Important QuestionsClass 10 Mathematics Chapter 6 – Triangles

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You have already learnt about the basics of triangles in your lower classes. The chapter, Triangles is one of the most vital parts of the Class 10 Geometry syllabus. In this class, we would learn about triangles and their associated laws and theorems and their collinearity. The important topics covered in the chapter include Similar Figures, Similarity of Triangles and their associated theorems, Criteria for Similarity of Triangles and theorems related to it and Areas of Similar Triangles and the Pythagoras Theorem. Students can visit our Extramarks’ website and refer to Important Questions Class 10 Mathematics Chapter 6.

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Important Questions Class 10 Mathematics Chapter 6 – With Solutions

The following Important Questions and their solutions are included in the Class 10 Mathematics Chapter 2.

Question 1. In the figure, ∠BAC =90° and AD⊥BC. So,

(a) BD.CD = BZC² 

(b) AB.AC = BC² 

(c) BD.CD=AD² 

(d) AB.AC =AD²

Answer 1:

1. c) BD.CD=AD²

Explanation:

From the ∆ADB and ∆ADC,

As per the question, we get,

∠D = ∠D = 90° (∵ AD ⊥ BC)

∠DBA = ∠DAC [each angle = 90°- ∠C]

By using AAA similarity criteria,

∆ADB ∼ ∆ADC

BD/AD = AD/CD

BD.CD = AD2

Question 2. When the lengths for the diagonals of the rhombus are 16 cm and 12 cm, the length of the sides of the rhombus should be

(a) 9 cm 

(b) 10 cm 

(c) 8 cm 

(d) 20 cm

Answer 2:

(b) 10 cm

Explanation: Given,

A rhombus is a simple quadrilateral that has four sides that are of equal length, and diagonals that are perpendicular bisectors to each other.

From the question, we have,

From the question,

AC = 16 cm, BD = 12 cm

∠AOB = 90°

Since AC and BD bisects each other

AO = ½ AC and BO = ½ BD

Thus, we have,

AO = 8 cm, BO = 6 cm

In the right angled ∆AOB,

By using the Pythagoras theorem,

We get,

AB2 = AO2 + OB 2

AB2 = 82 + 62 = 64 + 36 = 100

So, AB = √100 = 10 cm

The four sides of the rhombus are equal.

Hence, 

one side for the rhombus = 10 cm.

Question 3. When ΔABC ~ ΔEDF as well as ΔABC is not similar to ΔDEF, which among the following is not true?

(a) BC · EF = AC · FD 

(b) AB · EF = AC · DE

(c) BC · DE = AB · EF 

(d) BC · DE = AB · FD

Answer 3:

(c) BC · DE = AB · EF

Explanation:

The explanation for the question above is as follows,

When the sides of one of the triangle are proportional to the side towards the other triangle, and the corresponding angles are also the same, the triangles are similar by SSS similarity.

Thus, ∆ABC ∼ ∆EDF

By using similarity property,

AB/ED = BC/DF = AC/EF

By taking AB/ED = BC/DF, we have

AB/ED = BC/DF

AB.DF = ED.BC

Therefore, option (d) BC · DE = AB · FD is true

By taking BC/DF = AC/EF, we have,

BC/DF = AC/EF

 BC.EF = AC.DF

Therefore, option (a) BC · EF = AC · FD is true

By taking AB/ED = AC/EF, we get,

AB/ED = AC/EF

AB.EF = ED.AC

Therefore, option (b) AB · EF = AC · DE is true.

Question 4. When in two Δ QPR, given that AB/QR = BC/PR = CA/PQ, thus

(a)Δ PQR~Δ CAB 

(b) Δ QPR ~ Δ ABC

(c)Δ CBA ~ Δ QPR 

(d) Δ BCA ~ Δ QPR

Answer 4:

(a)Δ PQR~Δ CAB

Explanation:

From the ∆ABC and ∆QPR, we get,

AB/QR = BC/PR = CA/PQ

When sides of the one triangle are proportional to the side of the other triangle, and their corresponding angles are also similar, then both the triangles show similarity by SSS similarity.

Hence, we get,

Δ PQR~Δ CAB

Question 5. In the given figure, two line segments AC and BD bisect each other at the point P so that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Thus, ∠PBA is equal to

(a) 50° (b) 30° (c) 60° (d) 100°

Answer 5: (d) 100°

Explanation:

From the ∆APB and ∆CPD,

∠APB = ∠CPD = 50° (as they are vertically opposite angles)

AP/PD = 6/5 … (i)

And, BP/CP = 3/2.5

Also, BP/CP = 6/5 … (ii)

From the equations (i) and (ii),

We have,

AP/PD = BP/CP

Thus,, ∆APB ∼ ∆DPC [ by using SAS similarity criterion]

∠A = ∠D = 30° [as the corresponding angles of similar triangles]

As Sum of the angles for the triangle = 180°,

In the ∆APB,

∠A + ∠B + ∠APB = 180°

So, 30° + ∠B + 50° = 180°

So, ∠B = 180° – (50° + 30°)

∠B = 180 – 80° = 100°

Hence, ∠PBA = 100°

Question 6. It is given that the ΔDEF ~ ΔRPQ. Should it be true to say that ∠D = ∠R as well as ∠F = ∠P ? Why?

