Chapter-6 Triangles

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)

Explanation:

(i) similar

(ii) similar

(iii) equilateral

(iv) equal, proportional.

2. Give two different examples of pair of:

(i) similar figures

(ii) non-similar figures

Explanation:

(a) Any two rectangles

(b) Any two squares

(ii) The following are two distinct examples of a pair of non-similar figures:

(a) An equilateral triangle and a scalene

(b) A right-angled triangle and an equilateral triangle.

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

Explanation:

If the corresponding angles and sides of two figures are congruent (equal), then the figures are said to be comparable. The sides of the quadrilaterals PQRS and ABCD in the given figures are proportional, but their corresponding angles are not equal.

Hence, there is no similarity between the aforementioned quadrilaterals.

4. In Figures (i) and (ii), DE BC. Find EC in (i) and AD in (ii).

Explanation:

(i) Since DE BC,

EC =

EC = 2 cm

(ii)Since DE BC,

AD =

EC = 2.4 cm

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

(i) PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm

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

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

Explanation:

(i) Given: PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm

Now, = 0.97 cm

And = 1.2 cm

As a result, EF does not divide the PQ and PR sides of the PQR in an equal ratio.

EF is not parallel to QR.

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

Now, = cm

And cm

As a result, EF divides the PQ and PR sides of the PQR in the same proportion.

EF is parallel to QR.

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

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And ER = PR – PF = 2.56 – 0.36 = 2.20 cm

Now, = cm

And = cm

As a result, EF divides the PQ and PR sides of the PQR in the same proportion.

EF is parallel to QR.

6. In the figure, if LM CB and LN CD, prove that

Explanation:

In ABC, LM CB

[Basic Proportionality theorem] ……….(i)

And in ACD, LN CD

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

7. In the figure, DE AC and DF AE. Prove that

Explanation:

 In BCA, DE AC

[Basic Proportionality theorem] ……….(i)

And in BEA, DF AE

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

8. In the figure, DE OQ and DF OR. Show that EF QR.

Explanation:

In PQO, DE OQ

[Basic Proportionality theorem] ……….(i)

And in POR, DF OR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

EF QR [By the converse of BPT]

9. In the figure, A, B, and C are points on OP, OQ, and OR respectively such that AB PQ and AC PR. Show that BC QR.

Explanation:

Given: O is any point in PQR, in which AB PQ and AC PR.

To prove: BC QR

Construction: Join BC.

Proof: In OPQ, AB PQ

[Basic Proportionality theorem] ……….(i)

And in OPR, AC PR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

In OQR, B, and C are the locations that divide the sides OQ and OR in the same ratio.

By the converse of the Basic Proportionality theorem,

BC QR.

10. Using the 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).

Explanation:

Given: Given: A triangle ABC in which side AB's midpoint is represented by D.

Moreover, a line connecting AC and DE is drawn parallel to BC at E

To prove: AE = EC

Proof: Since DE BC

[Basic Proportionality theorem] ……….(i)

But AD = DB [Given]

[From eq. (i)]

AE = EC

As a result, E is where the third side AC splits in half.

11. Using the Converse of the basic proportionality theorem, 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).

Explanation:

A triangle ABC where the midpoints of D and E are AB and AC are the respective sides.

To Prove: DE BC

Proof: Since D and E are the mid-points of AB and AC

respectively.

AD = DB and AE = EC

Now, AD = DB

and AE = EC

= 1

As a result, in the triangle, ABC, the points D and E divide the sides AB and AC in the same proportion.

As a result, using the Basic Proportionality theorem opposite, we have

DE BC

12. ABCD is a trapezium in which AB DC and its diagonals intersect each other at point O. Show that 

Explanation:

An ABCD trapezium with AB DC and its diagonals is given.

At O, AC and BD come together.

To Prove:

Construction: Through O, draw OE AB, i.e. OE DC.

