Class 10 - Maths - Triangles

Exercise 6.1

Question 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)

(i) All circles are similar.

(ii) All squares are similar.

(iii) All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are

equal and (b) their corresponding sides are proportional.

Question 2:

Give two different examples of pair of

(i) similar figures.                           (ii) non-similar figures.

(i) similar figures.

Two equilateral triangles with sides

1 cm and 2 cm are given below:

Two squares with sides 1 cm and 2 cm are given below:

(ii) non-similar figures.

Trapezium and square

Triangle and parallelogram

Question 3:

State whether the following quadrilaterals are similar or not:

The sides of quadrilateral PQRS and ABCD are in the same ratio i.e. 1 : 2. Therefore, the

corresponding angles of the two quadrilaterals are not equal.

Hence, the two quadrilaterals are not similar.

Exercise 6.2

Question 1:

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

(i) Let EC = x cm

Given that DE || BC, therefore using Thales theorem, we get

=> 1.5/3 = 1/x

=> x = 3/1.5

=> x = 3/(15/10)

=> x = (3 * 10)/15

=> x = 30/15

=> x = 2

Hence, EC = 2 cm

(ii) Let AD = x cm

Given that DE || BC, therefore using Thales theorem, we get

=> x/7.2 = 1.8/5.4

=> x/7.2 = 1/3

=> x = 7.2/3

=> x = 2.4

Question 2:

E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR:

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

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

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

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm,

therefore

PE/EQ = 3.9/3 = 3 cm

PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5 cm

Since PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

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

PE/EQ = 4/4.5 = 40/45 = 8/9 cm

PF/FR = 8/9 cm

Since PE/EQ = PF/FR

Hence, according to converse of Thales theorem, EF || QR.

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

PE/PQ = 0.18/1.28 = 18/128 = 9/64 cm

PE/PR = 0.36/2.56 = 36/256 = 9/64 cm

Since PE/PQ = PE/PR

Hence, according to converse of Thales theorem, EF || QR.

Question 3:

In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

Given, in triangle ABC, LM || CB, therefore

According to Thales theorem, we have

AM/AB = AL/AC  ………………1

Similarly, in triangle ADC, LN || CD, therefore

According to Thales theorem, we have

From equation 1 and 2, we get

Question 4:

In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

Given, in triangle ABC, DE || AC, therefore

According to Thales theorem, we have

BD/DA = BE/EC  ………………1

Similarly, in triangle ABC, DE || AE, therefore

According to Thales theorem, we have

BD/DA = BF/FE  ………………2

From equation 1 and 2, we get

BF/FE = BE/EC

Question 5:

In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Given, in triangle POQ, DE || OQ, therefore

According to Thales theorem, we have

PE/EQ = PD/DO  ………………1

Similarly, in triangle POR, DF || OR, therefore

According to Thales theorem, we have

PF/FR = PD/DO  ………………2

From equation 1 and 2, we get

PE/EQ = PF/FR

In triangle PQR,

PE/EQ = PF/FR         [Proved]

Hence, according to converse of Thales theorem, EF || QR.

Question 6:

In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively

such that AB || PQ and AC || PR. Show that BC || QR.

Given, in triangle POQ, AB || PQ, therefore

According to Thales theorem, we have  OA/AP = OB/BQ  ………………1

Similarly, in triangle POR, AC || PR, therefore

According to Thales theorem, we have

OA/AP = OC/CR  ………………2

From equation 1 and 2, we get OB/AP = OC/CR

In triangle OQR,

OB/AP = OC/CR         [Proved]

Hence, according to converse of Thales theorem, BC || QR.

Question 7:

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

(Recall that you have proved it in Class IX).

Let PQ is a line through the mid-point of AB which is parallel to BC intersects Ac at Q

i.e. PQ || BC,

In triangle ABC,

According to converse of Thales theorem, we have

AP/PB = AQ/QC

=> 1 = AQ/QC                 [Since AP = PB]

=> AQ = QC

Hence, Q is the mid-point of AC.

Question 8:

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

(Recall that you have done it in Class IX).

Let PQ is a line which passes through the mid-point of AB and AC.

Therefore, AP = PB and AQ = QC

=>  AP/PB = 1 and AQ/QC = 1

=> AP/PB = AQ/QC = 1

Now, in triangle ABC,

AP/PB = AQ/QC

Hence, according to converse of Thales theorem, we have PQ || BC.

Question 9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

A line is drawn through the point O parallel to CD such that EO || AB.

In triangle ADC, EF || CD

According to Thales theorem, we have

AE/ED = AO/OC    ……….1

Similarly, in triangle ABD, EO || AB

According to Thales theorem, we have

AE/ED = BO/OD    ……….2

From equation 1 and 2, we get

AO/OC = BO/OD

=> AO/OB = OC/OD

Question 10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that

AO/BO = CO/DO. Show that ABCD is a trapezium.

A line is drawn through the point O parallel to AB such that EO || AB.

