Fundamental Concepts of Geometry

ICSE Class 6 Mathematics · Selina Solutions Concise Mathematics Class 6 ICSE · 97 solved questions

Exercise

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1Exercise 14(A)

State true or false, if false, correct the statement:

(i) A dot has width but no length.

(ii) A ray has an infinite length only on one side of it.

(iii) A line segment PQ is written as $\overleftrightarrow{PQ}$.

(iv) $\overleftrightarrow{PQ}$ represents a straight line.

(v) Three points are said to be collinear, if they lie in the same plane.

(vi) Three or more points, all lying in the same line, are called collinear points.

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(i) False. A dot has no length and no width.

(ii) True. A ray extends endlessly on one side of its end point, so it has an infinite length only on one side of it.

(iii) False. A line segment PQ is written as $\overline{PQ}$.

(iv) True. $\overleftrightarrow{PQ}$ has arrow heads on both sides and so it represents a straight line.

(v) False. Three points are said to be collinear if they lie in the same straight line.

(vi) True. Three or more points that all lie on the same line are called collinear points.

2Exercise 14(A)

Write how many lines can be drawn, passing through:

(i) a given point?

(ii) two given fixed points?

(iii) three collinear points?

(iv) three non-collinear points?

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(i) Through a single given point an unlimited number of lines can be drawn.

Hence, infinite lines can be drawn through a given point.

(ii) Through two given fixed points one and only one line can be drawn.

Hence, only one line can be drawn through two given fixed points.

(iii) Since three collinear points lie on the same straight line, that single line passes through all of them.

Hence, only one line can be drawn through three collinear points.

(iv) Three non-collinear points do not lie on one line, so no single line can pass through all three of them.

Hence, no line can be drawn through three non-collinear points.

3Exercise 14(A)

The shaded region of the given figure shows a plane:

(a) Name:

(i) three collinear points.

(ii) three non-collinear points.

(iii) a pair of intersecting lines.

(b) State whether true or false:

(i) Line DE is contained in the given plane P.

(ii) Lines AB and DE intersect at point C.

(iii) Points D, B and C are collinear.

(iv) Points D, B and E are collinear.

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(a) (i) A, B and C lie on the same straight line, so they are three collinear points.

(ii) A, D and C do not lie on the same straight line, so they are three non-collinear points.

(iii) AC and DC are a pair of intersecting lines.

(b) (i) True.

(ii) True.

(iii) False.

(iv) False.

4Exercise 14(A)

Correct the statement, if it is wrong:

(i) A ray can be extended infinitely on either side.

(ii) A ray has a definite length.

(iii) A line segment has a definite length.

(iv) A line has two end points.

(v) A ray has only one end point.

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(i) A ray can be extended infinitely on only one side of it.

(ii) A ray has an indefinite length.

(iii) Correct. A line segment has a definite length.

(iv) A line has no end point.

(v) Correct. A ray has only one end point.

5Exercise 14(A)

State true or false, if false, give the correct statement:

(i) A line has a countable number of points in it.

(ii) Only one line can pass through a given point.

(iii) The intersection of two planes is a straight line.

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(i) False. A line has an infinite number of points on it.

(ii) False. An infinite number of lines can pass through a given point.

(iii) True.

6Exercise 14(A)

State whether the following pairs of lines or rays appear to be parallel or intersecting.

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Two lines or rays are said to be parallel if they never meet however far they are produced, and intersecting if they meet (or would meet) at a point.

(i) The two rays in the figure appear that they will meet at a point.

Hence, the pair of lines are intersecting.

(ii) The two lines in the figure are the same distance apart everywhere and never meet.

Hence, the pair of lines are parallel.

(iii) The two lines in the figure remain the same distance apart and never meet however far they are extended.

Hence, the pair of lines are parallel.

(iv) The two lines in the figure appear that they will meet at a point.

Hence, the pair of lines are intersecting.

7Exercise 14(A)

Give two examples, from your surroundings, for each of the following:

(i) points

(ii) line segments

(iii) plane surfaces

(iv) curved surfaces.

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(i) Tip of your pencil and tip of a nail.

(ii) Lines printed on your note-book and edge of your table-top.

(iii) Top of your study table and a wall of your class-room.

(iv) Surface of a cricket-ball and surface of your pen.

8Exercise 14(A)

Under what condition will two straight lines in the same plane have:

(i) no point in common.

(ii) only one point in common.

(iii) an infinite number of points in common.

If possible, draw diagrams in support of your answer.

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(i) When the two lines are parallel to each other, they have no point in common.

Under what condition will two straight lines in the same plane have:. Fundamental Concepts, Mathematics Solutions ICSE Class 6.

(ii) When the two lines intersect each other, they have only one point in common.

(iii) When the two lines coincide (lie exactly one over the other), they have an infinite number of points in common.

9Exercise 14(A)

Mark two points A and B on a page of your exercise book. Mark a third point P such that:

(i) P lies between A and B and the three points A, P and B are collinear.

(ii) P does not lie between A and B yet the three points are collinear.

(iii) the three points do not lie in a line.

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(i)

Mark two points A and B on a page of your exercise book. Mark a third point P such that:. Fundamental Concepts, Mathematics Solutions ICSE Class 6.

(ii)

(iii)

10Exercise 14(A)

Mark two points P and Q on a piece of paper. How many lines can you draw:

(i) passing through both the points P and Q?

(ii) passing through the point P?

(iii) passing through the point Q?

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(i) Through two given points only one line can be drawn.

Mark two points P and Q on a piece of paper. How many lines can you draw:. Fundamental Concepts, Mathematics Solutions ICSE Class 6.

Hence, only one line can be drawn passing through both the points P and Q.

(ii) Through a single point an unlimited number of lines can be drawn.

Hence, infinite lines can be drawn passing through the point P.

(iii) Through a single point an unlimited number of lines can be drawn.

Hence, infinite lines can be drawn passing through the point Q.

11Exercise 14(A)

The adjoining figure shows a line AB. Draw figures to represent:

(i) ray AB, i.e. $\overrightarrow{AB}$

(ii) ray BA, i.e. $\overrightarrow{BA}$

(iii) line segment AB.

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(i)

(ii)

(iii)

12Exercise 14(A)

The adjoining figure shows a ray AB. Draw figures to show:

(i) ray BA, i.e. $\overrightarrow{BA}$

(ii) line AB

(iii) line segment BA.

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(i)

(ii)

(iii)

13Exercise 14(A)

The adjoining figure shows a line segment AB. Draw figures to represent:

(i) ray AB, i.e. $\overrightarrow{AB}$

(ii) line AB, i.e. $\overleftrightarrow{AB}$

(iii) ray BA.

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(i)

(ii)

The adjoining figure shows a line segment AB. Draw figures to represent:. Fundamental Concepts, Mathematics Solutions ICSE Class 6.

