Circle

Question 1:

Exercise 11A

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Question 1:

Exercise 11B

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Question 1:

Exercise 11C

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GeometricalConstructions

Exercise 12A

Question 1:

Steps of Construction:

1. Draw a line segment AB = 5 cm
2. With A as centre and radius equal to more than half of AB, draw two arcs, one above AB and the other below AB.
3. With B as a centre and the same radius draw two arcs which cuts the previously drawn arcs at C and D.
4. Join CD , intersecting AB at point P.

∴ CD is the perpendicular bisector of AB at the point P.

Question 2:

Step of Construction:

1. Draw a line segment OA.
2. AT A, draw ∠AOE=90 , using ruler and compass.
3. With B as centre and radius more than half of BD, draw an arc.
4. With D as centre and same radius draw another arc which cuts the previous arc at F.
5. Join OF. ∴ ∠AOF=45 .
6. Now with centre B and radius more than half of BC, draw an arc.
7. With centre C and same radius draw another arc which cuts the previously drawn arc at X.
8. Join OX. ∴ OX is the bisector of ∠AOF.

Question 3:

Step of Construction:

1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc, cutting OA at B.
3. With B as centre and the same radius cut the previously drawn arc at C.
4. With C as centre and the same radius cut the arc at D.
5. With C as centre and the radius more than half CD draw an arc.
6. With D as centre and the same radius draw another arc which cuts the previous arc at E.
7. Join E Now, ∠AOE =900
8. Now with B as centre and radius more than half of CB draw an arc.

(iv) With C as centre and same radius draw an arc which cuts the previous at F.

1. Join OF.
2. ∴ F is the bisector of right ∠AOE.

Question 4:

Step of construction:

1. Draw a line segment BC=5cm.
2. With B as centre and radius equal to BC draw an arc.
3. With C as centre and the same radius draw another arc which cuts the previous arc at A.
4. Join AB and AC.

Then ∆ABC is the required equilateral triangle.

Question 5:

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Question 13:

Steps of Construction:

• 1. Draw BC = 4.5 cm.
  2. Construct ∠CBX = 600
  3. Along BX set off BP =8cm.
  4. Join CP.
  5. Draw the perpendicular bisector of CP to intersecting BP at A.
  6. Join AC. ∴ ∆ABC is the required triangle.

Question 14:

Steps of Construction:

1. Draw BC = 5.2 cm.
2. Construct ∠CBX = 300
3. Set off BP = 3.5 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP produced at A.
6. Join AC. ∴ ∆ABC is the required triangle.

Probability

Question 1:

Exercise 15A

Question 2:

Question 3:

Question 4:

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Question 6:

Number of tests in which he gets more than 60% marks =2 Total numbers of tests =6

∴ Required probability

Question 7:

Question 8:

Question 9:

Question 10:

Total numbers of students =30

Numbers of students who lie in the interval21-30=6

∴ Required probability

Question 11:

Area

Question 1:

Exercise 10A

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Quadrilateralsand Parallelograms

Question 1:

Exercise 9A

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Since AB || DC

Question 4:

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Question 8:

Given: O is a point within a quadrilateral ABCD

Question 9:

Given: ABCD is a quadrilateral and AC is one of its disgonals.

Question 10:

Given: ABCD is a quadrilateral.

Question 1:

Exercise 9B

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Question 1:

Exercise 9C

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∴ PQRS is a rhombus.

Question 10:

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LinearEquations in Two Variables

Exercise 8A

Question 1:

1. The given equation is x = 5

Take two solutions of the given equation as x = 5, y = 1 and x = 5, y = -1 Thus we get the following table:

Plot points P(5,1) and Q(5,-1) on the graph paper. Join PQ. The line PQ is the required graph.

1. The given equation is y = -2

Take two solutions of the given equation as x = 1, y = -2 and x = 2, y = -2. Thus we have the following table:

Plot points P(1,-2) and Q(2,-2) on the graph paper. Join PQ. The line PQ is the required

graph.

1. The given equation is x + 6 = 0

⇒ x = -6

Let x = -6 & y = 1 x = -6 & y = -1

Plot points P(-6,1) and Q(-6,-1) on the graph paper. Join PQ. The line PQ is the required graph.

1. The given equation is x + 7 = 0

⇒ x = -7

Let x = -7, y = 2 and x = -7, y = 1 Thus we have the following table:

Plot points P(-7,2) and Q(-7,1) on the graph paper. Join PQ. The line PQ is the required graph.

1. y = 0 represents the x-axis
2. x = 0 represents the y-axis.

Question 2:

The given equation is y = 3x. Putting x = 1, y = 3 (1) = 3

Putting x = 2, y = 3 (2) = 6

Thus, we have the following table:

Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

So, y = ON = -6.

Question 3:

The given equation is, x + 2y – 3 = 0

⇒ x = 3 – 2y

Putting y = 1, x = 3 – (2 × 1) = 1

Putting y = 0, x = 3 – (2 × 0) = 3 Thus, we have the following table:

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P. Through P, draw PM parallel to x-axis cutting y-axis at M. So, y = OM = -1.

Question 4:

1. The given equation is y = x

Let x = 1, then y = 1 and let x = 2, then y = 2 Thus, we have the following table:

Plot points (1,1) and (2,2) on a graph paper and join them to get the required graph.

1. The given equation is y = -x

Now, if x = 1, y = -1 and if x = 2, y = -2 Thus, we have the following table:

Plot points (1,-1) and (2,-2) on a graph paper and join them to get the required graph.

1. The given equation is y + 3x = 0

⇒ y = -3x

Now, if x = -1, then y = -3 (-1) = 3

And, if x = 1, then y = -3 (1) = -3 Thus we have the following table:

Plot points (1,-3) and (-1,3) on a graph paper and join them to get the required graph.

1. The given equation is 2x + 3y = 0

⇒ y =

x

Now, if x = 3, then y =

× 3 = -2

And, if x = -3, then y =

× (-3) = 2

Thus, we have the following table

Plot points (3,-2) and (-3,2) on a graph paper and join them to get the required graph.

1. The given equation is 3x – 2y = 0

⇒ y =

x Now, if x = 2, y =

× 2 = 3

And, if x = -2,

y =

× (-2) = -3

Thus, we have the following table:

Plot points (2,3) and (-2,-3) on a graph paper and join them to get the required graph.

1. The given equation is 2x + y = 0

⇒ y = -2x

Now, if x = 1, then y = -2 1 = -2

And, if x = -1, then y = -2 (-1) = 2 Thus, we have the following table:

Plot points (1,-2) and (-1,2) on a graph paper and join them to get the required graph.

