Monday, December 18, 2023

Class – 7 CH- 7 CONGRUENCE OF TRIANGLES MATHS NCERT SOLUTIONS

Class – 7 CH- 7 CONGRUENCE OF TRIANGLES

MATHS NCERT SOLUTIONS

Exercise 7.1

Question 1:

Complete the following statements:

(a) Two line segments are congruent if _______________.

(b) Among two congruent angles, one has a measure of 70, the measure of other angle is _______________. (c) When we write  A =  B, we actually mean _______________.

SOLUTION 1:

(a) they have the same length

(b) 70

Question 2: Give any two real time examples for congruent shapes.

SOLUTION 2:

(i) Two footballs (ii) Two teacher’s tables

Question 3:

If  ABC  FED under the correspondence ABC  FED, write all the corresponding congruent parts of the triangles.

SOLUTION 3:

Given: ABC  FED.

The corresponding congruent parts of the triangles are:

(i)  A   F (ii)  B   E

(iii)  C   D

(iv) AB  FE (v) BC  ED

(vi) AC  FD

Question 4:

(i)  E

(ii) EF

(iii)  F

(iv) DF

If DEF BCA, write the part(s) of BCA that correspond to:

SOLUTION 4:

Given: DEF  BCA.

(i)  E   C

(ii) EF  CA

(iii)  F   A

(iv) DF  BA

Exercise 7.2

Question 1:

Which congruence criterion do you use in the following? (a) Given: AC = DF, AB = DE, BC = EF

So ABC DEF

(b) Given: RP = ZX, RQ = ZY,  PRQ =  XZY

So PQR XYZ

So LMN GFH

(d) Given: EB = BD, AE = CB,  A =  C = 90

So ABE CDB

SOLUTION 1:

(a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF

The three sides of one triangle are equal to the three corresponding sides of another triangle.

Therefore, ABC  DEF

(b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and  PRQ =  XZY

The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.

Therefore, PQR  XYZ

The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.

Therefore, LMN  GFH

(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB,  A =  C = 90

Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle. Therefore, ABE  CDB

Question 2:

You want to show that ART PEN:

(a) If you have to use SSS criterion, then you need to show:

(i) AR = (ii) RT = (iii) AT =

(b) If it is given that  T =  N and you are to use SAS criterion, you need to have:

(i) RT = and (ii) PN =

(i) ? (ii) ?

SOLUTION 2:

(a) Using SSS criterion, ART  PEN

(i) AR = PE (ii) RT = EN (iii) AT = PN

(b) Given:  T =  N

Using SAS criterion, ART  PEN

(i) RT = EN (ii) PN = AT

Using ASA criterion, ART  PEN

(i)  RAT =  EPN (ii)  RTA =  ENP

Question 3:

You have to show that AMP AMQ. In the following proof, supply the missing reasons:

Steps Reasons

(i) PM = QM

(ii)  PMA =  QMA

(iii) AM = AM

(iv) AMP AMQ (i)

(ii)

(iii)

(iv) __________

__________

SOLUTION 3:

Steps Reasons

(i) PM = QM

(ii)  PMA =  QMA

(iii) AM = AM

(iv) AMP AMQ (i) Given

(ii) Given

(iii) Common

(iv) SAS congruence rule

Question 4:

In ABC,  A = 30,  B = 40 and  C = 110 . In PQR,  P = 30,  Q = 40 and  R = 110.

A student says that ABC  PQR by AAA congruence criterion. Is he justified? Why or why not?

SOLUTION 4:

No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other.

Question 5:

In the figure, the two triangles are congruent. The corresponding parts are marked. We can write RAT  ?

SOLUTION 5:

In the figure, given two triangles are congruent. So, the corresponding parts are:

A  O, R  W, T  N.

We can write, RAT  WON [By SAS congruence rule]

Question 6:

Complete the congruence statement:

BCA  ? QRS  ?

SOLUTION 6:

In BAT and BAC, given triangles are congruent so the corresponding parts are:

B  B, A  A, T  C

Thus, BCA  BTA [By SSS congruence rule]

In QRS and TPQ, given triangles are congruent so the corresponding parts are:

P  R, T  Q, Q  S

Thus, QRS  TPQ [By SSS congruence rule]

Question 7:

In a squared sheet, draw two triangles of equal area such that:

(i) the triangles are congruent.