(iii)

14Exercise 14(A)

Use a ruler and find whether the points in the given figure are collinear or not:

(i) D, A and C

(ii) A, B and C

(iii) A, B and E

(iv) B, C and E

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(i) No, the points D, A and C are not collinear.

(ii) No, the points A, B and C are not collinear.

(iii) Yes, the points A, B and E are collinear.

(iv) No, the points B, C and E are not collinear.

15Exercise 14(A)

From the adjoining figure, write:

(i) all pairs of parallel lines.

(ii) all the lines which intersect EF.

(iii) lines whose point of intersection is G.

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(i) EF ∥ GH, EF ∥ IJ and GH ∥ IJ.

(ii) AB and CD are the lines which intersect EF.

(iii) AB and GH are the lines whose point of intersection is G.

1Exercise 14(B)

State which of the following is a plane closed figure:

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A plane closed figure is a figure drawn in a plane whose boundary starts and ends at the same point, leaving no gap.

(i) The boundary is completely closed with no gap, so it is a plane closed figure.

(ii) The boundary is not closed, as it has open ends, so it is not a plane closed figure.

(iii) The boundary starts and ends at the same point leaving no gap, so it is a plane closed figure.

(iv) The boundary starts and ends at the same point leaving no gap, so it is a plane closed figure.

(v) It is not a plane closed figure because it contains intersecting line segments inside the figure.

(vi) The boundary starts and ends at the same point leaving no gap, so it is a plane closed figure.

(vii) The boundary is not closed, as it has open ends, so it is not a plane closed figure.

(viii) The boundary starts and ends at the same point leaving no gap, so it is a plane closed figure.

Hence, the figures (i), (iii), (iv), (vi) and (viii) are plane closed figures.

2Exercise 14(B)

State which of the lines/line-segments are perpendicular to the line PQ:

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Two lines are perpendicular when they meet at an angle of 90°.

Hence, CD and MN are perpendicular to the line PQ.

3Exercise 14(B)

Which of the following figures shows two mutually perpendicular lines:

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Two lines are mutually perpendicular when they intersect each other at a right angle (90°).

Hence, the figures (i) and (iii) show two mutually perpendicular lines.

4Exercise 14(B)

For each figure given below name the line segment that is perpendicular bisector of the other:

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The perpendicular bisector of a line segment is a line that passes through its midpoint and is perpendicular to it.

(i) AB meets CD at a right angle and divides it into two equal parts.

AB is the perpendicular bisector of CD.

(ii) AB meets MN at a right angle and divides it into two equal parts.

AB is the perpendicular bisector of MN.

(iii) PQ and RS cut each other at a right angle and each divides the other into two equal parts.

PQ is the perpendicular bisector of RS and RS is the perpendicular bisector of PQ, i.e. PQ and RS are perpendicular bisectors of each other.

(iv) The two line segments meet at a right angle but do not divide each other into two equal parts, so neither is the perpendicular bisector of the other.

None.

(v) The two line segments do not meet at a right angle, and AB does not divide CD into two equal parts, so neither is the perpendicular bisector of the other.

None.

5Exercise 14(B)

Name three objects from your surroundings that contain perpendicular edges.

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Hence, a classroom, a textbook and a ruler contain perpendicular edges.

6Exercise 14(B)

Place a scale (ruler) on a sheet of paper and hold it firmly with one hand. Now draw two line segments AB and CD along the longer edges of the scale. State whether segment AB is parallel to or perpendicular to segment CD.

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The two longer edges of a scale are always at the same distance apart and never meet.

Hence, segment AB is parallel to segment CD.

7Exercise 14(B)

Check your textbook:

(i) How many pairs of its edges are parallel to each other?

(ii) How many pairs of its edges are perpendicular to each other?

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Check your textbook:. Fundamental Concepts, Mathematics Solutions ICSE Class 6.

A textbook is shaped like a cuboid and has 12 edges.

(i) Its edges form three groups:

4 edges along its length,

4 edges along its breadth,

4 edges along its height.

The four edges in each group are parallel to one another.

Among four parallel edges, the number of pairs is:

3 + 2 + 1 = 6

Therefore, the total number of pairs of parallel edges is:

6 + 6 + 6 = 18

Hence, a textbook has 18 pairs of parallel edges.

(ii) A cuboid has 8 corners. At each corner, three edges meet at right angles.

These three edges form 3 pairs of perpendicular edges.

Therefore,

8 × 3 = 24

Hence, a textbook has 24 pairs of perpendicular edges.

8Exercise 14(B)

Give two examples from your surroundings for each of the following:

(i) intersecting lines

(ii) parallel lines

(iii) perpendicular lines.

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(i) The two edges of a book meeting through any one corner of it.

The two roads crossing each other at a road crossing.

(ii) The opposite edges of a book.

The two rails of a railway track.

(iii) The two adjacent edges meeting at a corner of a book.

The two edges meeting at a corner of a window.

9Exercise 14(B)

State true or false, if false, give the correct statement:

(i) The maximum number of lines passing through three collinear points is three.

(ii) The maximum number of lines passing through three non-collinear points is three.

(iii) Two parallel lines always lie in the same plane.

(iv) Concurrent lines always meet at the same point.

(v) A surface can be plane or curved.

(vi) There are infinite number of points in a line segment of length 10 cm.

(vii) There are infinite number of points in a line.

(viii) A plane has infinite number of lines.

(ix) A plane has infinite number of points.

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(i) False. Only one line passes through three collinear points.

(ii) False. No line passes through all the three non-collinear points.

(iii) True.

(iv) True.

(v) True.

(vi) True.

(vii) True.

(viii) True.

(ix) True.

1Exercise 14(C)

For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.

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(i) From figure,

Vertex is O;

Arms are OA and OB;

The angle is written as angle AOB or ∠AOB or ∠O.

(ii) From figure,

Vertex is Q;

Arms are QP and QR;

The angle is written as angle PQR or ∠PQR or ∠Q.

(iii) From figure,

Vertex is M;

Arms are MN and ML;

The angle is written as angle NML or ∠LMN or ∠M.

2Exercise 14(C)

Name the points:

(i) in the interior of the angle PQR.

(ii) in the exterior of the angle PQR.

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(i) The points a, b and x lie in the interior of the angle PQR.

(ii) The points d, m, n, s and t lie in the exterior of the angle PQR.

3Exercise 14(C)

In the adjoining figure, write all possible angles formed. Neglect reflex angle(s).

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Taking the rays OA, OB, OC, OD and OE two at a time, the possible angles formed are :

∠AOB, ∠AOC, ∠AOD, ∠AOE, ∠BOC, ∠BOD, ∠BOE, ∠COD, ∠COE and ∠DOE.