Question 5:

The given equation is, 2x – 3y = 5

⇒ y =

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

1. When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

1. When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.

Question 6:

The given equation is 2x + y = 6

⇒ y = 6 – 2x

Now, if x = 1, then y = 6 – 2 (1) = 4

And, if x = 2, then y = 6 – 2 (2) = 2 Thus, we have the following table:

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).

Question 7:

The given equation is 3x + 2y = 6

⇒ 2y = 6 – 3x

⇒ y =

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x- axis.

So, co-ordinates of P are (0,3).

Areas

Question 1:

Exercise 7A

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Question 5:

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Question 7:

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Question 15:

(i) Area of an equilateral triangle =

Where a is the side of the equilateral triangle

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Question 24:

Angles, Linesand Triangles

Question 1:

Exercise 4A

1. Angle: Two rays having a common end point form an angle.
2. Interior of an angle: The interior of ∠AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.
3. Obtuse angle: An angle whose measure is more than 90° but less than 180°, is called an obtuse angle.
4. Reflex angle: An angle whose measure is more than 180° but less than 360° is called a reflex angle.
5. Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.
6. Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180°.

Question 2:

∠A = 36° 27′ 46″ and ∠B = 28° 43′ 39″

∴ Their sum = (36° 27′ 46″) + (28° 43′ 39″)

Therefore, the sum ∠A + ∠B = 65° 11′ 25″

Question 3:

Let ∠A = 36° and ∠B = 24° 28′ 30″ Their difference = 36° – 24° 28′ 30″

Thus the difference between two angles is ∠A – ∠B = 11° 31′ 30″

Question 4:

• 1. Complement of 58o = 90o – 58o = 32o
  2. Complement of 16o = 90 – 16o = 74o
  3. 
     of a right angle =
     
     × 90o = 60o Complement of 60o = 90o – 60o = 30o
  4. 1o = 60′

⇒ 90o = 89o 60′

Complement of 46o 30′ = 90o – 46o 30′ = 43o 30′ (v) 90o = 89o 59′ 60″

Complement of 52o 43′ 20″ = 90o – 52o 43′ 20″

= 37o 16′ 40″

(vi) 90o = 89o 59′ 60″

∴ Complement of (68o 35′ 45″)

= 90o – (68o 35′ 45″)

= 89o 59′ 60″ – (68o 35′ 45″)

= 21o 24′ 15″

Question 5:

1. Supplement of 63o = 180o – 63o = 117o
2. Supplement of 138o = 180o – 138o = 42o
3. 
   of a right angle =
   
   × 90o = 54o

∴ Supplement of 54o = 180o – 54o = 126o

1. 1o = 60′

⇒ 180o = 179o 60′

Supplement of 75o 36′ = 180o – 75o 36′ = 104o 24′ (v) 1o = 60′, 1′ = 60″

⇒ 180o = 179o 59′ 60″

Supplement of 124o 20′ 40″ = 180o – 124o 20′ 40″

= 55o 39′ 20″

(vi) 1o = 60′, 1′ = 60″

⇒ 180o = 179o 59′ 60″

∴ Supplement of 108o 48′ 32″ = 180o – 108o 48′ 32″

= 71o 11′ 28″.

Question 6:

• 1. 
     Let the required angle be xo Then, its complement = 90o – xo

∴ The measure of an angle which is equal to its complement is 45o.

• 1. 
     Let the required angle be xo Then, its supplement = 180o – xo

∴ The measure of an angle which is equal to its supplement is 90o.

Question 7:

Let the required angle be xo Then its complement is 90o – xo

∴ The measure of an angle which is 36o more than its complement is 63o.

Question 8:

Let the required angle be xo Then its supplement is 180o – xo

∴ The measure of an angle which is 25o less than its supplement is

Question 9:

Let the required angle be xo Then, its complement = 90o – xo

∴ The required angle is 72o.

Question 10:

Let the required angle be xo Then, its supplement is 180o – xo

∴ The required angle is 150o.

Question 11:

Let the required angle be xo

Then, its complement is 90o – xo and its supplement is 180o – xo That is we have,

∴ The required angle is 60o.

Question 12:

Let the required angle be xo

Then, its complement is 90o – xo and its supplement is 180o – xo

∴ The required angle is 45o.

Question 13:

Let the two required angles be xo and 180o – xo. Then,

⇒ 2x = 3(180 – x)

⇒ 2x = 540 – 3x

⇒ 3x + 2x = 540

⇒ 5x = 540

⇒ x = 108

Thus, the required angles are 108o and 180o – xo = 180 o – 108o = 72o.

Question 14:

Let the two required angles be xo and 90o – xo. Then

⇒ 5x = 4(90 – x)

⇒ 5x = 360 – 4x

⇒ 5x + 4x = 360

⇒ 9x = 360

⇒ x =

= 40

Thus, the required angles are 40o and 90o – xo = 90 o – 40o = 50o.

Question 15:

Let the required angle be xo.

Then, its complementary and supplementary angles are (90o – x) and (180o – x) respectively.

Then, 7(90o – x) = 3 (180o – x) – 10o

⇒ 630o – 7x = 540o – 3x – 10o

⇒ 7x – 3x = 630o – 530o

⇒ 4x = 100o

⇒ x = 25o

Thus, the required angle is 25o.

Question 1:

Exercise 4B

Since ∠BOC and ∠COA form a linear pair of angles, we have

∠BOC + ∠COA = 180o

⇒ xo + 62o = 180o

⇒ x = 180 – 62

∴ x = 118o

Question 2:

Since, ∠BOD and ∠DOA form a linear pair.

∠BOD + ∠DOA = 180o

∴ ∠BOD + ∠DOC + ∠COA = 180o

⇒ (x + 20)o + 55o + (3x – 5)o = 180o

⇒ x + 20 + 55 + 3x – 5 = 180

⇒ 4x + 70 = 180

⇒ 4x = 180 – 70 = 110

⇒ x =

= 27.5

∴ ∠AOC = (3 × 27.5 – 5)o = 82.5-5 = 77.5o And, ∠BOD = (x + 20)o = 27.5o + 20o = 47.5o.

Question 3:

Since ∠BOD and ∠DOA from a linear pair of angles.