(ii) the triangles are not congruent. What can you say about their perimeters? SOLUTION 7:

In a squared sheet, draw ABC and PQR.

When two triangles have equal areas and

(i) these triangles are congruent, i.e., ABC  PQR [By SSS congruence rule] Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congruence rule.

(ii) But, if the triangles are not congruent, then their perimeters are not same because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.

Question 8:

Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent. SOLUTION 8:

Let us draw two triangles PQR and ABC.

All angles are equal, two sides are equal except one side. Hence, PQR are not congruent to ABC.

Question 9:

If  ABC and  PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

ABC and PQR are congruent. Then one additional pair is BC = QR.

Given:  B =  Q = 90

 C =  R

BC = QR

Therefore, ABC  PQR [By ASA congruence rule]

Question 10:

Explain, why ABC  FED.

SOLUTION 10:

Given:  A =  F, BC = ED,  B =  E

In ABC and FED,

 B =  E = 90

 A =  F BC = ED

Therefore, ABC  FED [By RHS congruence rule]

Class – 7 CH-3 DATA HANDLING MATHS NCERT SOLUTIONS

Class – 7 CH-3 DATA HANDLING

MATHS NCERT SOLUTIONS

Exercise 3.1

Class – 7 CH-6 TRIANGLE AND ITS PROPERTIES MATHS NCERT SOLUTIONS

Class – 7 CH-6 TRIANGLE AND ITS PROPERTIES

MATHS NCERT SOLUTIONS

Exercise 6.1

Question 1:

In PQR, D is the mid-point of QR.

PM is _______________

PD is ________________ Is QM = MR?

SOLUTION 1:

Given: QD = DR

 PM is altitude.

PD is median.

No, QM  MR as D is the mid-point of QR.

Question 2:

Draw rough sketches for the following:

(a) In ABC, BE is a median.

(b) In PQR, PQ and PR are altitudes of the triangle.

SOLUTION 2:

(a) Here, BE is a median in ABC and AE = EC.

(b) Here, PQ and PR are the altitudes of the PQR and RP  QP.

Question 3: Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.

SOLUTION 3:

Isosceles triangle means any two sides are same. Take ABC and draw the median when AB = AC.

AL is the median and altitude of the given triangle.

Exercise 6.2

Question 1:

Find the value of the unknown exterior angle x in the following diagrams:

SOLUTION 1:

Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x  50 70 120

(ii) x  65 45 110

(iii) x   30 40 70

(iv) x  60 60 120

(v) x  50 50 100

(vi) x 60 30  90

Question 2:

Find the value of the unknown interior angle x in the following figures:

SOLUTION 2:

Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x 50 115 

(ii) 70 x 100 

(iii) x 90 125 

(iv) 60 x 120 

(v) 30  x 80 

(vi) x  35 75 

x115 50 65   x100 70 30   x120  90 35 x120  60 60 x   80 30 50 x   75 35 40

Exercise 6.3

Question 1:

Find the value of unknown x in the following diagrams:

(i) In ABC,

 BAC +  ACB +  ABC = 180 [By angle sum property of a triangle]

 x  50 60 180

 x110180

 x180 110 70  

(ii) In PQR,

 RPQ +  PQR +  RPQ = 180 [By angle sum property of a triangle]

 90 30  x 180

 x120180

 x180120 60

(iii) In XYZ,

 ZXY +  XYZ +  YZX = 180 [By angle sum property of a triangle]

    

 x140180

 x180140 40

30 110 x 180

(iv) In the given isosceles triangle,

x x  50 180 [By angle sum property of a triangle]

 2x 50 180

 2x180 50

 2x130

130

 x  65

(v) In the given equilateral triangle,

x x x  180 [By angle sum property of a triangle]

 3x180

180

 x  60

(vi) In the given right angled triangle,

x2x 90 180 [By angle sum property of a triangle]

 3x 90 180

 3 180 90x  

 3 90x 

90

 x  30

Question 2:

Find the values of the unknowns x and y in the following diagrams:

SOLUTION 2:

