Lines and Angles Class 6 NCERT Solutions Ganita Prakash Maths Chapter 2

Get the simplified Class 6 Maths NCERT Solutions of Ganita Prakash Chapter 2 Lines and Angles textbook exercise questions with complete explanation.

Ganita Prakash Class 6 Maths Chapter 2 Solutions Lines and Angles

NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 2 Lines and Angles

2.1 Point 2.2 Line Segment 2.3 Line 2.4 Ray Figure it Out (Page No. 15-17)

Question 1.

Can you help Rihan and Sheetal find their answers?
Solution:
Rihan can draw infinite number of lines that pass through the point.
Whereas Sheetal can draw only one line that passes through both of the points.

Question 2.
Name the line segments in given figure. Which of the five marked points are on exactly one of the line segments? Which are on two of the line segments?

Solution:
The line segments in the given figure are \(\overline{\mathrm{LM}}, \overline{\mathrm{MP}}, \overline{\mathrm{PQ}}, \overline{\mathrm{QR}}\). Points L and R are on exactly one of the line segments and points M, P and Q are on two of the line segments.

Question 3.
Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays?

Solution:
In the provided Fig. 2.5, there are two rays shown. Let’s identify and name them:

1. Ray \(\overrightarrow{T A}\): This ray starts from point T and passes through point A.
2. Ray \(\overrightarrow{T B}\): This ray starts from point T and passes through point B.

Yes, T is the starting point of both of these rays. In geometry, the starting point of a ray is called the ‘initial point’, and here, T serves as the initial point for both ray \(\overrightarrow{T A}\) and ray \(\overrightarrow{T B}\).

Question 4.
Draw a rough figure and write labels appropriately to illustrate each of the following:
(a) OP and OQ meet at O.
Solution:

(b) XY and PQ intersect at point M.
Solution:

(c) Line l contains joints E and F but not point D.
Solution:

(d) Point P lies on AB.
Solution:

Question 5.
In the figure, name
(a) five points
(b) a line
(c) four rays
(d) five line segments

Solution:
(a) Five points are D, E, O, C and B.
(b) In the given figure, a line is \(\overrightarrow{D B}\).
(c) Four rays are \(\overrightarrow{O C}, \overrightarrow{O B}, \overrightarrow{E B}\) and \(\overrightarrow{O D}\).
(d) Five line segments are \(\overline{D E}, \overline{E O}, \overline{O B}, \overline{D O}\) and \(\overline{E B}\).

Question 6.
Here is a ray \(\overrightarrow{OA}\) (Fig. 2.7). It starts at O and passes through the point A. It also passes through the point B.

(a) Can you also name it as \(\overrightarrow{OB}\)? Why?
(b) Can we write \(\overrightarrow{OA}\) as \(\overrightarrow{AO}\)? Why or why not?
Solution:
(a) No, we cannot name it as \(\overrightarrow{OB}\). A ray is defined as having a starting point and extending infinitely in one direction. In this case, the ray \(\overrightarrow{OA}\) starts at point O and passes through point A. If you name it \(\overrightarrow{OB}\), it would imply that O is the starting point and B is a point it passes through. Since \(\overrightarrow{OA}\) passes through A and not necessarily B (or if B is just a different point on the same line), \(\overrightarrow{OB}\) would not correctly describe the same ray unless B coincides with A.

(b) No, you cannot write \(\overrightarrow{OA}\) as \(\overrightarrow{AO}\). The notation \(\overrightarrow{OA}\) represents a ray that starts at O and passes through A, extending infinitely in the direction from O to A. In contrast, \(\overrightarrow{AO}\) would represent a ray starting at A and passing through O. The direction and starting point are different for \(\overrightarrow{AO}\), so it does not represent the same ray as \(\overrightarrow{OA}\).

2.5 Angle Figure it Out (Page No. 19-21)

Question 1.
Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.

Solution:
Yes, we can find the angles in the given pictures.

Question 2.
Draw and label an angle with arms ST and SR.
Solution:
An angle with arms ST and SR is ∠TSR.

