 m)²

BD² = (5.76 − 3.24) m² = 2.52 m²

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m