4(i)Exercise 14(C)

Add:

29° 16' 23" and 8° 27' 12"

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Adding the seconds, minutes and degrees separately :

$$\begin{array}{rrrr} &29^{\circ} &16' &23'' \\ + &8^{\circ} &27' &12'' \\ \hline &37^{\circ} &43' &35'' \\ \hline \end{array}$$

Hence, the sum is 37° 43' 35".

4(ii)Exercise 14(C)

Add:

9° 45' 56" and 73° 8' 15"

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Adding the seconds, minutes and degrees separately :

$$\begin{array}{rrrr} &9^{\circ} &45' &56'' \\ + &73^{\circ} &8' &15'' \\ \hline &82^{\circ} &53' &71'' \\ \hline \end{array}$$

As 60" = 1', we have 71" = 1' 11". Carrying 1' over to the minutes :

⇒ 82° 53' 71" = 82° 54' 11"

Hence, the sum is 82° 54' 11".

4(iii)Exercise 14(C)

Add:

56° 38' and 27° 42' 30"

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Adding the seconds, minutes and degrees separately :

$$\begin{array}{rrrr} &56^{\circ} &38' &0'' \\ + &27^{\circ} &42' &30'' \\ \hline &83^{\circ} &80' &30'' \\ \hline \end{array}$$

As 60' = 1°, we have 80' = 1° 20'. Carrying 1° over to the degrees :

⇒ 83° 80' 30" = 84° 20' 30"

Hence, the sum is 84° 20' 30".

4(iv)Exercise 14(C)

Add:

47° and 61° 17' 4"

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Adding the seconds, minutes and degrees separately :

$$\begin{array}{rrrr} &47^{\circ} &0' &0'' \\ + &61^{\circ} &17' &4'' \\ \hline &108^{\circ} &17' &4'' \\ \hline \end{array}$$

Hence, the sum is 108° 17' 4".

5Exercise 14(C)

In the figure given alongside, name:

(i) three pairs of adjacent angles.

(ii) two acute angles.

(iii) two obtuse angles.

(iv) two reflex angles.

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(i) ∠AOB and ∠BOC; ∠BOC and ∠COD; ∠COD and ∠DOA.

(ii) ∠AOB and ∠AOD.

(iii) ∠BOC and ∠COD.

(iv) Reflex ∠AOB and reflex ∠COD.

6Exercise 14(C)

In the given figure, PQR is a straight line. If:

(i) ∠SQR = 75°; find ∠PQS.

(ii) ∠PQS = 110°; find ∠RQS.

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Since PQR is a straight line, ∠PQS and ∠SQR form a linear pair.

⇒ ∠PQS + ∠SQR = 180°

(i) Given, ∠SQR = 75°

⇒ ∠PQS + 75° = 180°

⇒ ∠PQS = 180° − 75° = 105°

Hence, ∠PQS = 105°.

(ii) Given, ∠PQS = 110°

⇒ 110° + ∠RQS = 180°

⇒ ∠RQS = 180° − 110° = 70°

Hence, ∠RQS = 70°.

7Exercise 14(C)

In the given figure, AOC is a straight line. If angle AOB = 50°, angle AOE = 90° and angle COD = 25°, find the measure of:

(i) angle BOC

(ii) angle EOD

(iii) obtuse angle BOD

(iv) reflex angle BOD

(v) reflex angle COE.

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From the figure, AOC is a straight line. With ∠AOB = 50°, ∠AOE = 90° and ∠COD = 25°.

(i) Since AOC is a straight line, ∠AOB and ∠BOC form a linear pair.

⇒ ∠AOB + ∠BOC = 180°

⇒ 50° + ∠BOC = 180°

⇒ ∠BOC = 180° − 50° = 130°

Hence, ∠BOC = 130°.

(ii) Since AOC is a straight line, ∠AOE and ∠COE form a linear pair.

⇒ ∠AOE + ∠COE = 180°

⇒ 90° + ∠COE = 180°

⇒ ∠COE = 180° − ∠AOE = 180° − 90° = 90°

⇒ ∠EOD = ∠EOC − ∠COD = 90° − 25° = 65° (given ∠COD = 25°)

Hence, ∠EOD = 65°.

(iii) As OB and OD lie on opposite sides of the line AOC,

⇒ ∠BOD = ∠BOC + ∠COD = 130° + 25° = 155°

Hence, obtuse ∠BOD = 155°.

(iv) Reflex ∠BOD = 360° − ∠BOD = 360° − 155° = 205°

Hence, reflex ∠BOD = 205°.

(v) ∠COE = ∠AOC − ∠AOE = 180° − 90° = 90°

Reflex ∠COE = 360° − ∠COE = 360° − 90° = 270°

Hence, reflex ∠COE = 270°.

8Exercise 14(C)

In the given figure, if:

(i) a = 130°, find b.

(ii) b = 200°, find a.

(iii) a = $\frac{5}{3}$ right angle, find b.

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From the figure, ∠a and ∠b are the angles at the point O, and the sum of the angles at a point is 360°.

⇒ a + b = 360°

(i) Given,

a = 130°

⇒ 130° + b = 360°

⇒ b = 360° − 130° = 230°

Hence, b = 230°.

(ii) Given,

b = 200°

⇒ a + 200° = 360°

⇒ a = 360° − 200° = 160°

Hence, a = 160°.

(iii) Given,

a = $\frac{5}{3}$ right angle = $\frac{5}{3} \times 90^{\circ} = 150^{\circ}$

⇒ 150° + b = 360°

⇒ b = 360° − 150° = 210°

Hence, b = 210°.

9Exercise 14(C)

In the figure given alongside, ABC is a straight line.

(i) If x = 53°, find y.

(ii) If y = $1\frac{1}{2}$ right angles, find x.

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Since ABC is a straight line, x and y form a linear pair.

⇒ x + y = 180°

(i) Given,

x = 53°

⇒ 53° + y = 180°

⇒ y = 180° − 53° = 127°

Hence, y = 127°.

(ii) Given,

y = $1\frac{1}{2}$ right angles = $\frac{3}{2} \times 90^{\circ} = 135^{\circ}$

⇒ x + 135° = 180°

⇒ x = 180° − 135° = 45°

Hence, x = 45°.

10Exercise 14(C)

In the figure given alongside, AOB is a straight line. Find the value of x and also answer each of the following:

(i) ∠AOP = ..............

(ii) ∠BOP = ..............

(iii) which angle is obtuse?

(iv) which angle is acute?

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From the figure, ∠AOP = (x + 30)° and ∠BOP = (x − 30)°.

Since AOB is a straight line, ∠AOP and ∠BOP form a linear pair.