Answer 6:

False

We have,

The corresponding angles are equal in similar triangles.

Thus, we have,

∠D = ∠R

∠E = ∠P

∠F = ∠Q

Question 7. Is the following sentence true? Why?

“Two quadrilaterals are equal if their corresponding angles are the same”.

Answer 7:

False

Two quadrilaterals could not be similar, only if their corresponding angles are equal

Question 8. In the given figure, when AB || DC and AC, PQ bisect each other at the point O. Prove that the OA.CQ = 0C.AP.

Source: NCERT

Answer 8:

As per the question,

AC and PQ bisect each other at the point O and AB||DC.

From the ∆AOP and ∆COQ,

∠AOP = ∠COQ [As they are vertically opposite angles]

∠APO = ∠CQO [ As AB||DC and PQ is transversal, the angles are alternate angles]

∆AOP ∼ ∆COQ [ by using AAA similarity criterion]

As the corresponding sides are proportional

We get,

OA/OC = AP/CQ

OA × CQ = OC × AP

So Proved.

Question 9: In the given figure, we have PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove thatQPR is an isosceles triangle.

Answer 9:

Given that,

For PS/SQ = PT/TR

We are aware that if a line divides any two sides of the triangle in an equal ratio, the line will be parallel to the third side.

Hence, ST ||QR

Also,

 ∠ PST = ∠ QPR (Corresponding angles) ……..(i)

And

∠ PST = ∠ PRQ………(ii)

From the (i) and (ii),

∠ PRQ = ∠ QPR

So, PQ = PR (sides opposite the equal angles)

Thus, QPR is an Isosceles triangle.

Question  10: In the figure, we have DE || AC and DF || AE. Prove that the sides BF/FE = BE/EC.

Answer 10:

Given as

In the triangle ABC, DE || AC.

By the Basic Proportionality Theorem,

BD/DA = BE/EC……….(i)

And given as DF || AE.

Also, by the Basic Proportionality Theorem,

BD/DA = BF/FE……….(ii)

From the (i) and (ii),

BE/EC = BF/FE

Thus proved.

Question 11: In the given figure, the altitudes of AD and CE of the ∆ ABC cut each other at point P. Prove that:

(i) ∆AEP ~ ∆ CDP

(ii) ∆ABD ~ ∆ CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆ PDC ~ ∆ BEC

Answer 11:

Given as,

Side AD and CE are the altitudes of the triangle ABC, and these altitudes cut each other at P.

(i) In the ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

So, by the AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In the ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

So, by the AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In the ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

So, by the AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In the ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

So, by the AA similarity criterion,

ΔPDC ~ ΔBEC

Question 12: A vertical pole having a length of 6 m casts a 4m long shadow of 4 m long on the ground, and at  the same time, a tower casts a 28m long  shadow. Find out the height of the tower

Answer 12:

Given that,

length of the vertical pole = 6 m

Shadow of the pole = 4 m

Let the height of the tower be h m.

Length of the shadow of the tower = 28 m

In the ΔABC and ΔDFE,

∠C = ∠E (angle of elevation)

∠B = ∠F = 90°

By the AA similarity criterion,

ΔABC ~ ΔDFE

We have that the corresponding sides of the two similar triangles are proportional.

AB/DF = BC/EF

6/h = 4/28

h = (6 ×28)/4

h = 6 × 7

h = 42

Hence, the height of the tower = 42 m.

Question 13: When ΔABC ~ ΔQRP, area of (ΔABC) / area of (ΔPQR) =9/4 , AB = 18 cm and BC = 15 cm. Find out PR.

Answer 13:

Given as, 

ΔABC ~ ΔQRP.

area of (ΔABC) / area of (ΔQRP) =9/4

AB = 18 cm and BC = 15 cm

We have the ratio of the areas of two similar triangles is the same to the square of the ratio of their corresponding sides.

area of (ΔABC) / area of (ΔQRP) = BC2/RP2

9/4 = (15)2/RP2

RP2 = (4/9) × 225

PR2 = 100

Hence, PR = 10 cm

Question 14: When the areas of the two similar triangles are equal, prove that they are also congruent.

Answer 14:

Assume ΔABC and ΔPQR are the two similar triangles with the equal area.

To prove that

 ΔABC ≅ ΔPQR.