⇒ ∠BOD + ∠DOA = 180o

⇒ ∠BOD + ∠DOC + ∠COA = 180o

⇒ xo + (2x – 19)o + (3x + 7)o = 180o

⇒ 6x – 12 = 180

⇒ 6x = 180 + 12 = 192

⇒ x =

= 32

⇒ x = 32

⇒ ∠AOC = (3x + 7)o = (3 32 + 7)o = 103o

⇒ ∠COD = (2x – 19)o = (2 32 – 19)o = 45o

and ∠BOD = xo = 32o

Question 4:

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15 But x + y + z = 180o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5. If the sum of angles is 180o, then, measure of x =

× 180 = 60

And, if the total sum of the measures is 15, then the measure of y is 4. If the sum of the angles is 180o, then, measure of y =

× 180 = 48 And ∠z = 180o – ∠x – ∠y

= 180o – 60o – 48o

= 180o – 108o = 72o

∴ x = 60, y = 48 and z = 72.

Question 5:

AOB will be a straight line, if two adjacent angles form a linear pair.

∴ ∠BOC + ∠AOC = 180o

⇒ (4x – 36)o + (3x + 20)o = 180o

⇒ 4x – 36 + 3x + 20 = 180

⇒ 7x – 16 = 180o

⇒ 7x = 180 + 16 = 196

⇒ x =

= 28

∴ The value of x = 28.

Question 6:

Since ∠AOC and ∠AOD form a linear pair.

∴ ∠AOC + ∠AOD = 180o

⇒ 50o + ∠AOD = 180o

⇒ ∠AOD = 180o – 50o = 130o

∠AOD and ∠BOC are vertically opposite angles.

∠AOD = ∠BOC

⇒ ∠BOC = 130o

∠BOD and ∠AOC are vertically opposite angles.

∴ ∠BOD = ∠AOC

⇒ ∠BOD = 50o

Question 7:

Since ∠COE and ∠DOF are vertically opposite angles, we have,

∠COE = ∠DOF

⇒ ∠z = 50o

Also ∠BOD and ∠COA are vertically opposite angles. So, ∠BOD = ∠COA

⇒ ∠t = 90o

As ∠COA and ∠AOD form a linear pair,

∠COA + ∠AOD = 180o

⇒ ∠COA + ∠AOF + ∠FOD = 180o [∠t = 90o]

⇒ t + x + 50o = 180o

⇒ 90o + xo + 50o = 180o

⇒ x + 140 = 180

⇒ x = 180 – 140 = 40

Since ∠EOB and ∠AOF are vertically opposite angles So, ∠EOB = ∠AOF

⇒ y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90

Question 8:

Since ∠COE and ∠EOD form a linear pair of angles.

⇒ ∠COE + ∠EOD = 180o

⇒ ∠COE + ∠EOA + ∠AOD = 180o

⇒ 5x + ∠EOA + 2x = 180

⇒ 5x + ∠BOF + 2x = 180

[∴ ∠EOA and BOF are vertically opposite angles so, ∠EOA = ∠BOF]

⇒ 5x + 3x + 2x = 180

⇒ 10x = 180

⇒ x = 18

Now ∠AOD = 2xo = 2 × 18o = 36o

∠COE = 5xo = 5 × 18o = 90o

and, ∠EOA = ∠BOF = 3xo = 3 × 18o = 54o

Question 9:

Let the two adjacent angles be 5x and 4x. Now, since these angles form a linear pair. So, 5x + 4x = 180o

⇒ 9x = 180o

⇒ x =

= 20

∴ The required angles are 5x = 5x = 5 20o = 100o and 4x = 4 × 20o = 80o

Question 10:

Let two straight lines AB and CD intersect at O and let ∠AOC = 90o.

Now, ∠AOC = ∠BOD [Vertically opposite angles]

⇒ ∠BOD = 90o

Also, as ∠AOC and ∠AOD form a linear pair.

⇒ 90o + ∠AOD = 180o

⇒ ∠AOD = 180o – 90o = 90o

Since, ∠BOC = ∠AOD [Verticallty opposite angles]

⇒ ∠BOC = 90o

Thus, each of the remaining angles is 90o.

Question 11:

Since, ∠AOD and ∠BOC are vertically opposite angles.

∴ ∠AOD = ∠BOC

Now, ∠AOD + ∠BOC = 280o [Given]

⇒ ∠AOD + ∠AOD = 280o

⇒ 2∠AOD = 280o

⇒ ∠AOD =

= 140o

⇒ ∠BOC = ∠AOD = 140o

As, ∠AOC and ∠AOD form a linear pair. So, ∠AOC + ∠AOD = 180o

⇒ ∠AOC + 140o = 180o

⇒ ∠AOC = 180o – 140o = 40o

Since, ∠AOC and ∠BOD are vertically opposite angles.

∴ ∠AOC = ∠BOD

⇒ ∠BOD = 40o

∴ ∠BOC = 140o, ∠AOC = 40o , ∠AOD = 140o and ∠BOD = 40o.

Question 12:

Since ∠COB and ∠BOD form a linear pair So, ∠COB + ∠BOD = 180o

⇒ ∠BOD = 180o – ∠COB …. (1)

Also, as ∠COA and ∠AOD form a linear pair. So, ∠COA + ∠AOD = 180o

⇒ ∠AOD = 180o – ∠COA

⇒ ∠AOD = 180o – ∠COB …. (2)

[Since, OC is the bisector of ∠AOB, ∠BOC = ∠AOC] From (1) and (2), we get,

∠AOD = ∠BOD (Proved)

Question 13:

Let QS be a perpendicular to AB. Now, ∠PQS = ∠SQR

Because angle of incident = angle of reflection

⇒ ∠PQS = ∠SQR =

= 56o

Since QS is perpendicular to AB, ∠PQA and ∠PQS are complementary angles. Thus, ∠PQA + ∠PQS = 90o

⇒ ∠PQA + 56o = 90o

⇒ ∠PQA = 90o – 56o = 34o

Question 14:

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the

∠BOD. OF is a ray opposite to ray OE.

To Prove: ∠AOF = ∠COF

Proof : Since

and

are two opposite rays,

is a straight line passing through O.

∴ ∠AOF = ∠BOE and ∠COF = ∠DOE

[Vertically opposite angles] But ∠BOE = ∠DOE (Given)

∴ ∠AOF = ∠COF

Hence, proved.

Question 15:

Given:

is the bisector of ∠BCD and

is the bisector of ∠ACD. To Prove: ∠ECF = 90o

Proof: Since ∠ACD and ∠BCD forms a linear pair.