Question 3.
Explain why ∠APC cannot be labelled as ∠P.

Solution:
The angle ∠APC cannot be labelled just as ∠P because three different angles meet at point P.
If we just say ∠P, it won’t be clear which of these three angles we’re talking about.

• One angle is between the lines of PA and PC (this is ∠APC).
• Another angle is between the lines PA and PB (this is ∠APB).
• The third angle is between the lines PB and PC (this is ∠BPC).

So, to make it clear, we use all three letters: A, P, and C, to describe the exact angle we mean.

Question 4.
Name the angles marked in the given figure.

Solution:
In the given figure, marked angles are ∠RTQ and ∠RTP.

Question 5.
Mark any three points on your paper that are not on one line. Label them A, B, C. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C? Write them down, and mark each of them with a curve.
Solution:
We can draw the figure as shown below. We get 3 lines in all. These are \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}\) and \(\overrightarrow{\mathrm{CA}}\).

We get 3 angles in all. These are ∠ABC, ∠BCA and ∠CAB.

Question 6.
Now, mark any four points on your paper so that no three of them are on one line. Label them A, B, C and D. Draw all possible lines going through pairs of these points. How many lines do you get? Name them. How many angles can you name using A, B, C and D? Write them all down and mark each of them with a curve as in given figure.
Solution:
Point A, B, C and D are as follow
Possible lines are AB, BC, CD, AD, AC and BD.
Thus, there are six lines formed and angles are ∠BAC, ∠BAD, ∠ADB, ∠ADC, ∠DCA, ∠DCB, ∠ABD, ∠ABC, ∠CAD, ∠BDC, ∠ACB and ∠DBC.
Thus, there are total twelve angles formed.

2.6 Comparing Angles Figure it Out (Page No. 23)

Question 1.
Fold a rectangular sheet of paper, then draw a line along the fold created. Name and compare the angles formed between the fold and the sides of the paper. Make different angles by folding a rectangular sheet of paper and compare the angles. Which is the largest and smallest angle you made?

Solution:

Do it yourself. For your reference, ∠5 is the largest and ∠2 is the smallest among all the angles in the figure shown below.

Question 2.
In each case, determine which angle is greater and why.
(a) ∠AOB or ∠XOY
(b) ∠AOB or ∠XOB
(c) ∠XOB or ∠XOC
Discuss with your friends on how you decided which one is greater.

Solution:
On comparing the given angles in the figure by superimposition.
(a) ∠AOB is greater because size of ∠AOB is greater than size of ∠XOY.
(b) ∠AOB is greater because size of ∠AOB is greater than size of ∠XOB.
(c) ∠XOB and ∠XOC both are equal angle because vertex O and one ray \(\overrightarrow{O X}\) are common and arm \(\overrightarrow{O B}\) and \(\overrightarrow{O C}\) are overlaping.

Question 3.
Which angle is greater: ∠XOY or ∠AOB? Give reasons.

Solution:
On comparing by superimposition, the angles ∠XOY and ∠AOB in the given figure, we get ∠XOY is greater than ∠AOB because size of ∠XOY is greater.

2.7 Making Rotating Arms 2.8 Special Types of Angles Figure it Out (Page No. 29-31)

Question 1.
How many right angles do the windows of your classroom contain? Do you see other right angles in your classroom?
Solution:
Do it yourself.

Question 2.
Join A to other grid points in the figure by a straight line to get a straight angle. What are all the different ways of doing it?

Solution:

Question 3.
Now join A to other grid points in the figure by a straight line to get a right angle. What are all the different ways of doing it?

Solution:
We can produce BA beyond A to make a straight angle, then through A, draw the line DE, which is perpendicular to BC as shown in the figure below. Clearly, ∠DAB or ∠EAB is a right angle. There is only one way of doing it.