Proof: In ADC, we have OE DC

[By Basic Proportionality theorem]……….(i)

Again, in ABD, we have OE AB[Construction]

[By Basic Proportionality theorem]

……….(ii)

From eq. (i) and (ii), we get

 

13. The diagonals of a quadrilateral ABCD intersect each other at point O such that Such that ABCD is a trapezium.

Explanation:
Given: An ABCD quadrilateral with diagonals AC and BD cross at O in such a way that, i.e.

.

To Prove: Quadrilateral ABCD is a trapezium.

Construction: Through O, draw OE AB meeting AD at E.

Proof: In ADB, we have OE AB [By construction]

[By Basic Proportionality theorem]

=

=

As a result, in ADC, E and O are the points that divide AD and AC in the same proportion. As a result, using the Basic Proportionality theorem's opposite, we have

EO DC

But EO AB[By construction]

AB DC

Quadrilateral ABCD is a trapezium.

14. State which pairs of triangles in the figure, is similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form: 

Explanation:

(i) In s ABC and PQR, we observe that,

By the AAA criterion of similarity,

(ii) In s ABC and PQR, we observe that,

By the SSS criterion of similarity,

(iii) We note that the ratio of the sides of these triangles in s LMP and DEF is not equal.

Hence, there is no similarity between these two triangles.

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

But,

These two triangles are not comparable since they do not meet the SAS similarity requirement.

(v) In s ABC and FDE, we have,

But, [ AC is not given]

These two triangles are not comparable since they do not meet the SAS similarity requirement.

(vi) In s DEF and PQR, we have,

[ ]

And

By the AAA criterion of similarity, 

15. In figure 6.35, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO, and ∠ OAB.

Explanation:

Since BD is a line and OC is a ray on it.

In CDO, we have

It is given that ODC OBA

HenceDOC = DCO = and OAB = 

16. Diagonals AC and BD of a trapezium ABCD with AB CD intersect each other at the point O. Using a similarity criterion for two triangles, show that 

Explanation:

Given: ABCD is a trapezium in which AB DC.

To Prove:

Proof: In s OAB and OCD, we have,

5 = 6 [Vertically opposite angles]

1 = 2 [Alternate angles]

And 3 = 4 [Alternate angles]

By AAA criterion of similarity, OAB ODC

Hence, 

17. In the figure, 1 = 2. Show that PQS TQR.

Explanation:

We have,

……….(1)

Also, 1 = 2 [Given]

PR = PQ ……….(2) [Sides opposite to equal s are equal]

From eq.(1) and (2), we get

In s PQS and TQR, we have,

and PQS = TQR = Q

By the SAS criterion of similarity, PQS TQR

18. S and T are points on sides PR and QR of a PQR such that P = RTS. Show that RPQ RTS.

Explanation:

 In s RPQ and RTS, we have

RPQ = RTS [Given]

PRQ = TRS [Common]

By AA-criterion of similarity,

RPQ RTS

19. In the figure, if ABE ACD, show that ADE ABC.

Explanation:

It has been given that ABE ACD

AB = AC and AE = AD

……….(1)

In s ADE and ABC, we have,

[from eq.(1)]

And BAC = DAE [Common]

In light of the SAS similarity criterion, ADE ABC.

20. In the figure, altitude AD and CE of an ABC intersect each other at point P. Show that:

(i) AEP CDP

(ii)ABD CBE

(iii) AEP ADB

(iv) PDC BEC

Explanation:

(i) In s AEP and CDP, we have,

AEP = CDP = [ CEAB, ADBC]

AndAPE = CPD[ Vertically opposite]

As per the AA criterion of similarity, AEP CDP

(ii) In s ABD and CBE, we have,

ADB = CEB =

AndABD = CBE[ Common]

As per the AA criterion of similarity, ABD CBE

(iii) In s AEP and ADB, we have,

AEP = ADB = [ ADBC, CEAB]