In triangle ABD, EO || AB

According to Thales theorem, we have

AE/ED = BO/OD    ……….1

But given that

AO/OB = CO/OD

=> AO/CO = OB/OD   ………2

From equation 1 and 2, we get

AE/ED = AO/OC

According to converse of Thales theorem, we have

=> ED || DC

=> AB || OE || DC

=> AB || CD

Hence, ABCD is a trapezium.

Exercise 6.3

Question 1:

State which pairs of triangles in Fig. 6.34 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) ∠A =∠P = 60°, ∠B =∠Q = 80°, ∠C =∠R = 40°

Therefore, Δ PQS ~ Δ TQR            [AAA similarity]

(ii) AB/QR = BC/RP = CA/PQ

Therefore, Δ ABC ~ Δ QRP            [SSS similarity]

(iii) Triangles are not similar because the corresponding sides are not equal.

(iv) Triangles are not similar because the corresponding sides are not equal.

(v) Triangles are not similar because the corresponding sides are not equal.

(vi) In Δ DEF,

∠D + ∠E + ∠F = 180°

=> 70° + 80° + ∠F = 180°

=> 150° + ∠F = 180°

=> ∠F = 180° – 150°

=> ∠F = 30°

Similarly, in Δ PQR,

∠D + ∠E + ∠F = 180°

=> ∠P + 80° + 30° = 180°

=> ∠P + 110° = 180°

=> ∠P = 180° – 110°

=> ∠P = 70°

Now, in Δ DEF and Δ PQR,

∠D = ∠P           [Each 70°]

∠E = ∠Q           [Each 70°]

∠F = ∠R            [Each 70°]

So, Δ DEF ~ Δ PQR             [AAA similarity]

Question 2:

In Fig. 6.35, Δ ODC ~ Δ OBA, ∠ BOC = 125°

and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Since DOB is a straight line

So, ∠ DOC + ∠ COB = 180°

=> ∠ DOC + 125° = 180°

=> ∠ DOC = 180° – 125°

=> ∠ DOC = 55°

In Δ DOC,

∠ DCO + ∠ DCO + ∠ DOC = 180°

=> ∠ DCO + 70° + 55° = 180°

=> ∠ DCO + 125° = 180°

=> ∠ DCO = 180° – 125°

=> ∠ DCO = 55°

Given that, Δ ODC ~ Δ OBA

So, ∠ OAB = ∠ DCO                  [Corresponding angles of similar triangles]

=> ∠ OAB = 55°

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 OA/OC = OB/OD.

In Δ DOC and Δ BOA,

∠ DOC = ∠ BOA                  [Vertically opposite angles]

∠ CDO = ∠ ABO                  [Alternate angles]

∠ DCO = ∠ BAO                  [Alternate angles]

So, Δ DOC ~ Δ BOA            [AA similarity]

=> DO/BO = CO/AO

=> BO/DO = AO/CO

Question 4:

In Fig. 6.36, QR/QS = QT/PR and ∠ 1 = ∠ 2. Show that Δ PQS ~ Δ TQR.

In Δ PQR, ∠ PQR = ∠ PRQ

=> PQ = PR  …………..1

Given that QR/QS = QT/PR

=> QR/QS = QT/PQ ………2    [From equation 2]

In Δ PQS and Δ TQR,

QR/QS = QT/PQ            [From equation 2]

∠Q = ∠Q                         [Common]

So, Δ PQS ~ Δ TQR        [SAS similarity]

Question 5:

S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS.

Show that:   Δ RPQ ~ Δ RTS.

Given, S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS

Now, in Δ RPQ ~ Δ RTS

∠ RTS= ∠ QPS         [Given]

∠ R = ∠ R                 [Common]

Hence, Δ RPQ ~ Δ RTS    [By AA similarity]

Question 6:

In Fig. 6.37, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC.

Given, Δ ABE ≅ Δ ACD

So, AB = AC  ………1        [By CPCT]

and AD = AE  ………2      [By CPCT]

Now in Δ ADE and Δ ABC,

AD/AB = AE/AC                  [From equation 1 and 2]

∠ A = ∠ A                            [Common]

Hence, Δ ADE ~ Δ ABC     [By SAS similarity]

Question 7:

In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P.

Show that:

(i) Δ AEP ~ Δ CDP

(ii) Δ ABD ~ Δ CBE

(iii) Δ AEP ~ Δ ADB

(iv) Δ PDC ~ Δ BEC

(i) In Δ AEP and Δ CDP,

∠ APE = ∠ CPD            [Vertically opposite angles]

∠ AEP = ∠ CDP            [Common]

So, Δ AEP ~ Δ CDP       [AA similarity]

(ii) Δ ABD ~ Δ CBE

∠ ADB = ∠ CEB            [Each 90°]

∠ ABD = ∠ CBE            [Common]

So, Δ ABD ~ Δ CBE       [AA similarity]

(iii) Δ AEP ~ Δ ADB

∠ AEP = ∠ ADB            [Each 90°]