Question 4.
Get a slanting crease on the paper. Now, try to get another crease that is perpendicular to the slanting crease.
(a) How many right angles do you have now? Justify why the angles are exact right angles.
(b) Describe how you folded the paper so that any other person who doesn’t know the process can simply follow your description to get the right angle.
Solution:
(a) By folding a second crease that is perpendicular to the slanting crease, we’ll get 4 right angles.
(b) To explain some other person who doesn’t know the process, he can follow the given steps:

• Fold Diagonally: Fold a rectangular sheet of paper by bringing one corner to the opposite corner. Press to create a diagonal crease. Unfold.
• Fold Perpendicular: Lay the paper flat. Fold one edge to meet the diagonal crease, aligning it perfectly. Press to create a perpendicular crease. Unfold.
• Check: The intersection of the two creases should create four right angles (90°) at the crossing point.

2.7 Making Rotating Arms 2.8 Special Types of Angles Figure it Out (Page No. 31-32)

Question 1.
Identify acute, right, obtuse and straight angles in the previous figures.
Solution:

Question 2.
Make a few acute angles and a few obtuse angles. Draw them in different orientations.
Solution:
Do it yourself.

Question 3.
Do you know what the words acute and obtuse mean? Acute means sharp and obtuse means blunt. Why do you think these words have been chosen?
Solution:
Word ‘acute’ means ‘sharp’. The vertex of the angle appears as a sharp tip.
Word ‘obtuse’ means ‘blunt’. The vertex of the angle appears as a blunt tip.

Question 4.
Find out the number of acute angles in each of the figures below.

What will be the next figure and how many acute angles will it have? Do you notice any pattern in the numbers?
Solution:

3 + 9=12
12 + 9 = 21
In every step, the numbers of angles increases by 9.

Next figure will be as follows:

Number of acute angles = 21 + 9 = 30

2.9 Measuring Angles Figure it Out (Page No. 35)

Question 1.
Write the measures of the following angles:

(a) ∠KAL
Notice that the vertex of this angle coincides with the centre of the protractor. So the A number of units of 1 degree angle between KA and AL gives the measure of ∠KAL. By counting, we get ∠KAL = 30°.
Making use of the medium sized and large sized marks, is it possible to count the number of units in 5 s or 10s?
(b) ∠WAL
(c) ∠TAK
Solution:
(a) Yes it is possible to count the number of units in 5s or 10s because there are also marking for each 5° and 10° on the protractor and the arms \(\overrightarrow{\mathrm{AK}}\) and \(\overrightarrow{\mathrm{AL}}\) pass through the 10° markings.

(b) In the given figure, the arms \(\overrightarrow{\mathrm{AL}}\) and \(\overrightarrow{\mathrm{AW}}\) pass through the 10° markings. There are five 10° markings of scale where \(\overrightarrow{\mathrm{AL}}\) and \(\overrightarrow{\mathrm{AW}}\) pass. So, ∠WAL = 50°.

(c) In the given figure, the arms \(\overrightarrow{\mathrm{AK}}\) and \(\overrightarrow{\mathrm{AT}}\) pass through the 10° markings. There are twelve 10° markings of scale where \(\overrightarrow{\mathrm{AK}}\) and \(\overrightarrow{\mathrm{AT}}\) pass. So, ∠TAK = 120°.

2.9 Measuring Angles Figure it Out (Page No. 40-43)

Question 1.
Find the degree measures of the following angles using your protractor.

Solution:
(a) ∠IHJ = 47°
(b) ∠IHJ = 24°
(c) ∠IHJ =110°

Question 2.
Find the degree measures of different angles in your classroom using your protractor. .
Solution:
Angle at comer of blackboard = 90°
Angle at comer of desk = 90°

Question 3.
Find the degree measures for the angles given below. Check if your paper protractor can be used here!

Solution:
(a) ∠IHJ = 42°
(b) ∠IHJ =116°
Paper protractor cannot be used here.

Question 4.
How can you find the degree measure of the angle given below using a protractor?

Solution:
We require measure of reflex ∠AOB.
Step 1. We find measure of ∠AOB.
Step 2. We find 360° ∠AOB.
This is the required measure.

Question 5.
Measure and write the degree measures for each of the following angles:

Solution:
(a) Measure of given angle is 80°
(b) Measure of given angle is 120°
(c) Measure of given angle is 60°
(d) Measure of given angle is 130°
(e) Measure of given angle is 130°
(f) Measure of given angle is 60°

Question 6.
Find the degree measures of ∠BXE, ∠CXE, ∠AXB and ∠BXC.