(i) 50 x 120 [Exterior angle property of a  ]

 x120  50 70

Now, 50  x y 180 [Angle sum property of a  ]

 50 70  y 180

 120 y 180

 y180120 60

(ii) y80 ……….(i) [Vertically opposite angle]

Now, 50  x y 180 [Angle sum property of a  ]

 50  80 y 180 [From equation (i)]

 130 y 180

 y180 130 50  

(iii) 50 60 x [Exterior angle property of a  ]

 x110

Now 50 60  y 180 [Angle sum property of a  ]

 110 y 180

 y180110

 y70

(iv) x 60 ……….(i) [Vertically opposite angle]

Now, 30  x y 180 [Angle sum property of a  ]

 50  60 y 180 [From equation (i)]

 90 y 180

 y180 90 90  

(v) y 90 ……….(i) [Vertically opposite angle]

Now, y x x  180 [Angle sum property of a  ]

 902x180 [From equation (i)]

 2 180 90x  

 2 90x 

 x  45

(vi)

x y

Now,



 ……….(i) [Vertically opposite angle] x x y  180 [Angle sum property of a  ]

2x x 180 [From equation (i)]

3x180

180

x  60

Exercise 6.4

Question 1:

Is it possible to have a triangle with the following sides?

(i) 2 cm, 3 cm, 5 cm (ii) 3 cm, 6 cm, 7 cm (iii) 6 cm, 3 cm, 2 cm SOLUTION 1:

Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.

(i) 2 cm, 3 cm, 5 cm (ii) 3 cm, 6 cm, 7 cm

2 + 3 > 5 No 3 + 6 > 7 Yes

2 + 5 > 3 Yes 6 + 7 > 3 Yes

3 + 5 > 2 Yes 3 + 7 > 6 Yes

This triangle is not possible. This triangle is possible.

(iii) 6 cm, 3 cm, 2 cm

6 + 3 > 2 Yes

6 + 2 > 3 Yes

2 + 3 > 6 No

This triangle is not possible.

Question 2:

Take any point O in the interior of a triangle PQR. Is: R (i) OP + OQ > PQ ?

(ii) OQ + OR > QR ? (iii) OR + OP > RP ?

SOLUTION 2:

Join OR, OQ and OP.

(i) Is OP + OQ > PQ ?

Yes, POQ form a triangle. (ii) Is OQ + OR > QR ?

Yes, RQO form a triangle.

(iii) Is OR + OP > RP ? P

Yes, ROP form a triangle.

AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles ABM and AMC.)

SOLUTION 3:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In ABM, AB + BM > AM ... (i)

In AMC, AC + MC > AM

Adding eq. (i) and (ii),

AB + BM + AC + MC > AM + AM

 AB + AC + (BM + MC) > 2AM

 AB + AC + BC > 2AM

Hence, it is true. ... (ii)

Question 4:

ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

SOLUTION 4:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In ABC, AB + BC > AC ……….(i)

In ADC, AD + DC > AC ……….(ii)

In DCB, DC + CB > DB ……….(iii)

In ADB, AD + AB > DB ……….(iv)

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

AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB

 (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB

 2AB + 2BC + 2AD + 2DC > 2(AC + DB)

 2(AB + BC + AD + DC) > 2(AC + DB)

 AB + BC + AD + DC > AC + DB

 AB + BC + CD + DA > AC + DB

Hence, it is true.

Question 5:

ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

SOLUTION 5:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In AOB, AB < OA + OB ……….(i)

In BOC, BC < OB + OC ……….(ii)

In COD, CD < OC + OD ……….(iii)

In AOD, DA < OD + OA ……….(iv)

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

AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA

 AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD

 AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]

 AB + BC + CD + DA < 2(AC + BD)

Hence, it is proved.

The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

SOLUTION 6:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

It is given that two sides of triangle are 12 cm and 15 cm.

Therefore, the third side should be less than 12 + 15 = 27 cm.

And also the third side cannot be less than the difference of the two sides. Therefore, the third side has to be more than 15 – 12 = 3 cm.

Hence, the third side could be the length more than 3 cm and less than 27 cm.

Exercise 6.5

Question 1: PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

SOLUTION 1:

Given: PQ = 10 cm, PR = 24 cm Let QR be x cm.