Proof:

ΔABC ~ ΔPQR

Area for the (ΔABC)/Area for the (ΔPQR) = BC2/QR2 = AB2/PQ2 = AC2/PR2

 BC2/QR2 = AB2/PQ2 = AC2/PR2 = 1 [hence, ar (ΔABC) = ar (ΔPQR)] ⇒ BC2/QR2 = 1

 AB2/PQ2 = 1

AC2/PR2 = 1

BC = QR

AB = PQ

AC = PR

Hence, ΔABC ≅ ΔPQR [ By SSS criterion of congruence]

Question 15: O is any point inside the rectangle ABCD as shown in the figure. Prove that the OB2 + OD2 = OA2 + OC2.

Answer 15:

Through O, draw PQ || BC such that P lies on the AB and Q lies on the DC.

PQ || BC

Hence, PQ ⊥ AB and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)

Thus, ∠ BPQ = 90° and ∠ CQP = 90°

So, BPQC and APQD are both the rectangles.

By the Pythagoras theorem,

In the ∆ OPB,

OB2 = BP2 + OP2…..(1)

Same as,

In the ∆ OQD,

OD2= OQ2 + DQ2…..(2)

In the ∆ OQC,

OC2 = OQ2 + CQ2…..(3)

In the ∆ OAP,

OA2 = AP2 + OP2…..(4)

Adding the (1) and (2),

OB2+ OD2 = BP2 + OP2 + OQ2 + DQ2

= CQ2 + OP2 + OQ2 + AP2

(As BP = CQ and DQ = AP)

= CQ2 + OQ2 + OP2 + AP2

= OC2 + OA2 [From (3) and (4)]

Thus proved that OB2 + OD2 = OA2 + OC2.

Question 16: Sides of the triangles are given below. Find out which of them are right triangles.

In the case of the right triangle, write the length of the hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

Answer 16:

(i) Given that, sides of the triangle are 7 cm, 24 cm, and 25 cm.

On squaring the lengths of the sides of the triangle, we find 49, 576, and 625.

49 + 576 = 625

(7)2 + (24)2 = (25)2

Hence, in the above equation it satisfies the Pythagoras theorem. So, it is the right-angled triangle.

length of the Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

On squaring the lengths of the sides, we find 9, 64, and 36.

Thus, 9 + 36 ≠ 64

And 32 + 62 ≠ 82

So, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

So, the given triangle does not satisfy the Pythagoras theorem.

Question 17. Find the altitude of the equilateral triangle of side 8 cm.

Answer 17:

Let ABC be an equilateral triangle having the side of 8 cm

AB = BC = CA = 8 cm. (all sides of the equilateral triangle is equal)

Draw the altitude AD that is perpendicular to BC.

So, D is the midpoint of BC.

BD = CD = ½

BC = 8/2 = 4 cm

 By the Pythagoras theorem

AB2 = AD2 + BD2

 (8)2 = AD2 + (4)2

64 = AD2 + 16

AD = √48 = 4√3 cm.

Thus, the altitude of the equilateral triangle is 4√3 cm.

Question 18. Fill in the blanks by using the right word given below in the brackets:-

(i) All circles are __________. (congruent, similar)

Answer: Similar

(ii) All squares are __________. (similar, congruent)

Answer: Similar

(iii) All __________ triangles are similar. (isosceles, equilateral)

Answer: Equilateral

(iv) Two polygons having equal number of sides are equal, when (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Answer: (a) Equal

(b) Proportional

Question 19. Given that two different examples of the pair of

(i) Similar figures

(ii) Non-similar figures

Answer 19:

Question 20. State if the following quadrilaterals are similar or not:

Source: NCERT

Answer 20:

From the above given two figures, we can see their corresponding angles are either different or unequal. Hence they are not similar.

Question 21. In the figure. (i) and (ii), DE || BC. Find out EC in the (i) and AD in the (ii).

Source: NCERT

Answer 21:

(i) Given that,

 In the △ ABC, DE∥BC

AD/DB = AE/EC  [ By using the Basic proportionality theorem]

1.5/3 = 1/EC

EC = 3/1.5

EC = 3×10/15 = 2 cm

Thus, EC = 2 cm.

(ii) Given that, 

In △ ABC, DE∥BC

AD/DB = AE/EC [By using Basic proportionality theorem]

AD/7.2 = 1.8 / 5.4

AD = 1.8 ×7.2/5.4 

      = (18/10)×(72/10)×(10/54) 

      = 24/10

 AD = 2.4

Thus, AD = 2.4 cm.

Question 22. In the figure, When LM || CB and LN || CD, prove that the side AM/AB = AN/AD

Source: NCERT

Answer 22:

For the given figure, we know LM || CB,

By using the basic proportionality theorem, we find,

AM/AB = AL/AC……………………..(i)

In the same way, LN || CD and using the basic proportionality theorem,

∴AN/AD = AL/AC……………………………(ii)

From the equation (i) and (ii), we get,

AM/AB = AN/AD

Thus, proved.