Solution:
(a) ∠BXE =115°
(b) ∠CXE = 85°
(c) ∠AXB = 65°
(d) ∠BXC = 30°

Question 7.
Find the degree measures of ∠PQR, ∠PQS and ∠PQT.

Solution:
(a) ∠PQR = 45°
(b) ∠PQS = 105°
(c) ∠PQT = 150°

Question 8.
Make the paper craft as per the given instructions. Then, unfold and open the paper fully. Draw lines on the creases made and measure the angles formed.

Solution:
Do it yourself.

Question 9.
Measure all three angles of the triangle shown in Fig. (a), and write the measures down near the respective angles. Now add up the three measures. What do you get? Do the same for the triangles in Fig. (b) and (c). Try it for other triangles as well, and then make a conjecture for what happens in general! We will come back to why this happens in a later year.

Solution:
∠A + ∠B + ∠C = 180°

2.9 Measuring Angles Figure it Out (Page No. 45-46)

Question 1.
Angles in a clock
(a) The hands of a clock make different angles at different times. At 1 O’clock, the angle between the hands is 30°. Why?
(b) What will be the angle at 2 O’clock? And at 4 O’clock? 6 O’clock?
(c) Explore other angles made by the hands of a clock.

Solution:
(a) The clock is divided into 12 h, so each hour mark is 30° apart (360°- 12 = 30°).
Therefore, at 1 O’clock the hour hand is at 1 and the minute hand is at 12, forming a 30° angle.

(b) At 2 O’clock, it is 60° (i.e. 30° × 2 = 60°), at 4 O’clock, it is 120° (i.e. 30° × 4 = 120°) and at 6 O’clock, it is 180° (i.e. 30° × 6 = 180°)

(c) The angle increases by 30° for each hour. Other angle includes 90° at 3 O’clock, 150° at 5 O’clock and so on. Thus, on multiplying the hour by 30°, we can find the angle at any hour.

Question 2.
The angle of a door.
Is it possible to express the amount by which a door is opened using an angle? What will be the vertex of the angle and what will be the arms of the angle?

Solution:
Yes, it is possible to express the amount by which a door is opened using an angle. The hinge of the door will be the vertex of the angle. The wall and the door will be the arms of the angle.

Solution:
Yes, it is possible.

Here, vertex is B, and arms are AB and BC.

Question 3.
Vidya is enjoying her time on the swing. She notices that the greater the angle with which she starts the swinging, the greater is the speed she achieves on her swing. But where is the angle? Are you able to see any angle?

Solution:
Yes, an angle can be seen.

Question 4.
Here is a toy with slanting slabs attached to its sides; the greater the angles or slopes of the slabs, the faster the balls roll. Can angles be used to describe the slopes of the slabs?

Solution:
Greater the angle, greater the slope.
For each angle one arm is a side and one arm is the slope.

Question 5.
Observe the images below where there is an insect and its rotated version, fan angles be used to describe the amount of rotation? How? What will be the arms of the angle and the vertex?
Hint: Observe the horizontal line touching the insects.

Solution:
Both insects are rotated 90° clockwise.

2.10 Drawing Angles Figure it Out (Page No. 49-50)

Question 1.
In the given figure below, list all the angles possible. Did you find them all? Now, guess the measures of all the angles. Then, measure the angles with a protractor. Record all your numbers in a table. See how close your guesses are to the actual measures.

Solution:

Name of Angles	Estimated measure of Angles	Actual measures of Angles
∠PAC	100°	107°
∠ACD	80°	72°
∠CDL	180°	180°
∠DLP	95°	97°
∠LPR	95°	98°
∠PLS	85°	82°
∠LSR	75°	78°
∠PRS	105°	102°
∠BRS	75°	79°

Question 2.
Use a protractor to draw angles having the following degree measures:
(a) 110°
(b) 40°
(c) 75°
(d) 112°
(e) 134°
Solution:
(a) We can construct an angle of measure 110° using a protractor step by step as discussed below.
Step I: Draw a ray \(\overrightarrow{\mathrm{AB}}\).