∠ PAE = ∠ DAP            [Common]

So, Δ AEP ~ Δ ADB       [AA similarity]

(iv) Δ PDC ~ Δ BEC

∠ PDC = ∠ BEC            [Each 90°]

∠ PCD = ∠ BCE            [Common]

So, Δ PDC ~ Δ BEC       [AA similarity]

Question 8:

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

In Δ ABE and Δ CFB,

∠ A = ∠ C                       [Opposite angles of parallelogram]

∠ AEB = ∠ CBF              [Alternate angles as AE || BC]

So, Δ ABE ~ Δ CFB        [AA similarity]

Question 9:

In Fig. 6.39, ABC and AMP are two right triangles, right angled   at B and M respectively. Prove that:

(i) Δ ABC ~ Δ AMP

(ii) CA/PA = BC/MP

(i) In Δ ABC and Δ AMP,

∠ ABC = ∠ AMP              [Each 90°]

∠ A = ∠ A                         [Common]

So, Δ ABC ~ Δ AMP        [AA similarity]

(ii)  Since Δ ABC ~ Δ AMP     [Proved]

So, the corresponding sides of the similar triangles are proportional

=> AB/AM = BC/PM = AC/AP

=> CA/PA = BC/MP

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 Δ ABC and Δ EFG respectively.

If Δ ABC ~ Δ FEG, show that:

(i) CD/GH = AC/FG

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF

(i) Given, Δ ABC ~ Δ FEG

So, ∠ A = ∠ F, ∠ B = ∠ E and ∠ ACB = ∠ FGE

=> ∠ ACD = ∠ FGH    [CD and GH are the bisectors of equal angles]

and ∠ DCB = ∠ HGE    [CD and GH are the bisectors of equal angles]

In Δ ACD and Δ FGH,

∠ A = ∠ F and ∠ ACD = ∠ FGH   [Proved]

So, Δ ACD ~ Δ FGH     [AA similarity]

=> CD/GH = AC/FG

(ii) In Δ DCB and Δ HGE,

∠ B = ∠ E                     [Proved]

∠ DCB = ∠ HGE           [Proved]

So, Δ DCB ~ Δ HGE     [AA similarity]

(iii) In Δ DCA and Δ HGF,

∠ A = ∠ F                       [Proved]

∠ ACD = ∠ HGF            [Proved]

So, Δ DCA ~ Δ HGF      [AA similarity]

Question 11:

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC.

If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.

Given, ABC is an isosceles triangle.

So, AB = BC

=> ∠ ABD = ∠ ECF

In Δ ABD and Δ ECF,

∠ ADB = ∠ EFC                       [Each 90]

∠ ABD = ∠ ECF                       [Proved]

So, Δ ABD ~ Δ ECF                 [AA similarity]

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 Δ PQR (see Fig. 6.41). Show that: Δ ABC ~ Δ PQR.

Given, AD and Pm are the median of triangle, Therefore,

BD = BC/2 and QM = QR/2

Given that

=> AB/PQ = (BC/2)/(QR/2) = AD/PM

=> AB/PQ = BD/QM = AD/PM

In Δ ABD and Δ PQM,

AB/PQ = BD/QM = AD/PM       [Proved]

So, Δ ABD ~ Δ PQM                    [SSS similarity]

Hence, ∠ ABD = ∠ PQM            [Corresponding angles of similar triangles]

In Δ ABC and Δ PQR,

∠ ABD = ∠ PQM            [Proved]

AB/PQ = BC/QR

So, Δ ABC ~ Δ PQR      [SAS similarity]

Question 13:

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

In Δ ADC and Δ BAC,

∠ ADC = ∠ BAC            [Proved]

∠ ACD = ∠ BCA            [Proved]

So, Δ ADC ~ Δ BAC      [AA similarity]

We know that the corresponding sides of similar

triangles are proportional. Therefore,

CA/CB = CD/CA

=> CA2 = 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: Δ ABC ~ Δ PQR.

Given that,

Produce AD and PM to E and L such that AD = DE and PM = DE

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

AD and Pm are medians of triangles, therefore

BD and Dc and QM = MR

AD = DE and PM = ML    [By construction]

So, the diagonals of ABEC bisecting each other at D,

Therefore, ABEC is a parallelogram.

So, AC = BE, AB = EC, PR = QL and PQ = LR

Given that, AB/PQ = AC/PR = AD/PM

=> AB/PQ = BE/QL = 2AD/2PM

=> AB/PQ = BE/QL = AE/PL

Hence, Δ ABE ~ Δ PQL      [SSS similarity]

We know that the corresponding angles of similar triangles are equal.

Therefore, ∠ BAE = ∠ QPL   ……….1

Similarly, Δ AEC ~ Δ PLR

and ∠ CAE = ∠ RPL   ……….2

Add equation 1 and 2, we get

∠ BAE + ∠ CAE = ∠ QPL + ∠ RPL

=> ∠ CAB = ∠ RPQ   ……….3

In Δ ABC and Δ PQR,

AB/PQ = AC/PR           [Given]

∠ CAB = ∠ RPQ            [Proved]

So, Δ ABC ~ Δ PQR      [SAS similarity]

Question 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.

