# NCERT Solutions for Class 6 Maths Chapter 14

## NCERT Solutions for Class 6 Maths Chapter 14 – Practical Geometry

NCERT Solutions for Class 6 Maths Chapter 14 are reliable study materials to get answers to the questions given in the Chapter 14 Practical Geometry of NCERT book. These are prepared by the subject experts in a systematic manner, so students can go through them easily.

With NCERT Solutions for Class 6 Maths  Chapter 14 Practical Geometry, students will be able to understand how to solve problems related to shapes and their construction, which are important from the exam point of view. The NCERT Solutions Class 6 Chapter 14 are prepared in very simple language so that students can grasp the concepts easily. They can use the solutions for both preparation and revision of the chapter.

## Access NCERT Solutions for Class 6 Maths  Chapter 14 – Practical Geometry

### NCERT Solutions for Class 6 Maths

The NCERT Solutions for Class 6 Maths prove to be a help to those students who are trying to score better marks in Maths. Although this subject might be a bit tough to understand, practising more will help you gain confidence in the subject.

### Chapter 14 – Practical Geometry

Practical Geometry is an important branch of geometry dealing with the study of the shapes, size, shapes, positions and dimensions of various objects. Geometrical tools are very helpful in Practical Geometry as they are used in the construction of different shapes. For example, the divider is used for measuring lengths, the protractor is used for angles, set squares, compasses, rulers, etc. of a different use.

### NCERT Solutions for Class 6 Maths Chapter 14 Exercises

 Chapter 14 – Practical Geometry Exercises Exercise 14.1 5 Questions & Solutions Exercise 14.2 5 Questions & Solutions Exercise 14.3 2 Questions & Solutions Exercise 14.4 3 Questions & Solutions Exercise 14.5 9 Questions & Solutions Exercise 14.6 9 Questions & Solutions

### NCERT Solutions for Class 6 Maths Chapters

Students must go through the NCERT Solutions Class 6 Maths to practise better for the exams. The NCERT Solutions for Class 6 Maths will prove beneficial to all students. It is accurate and reliable practise material to help students master the subject.

#### Introduction

This section deals with how to construct shapes of different measurements. Class 6 chapter 14 Maths includes various practice questions at the end of the chapter and our NCERT solutions dealing with this chapter provide the required solutions for the same. The problems given in Practical Geometry are based on various difficulty levels that will help students develop a conceptual understanding.

#### The Circle

A circle is a body with no corners. The distance of any point on a circle is the same from the centre of the circle. The distance from the centre of a circle to any point on the circle is known as the radius. If you know the radius of a circle; it becomes easier to draw the circle. You will get the other details of the steps of constructing a circle with a known radius by going through the NCERT Solutions for Class 6 Maths Chapter 14.

#### Line Segment

There is a difference between a line and a line segment. A line has no endpoints whereas a part of a line is known as a line segment. In other words, it can be said that every line segment has two endpoints and so there is a defined length of a line segment. You can learn about the process of drawing a line segment of a defined length by going through the NCERT Solutions for Class 6 Maths chapter 14.

#### Perpendiculars

Constructing a perpendicular on a definite point of a line is easy. You can do so by choosing the definite point and then drawing a perpendicular using the set square on that point. The set square will help you draw a perfect perpendicular. If the point is not on the line, you can draw a perpendicular by using a compass and drawing an arc on the line to cross two points using it.

Another way of drawing the perpendiculars is by the paper folding technique discussed in Chapter 14 Maths Class 6.

#### Angles

You can construct an angle of any measure by placing a protractor on a line segment and marking the angle to be drawn.

Draw a line from the left end of the line segment to the marked point. You can also draw an angle using a compass. For this, all you need to do is use the compass to make an arc at the given angle from one end of the line segment. Then without disturbing the compass angle, draw another arc on the line segment with another point as the centre. Having done so, go back to the previous angle, and put the compass point and the pencil on the points of intersection of the arc on the angle, respectively. At last, all you need to do is draw another arc on the new line using the same measurement to intersect the previous arc. For more details on how to draw angles using different techniques, you can refer to Practical Geometry Class 6 NCERT Solutions.