AndPAE = DAB[ Common]

As per  AA-criterion of similarity, AEP ADB

(iv) In s PDC and BEC, we have,

PDC = BEC = [ CEAB, ADBC]

AndPCD = BEC[ Common]

As per the AA criterion of similarity, PDC BEC

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

Explanation:

In s ABE and CFB, we have,

AEB = CBF[Alt. s]

A = C [opp. s of a gm]

As per AA-criterion of similarity, we have

ABE CFB

22. In the figure, ABC and AMP are two right triangles, right angles at B and M respectively. Prove that:

(i) ABC AMP

(ii) 

Explanation:

(i) In s ABC and AMP, we have,

ABC = AMP = [Given]

BAC = MAP [Common angles]

As per AA-criterion of similarity, we have

ABC AMP

(ii) We have ABC AMP [As proven above]

 

23. 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) CD/GH = AC/FG

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

Explanation:

We have, ABC FEG

A = F………(1)

And C = G

C = G

1 = 3 and 2 = 4 ……….(2)

[CD and GH are bisectors of C and G respectively]

In s DCA and HGF, we have

A = F[From eq.(1)]

2 = 4[From eq.(2)]

As per AA-criterion of similarity, we have

DCA HGF

That supports the (iii) part

We have, DCA HGF

That supports the (i) part

In s DCA and HGF, we have

1 = 3[From eq.(2)]

B = E[ DCB HGE]

That supports the (ii) part

24. In the figure, E is a point on side CB produced by an isosceles triangle ABC with AB = AC. If AD BC and EF AC, prove that ABD ECF.

Explanation:

Here ABC is isosceles with AB = AC

B = C

In s ABD and ECF, we have

ABD = ECF[ B = C]

ABD = ECF = [ ADBC and EFAC]

As per  AA-criterion of similarity, we have

ABD ECF.

25. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of a PQR (see figure). Show that ABC PQR.

Explanation:

Given: The median of ABC is AD, whereas the median of PQR is PM.

To prove: ABC PQR

Proof: BD = BC [Given]

And QM = QR [Given]

Also [Given]

ABD PQM[As per SSS-criterion of similarity]

B = Q[Angles in similar triangles are equal to one another.]

And [Given]

As per SAS-criterion of similarity, we have

ABC PQR

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

Explanation:

In triangles ABC and DAC,

ADC = BAC [Given]

and C = C[Common]

As per the AA-similarity criterion,

ABC DAC

= CB.CD.

27. 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 ABC PQR.

Explanation:

Given: AD is the median of ABC and PM is the median of PQR such that

To prove: ABC PQR

Proof: BD = BC [Given]

And QM = QR and [Given]

ABD PQM[As per SSS-criterion of similarity]

B = Q[Angles in similar triangles are equal to one another.]

And [Given]

By SAS-criterion of similarity, we have

ABC PQR

28. 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.

Explanation:

Let AC be the shadow of AB and AB be the vertical pole. Let DF be the shadow of the vertical tower and DE be it. a member of EF and BC.

Let DE = meters

Here, AB = 6 m, AC = 4 m and DF = 28 m

In the triangles ABC and DEF,

A = D =

And C = F[Each is the angular elevation of the sun]

As per the AA-similarity criterion,

ABC DEF

= 42 m

Hence, the tower is 42 metres tall.

29. If AD and PM are medians of triangles ABC and PQR respectively, where ABC PQR, proves that 

Explanation:

Given: The medians of triangles are AD and PM.

While in ABC and PQR, respectively

ABC PQR

To prove:

Proof: In triangles ABD and PQM,

B = Q [Given]

And [ BC and QR are having  AD and PM as medians respectively]

As per SAS-criterion of similarity,

ABD PQM

=

= 

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

Explanation:

We consider,

BC = cm = 11.2 cm.