In right angled triangle QPR,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

 (QR)2 = (PQ)2 + (PR)2

 x2 102 242

 x2 = 100 + 576 = 676

 x 676 = 26 cm

Thus, the length of QR is 26 cm.

Question 2: ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

SOLUTION 2:

Given: AB = 25 cm, AC = 7 cm Let BC be x cm.

In right angled triangle ACB,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

 (AB)2 = (AC)2 + (BC)2

 252 72 x2

 625 = 49 + x2

 x2 = 625 – 49 = 576

 x 576 = 24 cm

Thus, the length of BC is 24 cm.

A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

SOLUTION 3:

Let AC be the ladder and A be the window.

Given: AC = 15 m, AB = 12 m, CB = a m

In right angled triangle ACB,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

 (AC)2 = (CB)2 + (AB)2

 152 a2 122

 225 = a2 + 144

 a2 = 225 – 144 = 81

 a 81 = 9 cm

Thus, the distance of the foot of the ladder from the wall is 9 m.

Question 4:

Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm (ii) 2 cm, 2 cm, 5 cm

(iii) 1.5 cm, 2 cm, 2.5 cm In the case of right angled triangles, identify the right angles.

SOLUTION 4:

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

(i)

(ii)

(iii)

2.5 cm, 6.5 cm, 6 cm

In ABC, AC2 AB2 BC2

L.H.S. = 6.52 = 42.25 cm

R.H.S. = 62 2.52 = 36 + 6.25 = 42.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

2 cm, 2 cm, 5 cm

In the given triangle, 52 22 22

L.H.S. = 52 = 25

R.H.S. = 22 22 = 4 + 4 = 8

Since, L.H.S.  R.H.S.

Therefore, the given sides are not of the right angled triangle.

1.5 cm, 2 cm, 2.5 cm

In PQR, PR2 PQ2 RQ2

L.H.S. = 2.52 = 6.25 cm

R.H.S. = 1.52 22 = 2.25 + 4 = 6.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

SOLUTION 5:

Let A’CB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then ABC is a right angled triangle, right angled at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem, In ABC

AC2 AB2 BC2

 AC2 122 52

 AC2 14425

 AC2 169

 AC = 13 m

Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.

Question 6:

Angles Q and R of a PQR are 25 and 65 .

Write which of the following is true: P

(i) PQ2 + QR2 = RP2 (ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

65 R

SOLUTION 6:

In PQR,

 PQR +  QRP +  RPQ = 180 [By Angle sum property of a  ]

 25 65 RPQ = 180

 90RPQ=180

  RPQ = 180 90 90  

Thus, PQR is a right angled triangle, right angled at P.

 (Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

 QR2 PR2 QP2

Hence, Option (ii) is correct.

Question 7:

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

SOLUTION 7:

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be x cm.

Now, in right angled triangle PQR,

PR2 RQ2 PQ2 [By Pythagoras theorem]

2 2

 41  x2 40

 1681 = x2 + 1600

 x2 = 1681 – 1600

 x2 = 81

 x 81 9 cm

Therefore the breadth of the rectangle is 9 cm.

Perimeter of rectangle = 2(length + breadth)

= 2 (9 + 49)

= 2 x 49 = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

Question 8:

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

SOLUTION 8:

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

DB 16

Therefore, OD =  = 8 cm

2 2

AC 30 And OC =  = 15 cm

2 2

Now, In right angle triangle DOC,

DC2 OD2 OC2 [By Pythagoras theorem]

2 2 2

 DC 8 15

 DC2 = 64 + 225 = 289

 DC = 289 = 17 cm

Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm Thus, the perimeter of rhombus is 68 cm.

Class – 7 CH-5 LINES AND ANGLES MATHS NCERT SOLUTIONS

Class – 7 CH-5 LINES AND ANGLES

MATHS NCERT SOLUTIONS

Exercise 5.1

Question 1:

Find the complement of each of the following angles:

Question 2:

Find the supplement of each of the following angles:

Identify which of the following pairs of angles are complementary and which are supplementary:

(i) 65 ,115  (ii) 63,27 (iii) 112 ,68 

(iv) 130 ,50  (v) 45,45 (vi) 80,10

SOLUTION 3:

If sum of two angles is 180, then they are called supplementary angles.