⇒ ∠AOP + ∠BOP = 180°

⇒ (x + 30°) + (x − 30°) = 180°

⇒ 2x = 180°

⇒ x = $\frac{180^{\circ}}{2}$

⇒ x = 90°

Hence, x = 90°.

(i) ∠AOP = x + 30° = 90° + 30° = 120°

Hence, ∠AOP = 120°.

(ii) ∠BOP = x − 30° = 90° − 30° = 60°

Hence, ∠BOP = 60°.

(iii) Since ∠AOP = 120° lies between 90° and 180°, it is obtuse.

Hence, ∠AOP is the obtuse angle.

(iv) Since ∠BOP = 60° is less than 90°, it is acute.

Hence, ∠BOP is the acute angle.

11Exercise 14(C)

In the figure given alongside, PQR is a straight line. Find x. Then complete the following:

(i) ∠AQB = ..............

(ii) ∠BQP = ..............

(iii) ∠AQR = ..............

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From the figure,

∠PQA = (x + 20°), ∠AQB = (2x + 10°) and ∠BQR = (x − 10°).

Since PQR is a straight line, these three angles lie on one side of it.

⇒ ∠PQA + ∠AQB + ∠BQR = 180°

⇒ (x + 20°) + (2x + 10°) + (x − 10°) = 180°

⇒ 4x + 20° = 180°

⇒ 4x = 160°

⇒ x = $\frac{160^{\circ}}{4}$

⇒ x = 40°

Hence, x = 40°.

(i) ∠AQB = 2x + 10°

= 2 × 40° + 10°

= 80° + 10° = 90°

Hence, ∠AQB = 90°.

(ii) ∠BQP = ∠PQA + ∠AQB

= (x + 20°) + 90° = (40° + 20°) + 90°

= 60° + 90° = 150°

Hence, ∠BQP = 150°.

(iii) ∠AQR = ∠AQB + ∠BQR

= 90° + (x − 10°) = 90° + (40° − 10°)

= 90° + 30° = 120°

Hence, ∠AQR = 120°.

12Exercise 14(C)

In the figure given alongside, lines AB and CD intersect at point O.

(i) Find the value of ∠a.

(ii) Name all the pairs of vertically opposite angles.

(iii) Name all the pairs of adjacent angles.

(iv) Name all the reflex angles formed and write the measure of each.

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(i) From the figure, ∠AOC = 68° and ∠a = ∠BOC.

Since AB is a straight line, ∠AOC and ∠BOC form a linear pair.

⇒ ∠AOC + ∠BOC = 180°

⇒ 68° + ∠a = 180°

⇒ ∠a = 180° − 68° = 112°

Hence, ∠a = 112°.

(ii) When two lines intersect, the angles opposite to each other at the point of intersection are vertically opposite angles.

Hence, the pairs of vertically opposite angles are ∠AOC and ∠BOD; ∠AOD and ∠BOC.

(iii) Two angles with a common vertex and a common arm, lying on opposite sides of the common arm, are adjacent angles.

Hence, the pairs of adjacent angles are ∠AOC and ∠BOC; ∠BOC and ∠BOD; ∠BOD and ∠DOA; ∠DOA and ∠AOC.

(iv) Here, ∠AOC = ∠BOD = 68° and ∠BOC = ∠AOD = 112°.

Reflex ∠BOC = 360° − ∠BOC = 360° − 112° = 248°

Reflex ∠AOC = 360° − ∠AOC = 360° − 68° = 292°

Reflex ∠AOD = 360° − ∠AOD = 360° − 112° = 248°

Reflex ∠BOD = 360° − ∠BOD = 360° − 68° = 292°

Hence, reflex ∠BOC = 248°, reflex ∠AOC = 292°, reflex ∠AOD = 248° and reflex ∠BOD = 292°.

13Exercise 14(C)

In the figure given alongside:

(i) if ∠AOB = 45°, ∠BOC = 30° and ∠AOD = 110°; find angles COD and BOD.

(ii) if ∠BOC = ∠DOC = 34° and ∠AOD = 120°; find angle AOB and angle AOC.

(iii) if ∠AOB = ∠BOC = ∠COD = 38°; find reflex angle AOC and reflex angle AOD.

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(i) From the figure,

∠AOD = ∠AOB + ∠BOC + ∠COD.

⇒ 110° = 45° + 30° + ∠COD

⇒ 110° = 75° + ∠COD

⇒ ∠COD = 110° − 75° = 35°

Now, ∠BOD = ∠BOC + ∠COD = 30° + 35° = 65°

Hence, ∠COD = 35° and ∠BOD = 65°.

(ii) From the figure,

∠AOC = ∠AOD − ∠DOC.

⇒ ∠AOC = 120° − 34° = 86°

Now, ∠AOB = ∠AOC − ∠BOC = 86° − 34° = 52°

Hence, ∠AOB = 52° and ∠AOC = 86°.

(iii) From the figure,

∠AOC = ∠AOB + ∠BOC = 38° + 38° = 76°.

Reflex ∠AOC = 360° − ∠AOC = 360° − 76° = 284°

Also, ∠AOD = ∠AOB + ∠BOC + ∠COD = 38° + 38° + 38° = 114°

Reflex ∠AOD = 360° − ∠AOD = 360° − 114° = 246°

Hence, reflex ∠AOC = 284° and reflex ∠AOD = 246°.

1(i)Exercise 14(D)

Write the complement angle of:

45°

Show SolutionHide Solution↓

The complement of an angle = 90° − the angle.

Complement of 45° = 90° − 45° = 45°

Hence, the complement angle of 45° is 45°.

1(ii)Exercise 14(D)

Write the complement angle of:

x°

Show SolutionHide Solution↓

The complement of an angle = 90° − the angle.

Complement of x° = (90 − x)°

Hence, the complement angle of x° is (90 − x)°.

1(iii)Exercise 14(D)

Write the complement angle of:

(x - 10)°

Show SolutionHide Solution↓

The complement of an angle = 90° − the angle.

Complement of (x − 10)° = 90° − (x − 10)° = 90° − x° + 10° = (100 − x)°

Hence, the complement angle of (x − 10)° is (100 − x)°.

1(iv)Exercise 14(D)

Write the complement angle of:

20° + y°

Show SolutionHide Solution↓

The complement of an angle = 90° − the angle.

Complement of (20° + y°) = 90° − (20° + y°) = 90° − 20° − y° = (70 − y)°

Hence, the complement angle of (20° + y°) is (70 − y)°.

2(i)Exercise 14(D)

Write the supplement angle of:

49°

Show SolutionHide Solution↓

The supplement of an angle = 180° − the angle.

Supplement of 49° = 180° − 49° = 131°

Hence, the supplement angle of 49° is 131°.

2(ii)Exercise 14(D)

Write the supplement angle of:

111°

Show SolutionHide Solution↓

The supplement of an angle = 180° − the angle.