31. Diagonals of a trapezium ABCD with AB DC intersect each other at point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Explanation:

In s AOB and COD, we have,

AOB = COD[Vertically opposite angles]

OAB = OCD[Alternate angles]

As per AA-criterion of similarity,

AOBCOD

Therefore, Area (AOB): Area (COD) = 4:1

32. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Explanation:

Given: Two s ABC and DBC which stand on the same base but on opposite sides of BC.

To Prove:

Construction: Draw AEBC and DFBC.

Proof: In s AOE and DOF, we have, AEO = DFO =

and AOE = DOF[They are Vertically opposite)

AOEDOF[As per AA-criterion]

……….(i)

Now,

[using eq. (i)]

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

Explanation:

Given: Two s ABC and DEF such that ABCDEF

And Area(ABC) = Area (DEF)

To Prove: ABCDEF

Proof: ABCDEF

A = D, B = E, C = F

And

To establish ABCDEF, it is enough to prove that, AB = DE, BC = EF and AC = DF

Now, Area(ABC) = Area (DEF)

= 1

= 1

AB = DE, BC = EF, AC = DF

Hence, ABCDEF

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

Explanation:

Thus  D and E are, respectively, the midpoints of sides BC and CA of ABC.

DEBA DEFA ……….(i)

thus, D and F are respectively the mid-points of the sides BC and AB of ABC respectively.

DFCA DEAE ……….(ii)

From (i) and (ii), we can assure that AFDE is a parallelogram.

Therefore, BDEF is a parallelogram.

Now, in s DEF and ABC, we have

FDE = A[opposite angles of gm AFDE]

And DEF = B[opposite angles of gm BDEF]

As per the AA criterion of similarity, we have DEF ABC

[ DE = AB]

Hence, Area (DEF): Area (ABC) = 1:4

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

Explanation:

Given: ABCPQR, AD and PM are respectively the medians of s ABC and PQR.

To Prove:

Proof: Since ABCPQR

……….(1)

But, ……….(2)

From eq. (1) and (2), we have,

36. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of the diagonals.

Explanation:

Given: A square ABCD,

Equilateral s BCE and ACF have been drawn respectively on side BC and the diagonal AC.

To Prove: Area (BCE) = Area (ACF)

Proof: BCEACF

[As equilateral so similar as per AAA criterion of similarity]

[ Diagonal = side AC = BC]

 

37. Tick the correct answer and justify:

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

(A) 2: 1

(B) 1: 2

(C) 4: 1

(D) 1: 4

Explanation:

(C) AsABC and BDE were equilateral, they are equiangular and therefore,

ABCBDE

[BC has mid point as D]


(C) is the correct answer.

38. 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

Explanation:

(D) So the ratio of the squares of any two related sides is equal to the ratio of the areas of two similar triangles. Therefore,

The ratio of areas =

(D) is the correct answer.

39. Sides of triangles are given below. Determine which of them are right triangles. In the 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

Explanation:

(i) Let us take  = 7 cm, = 24 cm and = 25 cm

In this, the larger side is = 25 cm.

We consider,

Therefore, the triangle as per the given sides is a right triangle. Its hypotenuse is 25 cm

(ii) Let = 3 cm, = 8 cm and = 6 cm

In this larger side is = 8 cm.

We consider,

So, the triangle as per the given sides is not a right triangle.

(iii) Let = 50 cm, = 80 cm and = 100 cm

In this larger side is= 100 cm.

We consider,

So, the triangle as per the given sides is not a right triangle.

(iv) Let = 13 cm, = 12 cm and = 5 cm

In this, the larger side is = 13 cm.

We consider,

So, the triangle as per the given sides is a right triangle. Its hypotenuse = 13 cm

40. PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that PM2 = QM x MR.

Explanation:

Given: PQR is a triangle with right angles at P and PMQR

To Prove: PM2 = QM.MR

Proof: Since PMQR
 QMPPMR
 

41. In the figure, ABD is a triangle right-angled at A and ACBD. Show that:

(i) AB2 = BC.BD

(ii) AC2 = BC.DC

(iii) AD2 = BD.CD

Explanation:

Given that: ABD is a triangle right angled at A and ACBD.