Let CD be the pole and AB is tower.

Therefore, DE and BE are the shadows of pole and tower.

In Δ ABE and Δ CDF,

∠ DFC = ∠ BAE            [Angle of sun at same place]

∠ CDF = ∠ ABE            [Each 90]

So, Δ ABE ~ Δ CDF      [AA similarity]

=> AB/CD = BE/QL

=> AB/6 = 28/4

=> AB/6 = 7

=> AB = 6 * 7 = 42

Hence, the height of tower is 42 m.

Question 16:

If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR,

Given, Δ ABC ~ Δ PQR

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

Therefore, AB/PQ = AC/PR = BC/QR  …………1

and ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R  …………2

AD and PM are medians of triangles. Therefore,

BD = BC/2 and QM = QR/2   ………3

From equation 1 and 3, we get

AB/PQ = BD/QM    ……….4

In Δ ABD and Δ PQM,

∠ B = ∠ Q                        [From equation 2]

AB/PQ = BD/QM            [From equation 4]

So, Δ ABD ~ Δ PQM       [SAS similarity]

=> AB/PQ = BD/QM = AD/PM

Exercise 6.4

Question 1:

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

Given, Δ ABC ~ Δ DEF, therefore

ar(ABC)/ar (DEF) = AB2 /DE2 = BC2 /EF2 = AC2 /DF2

Given, ar(ABC) = 64 cm2, ar(DEF) = 64 cm2, EF = 15.4 cm

=> ar(ABC)/ar (DEF) = BC2 /EF2

=> 64/121 = BC2 /(15.4)2

Take square root on both sides, we get

=> 8/11 = BC/15.4

=> BC = (15.4 * 8)/11

=> BC = 1.4 * 8

=> BC = 11.2 cm

Question 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.

Given, AB || DC

So, ∠ OAB = ∠ OCD and ∠ OBA = ∠ ODC          [Alternate angles]

In Δ AOB and Δ COD,

∠ AOB = ∠ COD                        [Vertically opposite angles]

∠ OAB = ∠ OCD                        [Alternate angles]

∠ OBA = ∠ ODC                        [Alternate angles]

So, Δ AOB ~ Δ COD                  [AAA similarity]

Therefore, ar(AOB)/ar (COD) = AB2 /CD2

=> ar(AOB)/ar (COD) = (2CD)2/CD2              [Since AB = 2CD]

=> ar(AOB)/ar (COD) = 4CD2/CD2

=> ar(AOB)/ar (COD) = 4/1

=> ar(AOB) : ar (COD) = 4 : 1

Question 3:

In Fig. 6.44, ABC and DBC are two triangles on the same base BC.  If AD intersects BC at O, show that ar(ABC)/ar (DBC) = AO/DO.

Let AP, DM are the perpendiculars on BC.

We know that,

Area of triangle = (1/2) * base * perpendicular

ar(ABC)/ar (DBC) = {(1/2) * BC * AP}/{(1/2) * BC * DM}  ……..1

In Δ APO and Δ DMO,

∠ APO = ∠ DMO                          [Each 90]

∠ AOP = ∠ DOM                          [Vertically opposite angles]

So, Δ APO ~ Δ DMO                    [AA similarity]

=> ar(ABC)/ar (DBC) = AO/DO  [From equation 1]

Question 4:

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

Let Δ ABC ~ Δ DEF, therefore,

ar(ABC)/ar (DEF) = AB2 /DE2 = BC2 /EF2 = AC2 /DF2   …….1

Given that ar(ABC) = ar (DEF)

=> ar(ABC)/ar (DEF) = 1

From equation 1, we have

AB2 /DE2 = BC2 /EF2 = AC2 /DF2 = 1

=> AB = DE, BC = EF and AC = DF

So, Δ ABC ≡ Δ DEF     [SSS congruency theorem]

Question 5:

D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC.

D and E are the mid-points of sides AB and CA respectively.

Therefore, DE || AC and DE = AC/2

In Δ BED and Δ BCA,

∠ BED = ∠ BCA                          [Corresponding angles]

∠ BDE = ∠ BAC                          [Corresponding angles]

So, Δ BED ~ Δ BCA                    [AA similarity]

ar(Δ BED)/ar (Δ BCA) = DE2 /AC2

=> ar(Δ BED)/ar (Δ BCA) = (AC/2)2 /AC2

=> ar(Δ BED)/ar (Δ BCA) = (1/4)/1

=> ar(Δ BED)/ar (Δ BCA) = 1/4

=> ar(Δ BED) =ar (Δ BCA)/4

Let ar(Δ BCA) = x

So, ar(Δ BED) = x/4

Similarly, ar(Δ CEF) = x/4 and ar(Δ ADF) = x/4

Now, ar(Δ DEF) = ar(Δ ABC) - ar(Δ BED) - ar(Δ CEF) - ar(Δ ADF)

=> ar(Δ DEF) = x – x/4 – x/4 – x/4

=> ar(Δ DEF) = x – 3x/4

=> ar(Δ DEF) = x/4

=> ar(Δ DEF) = ar(Δ ABC)/4               [Since ar(Δ ABC) = x]

=> ar(Δ DEF)/ar(Δ ABC) = 1/4

=> ar(Δ DEF) : ar(Δ ABC) = 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.