Supplement of 111° = 180° − 111° = 69°

Hence, the supplement angle of 111° is 69°.

2(iii)Exercise 14(D)

Write the supplement angle of:

(x - 30)°

Show SolutionHide Solution↓

The supplement of an angle = 180° − the angle.

Supplement of (x − 30)° = 180° − (x − 30)° = 180° − x° + 30° = (210 − x)°

Hence, the supplement angle of (x − 30)° is (210 − x)°.

2(iv)Exercise 14(D)

Write the supplement angle of:

20° + y°

Show SolutionHide Solution↓

The supplement of an angle = 180° − the angle.

Supplement of (20° + y°) = 180° − (20° + y°) = 180° − 20° − y° = (160 − y)°

Hence, the supplement angle of (20° + y°) is (160 − y)°.

3(i)Exercise 14(D)

Find the complement angle of:

$\frac{1}{2}$ of 60°

Show SolutionHide Solution↓

As, $\frac{1}{2} \text{ of } 60^{\circ} = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 30° = 90° − 30° = 60°

Hence, the complement of 30° is 60°.

3(ii)Exercise 14(D)

Find the complement angle of:

$\frac{1}{5}$ of 160°

Show SolutionHide Solution↓

As, $\frac{1}{5} \text{ of } 160^{\circ} = \frac{1}{5} \times 160^{\circ} = 32^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 32° = 90° − 32° = 58°

Hence, the complement of 32° is 58°.

3(iii)Exercise 14(D)

Find the complement angle of:

$\frac{2}{5}$ of 70°

Show SolutionHide Solution↓

As, $\frac{2}{5} \text{ of } 70^{\circ} = \frac{2}{5} \times 70^{\circ} = 28^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 28° = 90° − 28° = 62°

Hence, the complement of 28° is 62°.

3(iv)Exercise 14(D)

Find the complement angle of:

$\frac{1}{6}$ of 90°

Show SolutionHide Solution↓

As, $\frac{1}{6} \text{ of } 90^{\circ} = \frac{1}{6} \times 90^{\circ} = 15^{\circ}$

The complement of an angle = 90° − the angle.

Complement of 15° = 90° − 15° = 75°

Hence, the complement of 15° is 75°.

4(i)Exercise 14(D)

Find the supplement angle of:

50% of 120°

Show SolutionHide Solution↓

As, 50% of 120° = $\frac{50}{100} \times 120^{\circ} = 60^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 60° = 180° − 60° = 120°

Hence, the supplement of 60° is 120°.

4(ii)Exercise 14(D)

Find the supplement angle of:

$\frac{1}{3}$ of 150°

Show SolutionHide Solution↓

As, $\frac{1}{3} \text{ of } 150^{\circ} = \frac{1}{3} \times 150^{\circ} = 50^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 50° = 180° − 50° = 130°

Hence, the supplement of 50° is 130°.

4(iii)Exercise 14(D)

Find the supplement angle of:

60% of 100°

Show SolutionHide Solution↓

As, 60% of 100° = $\frac{60}{100} \times 100^{\circ} = 60^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 60° = 180° − 60° = 120°

Hence, the supplement of 60° is 120°.

4(iv)Exercise 14(D)

Find the supplement angle of:

$\frac{3}{4}$ of 160°

Show SolutionHide Solution↓

As, $\frac{3}{4} \text{ of } 160^{\circ} = \frac{3}{4} \times 160^{\circ} = 120^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 120° = 180° − 120° = 60°

Hence, the supplement of 120° is 60°.

5(i)Exercise 14(D)

Find the angle that is equal to its complement?

Show SolutionHide Solution↓

Let the required angle be x.

As the angle is equal to its complement,

⇒ x = 90° − x

⇒ x + x = 90°

⇒ 2x = 90°

⇒ x = $\frac{90^{\circ}}{2}$

⇒ x = 45°

Hence, the required angle is 45°.

5(ii)Exercise 14(D)

Find the angle that is equal to its supplement?

Show SolutionHide Solution↓

Let the required angle be x.

As the angle is equal to its supplement,

⇒ x = 180° − x

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\frac{180^{\circ}}{2}$

⇒ x = 90°

Hence, the required angle is 90°.

6Exercise 14(D)

Two complementary angles are in the ratio 7 : 8. Find the angles.

Show SolutionHide Solution↓

As the angles are in the ratio 7 : 8.

Let the angles be 7x and 8x.

As the given angles are complementary, their sum is 90°.

⇒ 7x + 8x = 90°

⇒ 15x = 90°

⇒ x = $\frac{90^{\circ}}{15}$

⇒ x = 6°

So, the angles are 7x = 7 × 6° = 42° and 8x = 8 × 6° = 48°.

Hence, the angles are 42° and 48°.

7Exercise 14(D)

Two supplementary angles are in the ratio 7 : 11. Find the angles.

Show SolutionHide Solution↓

As the angles are in the ratio 7 : 11.

Let the angles be 7x and 11x.

As the given angles are supplementary, their sum is 180°.

⇒ 7x + 11x = 180°

⇒ 18x = 180°

⇒ x = $\frac{180^{\circ}}{18}$

⇒ x = 10°

So, the angles are 7x = 7 × 10° = 70° and 11x = 11 × 10° = 110°.

Hence, the angles are 70° and 110°.

8Exercise 14(D)

The measures of two complementary angles are (2x - 7)° and (x + 4)°. Find x.

Show SolutionHide Solution↓

As the given angles are complementary, their sum is 90°.

⇒ (2x − 7)° + (x + 4)° = 90°

⇒ 2x° - 7° + x° + 4° = 90°

⇒ 3x° − 3° = 90°

⇒ 3x° = 90° + 3°

⇒ 3x° = 93°

⇒ x° = $\frac{93^{\circ}}{3}$

⇒ x° = 31°

⇒ x = 31

Hence, x = 31.

9Exercise 14(D)

The measures of two supplementary angles are (3x + 15)° and (2x + 5)°. Find x.

Show SolutionHide Solution↓

As the given angles are supplementary, their sum is 180°.

⇒ (3x + 15)° + (2x + 5)° = 180°

⇒ 3x° + 15° + 2x° + 5° = 180°

⇒ 5x° + 20° = 180°

⇒ 5x° = 180° − 20°

⇒ 5x° = 160°

⇒ x° = $\frac{160^{\circ}}{5}$

⇒ x° = 32°

⇒ x = 32

Hence, x = 32.

10Exercise 14(D)

For an angle x°, find:

(i) the complementary angle

(ii) the supplementary angle.

(iii) the value of x° if its supplementary angle is three times its complementary angle.

Show SolutionHide Solution↓

(i) Two angles are said to be complementary if their sum is 90°, so the complement of an angle = 90° − the angle.