Question 23. In the figure, side DE||AC and DF||AE. Prove that the sides BF/FE = BE/EC

Source: NCERT

Answer 23:

In the ΔABC, given that DE || AC

Therefore, by using the Basic Proportionality Theorem, we find,

∴BD/DA = BE/EC ………………………………………………(i)

In the ΔBAE, given that, DF || AE

So, by using the Basic Proportionality Theorem, we find,

Side BD/DA = BF/FE ………………………………………………(ii)

From the equation (i) and (ii), we get

Side BE/EC = BF/FE

Thus, proved.

Question 24. In the figure, side DE||OQ and PDF||OR show that the EF||QR.

Source: NCERT

Answer 24:

Given that,

In the ΔPQO, DE || OQ

Thus, by using the Basic Proportionality Theorem,

PD/DO = PE/EQ……………… ..(i)

Given, in the ΔPOR, DF || OR,

Thus, by using the Basic Proportionality Theorem,

PD/DO = PF/FR………………… (ii)

From the equation (i) and (ii), we find,

PE/EQ = PF/FR

Hence, by the converse of the Basic Proportionality Theorem,

EF || QR, in the ΔPQR.

Question 25: In the figure, Side DE || BC. Find out the length of the side AD, given as AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

Answer 25:

Given as,

DE || BC

AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm

By the basic proportionality theorem,

AD/DB = AE/EC

AD/7.2 = 1.8/5.4

AD = (1.8 × 7.2)/5.4

= 7.2/4

= 2.4

Hence, AD = 2.4 cm.

Question 26: Given as ΔABC ~ ΔPQR, when AB/PQ = ⅓, then find (area of ΔABC)/(area of ΔPQR).

Answer 26:

Given as,

ΔABC ~ ΔPQR

Also,

AB/PQ = ⅓, 

Given that the ratio of the areas of the two similar triangles is equal to the square of the ratio for the corresponding sides.

(area of ΔABC)/(area of ΔPQR) = AB2/PQ2 = (AB/PQ)2 = (⅓)2 = 1/9

Hence, (area of ΔABC)/(area of ΔPQR) = 1/9

And

(area of ΔABC) : (area of ΔPQR) = 1 : 9

Question 27: The sides of the two similar triangles are in the ratio 7 : 10. Find out the ratio of areas of these triangles.

Answer 27:

Given as

The ratio of the sides of the two similar triangles = 7 : 10

We get that the ratio of the areas of the two similar triangles is equal to the square of the ratio for the corresponding sides.

The ratio of areas of the triangles = (Ratio of the sides of the two similar triangles)2

= (7)2 : (10)2

= 49 : 100

Hence, the ratio for the areas of the given similar triangles is 49 : 100.

Question 28: In the equilateral ΔABC, D is a point on the side BC so that BD = (⅓) BC. Prove that the 

9(AD)2 = 7(AB)2.

Answer 28:

Given as 

ABC is the equilateral triangle.

Also, D is the point on side BC such that BD = (1/3)BC.

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

 BE = EC = BC/2 = a/2

Also, AE = a√3/2

Given as, BD = 1/3BC

BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

In given ΔADE, by the Pythagoras theorem,

AD2 = AE2 + DE2

= [(a√3)/2]2 + (a/6)2

= (3a2/4) + (a2/36)

= (37a2 + a2)/36

= (28a2)/36

= (7/9)a2

= (7/9) (AB)2

Hence, 9(AD)2 = 7(AB)2.

Question 29: Prove that in the right triangle, the square of the hypotenuse is same as the addition of the squares of the other two sides.

Answer 29:

Given as,

In the right-angled triangle ABC, the right-angled is at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw the perpendicular BD meeting AC at D.

Proof:

In the ΔABC and ΔADB,

∠ABC = ∠ADB = 90°

  ∠A = ∠A → common

By using the AA criterion for the similarity of triangles, 

ΔABC ~ ΔADB 

Hence,, AD/AB = AB/AC

 AB2 = AC x AD ……(1)

On Considering the  ΔABC and ΔBDC from the figure.

C = ∠C   (common)

 ∠CDB = ∠ABC = 90°

By using the Angle Angle(AA) criterion for the similarity of the triangles, we conclude as,

ΔBDC ~ ΔABC

Hence, CD/BC = BC/AC

BC2 = AC x CD …..(2)

By adding the equation (1) and equation (2), we find:

AB2 + BC2 = (AC x AD) + (AC x CD)

AB2+ BC2 = AC (AD + CD) …..(3)

AB2 + BC2 = AC (AC) { as AD + CD = AC}

AB2+ BC2 = AC2

Thus, proved.

Question 30: In the figure, when PQ || RS, prove that the ∆ POQ ~ ∆ SOR.

Source: NCERT

Answer 30:

Given as,

PQ || RS

∠P = ∠S (Alternate angles) 

also ∠Q = ∠R 

And ∠POQ = ∠SOR (Vertically opposite angles) 

Hence, ∆ POQ ~ ∆ SOR (by AAA similarity criterion)

Thus, proved.

Question 31. In the given figure, A, B and C are the points on the OP, OQ and OR, respectively so that AB || PQ and AC || PR. Show that BC || QR.