Let Δ ABC ~ Δ PQR.

Let AD and PS are the medians of triangle.

Given, Δ ABC ~ Δ PQR

=> AB/PQ = BC/QR = AC/PR       …………1

and ∠ A =∠ P, ∠ B = ∠ Q, ∠ C = ∠ R  …..2

AD and PS are the medians of the triangle.

Therefore, BD = DC = BC/2 and QS = SR = QR/2

From the equation 1, we have

AB/PQ = BD/QS = AC/PR       …………1

In Δ ABD and Δ PQS,

∠ B =∠ Q                  [From equation 2]

AB/PQ = BD/QS      [From equation 3]

So, Δ ABD ~ Δ PQS

Therefore, AB/PQ = BD/QS = AD/PS       …………4

and ar(ΔABC)/ar (ΔPQR) = AB2 /PQ2 = BC2 /QR2 = AC2 /PR2

From equation 1 and 4, we get

AB/PQ = BC/QR = AC/PR = AD/PS

Hence, ar(Δ ABC)/ar (Δ PQR) = AD2 /PS2

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.

Let ABCD is a square of side a unit.

Therefore, diagonal = a√2 units

The triangles form on side and diagonal are Δ ABE and Δ DBF respectively.

The length of each side of triangle ABE = a units

and the length of each side of triangle DBF = a√2 units

Both the triangles are equilateral and each angle of both the triangles are 60.

Therefore, by AAA similarity,

Δ ABE ~ Δ DBF

Now using the area theorem, we get

ar(Δ ABE)/ar (Δ DBF) = (a/a√2)2

= a2/2a2

= 1/2

=> ar(Δ ABE)/ar (Δ DBF) = 1/2

Tick the correct answer and justify:

Question 8:

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is

(A) 2 : 1                        (B) 1 : 2                         (C) 4 : 1                              (D) 1 : 4

Both the triangles are equilateral and each angle of both the triangles are 60.

Therefore, by AAA similarity,

Δ BCA ~ Δ BDE

Therefore, the side of Δ BDE = x/2

Now, ar(Δ ABC)/ar (Δ BDE) = (x/x/2)2

= x2/(x2/4)

= 4/1

=> ar(Δ ABE) : ar (Δ BDE) = 4 : 1

Hence, option (C) is the correct answer.

Question 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

We know that the ration of similar triangles is equal to the ratio of square of their

corresponding sides.

Therefore, the ratio of areas of two triangles = (4/9)2 = 16/81 = 16 : 81

Hence, the option (D) is the correct answer.

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

(i) Sides of triangle are: 7 cm, 24 cm, 25 cm

Square these sides, we get 49, 576 and 625

Now, 49 + 576 = 625

=> 72 + 242 = 252

These sides satisfy the Pythagoras theorem; hence these are sides of right angled triangle.

Since hypotenuse is the longest side in right angle triangle. So, its length is 25 cm.

(ii) Sides of triangle are: 3 cm, 8 cm, 6 cm

Square these sides, we get 9, 64 and 36

Now, 9 + 36 ≠ 64

=> 32 + 62 ≠ 82

These sides do not satisfy the Pythagoras theorem; hence these are not sides of right angled

triangle.

(iii) Sides of triangle are: 50 cm, 80 cm, 100 cm

Square these sides, we get 2500, 6400 and 10000

Now, 2500 + 6400 ≠ 10000

=> 502 + 802 ≠ 1002

These sides do not satisfy the Pythagoras theorem; hence these are not sides of right angled

triangle.

(iv) Sides of triangle are: 13 cm, 12 cm, 5 cm

Square these sides, we get 169, 144 and 25

Now, 25 + 144 = 169

=> 52 + 122 = 132

These sides satisfy the Pythagoras theorem; hence these are sides of right angled triangle.

Since hypotenuse is the longest side in right angle triangle. So, its length 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   PM2 = QM . MR.