If sum of two angles is 90, then they are called complementary angles.

(i) 65115180 These are supplementary angles.

(ii) 63  27 90 These are complementary angles.

(iii) 112 68 180 These are supplementary angles.

(iv) 130 50 180 These are supplementary angles.

(v) 45  45  90 These are complementary angles. (vi) 80  10 90 These are complementary angles.

Question 4: Find the angle which is equal to its complement.

SOLUTION 4:

Let one of the two equal complementary angles be x.

 x x 90

 2x  90

 x   45

Thus, 45 is equal to its complement.

Question 5: Find the angle which is equal to its supplement. SOLUTION 5:

Let x be two equal angles of its supplement.

Therefore, x x 180 [Supplementary angles]

 2x 180

180

 x 90

Thus, 90 is equal to its supplement.

In the given figure,  1 and  2 are supplementary angles. If  1 is decreased, what changes should take place in  2 so that both the angles still remain supplementary?

SOLUTION 6:

If 1 is decreased then,  2 will increase with the same measure, so that both the angles still remain supplementary.

Question 7:

Can two angles be supplementary if both of them are:

(i) acute (ii) obtuse (iii) right?

SOLUTION 7:

(i) No, because sum of two acute angles is less than 180 .

(ii) No, because sum of two obtuse angles is more than 180 .

(iii) Yes, because sum of two right angles is 180 .

Question 8:

An angle is greater than 45 . Is its complementary angle greater than 45 or equal to 45 or less than 45 ?

SOLUTION 8:

Let the complementary angles be x and y, i.e., x y  90

It is given that x  45

Adding y both sides, x y  45 y

 90  45 y

 90 45 y

 y 45

Thus, its complementary angle is less than 45 .

In the adjoining figure:

(i) Is  1 adjacent to  2?

(ii) Is  AOC adjacent to  AOE?

(iii) Do  COE and  EOD form a linear pair?

(iv) Are  BOD and  DOA supplementary?

(v) Is  1 vertically opposite to  4? (vi) What is the vertically opposite angle of  5?

SOLUTION 9:

(i) Yes, in  AOE, OC is common arm.

(ii) No, they have no non-common arms on opposite side of common arm.

(iii) Yes, they form linear pair.

(iv) Yes, they are supplementary.

(v) Yes, they are vertically opposite angles.

(vi) Vertically opposite angles of  5 is  COB.

Question 10:

Indicate which pairs of angles are:

(i) Vertically opposite angles? (ii) Linear pairs?

SOLUTION 10:

(i) Vertically opposite angles,  1 and 4;  5 and 2 +  3. (ii) Linear pairs  1 and 5;  5 and 4.

Question 11:

In the following figure, is  1 adjacent to  2? Give reasons.

SOLUTION 11:

 1 and  2 are not adjacent angles because their vertex is not common.

Question 12:

Find the values of the angles x y, and z in each of the following:

(i) (ii)

SOLUTION 12:

(i) x  55 [Vertically opposite angles]

Now 55 y 180 [Linear pair]

 y 180 55 125

Also y  z 125

Thus, x  55 , y 125 and z 125 .

[Vertically opposite angles]

(ii) 40  x 25 180 [Angles on straight line]

 65 x 180

 x 180 65 = 115

Now 40 y 180 [Linear pair]

 y 180 40 140 ……….(i)

Also y  z 180 [Linear pair]

 140 z 180 [From equation (i)]

 z 180140 40

Thus, x 115 , y 140 and z 40 .

Question 13:

Fill in the blanks:

(i) If two angles are complementary, then the sum of their measures is

_______________.

(ii) If two angles are supplementary, then the sum of their measures is

_______________.

(iii) Two angles forming a linear pair are _______________.

(iv) If two adjacent angles are supplementary, they form a _______________.

(v) If two lines intersect a point, then the vertically opposite angles are always _______________.

(vi) If two lines intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are

_______________.

SOLUTION 13:

(i) 90 (ii) 180 (iii) supplementary

(iv) linear pair (v) equal (vi) obtuse angles

Question 14:

In the adjoining figure, name the following pairs of angles:

(i) Obtuse vertically opposite angles.