Source: NCERT

Answer 31:

Given as

In the ΔOPQ, AB || PQ

By using the Basic Proportionality Theorem,

OA/AP = OB/BQ…………….(i)

And

In ΔOPR, AC || PR

By using the Basic Proportionality Theorem

OA/AP = OC/CR……………(ii)

From the equation (i) and (ii), we get,

OB/BQ = OC/CR

Hence, by the converse of Basic Proportionality Theorem,

In the ΔOQR, BC || QR.

Question 32. Using the Basic proportionality theorem, prove that the line drawn through the midpoints of one side of the triangle parallel to another side cuts the third side.

Answer 32:

Given as 

In the ΔABC, D is the midpoint of the side AB so that AD=DB.

A line parallel to BC cuts AC at E as shown in the above figure, so that DE || BC.

We should prove that E is the midpoint for the AC.

As D is the midpoint for the AB.

 AD=DB

AD/DB = 1 …………………………. (i)

In the ΔABC, DE || BC,

By using the Basic Proportionality Theorem,

Hence, the side AD/DB = AE/EC

From the equation (i), we can write,

1 = AE/EC

AE = EC

So, proved, E is the midpoint of the AC.

Question 33. By using the Converse of the basic proportionality theorem, prove that the line joining the midpoints for any two sides of the triangle is parallel to the third side. 

Answer 33:

Given as, in the ΔABC, D and E are the midpoints for the AB and AC respectively, so that,

AD=BD and AE=EC.

To prove that: DE || BC.

As D is the midpoint of AB

 AD=DB

AD/BD = 1……………………………….. (i)

And E is the midpoint for the AC.

AE=EC

AE/EC = 1

From the equation (i) and (ii), we get,

AD/BD = AE/EC

By the converse of the Basic of Proportionality Theorem,

DE || BC

So, proved.

Question 34. ABCD is the trapezium for which AB || DC and its diagonals cut each other at point O. Show that the AO/BO = CO/DO.

Answer 34:

Given as, ABCD is the trapezium here AB || DC and diagonals AC and BD cut each other at O.

To prove, side AO/BO = CO/DO

From the point O, draw the line EO touching AD at E in so that,

EO || DC || AB

In the ΔADC, we get OE || DC

Hence, By using the Basic Proportionality Theorem

AE/ED = AO/CO ……………..(i)

In the ΔABD, OE || AB

Hence, By using the Basic Proportionality Theorem

DE/EA = DO/BO…………….(ii)

From the equation (i) and (ii), we get,

AO/CO = BO/DO

AO/BO = CO/DO

Hence proved.

Question 35. The diagonals of the quadrilateral ABCD cut each other at the point O so that AO/BO = CO/DO. Show that ABCD is the trapezium.

Answer 35:

Given as Quadrilateral ABCD here, AC and BD intersect each other at O so that,

AO/BO = CO/DO.

To prove, ABCD is the trapezium

From the point O, draw the line EO touching AD at E so that,

EO || DC || AB

In the ΔDAB, EO || AB

Hence, By using the Basic Proportionality Theorem

DE/EA = DO/OB ……………………(i)

And

AO/BO = CO/DO

AO/CO = BO/DO

CO/AO = DO/BO

DO/OB = CO/AO …………………………..(ii)

From the equation (i) and (ii), we get

DE/EA = CO/AO

Hence, 

By using the converse of the Basic Proportionality Theorem,

EO || DC also EO || AB

AB || DC.

Therefore, quadrilateral ABCD is the trapezium with AB || CD.

Question 36. Is a triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Justify your answer.

Answer 36:

No it is not a right triangle

As per the question,

Assume that,

A = 25 cm

B = 5 cm

C = 24 cm

By using Pythagoras Theorem,

We get,

A2 = B2 + C2

B2 + C2 = (5)2 + (24)2

B2 + C2 = 25 + 576

B2 + C2 = 601

A2 = 600

600 ≠ 601

A2 ≠ B2 + C2

As the sides do not satisfy the property of the Pythagoras theorem, a triangle withsides 25cm, 5cm and 24cm is not the right triangle.

Question 37. A as well as B are respectively the points at the sides PQ and PR for the ΔPQR so that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Justify your answer.

Answer 37:

True

As per the question,

PQ = 12.5 cm

PA = 5 cm

BR = 6 cm

PB = 4 cm

Now,

QA = QP – PA = 12.5 – 5 = 7.5 cm

Thus,

PA/AQ = 5/7.5 = 50/75 = 2/3 … (i)

PB/BR = 4/6 = 2/3 … (ii)

Form the Equations (i) and (ii).

PA/AQ = PB/BR

We have, when the line divides any two sides of the triangle in the same ratio, the lines are parallel to the third side.

Hence,

AB || QR.

Question 38. In the figure, BD and CE cut each other at the point P. Is ΔPBC ~ ΔPDE? Why?