Similarly, you will learn the method to bisect an angle and also create 30, 60, 90 and 120 degrees angles.

### Key Features of NCERT Solutions for Class 6 Maths Chapter 14

The key features of NCERT Solutions for Class 6 Maths Chapter 14 are:

• They are easily accessible and can be easily understood by the students as it includes stepwise solutions.
• Solutions to all the textbook problems are covered according to the latest CBSE guidelines and syllabus to ease exam preparation.
• NCERT Solutions for Class 6 Maths Chapter 14 will help you clear all the doubts that you might have in revision.

Q.1 With the same centre O, draw two circles of radii 4 cm and 2.5 cm.

Ans-

Steps of construction:

1. Take a measurement of 4 cm by the compass on a scale.

2. Draw a circle with centre O and radius 4 cm.

3. Now take a measurement of 2.5 cm by the compass on a scale.

4. Draw a circle with centre O and radius 2.5 cm. Q.2 Draw a circle of radius 3.2 cm.

Ans-

Steps of construction:

1. Take a measurement of 3.2 cm by the compass on a scale.

2. Draw a circle with centre O and radius 3.2 cm. Q.3 Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?

Ans-

When we join the end points of diameters that are not perpendicular to each other, we get a rectangle. When diameters are at right angle and we join end points of these, we get square. Since diagonals of a rectangle are equal and bisect each other at a specific angle. and diagonals of square are equal and bisect each other at right angle.

Q.4 Draw any circle and mark points A, B and C such that

(a) A is on the circle.

(b) B is in the interior of the circle.

(c) C is in the exterior of the circle.

Ans-

(a) A is on the circle with centre O. (b) B is in the interior of the circle with centre O. (c) C is in the exterior of the circle with centre O. $\begin{array}{l}\text{Q.5 Let A, B be the centres of two circles of equal radii; draw}\\ \text{them so that each one of them passes through the centre}\\ \text{of the other. Let them intersect at C and D. Examine}\\ \text{whether }\overline{\text{AB}}\text{ and }\overline{\text{CD}}\text{ are at right angles.}\end{array}$

Ans-

$\begin{array}{l}\text{Draw two circles of same radius which are passing through the centres of the other circle.}\\ \text{Here, point A and B are the centres of these circles and these circles are intersecting each other}\\ \text{at point C and D}.\\ \text{In quadrilateral ADBC,}\\ \text{AD}=\text{AC (Radius of circle centered at A)}\\ \text{BC}=\text{BD (Radius of circle centered at B)}\\ \text{As radius of both circles are equal, therefore, AD = AC = BC = BD}\\ \text{Hence, ADBC is a rhombus and in rhombus, the diagonals bisect each other at}90°\text{.}\\ \text{Hence,}\overline{\text{AB}}\text{and}\overline{\text{CD}}\text{are at right angles.}\end{array}$ Q.6 Draw a line segment of length 7.3 cm using a ruler.

Ans-

Steps of construction:

1. Draw a line l. Mark a point A on line l.

2. Take a measurement of 7.3 cm from the scale with the help of compass.

3. Mark an arc of radius 7.3 with centre A on line l which intersect at point B on line l.

4. AB is the required line segment. Q.7 Construct a line segment of length 5.6 cm using ruler and compasses.

Ans-

Steps of construction:

1. Draw a line l. Mark a point A on line l.

2. Take a measurement of 5.6 cm from the scale with the help of compass.

3. Mark an arc of radius 5.6 with centre A on line l which intersect at point B on line l.

4. AB is the required line segment. $\begin{array}{l}\text{Q.8 Construct }\overline{\text{AB}}\text{ of length 7.8 cm. From this, cut off }\overline{\text{AC}}\text{ of}\\ \text{length 4.7 cm. Measure }\overline{\text{BC}}\text{.}\end{array}$

Ans-

Steps of construction:
1. Draw a line l. Mark a point A on line l.
2. Take a measurement of 7.8 cm from the scale with the help of compass.
3. Mark an arc of radius 7.8 with centre A on line l which intersect at point B on line l.
4. AB is the required line segment.
5. Similarly, we cut line segment AC = 4.7 from line segment AB. The length of the remaining line segment BC is 3.1 cm.