To Prove (i) AB2 = BC.BD, (ii) (iii)

Proof:(i) Since ACBD

ABCADC and each triangle is similar to ABD

ABCABD

(ii)Since ABCADC

(iii) Since ACDABD

 

42. ABC is an isosceles triangle right angled at C. Prove that.

Explanation:

 As ABC is an isosceles right triangle, right angled at C.

[BC = AC, given]

 

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

Explanation:

As ABC is an isosceles right triangle with AC = BC and

[BC = AC, given]

ABC is right-angled at C.

44. ABC is an equilateral triangle of sides Find each of its altitudes.

Explanation:

Let us consider ABC to be an equilateral triangle of side units.

Draw ADBC. Then, BC has a midpoint as  D.

BD = BC =

Therefore, ABD is a right triangle, the right triangle at D.

= AD2 +

=

Each of its altitude = 

45. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of squares of its diagonals.

Explanation:

Let O be the point where the rhombus ABCD's diagonals AC and BD meet. Considering that a rhombus' diagonals cut each other at an angle.

and AO = CO, BO = OD

Therefore AOB is a right triangle, right-angled at O.

=

[OA = OC and OB = OD]

……….(1)

In the same way, we have ……….(2)

……….(3)

……….(4)

Combining all these results, we get

=

= 

46. In the figure, O is a point in the interior of a triangle ABC, OD BC,

OE AC and OF AB. Show that:

(i) =

(ii) = 

Explanation:

Combine AO, BO and CO.

(i) In right s OFA, ODB and OEC, we have

, and

Combining all these, we get

=

=

(ii) In right s ODB and ODC, we have

and

……….(1)

In the same way, we have ……….(2)

and = ……….(3)

Combining equations (1), (2) and (3), we get

=

=

–= 0

= 

47. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Explanation:

Assume that AB represents the ladder, B the window, and CB the wall. ABC then is a

A right triangle with C at the right angle.

= 100 – 64

= 36

AC = 6

As a result, the ladder's foot is 6 metres away from the wall's foundation.

48. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other hand. How far from the base of the pole should the stake be driven so that the wire will be taut?

Explanation:

The guy wire AB (= 24m) should be fastened to a vertical pole. BC of 18 m height. Let the wire be secured to a stake at A to keep it taut. ABC is then a right triangle with a right angle at C.

= 576 – 324

= 252

AC = 6

Hence, the stake may be positioned 6 metres from the pole's base.

49. An airplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another airplane 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 hours?

Explanation:

Let the first aircraft take off from O and fly up to A in the direction of the north, where OA = km = 1500 km

Allow the second aircraft to take off from O simultaneously and travel up to 1500 km 

B in the direction of the west, where

OB = km = 1800 km

The required distance, as stated in the question, is BA.

In the right-angled triangle ABC, using Pythagoras' theorem, we obtain,

=

= 2250000 + 3240000

= 5490000 =

AB = km

50. 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.

Explanation:

Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m 

Draw CEAB and join AC.

CE = DB = 12 m

AE = AB – BE = AB – CD = (11 – 6)m = 5 m

By using Pythagoras' Theorem to right-angled triangle ACE, we have

=

= 144 + 25 = 169

AC = 13 m

As a result, there are 13 metres between the tops of the two poles.

51. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that =.