Answer 38:

True

In the ∆PBC and ∆PDE,

∠BPC = ∠EPD [vertically opposite angles]

PB/PD = 5/10 = ½ … (i)

PC/PE = 6/12 = ½ … (ii)

From the equation (i) and (ii),

We find,

PB/PD = PC/PE

As the ∠BPC of ∆PBC = ∠EPD of ∆PDE and the sides including these.

By the SAS similarity criteria

∆PBC ∼ ∆PDE

Question 39. In the ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?

Answer 39:

We have,

The sum of the three angles of a triangle = 180°.

From the ∆PQR,

∠P + ∠Q + ∠R = 180°

55° + 25° + ∠R = 180°

Hence, we have,

∠R = 180° – (55° + 25°) = 180°- 80° = 100°

From the ∆TSM,

∠T + ∠S + ∠M = 180°

∠T + ∠25° + 100° = 180°

Hence, we have,

∠T = 180°- (∠25° + 100°)

∠T = 180° – 125° = 55°

In the ∆PQR and ∆TSM,

We get,

∠P = ∠T,

∠Q = ∠S

∠R = ∠M

So, the ∆PQR ∼ ∆TSM

As all the corresponding angles are equal,

The ∆QPR is similar to the ∆TSM,

Question 40. In the figure, when ∠1 =∠2 and ΔNSQ = ΔMTR, prove that ΔPTS ~ ΔPRQ.

Answer 40:

As per the question,

∆ NSQ ≅ ∆MTR

∠1 = ∠2

As

∆NSQ = ∆MTR

Thus,

SQ = TR ….(i)

And,

∠1= ∠2 PT = PS….(ii)

[As the sides opposite to the equal angles are also equal]

From the Equation (i) and (ii).

PS/SQ = PT/TR

 ST || QR

By the converse of the basic proportionality theorem, when a line is drawn parallel to one side of the triangle to cut the other sides at distinct points, the other two sides are divided in a similar ratio.

 ∠1 = QPR

Also,

∠2 = ∠PRQ

In the ∆PTS and ∆PRQ.

∠P = ∠P [Common angles]

∠1 = ∠QPR (proved)

∠2 = ∠PRQ (proved)

∆PTS – ∆PRQ

[By the AAA similarity criteria]

Thus proved.

Question 41. In the figure 6.35, ΔODC ~ ΔOBA, for which ∠ BOC = 125° and ∠ CDO = 70°. Find out ∠ DOC, ∠ DCO and ∠ OAB.

Source: NCERT

Answer 41:

We have from the given figure, DOB is the straight line.

Hence, ∠DOC + ∠ COB = 180°

∠DOC = 180° – 125° (Given, ∠ BOC = 125°)

= 55°

In the ΔDOC, sum of the measures of the angles for the triangle is 180º

Hence, ∠DCO + ∠ CDO + ∠ DOC = 180°

 ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)

 ∠DCO = 55°

We have, ΔODC ~ ΔOBA,

Hence, ΔODC ~ ΔOBA.

Thus, the corresponding angles are equal in similar triangles

∠OAB = ∠OCD

 ∠ OAB = 55°

∠OAB = ∠OCD

 ∠OAB = 55°

Question 42. Diagonals AC and BD of the trapezium ABCD having AB || DC cut each other at the point O. By using the similarity criterion of two triangles, show that side AO/OC = OB/OD

Answer 42:

In the ΔDOC and ΔBOA,

AB || CD, hence alternate interior angles will be equal,

∠CDO = ∠ABO

In the same way,

∠DCO = ∠BAO

And, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be same;

∠DOC = ∠BOA

So, by the AAA similarity criterion,

ΔDOC ~ ΔBOA

Hence, the corresponding sides are proportional.

DO/BO = OC/OA

OA/OC = OB/OD

Therefore, proved.

Question 43. In the given fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Source: NCERT

Answer 43:

In ΔPQR,

∠QPR = ∠PRQ

∴ PQ = PR ………………………(i)

Given,

QR/QS = QT/PR Using equation (i), we observe,

QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

∠Q = ∠Q

therefore, ΔPQS ~ ΔTQR [By SAS similarity criterion]

Question 44. S and T are points on the sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer 44:

Given that, S and T are points on sides PR and QR of ΔPQR

And ∠P = ∠RTS.

In ΔRPQ and ΔRTS,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

therefore, ΔRPQ ~ ΔRTS (AA similarity criterion)

Question 45. In the given figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Source: NCERT

Answer 45:

Given, ΔABE ≅ ΔACD.

therefore, AB = AC [By CPCT] ……………………………….(i)

And, AD = AE [By CPCT] ……………………………(ii)

In ΔADE and ΔABC, dividing eq.(ii) with eq(i),

AD/AB = AE/AC

∠A = ∠A [Common angle]

therefore, ΔADE ~ ΔABC [SAS similarity criterion]

Question 46. In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

Source: NCERT

(i) ΔAEP ~ ΔCDP

(ii) ΔABD ~ ΔCBE

(iii) ΔAEP ~ ΔADB

(iv) ΔPDC ~ ΔBEC

Answer 46 :

Given that, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

therefore, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

therefore, by AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

therefore, by AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

therefore, by AA similarity criterion,

ΔPDC ~ ΔBEC

Question 47. E is the point on the side AD produced of the parallelogram ABCD and BE intersects CD at F. so Show that ΔABE ~ ΔCFB.