∠ACD + ∠BCD = 180o

∠ACE + ∠ECD + ∠DCF + ∠FCB = 180o

∠ECD + ∠ECD + ∠DCF + ∠DCF = 180o

because ∠ACE = ∠ECD and ∠DCF = ∠FCB

2(∠ECD) + 2 (∠CDF) = 180o

2(∠ECD + ∠DCF) = 180o

∠ECD + ∠DCF =

= 90o

∠ECF = 90o (Proved)

Question 1:

Exercise 4C

Since AB and CD are given to be parallel lines and t is a transversal. So, ∠5 = ∠1 = 70o [Corresponding angles are equal]

∠3 = ∠1 = 70o [Vertically opp. Angles]

∠3 + ∠6 = 180o [Co-interior angles on same side]

∴ ∠6 = 180o – ∠3

= 180o – 70o = 110o

∠6 = ∠8 [Vertically opp. Angles]

⇒ ∠8 = 110o

⇒ ∠4 + ∠5 = 180o [Co-interior angles on same side]

∠4 = 180o – 70o = 110o

∠2 = ∠4 = 110o [ Vertically opposite angles]

∠5 = ∠7 [Vertically opposite angles] So, ∠7 = 70o

∴ ∠2 = 110o, ∠3 = 70o , ∠4 = 110o, ∠5 = 70o, ∠6 = 110o, ∠7 = 70o and ∠8 = 110o.

Question 2:

Since ∠2 : ∠1 = 5 : 4.

Let ∠2 and ∠1 be 5x and 4x respectively.

Now, ∠2 + ∠1 = 180o , because ∠2 and ∠1 form a linear pair. So, 5x + 4x = 180o

⇒ 9x = 180o

⇒ x = 20o

∴ ∠1 = 4x = 4 × 20o = 80o And ∠2 = 5x = 5 × 20o = 100o

∠3 = ∠1 = 80o [Vertically opposite angles]

And ∠4 = ∠2 = 100o [Vertically opposite angles]

∠1 = ∠5 and ∠2 = ∠6 [Corresponding angles] So, ∠5 = 80o and ∠6 = 100o

∠8 = ∠6 = 100o [Vertically opposite angles] And ∠7 = ∠5 = 80o [Vertically opposite angles]

Thus, ∠1 = 80o, ∠2 = 100o, ∠3 = ∠80o, ∠4 = 100o, ∠5 = 80o, ∠6 = 100o, ∠7 = 80o and

∠8 = 100o.

Question 3:

Given: AB || CD and AD || BC To Prove: ∠ADC = ∠ABC

Proof: Since AB || CD and AD is a transversal. So sum of consecutive interior angles is

180o.

⇒ ∠BAD + ∠ADC = 180o ….(i)

Also, AD || BC and AB is transversal. So, ∠BAD + ∠ABC = 180o ….(ii) From (i) and (ii) we get:

∠BAD + ∠ADC = ∠BAD + ∠ABC

⇒ ∠ADC = ∠ABC (Proved)

Question 4:

1. 
   Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So, ∠GED = ∠EDC = 65o [Alternate interior angles] Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So, ∠BEG = ∠ABE = 35o [Alternate interior angles] So, ∠DEB = xo

⇒ ∠BEG + ∠GED = 35o + 65o = 100o.

Hence, x = 100.

1. 
   Through O draw OF||CD.

Now since OF || CD and OD is transversal.

∠CDO + ∠FOD = 180o

[sum of consecutive interior angles is 180o]

⇒ 25o + ∠FOD = 180o

⇒ ∠FOD = 180o – 25o = 155o

As OF || CD and AB || CD [Given] Thus, OF || AB and OB is a transversal.

So, ∠ABO + ∠FOB = 180o [sum of consecutive interior angles is 180o]

⇒ 55o + ∠FOB = 180o

⇒ ∠FOB = 180o – 55o = 125o

Now, xo = ∠FOB + ∠FOD = 125o + 155o = 280o.

Hence, x = 280.

1. 
   Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

∠FEC + ∠ECD = 180o

[sum of consecutive interior angles is 180o]

⇒ ∠FEC + 124o = 180o

⇒ ∠FEC = 180o – 124o = 56o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So, ∠BAE + ∠FEA = 180o

[sum of consecutive interior angles is 180o]

∴ 116o + ∠FEA = 180o

⇒ ∠FEA = 180o – 116o = 64o

Thus, xo = ∠FEA + ∠FEC

= 64o + 56o = 120o.

Hence, x = 120.

Question 5:

Since AB || CD and BC is a transversal.

So, ∠ABC = ∠BCD [atternate interior angles]

⇒ 70o = xo + ∠ECD ….(i)

Now, CD || EF and CE is transversal.

So, ∠ECD + ∠CEF = 180o

∴ ∠ECD + 130o = 180o

[sum of consecutive interior angles is 180o]

⇒ ∠ECD = 180o – 130o = 50o

Putting ∠ECD = 50o in (i) we get, 70o = xo + 50o

⇒ x = 70 – 50 = 20

Question 6:

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, ∠GCE = ∠CEA = 20o

∴ ∠DCG = 130o – ∠GCE

= 130o – 20o = 110o

[Alternate angles]

Also, we have AB || CD and FG is a transversal.

So, ∠BFC = ∠DCG = 110o

As, FG || AE, AF is a transversal.

[Corresponding angles]

∠BFG = ∠FAE [Corresponding angles]

∴ xo = ∠FAE = 110o. Hence, x = 110

Question 7:

Given: AB || CD

To Prove: ∠BAE – ∠DCE = ∠AEC

Construction : Through E draw EF || AB Proof : Since EF || AB, AE is a transversal. So, ∠BAE + ∠AEF = 180O ….(i)

[sum of consecutive interior angles is 180o] As EF || AB and AB || CD [Given]

So, EF || CD and EC is a transversal.

So, ∠FEC + ∠DCE = 180o ….(ii)

[sum of consecutive interior angles is 180o] From (i) and (ii) we get,

∠BAE + ∠AEF = ∠FEC + ∠DCE

⇒ ∠BAE – ∠DCE = ∠FEC – ∠AEF = ∠AEC [Proved]

Question 8:

Since AB || CD and BC is a transversal.

So, ∠BCD = ∠ABC = xo

[Alternate angles]

As BC || ED and CD is a transversal.

∠BCD + ∠EDC = 180o

⇒ ∠BCD + 75o =180o

⇒ ∠BCD = 180o – 75o = 105o

∠ABC = 105o

∴ xo = ∠ABC = 105o Hence, x = 105.

[since ∠BCD = ∠ABC]

Question 9:

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

⇒ ∠KFG = ∠FGD = ro …. (i) [alternate angles]

Again AE || KF, and EF is a transversal.