Let ∠ MPR = x

Now, in Δ MPR,

∠ MPQ = 90 - ∠ MPR = 90 – x

∠ MQP = 180 – 90 – (90 – x) = x

In Δ QMP and Δ PMR,

∠ MPQ = ∠ MRP

∠ PMQ = ∠ RMP

∠ MQP = ∠ MPR

So, Δ QMP ~ Δ PMR            [AAA similarity]

We know that the corresponding sides of similar triangle are proportional

Therefore, QM/PM = MP/MR

=> PM2 = MQ * MR

Question 3:

In Fig. 6.53, 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

(i) In Δ ADB and Δ CAB,

∠ DAB = ∠ ACB          [Each 90]

∠ ABD = ∠ CBA           [common]

So, Δ ADB ~ Δ CAB     [AA similarity]

=> AB/CB = BD/AB

=> AB2 = CB * BD

(ii) Let ∠ CAB = x

Now, in Δ CBA,

∠ CBA = 180 - 90 – x

=> ∠ CBA = 90 – x

∠ CAD = 90 – ∠ CBA

=> ∠ CAD = 90 – x

∠ CDA = 180 – 90 – (90 - x)

=> ∠ CDA = x

In Δ CBA and Δ CAD,

∠ CBA = ∠ CAD          [Proved]

∠ CAB = ∠ CDA           [Proved]

∠ ACB = ∠ DCA           [Each 90]

So, Δ CBA ~ Δ DCA     [AAA similarity]

=> AC/DC = BC/AC

=> AC2 = BC * DC

(iii) In Δ DCA and Δ DAB,

∠ DCA = ∠ DAB           [Each 90]

∠ CDA = ∠ ADB           [Common]

So, Δ DCA ~ Δ DAB     [AA similarity]

=> DC/DA = DA/DB

=> AD2 = BD * CD

Question 4:

ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Given that the triangle ABC is an isosceles triangle such that

AC = BC and ∠ C = 90

In ΔABC, by Pythagoras theorem,

AB2 = AC2 + BC2

=> AB2 = AC2 + AC2     [Since AC = BC]

=> AB2 = 2AC2

Question 5:

ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.

Given, AB2 = 2 AC2,

=> AB2 = AC2 + AC2

=> AB2 = AC2 + BC2      [Since AC = BC]

These sides satisfy the Pythagoras theorem.

Hence, the triangle ABC is a right angled triangle.

Question 6:

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Let ABC be any equilateral triangle with each side of length 2a.

Now, draw perpendicular AD from A to BC.

We know that the altitude in equilateral triangle bisects the opposite sides.

So, BD = DC = a

In Δ ADB, by Pythagoras theorem,

=> (2a)2 = AD2 + a2         [Since AB = 2a]

=> 4a2 = AD2 + a2

=> AD2 = 4a2 - a2

Hence, the length of each altitude is a√3 unit.

Question 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.

In Δ AOB, by Pythagoras theorem,

AB2 = AO2 + OB2  …………1

In Δ BOC, by Pythagoras theorem,

BC2 = BO2 + OC2  …………2

In Δ COD, by Pythagoras theorem,

CD2 = CO2 + OD2  …………3

In Δ AOD, by Pythagoras theorem,

AD2 = AO2 + OD2  …………4

Add equation 1, 2, 3 and 4, we get

AB2 + BC2 + CD2 + AD2 = AO2 + OB2 + BO2 + OC2 + CO2 + OD2 + AO2 + OD2

= 2[AO2 + OB2 + BO2 + OC2]

= 2[2AO2 + 2 OB2]                 [Since OA = OC, OB = OD]

= 4[AO2 + OB2]

= 4[(AC/2)2 + (BD/2)2]          [Since OA = AC/2, OB = BD/2]

= 4[AC2/4 + BD2/4]

= AC2 + BD2

Question 8:

In Fig. 6.54, 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.

Given, O is a point in the interior of a triangle ABC where OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC.

In Δ AOF, by Pythagoras theorem,

OA2 = OF2 + AF2  …………1

In Δ BOD, by Pythagoras theorem,

OB2 = OD2 + BD2  …………2

In Δ COE, by Pythagoras theorem,

OC2 = OE2 + EC2  …………3

Add equation 1 , 2 and 3, we get

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2

=> OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF2 + BD2 + EC2  ………4

(ii) From equation 4, we get

AF2 + BD2 + EC2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2

=> AF2 + BD2 + EC2 = (OA2 - OE2) + (OC2 - OD2)- (OB2 - OF2)

=> AF2 + BD2 + EC2 = AE2 + CD2 + BF2

Question 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.

Let OA is wall and AB is ladder in the given figure.

In Δ AOB, by Pythagoras theorem,

AB2 = OA2 + OB2

=> 102 = 82 + OB2

=> 102 = 64 + OB2

=> OB2 = 100 - 64

=> OB2 = 36

=> OB = 6

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

Question 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?

Let OB is vertical pole in the given figure.

In Δ AOB, by Pythagoras theorem,

AB2 = OA2 + OB2

=> 242 = OA2 + 182

=> 576 = OA2 + 324

=> OA2 = 576 - 324

=> OA2 = 252

=> OA = √252

=> OA = 6√7

Hence, the distance of the stake from the base of the pole is 6√7 m.

Question 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?

Distance travelled by first aeroplane(due north) in 1½ hours = 1000 * 3/2

= 3000/2

= 1500 km

Distance travelled by second aeroplane(due north) in 1½ hours = 1200 * 3/2

= 3600/2

= 1800 km

Now, OA and OB are the distance travelled.