Answer 47:

Given that, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD on F. Consider the figure given below,

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of the parallelogram)

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

therefore, ΔABE ~ ΔCFB (AA similarity criterion)

Question 48. In the given figure, ABC and AMP are the two right triangles, right-angled at B and M respectively. Prove that:

Source: NCERT

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

Answer 48:

Given that, ABC and AMP are the two right triangles, right-angled at B and M respectively.

(i) In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

therefore, ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

when two triangles are similar, the corresponding sides are always equal,

therefore, CA/PA = BC/MP

Question 49. CD and GH are bisectors of the angle ∠ACB and ∠EGF such that D and H lie on the sides AB and FE of ΔABC and ΔEFG respectively. When ΔABC ~ ΔFEG, show that:

(i) CD/GH = AC/FG

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

Answer 49:

Given that, CD and GH are bisectors of the angle of ∠ACB and ∠EGF such that D and H lie on the sides AB and FE of ΔABC and ΔEFG respectively.

(i) From the given condition,

ΔABC ~ ΔFEG.

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

Since, ∠ACB = ∠FGE

therefore, ∠ACD = ∠FGH (Angle bisector)

or, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

therefore, ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

therefore, ΔDCA ~ ΔHGF (AA similarity criterion)

Question 50. In the given 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.

Source: NCERT

Answer 50:

Given, ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (using AA similarity criterion)

Question 51. A 5 m long ladder has been placed leaning towards the vertical wall such that it reaches to the wall at a height of 4 m. If the foot of the ladder has moved 1.6 m towards the wall, find out the distance by which the top for the ladder would slide upwards on the wall.

Answer 51:

Let us assume the length of the ladder = AC = 5 m

Let us assume the height of the wall on which the ladder is placed = BC = 4m.

From right-angled ∆EBD,

Using the Pythagoras Theorem as given,

ED2 = EB2 + BD2

(5)2 = (EB)2 + (14)2 [ BD = 1.4]

25 = (EB)2 + 1.96

(EB)2 = 25 –1.96 = 23.04

EB = √23.04 = 4.8

then, we know,

EC = EB – BC = 4.8 – 4 = 0.8

Therefore, the top of the ladder would slide upwards on the wall by a distance of 0.8 m.

Question 52. For going to the city B from the city A, there is a route via city C such that perpendicular, AC⊥CB, AC = 2 x km as well as CB = 2 (x + 7) km. It is proposed to build a 26 km highway that directly connects the two cities A and B. Find out how much distance will be saved in reaching city B from city A after the construction of the highway.

Answer 52:

As per the question,

AC⊥CB,

AC = 2x km,

CB = 2 (x + 7) km as well as AB = 26 km

here, we observe ∆ ACB right angled at C.

then, from ∆ACB,

Using the Pythagoras Theorem,

AB2 = AC2 + BC2

⇒ (26)2 = (2x)2 + {2(x + 7)}2

⇒ 676 = 4x2 + 4(x2 + 196 + 14x)

⇒ 676 = 4x2 + 4x2 + 196 + 56x

⇒ 676 = 8x2 + 56x + 196

⇒ 8x2 + 56x – 480 = 0

Dividing the equation by 8, we observe,

x2 + 7x – 60 = 0

x2 + 12x – 5x – 60 = 0

x(x + 12)-5(x + 12) = 0

(x + 12)(x – 5) = 0

∴ x = -12 or x = 5

hence the distance can’t be negative, we should neglect x = -12

∴ x = 5

then,

AC = 2x = 10km

BC = 2(x + 7) = 2(5 + 7) = 24 km

therefore, the distance covered to city B from the city A via city C = AC + BC

AC + BC = 10 + 24

= 34 km

Distance covered to city B from city A after the highway was constructed = BA = 26 km

Hence, the distance saved = 34 – 26 = 8 km.

Question 53. The flag pole 18 m high casts a shadow with a height 9.6 m. Find out the distance of the top of the pole from the far end of the shadow.

Answer 53:

Let’s assume, MN = 18 m be the flag pole and its shadow will be LM = 9.6 m.

The distance for the top of the pole, N from the far end, L of the shadow is LN.