So, ∠AEF + ∠KFE = 180o

∠KFE = 180o – po …. (ii) Adding (i) and (ii) we get,

∠KFG + ∠KFE = 180 – p + r

⇒ ∠EFG = 180 – p + r

⇒ q = 180 – p + r i.e., p + q – r = 180

Question 10:

Since AB || PQ and EF is a transversal.

So, ∠CEB = ∠EFQ [Corresponding angles]

⇒ ∠EFQ = 75o

⇒ ∠EFG + ∠GFQ = 75o

⇒ 25o + yo = 75o

⇒ y = 75 – 25 = 50

Also, ∠BEF + ∠EFQ = 180o

∠BEF = 180o – ∠EFQ

= 180o – 75o

∠BEF = 105o

[sum of consecutive interior angles is 180o]

∴ ∠FEG + ∠GEB = ∠BEF = 105o

⇒ ∠FEG = 105o – ∠GEB = 105o – 20o = 85o

In ∆EFG we have,

xo + 25o + ∠FEG = 180o

Hence, x = 70.

Question 11:

Since AB || CD and AC is a transversal.

So, ∠BAC + ∠ACD = 180o

⇒ ∠ACD = 180o – ∠BAC

= 180o – 75o = 105o

[sum of consecutive interior angles is 180o]

⇒ ∠ECF = ∠ACD [Vertically opposite angles]

∠ECF = 105o

Now in ∆CEF,

∠ECF + ∠CEF + ∠EFC =180o

⇒ 105o + xo + 30o = 180o

⇒ x = 180 – 30 – 105 = 45

Hence, x = 45.

Question 12:

Since AB || CD and PQ a transversal.

So, ∠PEF = ∠EGH [Corresponding angles]

⇒ ∠EGH = 85o

∠EGH and ∠QGH form a linear pair. So, ∠EGH + ∠QGH = 180o

⇒ ∠QGH = 180o – 85o = 95o

Similarly, ∠GHQ + 115o = 180o

⇒ ∠GHQ = 180o – 115o = 65o

In ∆GHQ, we have, xo + 65o + 95o = 180o

⇒ x = 180 – 65 – 95 = 180 – 160

∴ x = 20

Question 13:

Since AB || CD and BC is a transversal. So, ∠ABC = ∠BCD

⇒ x = 35

Also, AB || CD and AD is a transversal. So, ∠BAD = ∠ADC

⇒ z = 75

In ∆ABO, we have,

∠AOB + ∠BAO + ∠BOA = 180o

⇒ xo + 75o + yo = 180o

⇒ 35 + 75 + y = 180

⇒ y = 180 – 110 = 70

∴ x = 35, y = 70 and z = 75.

Question 14:

Since AB || CD and PQ is a transversal. So, y = 75 [Alternate angle]

Since PQ is a transversal and AB || CD, so x + APQ = 180o

[Sum of consecutive interior angles]

⇒ xo = 180o – APQ

⇒ x = 180 – 75 = 105

Also, AB || CD and PR is a transversal.

So, ∠APR = ∠PRD [Alternate angle]

⇒ ∠APQ + ∠QPR = ∠PRD [Since ∠APR = ∠APQ + ∠QPR]

⇒ 75o + zo = 125o

⇒ z = 125 – 75 = 50

∴ x = 105, y = 75 and z = 50.

Question 15:

∠PRQ = xo = 60o

[vertically opposite angles]

Since EF || GH, and RQ is a transversal. So, ∠x = ∠y [Alternate angles]

⇒ y = 60

AB || CD and PR is a transversal.

So, ∠PRD = ∠APR [Alternate angles]

⇒ ∠PRQ + ∠QRD = ∠APR [since ∠PRD = ∠PRQ + ∠QRD]

⇒ x + ∠QRD = 110o

⇒ ∠QRD = 110o – 60o = 50o

In ∆QRS, we have,

∠QRD + to + yo = 180o

⇒ 50 + t + 60 = 180

⇒ t = 180 – 110 = 70

Since, AB || CD and GH is a transversal So, zo = to = 70o [Alternate angles]

∴ x = 60 , y = 60, z = 70 and t = 70

Question 16:

1. Lines l and m will be parallel if 3x – 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

⇒ 3x – 2x = 10 + 20

⇒ x = 30

1. Lines will be parallel if (3x + 5)o + 4xo = 180o

[if sum of pairs of consecutive interior angles is 180o, the lines are parallel] So, (3x + 5) + 4x = 180

⇒ 3x + 5 + 4x = 180

⇒ 7x = 180 – 5 = 175

⇒ x =

= 25

Question 17:

Given: Two lines m and n are perpendicular to a given line l.

To Prove: m || n Proof : Since m ⊥ l So, ∠1 = 90o Again, since n ⊥ l

∠2 = 90o

∴ ∠1 = ∠2 = 90o

But ∠1 and ∠2 are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.

Thus, m || n.

Question 1:

Exercise 4D

Since, sum of the angles of a triangle is 180o

∠A + ∠B + ∠C = 180o

⇒ ∠A + 76o + 48o = 180o

⇒ ∠A = 180o – 124o = 56o

∴ ∠A = 56o

Question 2:

Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o. Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180o ]

⇒ 9x = 180

⇒ x =

= 20

∴ The measures of the required angles are: 2x = (2 × 20)o = 40o

3x = (3 × 20)o = 60o

4x = (4 × 20)o = 80o

Question 3:

Let 3∠A = 4∠B = 6∠C = x (say) Then, 3∠A = x

⇒ ∠A =

4∠B = x

⇒ ∠B =

and 6∠C = x

⇒ ∠C =

As ∠A + ∠B + ∠C = 180o

Question 4:

∠A + ∠B = 108o [Given]

But as ∠A, ∠B and ∠C are the angles of a triangle,

∠A + ∠B + ∠C = 180o

⇒ 108o + ∠C = 180o

⇒ C = 180o – 108o = 72o

Also, ∠B + ∠C = 130o [Given]

⇒ ∠B + 72o = 130o

⇒ ∠B = 130o – 72o = 58o Now as, ∠A + ∠B = 108o

⇒ ∠A + 58o = 108o

⇒ ∠A = 108o – 58o = 50o

∴ ∠A = 50o, ∠B = 58o and ∠C = 72o.