Complement of x° = 90° − x° = (90 − x)°

Hence, the complementary angle is (90 − x)°.

(ii) Two angles are said to be supplementary if their sum is 180°, so the supplement of an angle = 180° − the angle.

Supplement of x° = 180° − x° = (180 − x)°

Hence, the supplementary angle is (180 − x)°.

(iii) As the supplementary angle is three times the complementary angle,

⇒ (180 − x)° = 3(90 − x)°

⇒ 180° − x° = 270° − 3x°

⇒ 3x° − x° = 270° − 180°

⇒ 2x° = 90°

⇒ x° = $\frac{90^{\circ}}{2}$

⇒ x° = 45°

Hence, x° = 45°.

1Exercise 14(E)

Two straight lines AB and CD intersect each other at a point O and angle AOC = 50°; find:

(i) angle BOD

(ii) ∠AOD

(iii) ∠BOC

Show SolutionHide Solution↓

Given, ∠AOC = 50°.

(i) As, AB and CD intersect each other at point O.

⇒ ∠BOD and ∠AOC are vertically opposite angles.

⇒ ∠BOD = ∠AOC = 50° [vertically opposite angles are equal]

Hence, ∠BOD = 50°.

(ii) Since CD is a straight line, ∠AOC and ∠AOD form a linear pair.

⇒ ∠AOC + ∠AOD = 180°

⇒ 50° + ∠AOD = 180°

⇒ ∠AOD = 180° − 50° = 130°

Hence, ∠AOD = 130°.

(iii) Since AB is a straight line, ∠AOC and ∠BOC form a linear pair.

⇒ ∠AOC + ∠BOC = 180°

⇒ 50° + ∠BOC = 180°

⇒ ∠BOC = 180° − 50° = 130°

Hence, ∠BOC = 130°.

2Exercise 14(E)

The adjoining figure shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x - 5° and ∠APD = 3x + 15°, find:

(i) the value of x.

(ii) ∠APD

(iii) ∠BPD

(iv) ∠BPC.

Show SolutionHide Solution↓

(i) As,

AB and CD are intersecting lines at point P.

⇒ ∠BPC and ∠APD are vertically opposite angles.

⇒ ∠BPC = ∠APD [vertically opposite angles are equal]

⇒ 4x − 5° = 3x + 15°

⇒ 4x − 3x = 15° + 5°

⇒ x = 20°

Hence, x = 20°.

(ii) ∠APD = 3x + 15° = 3 × 20° + 15° = 60° + 15° = 75°

Hence, ∠APD = 75°.

(iii) Since CD is a straight line, ∠APD and ∠BPD form a linear pair.

⇒ ∠APD + ∠BPD = 180°

⇒ 75° + ∠BPD = 180°

⇒ ∠BPD = 180° − 75° = 105°

Hence, ∠BPD = 105°.

(iv) ∠BPC = 4x − 5° = 4 × 20° − 5° = 80° − 5° = 75°

Hence, ∠BPC = 75°.

3Exercise 14(E)

The figure, given alongside, shows two adjacent angles AOB and AOC whose exterior sides are along the same straight line. Find the value of x.

Show SolutionHide Solution↓

From the figure,

∠AOB = 68° and ∠AOC = (3x − 20°).

Since the exterior arms OB and OC of the adjacent angles are in the same straight line, the adjacent angles are supplementary.

⇒ ∠AOB + ∠AOC = 180°

⇒ 68° + (3x − 20°) = 180°

⇒ 3x + 48° = 180°

⇒ 3x = 180° − 48°

⇒ 3x = 132°

⇒ x = $\frac{132^{\circ}}{3}$

⇒ x = 44°

Hence, x = 44°.

4Exercise 14(E)

Each figure, given below, shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

Show SolutionHide Solution↓

The exterior arms OA and OC are in the same straight line only if the sum of the adjacent angles ∠AOB and ∠BOC is 180°.

(i) ∠AOB + ∠BOC = (90° − x) + (90° + x) = 180°

Since the sum is 180°, the exterior arms OA and OC are in the same straight line.

Hence, Yes, OA and OC are in the same straight line.

(ii) ∠AOB + ∠BOC = 97° + 83° = 180°

Since the sum is 180°, the exterior arms OA and OC are in the same straight line.

Hence, Yes, OA and OC are in the same straight line.

(iii) ∠AOB + ∠BOC = 88° + 112° = 200° ≠ 180°

Since the sum is not 180°, the exterior arms OA and OC are not in the same straight line.

Hence, No, OA and OC are not in the same straight line.

5Exercise 14(E)

A line segment AP stands at point P of a straight line BC such that ∠APB = 5x - 40° and ∠APC = x + 10°; find the values of x and angle APB.

Show SolutionHide Solution↓

Since BC is a straight line and AP stands on it at P, ∠APB and ∠APC form a linear pair.

⇒ ∠APB + ∠APC = 180°

⇒ (5x − 40°) + (x + 10°) = 180°

⇒ 6x − 30° = 180°

⇒ 6x = 180° + 30°

⇒ 6x = 210°

⇒ x = $\frac{210^{\circ}}{6}$

⇒ x = 35°

Now,

∠APB = 5x − 40° = 5 × 35° − 40° = 175° − 40° = 135°

Hence, x = 35° and ∠APB = 135°.

1(i)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

The angle 140° and ∠a form a linear pair.

⇒ a + 140° = 180° [linear pair]

⇒ a = 40°

⇒ b = a = 40° [alternate interior angles are equal]

Hence, a = 40° and b = 40°.

1(ii)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠a and the 60° angle are corresponding angles.

⇒ a = 60° [corresponding angles are equal]

The 60° angle and ∠b form a linear pair.

⇒ b + 60° = 180° [linear pair]

⇒ b = 120°

Hence, a = 60° and b = 120°.

1(iii)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠a and the 110° angle are vertically opposite angles.

⇒ a = 110° [vertically opposite angles are equal]

∠a and ∠b are co-interior angles.

⇒ a + b = 180° [co-interior angles are supplementary]

⇒ 110° + b = 180°

⇒ b = 70°

Hence, a = 110° and b = 70°.

1(iv)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠a and the 60° angle are alternate angles.

⇒ a = 60° [alternate interior angles are equal]

As, ∠a and ∠b form a linear pair.

⇒ a + b = 180° [linear pair]

⇒ 60° + b = 180°

⇒ b = 120°

Hence, a = 60° and b = 120°.

1(v)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠a and the 72° angle are alternate interior angles.

⇒ a = 72° [alternate interior angles are equal]

∠b and ∠a are vertically opposite angles.

⇒ b = a = 72° [vertically opposite angles are equal]

Hence, a = 72° and b = 72°.

1(vi)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

From figure,

b = 100° [corresponding angles are equal]

a and b lie on a straight line and forms a linear pair.