In the right angled ∆LMN,

LN2 = LM2 + MN2 [by the Pythagoras theorem]

⇒ LN2 = (9.6)2 + (18)2

⇒ LN2 = 9.216 + 324

⇒ LN2 = 416.16

∴ LN = √416.16 = 20.4 m

therefore, the required distance is 20.4 m

Question 54. In the given figure, AD is a median of the triangle ABC and AM ⊥ BC. Prove that :

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

(ii) AB2 = AD2 – BC.DM + 2 (BC/2)2

(iii) AC2 + AB2 = 2 AD2 + ½ BC2

Source: NCERT

Answer 54:

(i) By applying the Pythagoras Theorem in ∆AMD, we get,

AM2 + MD2 = AD2 ………………. (i)

Again, by applying the Pythagoras Theorem in ∆AMC, we observe,

AM2 + MC2 = AC2

AM2 + (MD + DC)2 = AC2

(AM2 + MD2 ) + DC2 + 2MD.DC = AC2

From equation(i), we observe,

AD2 + DC2 + 2MD.DC = AC2

Since, DC=BC/2, thus, we observe,

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

AD2 + (BC/2)2 + 2MD × BC = AC2

therefore, proved.

(ii) By applying the Pythagoras Theorem in ∆ABM, we observe;

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

therefore, proved.

(iii) By applying the Pythagoras Theorem in ∆ABM, we get,

AM2 + MB2 = AB2 ………………….… (i)

By applying the Pythagoras Theorem in ∆AMC, we observe,

AM2 + MC2 = AC2…………………..… (ii)

sum of both the equations (i) and (ii), we observe,

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)2 = AB2 + AC2

2AD2 + BC2/2 = AB2 + AC2

Question 55. Prove that the sum of the squares of the diagonals of the parallelogram is similar  to the addition of the squares of its sides.

Answer 55:

Let us consider that, ABCD is the parallelogram. Now, draw a perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we observe,

DE2 + EA2 = DA2 ……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we observe,

DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2

(DE2 + EA2) + AB2 + 2EA × AB = DB2

DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we observe,

AD2 = AF2 + FD2

Again, applying Pythagoras theorem in ∆AFC, we observe,

AC2 = AF2 + FC2 = AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC × FD

= (AF2 + FD2) + DC2 − 2DC × FD AC2

AC2= AD2 + DC2 − 2DC × FD ………………… (iii)

hence ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

In ∆DEA and ∆ADF,

∠DEA = ∠AFD (Each 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common Angles)

therefore, ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF ……………… (vi)

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

DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2

From equation (iv) and (vi),

BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Question 56. In the given figure, two chords AB and CD intersect each other at point P. Prove that:

(i) ∆APC ~ ∆ DPB

(ii) AP . PB = CP . DP

Source: NCERT

Answer 56:

Firstly, let’s join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

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

hence,

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that the triangles ∆APC ∼ ∆DPB

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

∴ AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴AP. PB = PC. DP

therefore, proved.

Question 57. In the given figure., two chords AB and CD of the circle intersect each other at point P (when produced) outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB

(ii) PA . PB = PC . PD.

Source: NCERT

Answer 57:

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, the exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is the opposite interior angle, which are both the equal.

∠PAC = ∠PDB

so, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,

∆APC ∼ ∆DPB

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

thus,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

therefore, AP. PB = PC. DP

Question 58. In the given figure, D is a point on the side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of the angle ∠ BAC.

Source: NCERT

Answer 58:

It is given in the figure, assume that we extend BA to P such that;

AP = AC.

Now join PC.

We have, BD/CD = AB/AC

BD/CD = AP/AC

Using the converse of the basic proportionality theorem, we observe,

AD || PC

∠BAD = ∠APC (Corresponding angles) ……………….. (i)

Also, the ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

From the new figure, we get,

AP = AC

⇒ ∠APC = ∠ACP ……………………. (iii)

Comparing the equations (i), (ii), and (iii), we find,

∠BAD = ∠APC

Thus, AD is the bisector of the angle BAC.

So, proved.

Question 59. Nazima is fly fishing in the stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly by the end of the given string rests on the water 3.6 m away and 2.4 m from the point directly under the tip of the rod. If the string from the tip of the rod to the fly is taut, then how much string does she have (see Figure)? If she pulls the string at the rate of 5 cm per second, what would be the horizontal distance of the fly after 12 seconds?

Source: NCERT

Answer 59:

Consider AB as the height of the tip of the fishing rod from the water surface and BC as the

horizontal distance for the fly from the tip of the fishing rod. So, AC is now considered the length of the string.

To find AC, we need to use Pythagoras theorem in  ∆ABC in such a way;

AC2 = AB2+ BC2

AB2 = (1.8 m)2 + (2.4 m)2

AB2 = (3.24 + 5.76) m2

AB2 = 9.00 m2

AB = √9 m = 3m

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

 Given that, she pulls the string at the rate of 5 cm per second.

Thus, the string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Assume the fly is at the point D after 12 seconds.

The length of the 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 the ∆ADB, by the Pythagoras Theorem,

AB2 + BD2 = AD2

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

BD2 = (5.76 − 3.24) m2 = 2.52 m2

BD = 1.587 m

Horizontal distance for the fly = BD + 1.2 m

= (1.587 + 1.2) m = 2.787 m

= 2.79 m

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