Step II: Place the centre of the protractor at A and the zero edge along \(\overrightarrow{\mathrm{AB}}\).

Step III: Start counting from zero near B. Mark a point P at 110° and join AP.

(c) We can construct an angle of measure 75° using a protractor step by step as discussed below.
Step I: Draw a ray \(\overrightarrow{\mathrm{AB}}\).

Step II: Place the centre of the protractor at A and the zero edge along \(\overrightarrow{\mathrm{AB}}\).

Step III: Start counting from zero near B. Mark a point P at 75°.

Step IV: Join AP.
Thus, ∠BAP = 75°.
(b) , (d) and (e). Do it yourself as done above.

Question 3.
Draw an angle whose degree measure is the same as the angle given below:

Also, write down the steps you followed to draw the angle.
Solution:
Step 1. Measure the given angle (∠IHJ = 120°)
Step 2. Using a protractor draw ∠ABC = 120°

2.11 Types of Angles and their Measures Figure it Out (Page No. 51-52)

Question 1.
In each of the below grids, join A to other grid points in the figure by a straight line to get
(a) An acute angle

(b) An obtuse angle

(c) A reflex angle

Mark the intended angles with curves to specify the angles. One has been done for you.
Solution:

Question 2.
Use a protractor to find the measure of each angle. Then classify each angle as acute, obtuse, right, or reflex.
(a) ∠PTR
(b) ∠PTQ
(c) ∠PTW
(d) ∠WTP

Solution:
Using a protractor we fiipd the measure of each angle and then classify each angle as follows:
(a) ∠PTR = 31°, acute angle
(b) ∠PTQ = 60°, acute angle
(c) ∠PTW = 104°, obtuse angle
(d) ∠WTP = 360° – 104° = 256°, reflex angle.

2.11 Types of Angles and their Measures Figure it Out (Page No. 53-54)

Question 1.
Draw angles with the following degree measures
(a) 140°
(b) 82°
(c) 195°
(d) 70°
(d) 35°
Solution:

Question 2.
Estimate the size of each angle and then measure it with a protractor:

Classify these angles as acute, right, obtuse or reflex angles.
Solution:
(a) 45°, acute angle
(b) 169°, obtuse angle
(c) 120°, obtuse angle
(d) 33°, acute angle
(e) 99°, obtuse angle
(f) 348°, reflex angle

Question 3.
Make any figure with three acute angles, one right angle and two obtuse angles.
Solution:
Sample figure

Question 4.
Draw the letter !W such that the angles on the sides are 40° each and the angle in the middle is 60°.
Solution:

Question 5.
Draw the letter V such that the three angles formed are 150°, 60° and 150°.
Solution:

Question 6.
The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?

Solution:
The angle between two adjacent spokes:

• The Ashoka Chakra has 24 spokes, and a full circle is 360°.
• The angle between two adjacent spokes is \(\frac{360^{\circ}}{24}\) = 15°.

Largest Acute Angle Formed Between Two Spokes:

• The largest acute angle formed between any two spokes will be 15°, which is already the angle between two adjacent spokes.
• However, the largest angle may also be the reflex angle between any two adjacent spokes. The reflex angle will be 360° – 15° = 345°.

Question 7.
Puzzle: I am an acute angle. If you double my measure, you get an acute angle. If you triple my measure, you will get an acute angle again. If you quadruple (four times) my measure, you will get an acute angle yet again! But if you multiply my measure by 5, you will get an obtuse angle measure. What are the possibilities for my measure?
Solution:
As per given condition, we have 4 × acute angle < 90° and 5 × acute angle > 90°
⇒ acute angle < \(\frac{90^{\circ}}{4}\) = 22\(\frac{1}{2}^{\circ}\) and acute angle > \(\frac{90^{\circ}}{5}\) = 18°
So, possible measures of the acute angle would be 19°, 20°, 21° and 22° in whole numbers.