$\begin{array}{l}\text{Q.9 Given }\overline{\text{AB}}\text{ of length 3.9 cm, construct }\overline{\text{PQ}}\text{ such that the}\\ \text{length of }\overline{\text{PQ}}\text{ is twice that of }\overline{\text{AB}}\text{. Verify by measurement.}\end{array}$

Ans-

Steps of construction:
1. Draw a line l and cut a line segment AB of 3.9 cm from it.
2. Mark an arc of length 3.9 cm with centre B, which intersect given line l at the point C such that AC = 2AB. 3. Now, again draw a line l’.
4. Take distance AC in compass and mark an arc with centre P and radius AC on line l’. 5. PQ is the required line segment.

$\begin{array}{l}\text{Q.10 Given AB of length 7.3 cm and CD of length 3.4 cm, construct}\\ \text{a line segment XY such that the length of XY is equal to the}\\ \text{difference between the lengths of AB and CD.}\\ \text{Verify by measurement.}\end{array}$

Ans-

Steps of construction:

1. Draw a line l and take a point X on it.

2. Construct a line segment XM such that XM = AB = 7.3 cm

3. Then cut off a line segment YM such that YM = CD = 3.4 cm

4. Thus, length of XY = Length of AB – Length of CD By measurement, we find that XY = 3.9 cm = 7.3 cm –3.4 cm = AB –CD

$\begin{array}{l}\text{Q.11 Draw any line segment }\overline{\text{PQ}}\text{. Without measuring }\overline{\text{PQ}}\text{,}\\ \text{construct a copy of }\overline{\text{PQ}}\text{.}\end{array}$

Ans-

Steps of construction:
(i) Draw a line segment PQ of any length.
(ii) Fix the compass pointer at P and pencil pointer at Q. The opening of the instrument gives the length of PQ. (iii) Draw a line l’.
(iv) Choose a point R on l’. Mark an arc of radius equal to PQ and with centre R which intersects the line at point S. So, RS is a copy of line segment PQ.

$\begin{array}{l}\text{Q.12 Given some line segment }\overline{\text{AB}}\text{, whose length you do not}\\ \text{know, construct }\overline{\text{PQ}}\text{ such that the length of }\overline{\text{PQ}}\text{ is twice}\\ \text{that of }\overline{\text{AB}}\text{.}\end{array}$

Ans-

Step of construction:
(i) AB is a line segment of any length. (ii) Take the measurement of AB with the help of compass.
(iii) Draw a line l’.
(iv) Mark an arc of radius equal to AB and with centre P which intersects line at point R.
(v) Again mark another arc with radius AB and centre R which cuts line at point Q.
(vi) Therefore, PQ is required line segment. $\begin{array}{l}\text{Q.13 Draw any line segment }\overline{\text{AB}}\text{. Mark any point M on it. Through}\\ \text{M, draw a perpendicular to }\overline{\text{AB}}\text{. (use ruler and compasses)}\end{array}$

Ans-

Steps of construction:
1. Draw a line l and mark points A, B and M on it.
2. With the help of compasses mark an arc of any radius with centre M, which intersects AB at X and Y. 3. Mark arcs of same radius with centre X and Y. These arcs intersect each other at N.
4. Join MN. Therefore, TM is the required perpendicular to the line segment AB.