Question 5:

Since. ∠A , ∠B and ∠C are the angles of a triangle . So, ∠A + ∠B + ∠C = 180o

Now, ∠A + ∠B = 125o [Given]

∴ 125o + ∠C = 180o

⇒ ∠C = 180o – 125o = 55o

Also, ∠A + ∠C = 113o [Given]

⇒ ∠A + 55o = 113o

⇒ ∠A = 113o – 55o = 58o Now as ∠A + ∠B = 125o

⇒ 58o + ∠B = 125o

⇒ ∠B = 125o – 58o = 67o

∴ ∠A = 58o, ∠B = 67o and ∠C = 55o.

Question 6:

Since, ∠P, ∠Q and ∠R are the angles of a triangle. So, ∠P + ∠Q + ∠R = 180o ….(i)

Now, ∠P – ∠Q = 42o [Given]

⇒ ∠P = 42o + ∠Q ….(ii)

and ∠Q – ∠R = 21o [Given]

⇒ ∠R = ∠Q – 21o ….(iii)

Substituting the value of ∠P and ∠R from (ii) and (iii) in (i), we get,

⇒ 42o + ∠Q + ∠Q + ∠Q – 21o = 180o

⇒ 3∠Q + 21o = 180o

⇒ 3∠Q = 180o – 21o = 159o

∠Q =

= 53o

∴ ∠P = 42o + ∠Q

= 42o + 53o = 95o

∠R = ∠Q – 21o

= 53o – 21o = 32o

∴ ∠P = 95o, ∠Q = 53o and ∠R = 32o.

Question 7:

Given that the sum of the angles A and B of a ABC is 116o, i.e., ∠A + ∠B = 116o. Since, ∠A + ∠B + ∠C = 180o

So, 116o + ∠C = 180o

⇒ ∠C = 180o – 116o = 64o

Also, it is given that:

∠A – ∠B = 24o

⇒ ∠A = 24o + ∠B

Putting, ∠A = 24o + ∠B in ∠A + ∠B = 116o, we get,

⇒ 24o + ∠B + ∠B = 116o

⇒ 2∠B + 24o = 116o

⇒ 2∠B = 116o – 24o = 92o

∠B =

= 46o

Therefore, ∠A = 24o + 46o = 70o

∴ ∠A = 70o, ∠B = 46o and ∠C = 64o.

Question 8:

Let the two equal angles, A and B, of the triangle be xo each. We know,

∠A + ∠B + ∠C = 180o

⇒ xo + xo + ∠C = 180o

⇒ 2xo + ∠C = 180o ….(i) Also, it is given that,

∠C = xo + 18o ….(ii)

Substituting ∠C from (ii) in (i), we get,

⇒ 2xo + xo + 18o = 180o

⇒ 3xo = 180o – 18o = 162o x =

= 54o

Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.

Question 9:

Let ∠C be the smallest angle of ABC. Then, ∠A = 2∠C and B = 3∠C

Also, ∠A + ∠B + ∠C = 180o

⇒ 2∠C + 3∠C + ∠C = 180o

⇒ 6∠C = 180o

⇒ ∠C = 30o

So, ∠A = 2∠C = 2 (30o) = 60o

∠B = 3∠C = 3 (30o) = 90o

∴ The required angles of the triangle are 60o, 90o, 30o.

Question 10:

Let ABC be a right angled triangle and ∠C = 90o Since, ∠A + ∠B + ∠C = 180o

⇒ ∠A + ∠B = 180o – ∠C = 180o – 90o = 90o

Suppose ∠A = 53o

Then, 53o + ∠B = 90o

⇒ ∠B = 90o – 53o = 37o

∴ The required angles are 53o, 37o and 90o.

Question 11:

Let ABC be a triangle. Given, ∠A + ∠B = ∠C

We know, ∠A + ∠B + ∠C = 180o

⇒ ∠C + ∠C = 180o

⇒ 2∠C = 180o

⇒ ∠C =

= 90o

So, we find that ABC is a right triangle, right angled at C.

Question 12:

Given : ∆ABC in which ∠A = 90o, AL ⊥ BC To Prove: ∠BAL = ∠ACB

Proof :

In right triangle ∆ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180o

⇒ ∠ABC + 90o + ∠ACB = 180o

⇒ ∠ABC + ∠ACB = 180o – 90o

∴ ∠ABC + ∠ACB = 90o

⇒ ∠ ACB = 90o – ∠ABC ….(1)

Similarly since ∆ABL is a right triangle, we find that,

∠BAL = 90o – ∠ABC …(2)

Thus from (1) and (2), we have

∴ ∠BAL = ∠ACB (Proved)

Question 13:

Let ABC be a triangle. So, ∠A < ∠B + ∠C

Adding A to both sides of the inequality,

⇒ 2∠A < ∠A + ∠B + ∠C

⇒ 2∠A < 180o

⇒ ∠A <

= 90o

[Since ∠A + ∠B + ∠C = 180o]

Similarly, ∠B < ∠A + ∠C

⇒ ∠B < 90o

and ∠C < ∠A + ∠B

⇒ ∠C < 90o

∆ABC is an acute angled triangle.

Question 14:

Let ABC be a triangle and ∠B > ∠A + ∠C Since, ∠A + ∠B + ∠C = 180o

⇒ ∠A + ∠C = 180o – ∠B Therefore, we get

∠B > 180o – ∠B

Adding ∠B on both sides of the inequality, we get,

⇒ ∠B + ∠B > 180o – ∠B + ∠B

⇒ 2∠B > 180o

⇒ ∠B >

= 90o

i.e., ∠B > 90o which means ∠B is an obtuse angle.

∆ABC is an obtuse angled triangle.

Question 15:

Since ∠ACB and ∠ACD form a linear pair. So, ∠ACB + ∠ACD = 180o

⇒ ∠ACB + 128o = 180o

⇒ ∠ACB = 180o – 128 = 52o

Also, ∠ABC + ∠ACB + ∠BAC = 180o

⇒ 43o + 52o + ∠BAC = 180o

⇒ 95o + ∠BAC = 180o

⇒ ∠BAC = 180o – 95o = 85o

∴ ∠ACB = 52o and ∠BAC = 85o.