Explanation:

In right-angled s ACE and DCB, we have

and

=

=

=

[By Pythagoras theorem, and

52. The perpendicular from A on side BC of an ABC intersects BC at D such that DB = 3CD (see figure). Prove that.

Explanation:

We are given, DB = 3CD

consider,, BC = DB + CD

BC = 3CD + CD

BC = 4CD

CD = BC and DB = 3CD = BC ……….(1)

Therefore, ∆ ABD is a right triangle, right-angled at D. According to Pythagoras' theorem, we have,

……….(2)

In the same way, from ACD, we have, ……….(3)

From eq. (2) and (3)

= [Using eq.(1)]

=

=

 

53. In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that.

Explanation:

Let us consider ABC to be an equilateral triangle and let D be a point on BC such that BD = BC

Draw AEBC, Join AD.

In s AEB and AEC, we have,

AB = AC[ABC is equilateral]

AEB = AEC [ each ]

And AE = AE

As per SAS-criterion of similarity, we have

AEB AEC

BE = EC

Therefore, we have, BD = BC, DC = BC and BE = EC = BC ………(1)

Since, C =

ADC is an acute angle triangle.

= [using eq.(1)]

= [ AB = BC = AC]

=

 

54. 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.

Explanation:

Let us take ABC to be an equilateral triangle and let ADBC. In s ADB and ADC, we have,

AB = AC [Given]

B = C = [Given]

And ADB = ADC[Each = ]

ADB ADC[As per RHS criterion of congruence]

BD = DC

BD = DC = BC

AS ADB is a right triangle, right-angled at D, According to Pythagoras' theorem, we have,

[ BC = AB]

 

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

(a) 120°

(b) 60°

(c) 90°

(d) 45o

Explanation:

(C) In ABC, we have, AB = cm, AC = 12 cm, and BC = 6 cm.

consider,, = = 144 =

Hence, ABC is a right triangle, right angled at B.

B =

56. In the figure, PS is the bisector of QPR of PQR. Prove that

Explanation:

Given: PQR is a triangle, and PS is the internal bisector formed by QPR and QR's intersection at S.

QPS = SPR

To prove:

Construction: Draw RT SP to cut QP produced at T.

Proof: Since PS TR and PR cuts them, hence,

SPR = PRT ……….(i) [Alternate s]

And QPS = PTR ……….(ii)[Corresponding s]

But QPS = SPR [Given]

PRT = PTR[From eq. (i) & (ii)]

PT = PR……….(iii)

[Angles opposite to equal sides have equal sides.]

Now, in QRT,

RT SP[By construction]

[Thales theorem]

[From eq. (iii)]

57. In the figure, D is a point on hypotenuse AC of ABC, BD AC, DM BC, and DN AB. Prove that:

(i) = DN.MC

(ii) = DM.AN

Explanation:

Therefore, AB BC and DM BC

AB DM

In the same way, BC AB and DN AB

CB DN

quadrilateral BMDN is a rectangle.

BM = ND

(i) In BMD, 1 + BMD + 2 =

1 + + 2 =

1 + 2 =

Similarly, in DMC,3 + 4 =

Since BD AC,

2 + 3 =

Now, 1 + 2 = and 2 + 3 =

1 + 2 = 2 + 3

1 = 3

Also, 3 + 4 = and 2 + 3 =

3 + 4 = 2 + 3

4 = 2

Thus, in BMD and DMC,

1 = 3 and 4 = 2

BMD DMC

[BM = ND]

= DN.MC

(ii) Processing as in (i), we can prove that

BND DNA

[BN = DM]

= DM.AN

58. In the figure, ABC is a triangle in which ABC > and AD CB are produced. Prove that:

.BD

Explanation:

Given: ABC is considered as a triangle in which ABC > and AD CB are produced.

To prove:.BD

Proof: Therefore ADB is a right triangle, right-angled at D, therefore, by Pythagoras' theorem,

………(i)

We take again, ADB is a right triangle, right-angled at D, therefore, by Pythagoras theorem,

.BC

.BC

.BC [Using eq. (i)]

59. In the figure, ABC is a triangle in which ABC < and AD BC are produced. Prove that:

.BD

Explanation:

Given: ABC be a triangle in which ABC < and AD BC are produced.