$\begin{array}{l}\text{Q.14 Draw any line segment }\overline{\text{PQ}}\text{. Take any point R not on it.}\\ \text{Through R, draw a perpendicular to }\overline{\text{PQ}}\text{.}\\ \text{ (use ruler and set-square)}\end{array}$

Ans-

Steps of construction:
Step 1: Draw a line segment PQ and take a point R outside it. Step 2: Put a set square on line segment PQ such that one side of its right angle and line PQ should be in same direction. Step 3: Place a ruler along the edge opposite to the right angle of the set-square. Step 4: Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square. Step 5: Join RS along the edge through R meeting PQ at S. Then RS is perpendicular to PQ. $\begin{array}{l}\text{Q.15 Draw a line l and a point X on it. Through X, draw a line}\\ \text{segment XY perpendicular to l. Now draw a perpendicular}\\ \text{to XY at Y. (use ruler and compasses)}\end{array}$

Ans-

Steps of construction:
1. Draw a line l and mark any point X on it.
2. With the help of compass mark an arc of any radius on the line with centre X. It intersects the line at the points A and B.
3. Mark another arcs of same radius with centres A and B respectively which intersect each other at Y.
4. Join line segment XY.
5. Therefore, XY is required perpendicular to line l. 6. In the same way, line ZY is perpendicular on XY. Q.16 Draw

$\overline{\mathrm{AB}}$

of length 7.3 cm and find its axis of symmetry.

Ans-

Steps of construction:
1. Draw a line segment

$\overline{\mathrm{AB}}$

of length 7.3 cm.
2. Mark two arcs of same radii and centres A and B, which intersect each other at C and D respectively.
3. CD is the axis of symmetry i.e., perpendicular bisector of line segment AB. Q.17 Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

Ans-

Steps of construction:
1. Draw a line segment PQ of length 9.5 cm.
2. Mark two arcs of same radii and centres P and Q, which intersect each other at R and S respectively.
3. RS is the axis of symmetry i.e., perpendicular bisector of line segment PQ. $\begin{array}{l}\text{Q.18 Draw the perpendicular bisector of }\overline{\text{XY}}\text{ whose length is}\\ \text{10.3 cm. }\left(\text{a}\right)\text{ Take any point P on the bisector drawn. Examine}\\ \text{whether }\overline{\text{PX}}\text{ = }\overline{\text{PY}}\text{. }\left(\text{b}\right)\text{ If M is the midpoint of XY, what can}\\ \text{you say about the lengths }\overline{\text{MX}}\text{ and }\overline{\text{XY}}\text{?}\end{array}$

Ans- Steps of construction:
1. Draw a line segment XY = 10.3 cm.

2. Draw perpendicular bisector of line segment XY, which intersect each other at R and S.

3. Join RS. RS is the axis of symmetry.

(a) Here, by measuring we get: PX = PY (b) If M is the midpoint of line segment XY, then MX = MY. This means MX = (1/2) XY. Q.19 Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts.Verify by actual measurement.

Ans-

Step of construction:
1. Draw a line segment XY = 12.8 cm.
2. Mark two arcs with radius more than 6.4 cm and centres X and Y respectively which intersect each other at R and S.
3. Join RS which intersect XY at P.
4. Similarly draw perpendicular bisectors of XP and PY.
5. By measuring with scale, we find that XA= 3.2 cm, AP = 3.2 cm, PB = 3.2 cm and BY = 3.2 cm. $\mathrm{Q.20With}\stackrel{\to }{\mathrm{PQ}}\mathrm{of}\mathrm{length}6.1\mathrm{cm}\mathrm{as}\mathrm{diameter},\mathrm{draw}\mathrm{a}\mathrm{circle}.$

Ans-

Step of construction:
1. Draw a line segment PQ of length 6.1 cm.
2. Mark arcs of radius more than half of 6.1 cm with centre P and Q respectively. These arcs intersect each other at points C and D.
3. Join C and D. CD intersects PQ at the point O.
4. Draw a circle of radius OP or OQ with centre O.Thus, required circle passes through P and Q. $\begin{array}{l}\mathrm{Q.21Draw}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{C}\mathrm{and}\mathrm{radius}3.4\mathrm{cm}.\mathrm{Draw}\mathrm{any}\\ \mathrm{chord}\overline{\mathrm{AB}}.\mathrm{Construct}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\overline{\mathrm{AB}}\mathrm{and}\\ \mathrm{examine}\mathrm{if}\mathrm{it}\mathrm{passes}\mathrm{through}\mathrm{C}.\end{array}$

Ans-

Steps of construction:

1. Draw a circle with centre C and radius 3.4 cm.
2. Draw any chord AB.
3. Taking A and B as centres and radius more than half of AB, draw two arcs which cut each other at S and T.
4. Join ST. Then ST is the perpendicular bisector of AB.