Question 16:

As ∠DBA and ∠ABC form a linear pair. So, ∠DBA + ∠ABC = 180o

⇒ 106o + ∠ABC = 180o

⇒ ∠ABC = 180o – 106o = 74o

Also, ∠ACB and ∠ACE form a linear pair. So, ∠ACB + ∠ACE = 180o

⇒ ∠ACB + 118o = 180o

⇒ ∠ACB = 180o – 118o = 62o

In ∠ABC, we have,

∠ABC + ∠ACB + ∠BAC = 180o 74o + 62o + ∠BAC = 180o

⇒ 136o + ∠BAC = 180o

⇒ ∠BAC = 180o – 136o = 44o

∴ In triangle ABC, ∠A = 44o, ∠B = 74o and ∠C = 62o

Question 17:

(i) ∠EAB + ∠BAC = 180o [Linear pair angles]

110o + ∠BAC = 180o

⇒ ∠BAC = 180o – 110o = 70o

Again, ∠BCA + ∠ACD = 180o [Linear pair angles]

⇒ ∠BCA + 120o = 180o

⇒ ∠BCA = 180o – 120o = 60o

Now, in ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180o

xo + 70o + 60o = 180o

⇒ x + 130o = 180o

⇒ x = 180o – 130o = 50o

∴ x = 50 (ii)

In ∆ABC,

∠A + ∠B + ∠C = 180o

⇒ 30o + 40o + ∠C = 180o

⇒ 70o + ∠C = 180o

⇒ ∠C = 180o – 70o = 110o

Now ∠BCA + ∠ACD = 180o [Linear pair]

⇒ 110o + ∠ACD = 180o

⇒ ∠ACD = 180o – 110o = 70o In ∆ECD,

⇒ ∠ECD + ∠CDE + ∠CED = 180o

⇒ 70o + 50o + ∠CED = 180o

⇒ 120o + ∠CED = 180o

∠CED = 180o – 120o = 60o

Since ∠AED and ∠CED from a linear pair So, ∠AED + ∠CED = 180o

⇒ xo + 60o = 180o

⇒ xo = 180o – 60o = 120o

∴ x = 120 (iii)

∠EAF = ∠BAC [Vertically opposite angles]

⇒ ∠BAC = 60o

In ∆ABC, exterior ∠ACD is equal to the sum of two opposite interior angles. So, ∠ACD = ∠BAC + ∠ABC

⇒ 115o = 60o + xo

⇒ xo = 115o – 60o = 55o

∴ x = 55 (iv)

Since AB || CD and AD is a transversal. So, ∠BAD = ∠ADC

⇒ ∠ADC = 60o

In ∠ECD, we have,

∠E + ∠C + ∠D = 180o

⇒ xo + 45o + 60o = 180o

⇒ xo + 105o = 180o

⇒ xo = 180o – 105o = 75o

∴ x = 75 (v)

In ∆AEF,

Exterior ∠BED = ∠EAF + ∠EFA

⇒ 100o = 40o + ∠EFA

⇒ ∠EFA = 100o – 40o = 60o

Also, ∠CFD = ∠EFA [Vertically Opposite angles]

⇒ ∠CFD = 60o Now in ∆FCD,

Exterior ∠BCF = ∠CFD + ∠CDF

⇒ 90o = 60o + xo

⇒ xo = 90o – 60o = 30o

∴ x = 30 (vi)

In ∆ABE, we have,

∠A + ∠B + ∠E = 180o

⇒ 75o + 65o + ∠E = 180o

⇒ 140o + ∠E = 180o

⇒ ∠E = 180o – 140o = 40o

Now, ∠CED = ∠AEB [Vertically opposite angles]

⇒ ∠CED = 40o

Now, in ∆CED, we have,

∠C + ∠E + ∠D = 180o

⇒ 110o + 40o + xo = 180o

⇒ 150o + xo = 180o

⇒ xo = 180o – 150o = 30o

∴ x = 30

Question 18:

Produce CD to cut AB at E.

Now, in ∆BDE, we have,

Exterior ∠CDB = ∠CEB + ∠DBE

⇒ xo = ∠CEB + 45o In ∆AEC, we have,

…..(i)

Exterior ∠CEB = ∠CAB + ∠ACE

= 55o + 30o = 85o

Putting ∠CEB = 85o in (i), we get, xo = 85o + 45o = 130o

∴ x = 130

Question 19:

The angle ∠BAC is divided by AD in the ratio 1 : 3. Let ∠BAD and ∠DAC be y and 3y, respectively. As BAE is a straight line,

∠BAC + ∠CAE = 180o

[linear pair]

⇒ ∠BAD + ∠DAC + ∠CAE = 180o

⇒ y + 3y + 108o = 180o

⇒ 4y = 180o – 108o = 72o

⇒ y =

= 18o Now, in ∆ABC,

∠ABC + ∠BCA + ∠BAC = 180o

y + x + 4y = 180o

[Since, ∠ABC = ∠BAD (given AD = DB) and ∠BAC = y + 3y = 4y]

⇒ 5y + x = 180

⇒ 5 × 18 + x = 180

⇒ 90 + x = 180

∴ x = 180 – 90 = 90

Question 20:

Given : A ∆ABC in which BC, CA and AB are produced to D, E and F respectively. To prove : Exterior ∠DCA + Exterior ∠BAE + Exterior ∠FBD = 360o

Proof : Exterior ∠DCA = ∠A + ∠B ….(i) Exterior ∠FAE = ∠B + ∠C ….(ii) Exterior ∠FBD = ∠A + ∠C ….(iii) Adding (i), (ii) and (iii), we get,

Ext. ∠DCA + Ext. ∠FAE + Ext. ∠FBD

= ∠A + ∠B + ∠B + ∠C + ∠A + ∠C

= 2∠A + 2∠B + 2∠C

= 2 (∠A + ∠B + ∠C)

= 2 × 180o

[Since, in triangle the sum of all three angle is 180o]

= 360o

Hence, proved.

Question 21:

In ∆ACE, we have,

∠A + ∠C + ∠E = 180o ….(i) In ∆BDF, we have,

∠B + ∠D + ∠F = 180o ….(ii)

Adding both sides of (i) and (ii), we get,

∠A + ∠C +∠E + ∠B + ∠D + ∠F = 180o + 180o

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o.

Question 22:

Given : In ∆ABC, bisectors of ∠B and ∠C meet at O and ∠A = 70o In ∆BOC, we have,

∠BOC + ∠OBC + ∠OCB = 180o

= 180o – 55o = 125o

∴ ∠BOC = 125o.

Question 23:

We have a ∆ABC whose sides AB and AC have been procued to D and E. A = 40o and bisectors of ∠CBD and ∠BCE meet at O.