⇒ a + b = 180°

⇒ a + 100° = 180°

⇒ a = 180° - 100° = 80°

Hence, a = 80° and b = 100°.

1(vii)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠b and the 130° angle are vertically opposite angles.

⇒ ∠b = 130° [vertically opposite angles are equal]

∠a and the 130° angle are co-interior angles.

⇒ a + 130° = 180° [co-interior angles are supplementary]

⇒ a = 50°

Hence, a = 50° and b = 130°.

1(viii)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

From figure,

b = 62° [corresponding angles are equal]

a and b lie on a straight line and form a linear pair.

⇒ a + b = 180°

⇒ a + 62° = 180°

⇒ a = 180° - 62° = 118°

Hence, a = 118° and b = 62°.

1(ix)Exercise 14(F)

Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

Show SolutionHide Solution↓

∠a and the 90° angle form a linear pair.

⇒ a + 90° = 180°

⇒ a = 180° - 90°

⇒ a = 90°

∠b and the 90° angle are corresponding angles.

⇒ b = 90° [corresponding angles are equal]

Hence, a = 90° and b = 90°.

2Exercise 14(F)

If ∠1 = 120°, find the measures of ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8. Give reasons.

Show SolutionHide Solution↓

Given,

∠1 = 120°.

∠1 and ∠2 form a linear pair.

⇒ ∠2 = 180° − ∠1 = 180° − 120° = 60°

∠3 and ∠1 are vertically opposite angles.

⇒ ∠3 = ∠1 = 120° [vertically opposite angles are equal]

∠4 and ∠2 are vertically opposite angles.

⇒ ∠4 = ∠2 = 60° [vertically opposite angles are equal]

∠5 and ∠1 are corresponding angles.

⇒ ∠5 = ∠1 = 120° [corresponding angles are equal]

∠6 and ∠2 are corresponding angles.

⇒ ∠6 = ∠2 = 60° [corresponding angles are equal]

∠7 and ∠3 are corresponding angles.

⇒ ∠7 = ∠3 = 120° [corresponding angles are equal]

∠8 and ∠4 are corresponding angles.

⇒ ∠8 = ∠4 = 60° [corresponding angles are equal]

Hence, ∠2 = 60°, ∠3 = 120°, ∠4 = 60°, ∠5 = 120°, ∠6 = 60°, ∠7 = 120° and ∠8 = 60°.

3Exercise 14(F)

In the figure given alongside, find the measure of the angles denoted by x, y, z, p, q and r.

Show SolutionHide Solution↓

From the figure, the given angle is 100°.

The 100° angle and ∠x form a linear pair.

⇒ x + 100° = 180° [linear pair]

⇒ x = 180° - 100°

⇒ x = 80°

∠q and the 100° angle are vertically opposite angles.

⇒ q = 100° [vertically opposite angles are equal]

∠x and ∠p are vertically opposite angles.

⇒ ∠p = 80° [vertically opposite angles are equal]

∠y and ∠p are corresponding angles.

⇒ y = p = 80° [corresponding angles are equal]

∠z and the 100° angle are corresponding angles.

⇒ z = 100° [corresponding angles are equal]

∠r and ∠q are corresponding angles.

⇒ r = q = 100° [corresponding angles are equal]

Hence, x = 80°, y = 80°, z = 100°, p = 80°, q = 100° and r = 100°.

4Exercise 14(F)

Using the figure given alongside, fill in the blanks:

∠x = ..............; ∠z = ..............;

∠p = ..............; ∠q = ..............;

∠r = ..............; ∠s = ..............;

Show SolutionHide Solution↓

From the figure, the given angle is 60°.

∠x and the 60° angle are corresponding angles.

⇒ x = 60° [corresponding angles are equal]

∠z and ∠x are corresponding angles.

⇒ ∠z = 60° [corresponding angles are equal]

∠p and ∠z are vertically opposite angles.

⇒ ∠p = ∠z = 60° [vertically opposite angles are equal]

∠q and ∠p form a linear pair.

⇒ ∠q + ∠p = 180° [linear pair]

⇒ q + 60° = 180°

⇒ q = 120°

∠r and ∠x form linear pair.

⇒ ∠r + ∠x = 180° [linear pair]

⇒ ∠r + 60° = 180°

⇒ ∠r = 120°

∠s and ∠r are vertically opposite angles.

⇒ ∠s = ∠r = 120° [vertically opposite angles are equal]

Hence, ∠x = 60°, ∠z = 60°, ∠p = 60°, ∠q = 120°, ∠r = 120° and ∠s = 120°.

5Exercise 14(F)

In the figure given alongside, find the angles shown by x, y, z and w. Give reasons.

Show SolutionHide Solution↓

From the figure, the given angles are 115° and 70°.

∠x and the 115° angle are vertically opposite angles.

⇒ x = 115° [vertically opposite angles are equal]

∠w and ∠x are corresponding angles.

⇒ w = x = 115° [corresponding angles are equal]

∠y and the 70° angle are vertically opposite angles.

⇒ y = 70° [vertically opposite angles are equal]

∠z and ∠y are corresponding angles.

⇒ z = y = 70° [corresponding angles are equal]

Hence, x = 115°, y = 70°, z = 70° and w = 115°.

6Exercise 14(F)

Find a, b, c and d in the figure given below:

Show SolutionHide Solution↓

From the figure, the given angles are 130° and 150°.

∠a and the 130° angle are vertically opposite angles.

⇒ a = 130° [vertically opposite angles are equal]

∠d and the 130° angle are alternate interior angles.

⇒ d = 130° [alternate interior angles are equal]

∠b and the 150° angle are vertically opposite angles.

⇒ b = 150° [vertically opposite angles are equal]

∠c and the 150° angle are alternate interior angles.

⇒ c = 150° [alternate interior angles are equal]

Hence, a = 130°, b = 150°, c = 150° and d = 130°.

7Exercise 14(F)

Find x, y and z in the figure given below:

Show SolutionHide Solution↓

From the figure,

∠ABC = 75°.

As, BE is parallel to CD and BC is a transversal.

⇒ ∠BCD and ∠ABC are co-interior angles.

⇒ x + 75° = 180° [co-interior angles are supplementary]

⇒ x = 180° - 75°

⇒ x = 105°

Now, AD is parallel to BC and CD is a transversal.

⇒ ∠BCD and ∠ADC are co-interior angles.

⇒ ∠BCD + ∠ADC = 180° [co-interior angles are supplementary]

⇒ x + y = 180°

⇒ 105° + y = 180°

⇒ y = 180° - 105°

⇒ y = 75°

Since BE is parallel to CD and AD is a transversal.

⇒ z = y = 75° [alternate interior angles are equal]

Hence, x = 105°, y = 75° and z = 75°.