This perpendicular bisector of AB passes through the centre C of the circle. $\begin{array}{l}\mathrm{Q.22Draw}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{C}\mathrm{and}\mathrm{diameter}\overline{\mathrm{AB}}3.4\mathrm{cm}.\\ \mathrm{Construct}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\overline{\mathrm{AB}}\mathrm{and}\\ \mathrm{examine}\mathrm{if}\mathrm{it}\mathrm{passes}\mathrm{through}\mathrm{C}.\end{array}$

Ans-

Steps of construction:

1. Draw a circle with radius 3.4/2 = 1.7 cm with
centre C.
2. Draw diameter AB.

3. Draw perpendicular bisector of AB with centre A
and B and any radius more than 1.7 cm.
4. The bisector PQ of AB intersects AB at centre C. Therefore, it can be observed that perpendicular bisector of AB passes through centre C. Q.23 Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

Ans-

Step of construction:
1. Draw a circle of radius 4 cm with centre C.
2. Draw two chords AB and PQ.
3. Draw perpendicular bisectors of AB and PQ, which intersect at C. Therefore, it can be observed that bisectors of both chords intersect at the centre C of the circle. $\mathrm{Q.24Draw}\angle \mathrm{POQ}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{measure}\text{\hspace{0.17em}}75\mathrm{°}\mathrm{and}\text{\hspace{0.17em}}\mathrm{find}\text{\hspace{0.17em}}\mathrm{its}\text{\hspace{0.17em}}\mathrm{line}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{symmetry}.$

Ans-

Steps of construction:
1. Draw a ray OP.

2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Arc of same radius with centre B intersects the earliest arc at R.
5. The arcs drawn of any radius with centre B and R respectively intersect at the point S.
6. Draw ray OC through the point S.
6. Mark another arc of any radius with centre U (intersection point of arc and OC) and B which intersect at T.
7. Draw ray OQ through the point T
8. Draw two arcs of any radius with centre W (intersection point of arc and OQ) and A. These arcs intersect at the point V. Therefore, line OV is the line of symmetry for angle POQ. Q.25 Draw an angle of measure 147° and construct its bisector.

Ans-

Steps of construction:
1. Draw a ray OP.

2. Draw an angle of 147° with the help of protractor at point O. 3. Draw an arc of any radius with centre O which intersects OP at A and OQ at B.
4. Mark two arcs of any radius with centre A and B, which intersect each other at C.
5. Draw ray OR through C. Therefore, OR is bisector of angle POQ. Q.26 Draw a right angle and construct its bisector.

Ans-

Steps of construction:

1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Again mark two arcs of any radius with centre A and D (intersection point of arc and OQ), which intersect each other at E.
8. Draw ray OS through E. Therefore, OS is bisector of angle POQ. Q.27 Draw an angle of measure 153° and divide it into four equal parts.

Ans-

Steps of construction:
1. Draw a ray OP.

2. Draw an angle of 153° with the help of protractor at point O. 3. Taking O as centre draw an arc of any radius which cuts OP at A and OQ at B.
4. Mark two arcs of any radius with centre A and B which intersect each other at C.
5. Draw ray OR through C. Ray OR is bisector of angle POQ. 6. Mark two arcs of any radius with centre A and D which intersect at E.
7. Draw ray OS through the point E. OS is bisector of angle POR.
6. Draw two arcs of any radius with centre D and B, which intersect at F.
8. Draw ray OT through the point F. Therefore, by observation we see rays OS, OR and OT divide the angle POQ into four equal parts. Q.28 Construct with ruler and compasses, angles of following measures:
(a) 60° (b) 30° (c) 90°
(d) 120° (e) 45° (f) 135°