In ∆ABC, we have, Exterior ∠CBD = C + 40o

And exterior ∠BCE = B + 40o

Now, in ∆BCO, we have,

= 50o + 20o

= 70o

Thus, ∠BOC = 70o

Question 24:

In the given ∆ABC, we have,

∠A : ∠B : ∠C = 3 : 2 : 1

Let ∠A = 3x, ∠B = 2x, ∠C = x. Then,

∠A + ∠B + ∠C = 180o

⇒ 3x + 2x + x = 180o

⇒ 6x = 180o

⇒ x = 30o

∠A = 3x = 3 30o = 90o

∠B = 2x = 2 30o = 60o

and, ∠C = x = 30o

Now, in ∆ABC, we have,

Ext ∠ACE = ∠A + ∠B = 90o + 60o = 150o

∠ACD + ∠ECD = 150o

⇒ ∠ECD = 150o – ∠ACD

⇒ ∠ECD = 150o – 90o

⇒ ∠ECD= 60o

[since , AD ⊥ CD, ∠ACD = 90o]

Question 25:

In ∆ABC, AN is the bisector of ∠A and AM ⊥ BC. Now in ∆ABC we have;

∠A = 180o – ∠B – ∠C

⇒ ∠A = 180o – 65o – 30o

= 180o – 95o

= 85o

Now, in ∆ANC we have;

Thus, ∠MAN =

Question 26:

(i) False (ii) True (iii) False (iv) False (v) True (vi) True.

CoordinateGeometry

Question 1:

Exercise 6A

Draw the perpendiculars from the AF, BG, CH, DI and EJ on the x-axis.

1. The distance of A from the y-axis = OF = -6 units The distance of A from the x-axis = AF = 5 units Hence, the coordinate of A are (-6, 5)
2. The distance of B from the y-axis = OG = 5 units The distance of B from the x-axis = BG = 4 units Hence, the coordinate of B are (5, 4)
3. The distance of C from the y-axis = OH = -3 units The distance of C from the x-axis = HC = 2 units Hence, the coordinate of C are (-3, 2)
4. The distance of D from the y-axis = OI = 2 units The distance of D from the x-axis = ID = -2 units Hence, the coordinate of D are (2, -2)
5. The distance of E from the y-axis = OJ = -1 unit The distance of E from the x-axis = JE = -4 units Hence, the coordinate of E are (-1, -4)

Thus, the coordinates of A, B, C, D and E are respectively, A(-6,5), B(5,4), C(-3,2), D(2,-2) and E(-1,-4)

Question 2:

Let X’OX and Y’OY be the coordinate axes.

Fix the side of the small squares as one units.

• 1. Starting from O, take +7 units on the x-axis and then +4 units on the y-axis to obtain the point P(7, 4)
  2. Starting from O, take -5 units on the x-axis and then +3 units on the y-axis to obtain the point Q(-5, 3)
  3. Starting from O, take -6 units on the x-axis and then -3 units on the y-axis to obtain the point R(-6, -3)
  4. Starting from O, take +3 units on the x-axis and then -7 units on the y-axis to obtain the point S(3, -7)
  5. Starting from O, take 6 units on the x-axis to obtain the point A(6, 0)
  6. Starting from O, take 9 units on the y-axis to obtain the point B(0,9)
  7. Mark the point O as O(0, 0)
  8. Starting from O, take -3 units on the x-axis and then -3 units on the y-axis to obtain the point C(-3, -3)

These points are shown in the following graph:

Question 3:

1. In (7, 0), we have the ordinate = 0.

Therefore, (7,0) lies on the x-axis

1. In (0, -5), we have the abscissa = 0.

Therefore, (0,-5) lies on the y-axis

1. In (0,1), we have the abscissa = 0.

Therefore, (0,1) lies on the y-axis

1. In (-4,0), we have the ordinate = 0.

Therefore, (-4,0) lies on the x-axis

Question 4:

1. Points of the type (-, +) lie in the second quadrant. Therefore, the point (-6,5) lies in the II quadrant.
2. Points of the type (-, -) lie in the third quadrant. Therefore, the point (-3,-2) lies in the III quadrant.
3. Points of the type (+, -) lie in the fourth quadrant. Therefore, the point (2,-9) lies in the IV quadrant.

Question 5:

The given equation is y = x + 1 Putting x = 1, we get y = 1 + 1 = 2

Putting x = 2, we get y = 2 + 1 = 3 Thus, we have the following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Then, plot points P (1, 2) and Q (2, 3) on the graph paper. Join PQ and extend it to both sides.

Then, line PQ is the graph of the equation y = x + 1.

Question 6:

The give equation is y = 3x + 2 Putting x = 1, we get y = (3 1) + 2 = 5

Putting x = 2, we get y = (3 2) + 2 = 8 Thus, we have the following table:

On the graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot points P(1,5) and Q(2,8) on the graph paper.

Join PQ and extend it to both sides.

Then, line PQ is the graph of the equation y = 3x + 2.

Question 7:

The given equation is y = 5x – 3 Putting x = 0, we get y = (5 × 0) – 3 = -3

Putting x = 1, we get y = (5 × 1) – 3 = 2 Thus, we have following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now plot the points P(0,-3) and Q(1,2).

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation, y = 5x – 3.

Question 8:

The given equation is y = 3x Putting x = 1, we get y = (3 1) = 3

Putting x = 2, we get y = (3 2) = 6 Thus, we have the following table:

On a graph paper draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot points P(1,3) and Q(2,6).

Join PQ and extend it in both the directions. Then, line PQ is the graph of the equation y = 3x.

Question 9:

The given equation is y = -x

Putting x = 1, we get y = -1 Putting x = 2, we get y = -2

Thus, we have the following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot the points P(1,-1) and Q(2,-2).

Join PQ and extend it in both the directions. Then, line PQ is the graph of the equation y = -x.

Introduction toEuclid’s Geometry

Question 1:

Exercise 3A

A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.

Example of Theorem: Pythagoras Theorem

Example of axiom: A unique line can be drawn through any two points.

Question 2:

1. Line segment: The straight path between two points is called a line segment.
2. Ray: A line segment when extended indefinitely in one direction is called a ray.
3. Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.
4. Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.
5. Half-line: A ray without its initial point is called a half-line.
6. Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.
7. Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.
8. Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.

Question 3:

1. Six points: A,B,C,D,E,F
2. Five line segments: , , , ,

1. Four rays: , , ,

1. Four lines: , , ,
2. Four collinear points: M,E,G,B

Question 4:

• 1. and their corresponding point of intersection is R. and their corresponding point of intersection is P.
  2. , , and their point of intersection is R.
  3. Three rays are: , ,
  4. Two line segments are: ,

Question 5:

1. An infinite number of lines can be drawn to pass through a given point.
2. One and only one line can pass through two given points.
3. Two given lines can at the most intersect at one and only one point.
4. , ,

Question 6:

1. False
2. False
3. False
4. True
5. False
6. True
7. True
8. True
9. True
10. False
11. False
12. True