By Pythagoras theorem, the distance between two lanes

AB = √{OA2 + OB2}

=> AB = √{15002 + 18002}

=> AB = √{2250000 + 3240000}

=> AB = √5490000

=> AB = 300√61 km

Hence, in 1½ hours, the distance between two planes is 300√61 km.

Question 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.

Let AB and CD are the two poles with height 6 m and 11 m respectively.

Therefore, CP = 11 – 6 = 5 m and AP = 12 m

In Δ APC, by Pythagoras theorem,

AC2 = AP2 + AC2

=> AC2 = 122 + 52

=> AC2 = 144 + 25

=> AC2 = 169

=> OA = √169

=> AC = 13

Hence, the distance between the tops of two poles is 13 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 AE2 + BD2 = AB2 + DE2.

In Δ ACE, by Pythagoras theorem,

AC2 + CE2 = AE2   …………1

In Δ BCD, by Pythagoras theorem,

BC2 + CD2 = DB2   …………2

Add equation 1and 2, we get

AC2 + CE2 + BC2 + CD2 = AE2 + DB2 ………3

In Δ CDE, by Pythagoras theorem,

DE2 = CD2 + CE2   …………4

In Δ ABC, by Pythagoras theorem,

AB2 = AC2 + CB2   …………5

From equation 3, 4 and 5, we get

DE2 + AB2 = AE2 + DB2

Question 14:

The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD

(see Fig. 6.55). Prove that 2 AB2 = 2 AC2 + BC2.

In ΔACD, by Pythagoras theorem,

=> AC2 - CD2 = AD2   …………..1

In ΔABD, by Pythagoras theorem,

=> AB2 - BD2 = AD2   …………..2

From equation 1 and 2, we get

AC2 - CD2 = AB2 - BD2  …………3

Given, 3DC = DB, therefore

DC = BC/4 and BD = 3BC/4   …….4

From equation 3 and 4, we get

=> AC2 – (BC/4)2 = AB2 – (3BC/4)2

=> AC2 – BC2/16 = AB2 – 9BC2/16

=> AC2 = AB2 – 9BC2/16 – BC2/16

=> AC2 = AB2 – 8BC2/16

=> AC2 = AB2 – BC2/2

=> 2AC2 = 2AB2 – BC2

=> 2AB2 = 2AC2 + BC2

Question 15:

In an equilateral triangle ABC, D is a point on side BC such that BD = BC/3.   Prove that 9 AD2 = 7 AB2.

From the figure, AD = a√3/2 and DE = a/6

=> AD2 = 3a2/4 + a2/36

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.

Let a is the length of the side of equilateral triangle and AE is the altitude.

So BE = EC = BC/2 = a/2

Now in triangle ABC,

From Pythagoras Theorem

AB2 = AE2 + BE2

=> a2 = AE2 + (a/2) 2

=> AE2 = a2 - a2/4

=> AE2 = 3a2/4

=> 4AE2 = 3a2

=> 4*(square of altitude) = 3*(square of one side)

So, three times the square of one side is equal to four times the square of one of its altitudes.

Question 17:

Tick the correct answer and justify: In Δ ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:

(A) 120°                          (B) 60°                               (C) 90°                             (D) 45°

Given, AB = 6√3 cm, AC = 12 cm and BC = 6 cm

Therefore, AB2 = (6√3) 2 = 108

AC = 122 = 144

BC = 62 = 36

Now, AB2 + BC2 = 108 + 36

= 144  = AC2

The sides are satisfying the Pythagoras triplet in ΔABC.

Hence, these are the sides of a right angle triangle.

So, ∠ B = 90°      Hence, option (C) is the correct answer.

Exercise 6.6 (Optional)*

Question 1:

In Fig. 6.56, PS is the bisector of ∠ QPR of Δ PQR. Prove that QS/SR = PQ/PR.

Line RT is drawn parallel to SP which intersects QP produced at T.

Given that SP bisects angle QPR, therefore

∠ QPS = ∠ SPR    ……….1

By construction,

∠ SPR = ∠ PRT (As PS || TR)    ……….2

∠ QPS = ∠ QTR (As PS || TR)   ……….3

From the above equations, we have

∠ PRT = ∠ QTR

So, PT = PR

By construction, PS || TR

In ΔQTR, by Thales theorem,

QS/SR = QP/PT

=> QS/SR = PQ/QR                 [Since PT = TR]

Question 2:

In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:

(i) DM2 = DN . MC                             (ii) DN2 = DM . AN

(i) Join B and D.

Given that DN || CB, DM || AB and ∠ B = 90.

So, DMBN is a rectangle.