Ans-

(a) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw ray OQ through the point B.Therefore, angle POQ is required angle of 60°. (b) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw a ray OQ through the point B.
5. Mark two arcs of any radius with centre A and B, which intersect at point C.
6. Draw ray OR through point C. Therefore, angle POR is required angle of 30°. (c) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ. (d) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O. This arc intersects OP at the point A.
3. Mark another arc of the same radius with centre A, it intersects earlier arc at the point B.
4. Draw an arc with centre B of same radius. This arc intersects the earliest arc at the point C.
5. Draw ray OQ through the point C. Therefore, angle POQ is required angle of 120°. (e) Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Mark two arcs of any radius with centre A and D, which intersect at E.
8. Draw a ray OS through the point E. Therefore, POS is required angle of 45°. (f) Steps of construction:
1. Draw a line PR.
2. Take a point O on the line.
3. Draw a semi-circle of any radius with centre O.
4. Mark two arcs of any radius with centre A and B respectively which intersect at R.
5. Draw a ray OQ through the point D.
6. Mark two arcs with centre D and B which intersect at E.
7. Draw ray OS through the point E. Therefore, angle POS is required angle of 135°. Q.29 Draw an angle of measure 45° and bisect it.

Ans-

Steps of construction:
1. Draw a ray OP.
2. Mark an arc of any radius with centre O, which intersects ray OP at the point A.
3. Mark another arc of same radius with centre A which intersects earlier arc at B.
4. Mark an arc of same radius with centre B which intersects the earliest arc at the point C.
5. Mark two arcs of any radius with centre B and C respectively which intersect at the point R.
6. Draw ray OQ through the point R. In this way we get right angle POQ.
7. Mark two arcs of any radius with centre A and D (intersection point of arc and OQ), which intersect at E.
8. Draw a ray OS through the point E.
9. Mark two arcs of any radius with centre A and G (intersection point of arc and OS), which intersect at F.
10. Draw ray OT through the point F. Therefore, angle POS is required angle of measure 45° and OT is bisector of angle POS.

Q.30 Draw an angle of measure 135° and bisect it.

Ans-

Steps of construction:
1. Draw a line MN.
2. Take a point O on the line.
3. Draw an arc of any radius with centre O which cuts the line MN at A and B.
4. Mark two arcs of radius more than half of AB with centre A and B respectively which intersect at R.
5. Draw a ray OQ through the point R. Let the ray OQ intersects the arc in (3) at D.
6. Mark two arcs with centre D and B of radius more than half of DB which intersect at E.
7. Draw ray OS through the point E. Therefore, angle NOS is required angle of 135°.
8. Mark two arcs of radius more than half of AC with centre A and C respectively which intersect at F.
9. Draw ray OT through the intersection point F. Therefore, angle NOT is the bisector of angle NOS. Q.31 Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.

Ans-

Steps of construction:
1. Draw a ray OA.
2. Draw an angle of 70° at point O with the help of protractor.  3. Mark an arc with centre O which cuts OA and OB at C and D respectively.
4. Draw a ray QP and mark an arc with the same radius and centre Q on ray QP.
5. Set your compasses to the length CD with the same radius and mark an arc with centre T which cuts earlier arc at the point R.
6. Draw ray QS through the point R. Therefore, angle PQS is required angle of 70°. Q.32 Draw an angle of 40°. Copy its supplementary angle.

Ans-

Steps of construction:
Draw a line PA and mark a point O on it.
Draw an angle of 40° at point O with the help of protractor. 3. Angle POB is supplementary of 40°.
4. Taking O as centre, draw an arc between the rays of angle POB which cuts PO and OB at D and C respectively. 5. Now draw a line m and take a point S on it. Taking same radius and S as centre draw an arc. This arc cuts line m at point T.
6. Set your compasses to the length CD with the same radius and mark an arc with centre T which cuts earlier arc at the point R.
7. Join S and R. SR is the required ray which makes a supplementary angle of 40° with line m or makes an angle of 140°. 