11 Mark · Multiple Choice Questions

Which of the following has neither length nor width nor thickness?

(a)Line
(b)Surface
(c)Plane
(d)Point

Show SolutionHide Solution↓

A point is only a mark of position; it has neither length nor width nor thickness.

Hence, option 4 is the correct option.

21 Mark · Multiple Choice Questions

In which of the following figures, AB is perpendicular bisector of line segment PQ?

Show SolutionHide Solution↓

AB is the perpendicular bisector of PQ only when AB is perpendicular to PQ and also cuts it into two equal parts.

This condition is satisfied only in the second figure.

Hence, option 2 is the correct option.

31 Mark · Multiple Choice Questions

Which of the following statements is incorrect?

(a)Ray has one fixed point.
(b)Ray can be extended in only one direction.
(c)Ray has a definite length.
(d)All of these.

Show SolutionHide Solution↓

A ray extends endlessly in one direction, so it does not have a definite length.

Hence, option 3 is the correct option.

41 Mark · Multiple Choice Questions

The number of line(s) passing through two fixed points is:

(a)1
(b)2
(c)0
(d)3

Show SolutionHide Solution↓

Through two fixed points, one and only one line can be drawn.

Hence, option 1 is the correct option.

51 Mark · Multiple Choice Questions

The complement of (x - y)° is:

(a)(90 - x - y)°
(b)(90 - x + y)°
(c)(90 + x - y)°
(d)(x - y - 90)°

Show SolutionHide Solution↓

The complement of an angle = 90° − the angle.

Complement of (x − y)° = 90° − (x − y)°

= (90 − x + y)°

Hence, option 2 is the correct option.

61 Mark · Multiple Choice Questions

The supplement of $\frac{4}{5}$ of 60° is:

(a)52°
(b)132°
(c)40°
(d)80°

Show SolutionHide Solution↓

Since,

$\frac{4}{5} \text{ of } 60^{\circ} = \frac{4}{5} \times 60^{\circ} = 48^{\circ}$

The supplement of an angle = 180° − the angle.

Supplement of 48° = 180° − 48° = 132°

Hence, option 2 is the correct option.

71 Mark · Multiple Choice Questions

Reflex angle measures:

(a)more than 90° and less than 180°
(b)more than 180° and less than 270°
(c)more than 180° and less than 360°
(d)more than 90° and less than 360°

Show SolutionHide Solution↓

A reflex angle is an angle whose measure is more than 180° and less than 360°.

Hence, option 3 is the correct option.

81 Mark · Multiple Choice Questions

When two parallel lines are cut by a transversal then which one of following is not true?

(a)Interior alternate angles are equal.
(b)Exterior alternate angles are complementary.
(c)Co-interior angles are supplementary.
(d)Corresponding angles are equal.

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When two parallel lines are cut by a transversal, alternate exterior angles are equal. They are not necessarily complementary, so “exterior alternate angles are complementary” is not a general property.

Hence, option 2 is the correct option.

91 Mark · Multiple Choice Questions

In the given figure, lines AB and CD are parallel to each other and ∠a : ∠b = 7 : 11, then ∠b is equal to:

(a)70°
(b)110°
(c)100°
(d)none of these

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Given,

∠a : ∠b = 7 : 11

Let ∠a = 7x and ∠b = 11x.

Since AB ∥ CD,

∠BOE = ∠AOP = b [vertically opposite angles are equal]

∠BOE and ∠DEO are co-interior angles.

⇒ ∠BOE + ∠DEO = 180° [co-interior angles are supplementary]

⇒ ∠a + ∠b = 180°

⇒ 7x + 11x = 180°

⇒ 18x = 180°

⇒ x = $\frac{180^{\circ}}{18}$

⇒ x = 10°

⇒ ∠b = 11x = 11 × 10° = 110°

Hence, option 2 is the correct option.

101 Mark · Multiple Choice Questions

Two parallel lines are cut by a transversal, then which of the following can not be true?

(a)Corresponding angles are equal.
(b)Co-interior angles are complementary.
(c)Co-interior angles are supplementary.
(d)Alternate angles are equal.

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When two parallel lines are cut by a transversal, the co-interior angles are supplementary (sum = 180°); they can never be complementary (sum = 90°).

Hence, option 2 is the correct option.

11Statement I-II Type Questions

Statement 1: Two lines in a plane either intersect at a point or they are parallel to each other.

Statement 2: Exactly two lines in the same plane which do not intersect at any point even when further extended up to infinity are called parallel lines.

Which of the following options is correct?

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

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Two lines lying in the same plane either meet at a point or never meet; if they never meet, they are parallel. So, Statement 1 is true.

Two lines in the same plane which do not intersect at any point even when extended to infinity are called parallel lines. So, Statement 2 is true.

Hence, option 1 is the correct option.

12Statement I-II Type Questions

Statement 1: If an angle is such that its supplement is 5 times its complement then the angle is $67\frac{1}{2}°$.

Statement 2: When two adjacent angles form a linear pair then they are supplementary.

Which of the following options is correct?

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

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Let the angle be x. Then its supplement = (180° − x) and its complement = (90° − x).

As the supplement is 5 times the complement,

⇒ 180° − x = 5(90° − x)

⇒ 180° − x = 450° − 5x

⇒ 5x − x = 450° − 180°

⇒ 4x = 270°

⇒ x = $\frac{270^{\circ}}{4}$

⇒ x = $67\frac{1}{2}^{\circ}$

So, Statement 1 is true.

When two adjacent angles form a linear pair, their sum is 180°, i.e. they are supplementary. So, Statement 2 is true.

Hence, option 1 is the correct option.

13Assertion-Reason Type Questions

Assertion (A): There are some figures in geometry which have both curvilinear and linear boundary.

Reason (R): Curvilinear means having curved line, linear means straight line.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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Figures such as a semicircle have both a curved boundary and a straight boundary, so some geometrical figures do have both curvilinear and linear boundaries. So, Assertion (A) is true.

Curvilinear means having a curved line and linear means a straight line. So, Reason (R) is true.

Hence, option 3 is the correct option.

14Assertion-Reason Type Questions

Assertion (A): The angles around a point are: a right angle, x - 10° and 2x + 40°. Then the value of x is 40°.

Reason (R): The sum of all angles around a point is 360°.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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The sum of all the angles around a point is 360°.

⇒ 90° + (x − 10°) + (2x + 40°) = 360°

⇒ 3x + 120° = 360°

⇒ 3x = 360° − 120°

⇒ 3x = 240°

⇒ x = $\frac{240^{\circ}}{3}$

⇒ x = 80°

Since x = 80° and not 40°, Assertion (A) is false.

The sum of all angles around a point is 360°. So, Reason (R) is true.

Hence, option 2 is the correct option.

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