Given that BD Ʇ AC,

So, ∠ CDB = 90

=> ∠ 2 + ∠ 3 = 90  …………..1

In ΔCDM,

∠ 1 + ∠ 2 + ∠ DMC = 180

=> ∠ 1 + ∠ 2 = 180 - ∠ DMC

=> ∠ 1 + ∠ 2 = 180 – 90

=> ∠ 1 + ∠ 2 = 90  …….2

In ΔDMB,

∠ 2 + ∠ DMB + ∠ 4 = 180

=> ∠ 2 + ∠ 4 = 180 - ∠ DMB

=> ∠ 2 + ∠ 4 = 180 – 90

=> ∠ 2 + ∠ 4 = 90  …….3

From equation 1 and 2, we get

∠ 1 = ∠ 3

From equation 1 and 3, we get

∠ 2 = ∠ 4

In ΔDCM and ΔBDM,

∠ 1 = ∠ 3                        [Proved]

∠ 2 = ∠ 4                        [Proved]

So, Δ DCM ~ Δ BDM     [AAA similarity]

=> BM/DM = DM/MC

=> DN/DM = DM/MC      [Since BM = DN]

=> DM2 = DN * MC

(ii) In ΔDBN,

∠ 5 + ∠ 7 = 90  …………4

In ΔDAN,

∠ 6 + ∠ 8 = 90  …………5

But, BD Ʇ AC

=> ∠ 5 + ∠ 6 = 90  …………6

From equation 4 and 6, we have ∠ 6 = ∠ 7

From equation 4 and 6, we have ∠ 8 = ∠ 5

In ΔDNA and ΔBND,

∠ 6 = ∠ 7                        [Proved]

∠ 8 = ∠ 5                        [Proved]

So, Δ DNA ~ Δ BND      [AA similarity]

=> AN/DN = DN/NB

=> DN2 = AN * NB

=> DN2 = AN * DM              [Since NB = DM]

Question 3:

In Fig. 6.58, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that

AC2 = AB2 + BC2 + 2 BC . BD.

AB2 = AD2 + DB2  ………..1

In ΔACD, By Pythagoras theorem,

=> AC2 = AD2 + (DB + BC) 2

=> AC2 = AD2 + DB2 + BC2 + 2 * DB * BC

=> AC2 = AB2 + BC2 + 2DB * BC                 [From equation 1]

Question 4:

In Fig. 6.59, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced.

Prove that              AC2 = AB2 + BC2 + 2 BC . BD

=> AD2 = AB2 – DB2   ………..1

In ΔACD, By Pythagoras theorem,

=> AC2 = AB2 – DB2 + DC 2          [From equation 1]

=> AC2 = AB2 – DB2 + (BC - BD) 2

=> AC2 = AB2 – DB2 + BC2 + BD2 - 2BC * BD

=> AC2 = AB2 + BC2 - 2BC * BD

Question 5:

In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) AC2 = AD2 + BC . DM + (BC/2)2

(ii) AB2 = AD2 - BC . DM + (BC/2)2

(ii) AC2 + AB2 = 2AD2 + BC2/2

(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 DC = BC/2, we obtain

AD2 + (BC/2)2 + 2MD.(BC/2) = AC2

AD2 + (BC/2)2 + MD.BC = AC2

(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

= AD2 + (BC/2)2 – 2(BC/2) * MD

= AD2 + (BC/2)2 – BC * 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

=> 2(AM2 + MD2) + (BC/2)2 + (BC/2)2 + 2MD (− BC/2 + BC/2) = AB2 + AC2

=> 2AD2 + BC2/2 = AB2 + AC2

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.

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

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)

∠DEA = ∠AFD (Both 90°)

∴ 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

Question7:

In Fig. 6.61, two chords AB and CD intersect each other at the point P.

Prove that:                      (i) Δ APC ~ Δ DPB                                                       (ii) AP . PB = CP . DP

Join CB as shown in the figure.

(i) In ΔAPC and ΔDPB,

∠ APC = ∠ DPB            [Vertically opposite angles]

∠ CAP = ∠ BDP            [angles in the same segment]

So, Δ APC ~ Δ DPB      [AAA similarity]

(ii) Since Δ APC ~ Δ DPB       [Proved]

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

So, AP/DP = PC/PB = CA/BD

=> AP/DP = PC/PB

=> AP * PB = PC * DP

Question 8:

In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) Δ PAC ~ Δ PDB

(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)

So, ΔPAC ∼ ΔPDB         [AA Similarity]

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

So, PA/PD = AC/DB = PC/PB

=> PA/PD = PC/PB

=> PA * PB = PC * PD

Question 9:

In Fig. 6.63, D is a point on side BC of Δ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

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

It is given that,

BD/CD = AB/AC

=> BD/CD = AP/AC

By using the converse of basic proportionality theorem, we obtain

=> ∠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

=> 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, how 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?

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

=> AC2 = (1.8)2 + (2.4)2

=> AC2 = 3.24 + 5.76

=> AC2 = 9.00

=> AC = √9

=> AC = 3

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 = 60/100 m = 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

=> (1.8) 2 + BD2 = (2.4) 2

=> BD2 = 5.76 − 3.24

=> BD2 = 2.52

=> BD = √2.52

=> BD = 1.587

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m