(ii) Adjacent complementary angles.

(iii) Equal supplementary angles.

(iv) Unequal supplementary angles. (v) Adjacent angles that do not form a linear pair.

SOLUTION 14:

(i) Obtuse vertically opposite angles means greater than 90 and equal  AOD =  BOC.

(ii) Adjacent complementary angles means angles have common vertex, common arm, non-common arms are on either side of common arm and sum of angles is 90 .

(iii) Equal supplementary angles means sum of angles is 180 and supplement angles are equal.

(iv) Unequal supplementary angles means sum of angles is 180 and supplement angles are unequal.

i. e.,  AOE,  EOC;  AOD,  DOC and  AOB,  BOC

(v) Adjacent angles that do not form a linear pair mean, angles have common ray but the angles in a linear pair are not supplementary.

i. e.,  AOB,  AOE;  AOE,  EOD and  EOD,  COD

Exercise 5.2

Question 1:

State the property that is used in each of the following statements:

(i) If a||b, then  1 =  5. (ii) If  4 =  6, then a||b.

(iii) If  4 +  5 + 180, then a||b.

SOLUTION 1:

(i) Given, a||b, then  1 =  5 [Corresponding angles]

If two parallel lines are cut by a transversal, each pair of corresponding angles are equal in measure.

(ii) Given,  4 =  6, then a||b [Alternate interior angles] When a transversal cuts two lines such that pairs of alternate interior angles are equal, the lines have to be parallel.

(iii) Given,  4 +  5 = 180, then a||b [Co-interior Angles]

When a transversal cuts two lines, such that pairs of interior angles on the same side of transversal are supplementary, the lines have to be parallel.

Question 2:

In the adjoining figure, identify:

(i) the pairs of corresponding angles.

(ii) the pairs of alternate interior angles.

(iii) the pairs of interior angles on the same side of the transversal. (iv) the vertically opposite angles.

SOLUTION 2:

(i) The pairs of corresponding angles:

 1,  5;  2,  6;  4,  8 and  3,  7 (ii) The pairs of alternate interior angles are:

 3,  5 and  2,  8

(iii) The pair of interior angles on the same side of the transversal:

 3,  8 and  2,  5

(iv) The vertically opposite angles are:

 1,  3;  2,  4;  6,  8 and  5,  7

Question 3:

In the adjoining figure, p||q. Find the unknown angles.

SOLUTION 3:

Given, p||q and cut by a transversal line.

125 e 180 [Linear pair]

 e180125 55 ……….(i)

Now e  f 55 [Vertically opposite angles]

Also a  f 55 [Alternate interior angles]

a b 180 [Linear pair]

 55 b 180 [From equation (i)]

 b180 55 125

Now a   c 55 and b  d 125 [Vertically opposite angles]

Thus, a  55 ,b 125 ,c 55 ,d 125 ,e 55 and f  55 .

Question 4:

Find the values of x in each of the following figures if l||m

(i) Given, l||m and t is transversal line.

 Interior vertically opposite angle between lines l and t 110 .

 110 x 180 [Supplementary angles]

 x 180110 70

(ii) Given, l||m and t is transversal line.

x2x 180 [Interior opposite angles]

 3x 180

180

 x  60

(iii) Given, l||m and a||b.

x 100 [Corresponding angles]

Question 5:

In the given figure, the arms of two angles are parallel. If ABC = 70 , then find:

(i)  DGC (ii)  DEF

SOLUTION 5:

(i) Given, AB || DE and BC is a transversal line and ABC  70

 ABC =  DGC [Corresponding angles]

  DGC = 70 ……….(i)

(ii) Given, BC || EF and DE is a transversal line and DGC 70

 DGC =  DEF [Corresponding angles]

  DEF = 70 [From equation (i)]

Question 6:

In the given figures below, decide whether l is parallel to m.

SOLUTION 6:

(i) 126 44 170

l ||m because sum of interior opposite angles should be 180 .

(ii) 75 75 150

l ||m because sum of angles does not obey the property of parallel lines.

(iii) 57123180 l ||m due to supplementary angles property of parallel lines.

(iv) 98 72 170

l is not parallel to m because sum of angles does not obey the property of parallel lines.

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Monday, December 18, 2023