To prove:.BD

Proof: Therefore ADB is a right triangle, right-angled at D, therefore, by Pythagoras' theorem,

………(i)

We take again, ADB is a right triangle, right-angled at D, therefore, by Pythagoras theorem,

– 2BC.BD

– 2DB.BC

– 2DB.BC

[Using eq. (i)]

60. In the figure, AD is a median of a triangle ABC and AM BC. Prove that:

(i)

(ii)

(iii) BC

Explanation:

ThereforeAMD = therefore ADM < and ADC >

Since ADC is an acute angle and ADC is an obtuse angle.

(i) In ADC, ADC is an obtuse angle.

+ 2..DM

+ BC.DM

……….(i)

(ii) In ABD, ADM is an acute angle.

– 2..DM

……….(ii)

(iii) From eq. (i) and eq. (ii),

61. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Explanation:

If AD is a median of ABC, then

[See Q.5 (iii)]

BO and DO are the medians of triangles ABC and ADC, respectively, because the diagonals of a parallelogram cut each other in half.

AB2 + BC2 = 2BO2 + AC2 ……….(i)

And AD2 + CD2 = 2DO2 + AC2 ……….(ii)

Adding eq. (i) and (ii),

AB2 + BC2 + AD2 + CD2 = 2 (BO2 + DO2) + AC2

AB2 + BC2 + AD2 + CD2 = + AC2

AB2 + BC2 + AD2 + CD2 = AC2 + BD2

62. In the figure, two chords AB and CD intersect each other at point P. Prove that:

(i) APC DPB

(ii) AP.PB = CP.DP

Explanation:

(i) In the triangles APC and DPB,

APC = DPB [Vertically opposite angles]

CAP = BDP [Angles within the same circle segment are equal.l]

As per  AA-criterion of similarity,

APC DPB

(ii) Since APC DPB

AP x PB = CP x DP

63. 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) ∆PAC ~ ∆PDB

(ii)PA X PB = PC X PD

Explanation:

(i) In the triangles PAC and PDB,

APC = DPB [Common]

CAP = BDP [BAC = PAC and PDB = CDB]

= BAC = = PAC]

As per AA-criterion of similarity,

APC DPB

(ii) Since APC DPB

PA.PB = PC.PD

64. In the figure, D is appointing on side BC of ABC such that Prove that AD is the bisector of BAC.

Explanation:

 Given: ABC is a triangle and D is a point on BC such that

To prove: AD is the internal bisector of BAC.

Construction: Produce BA to E such that AE = AC. Join CE.

Proof: In AEC, since AE = AC

AEC = ACE ……….(i)

[Angles on a triangle's opposite side of an equal side are equal.]

Now, [Given]

[ AE = AC, by construction]

By opposite of the Basic Proportionality Theorem,

DA CE

Therefore CA is a transversal,

BAD = AEC ……….(ii) [Corresponding s]

And DAC = ACE ……….(iii) [Alternate s]

Also AEC = ACE [From eq. (i)]

Hence, BAD = DAC [From eq. (ii) and (iii)]

Thus, AD bisects BAC internally.

65. 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 figure)? 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?

Explanation:

 I. To find: The length of AC.

By Pythagoras' theorem,

AC2 = (2.4)2 + (1.8)2

AC2 = 5.76 + 3.24 = 9.00

AC = 3 m

She has a 3 m long string out.

length of the string after 12 seconds of 5 cm/sec pulling


= (5 x 12) cm = 60 cm = 0.60 m

Remaining string left out = 3 – 0.6 = 2.4 m

II. To find: The length of PB

PB2 = PC2 – BC2

= (2.4)2 – (1.8)2

= 5.76 – 3.24 = 2.52

PB = = 1.59 (approx.)

Hence, the fly's horizontal distance from Nazima after 12 seconds.

= 1.59 + 1.2 = 2.79 m (approx.)

Chapter-6 Triangles