NCERT Exemplar Class 6 Maths Chapter 9 Symmetry and Practical Geometry

NCERT Exemplar Class 6 Maths Chapter 9 Symmetry and Practical Geometry are part of NCERT Exemplar Class 6 Maths. Here we have given NCERT Exemplar Class 6 Maths Solutions Chapter 9 Symmetry and Practical Geometry.

NCERT Exemplar Class 6 Maths Chapter 9 Symmetry and Practical Geometry

Directions: In questions 1 to 17, out of the given four options only one is correct. Write the correct answers.

Question 1.
In the following figures, the figure that is not symmetric with respect to any line is

(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Solution:

Question 2.
The number of lines of symmetry in a scalene triangle is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(A)
A scalene triangle has no line of symmetry.

Question 3.
The number of lines of symmetry in a circle is
(A) 0
(B) 2
(C) 4
(D) more than 4
Solution:
(D)
Since, a circle is symmetrical about each of its diameters. Thus, each diameter of a circle is an axis of symmetry.

Question 4.
Which of the following letters does not have the vertical line of symmetry?
(A) M
(B) H
(C) E
(D) V
Solution:

Above letters show that E is the only letter which has no vertical line of symmetry.

Question 5.
Which of the following letters have both horizontal and vertical lines of symmetry?
(A) X
(B) E
(C) M
(D) K
Solution:

Above letters show that X has both horizontal and vertical line of symmetry, E and K has horizontal line of symmetry, M has vertical line of symmetry.

Question 6.
Which of following letters does not have any line of symmetry?
(A) M
(B) S
(C) K
(D) H
Solution:

Above letters show that S has no lines of symmetry.

Question 7.
Which of the following letters has only one line of symmetry?
(A) H
(B) X
(C) Z
(D) T
Solution:

The above letters show that H and X has both vertical and horizontal line of symmetry, Z has no line of symmetry and T has only one (horizontal) line of symmetry.

Question 8.
The instrument to measure an angle is a
(A) Ruler
(B) Protractor
(C) Divider
(D) Compasses
Solution:
(B)
The instrument protractor is used to measure an angle.

Question 9.
The instrument to draw a circle is
(A) Ruler
(B) Protractor
(C) Divider
(D) Compasses
Solution:
(D)
The instrument compasses is used to draw a circle.

Question 10.
Number of set squares in the geometry box is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(C)
We have two set squares in the geometry box.

Question 11.
The number of lines of symmetry in a ruler is
(A) 0
(B) 1
(C) 2
(D) 4
Solution:
(C)
The ruler is in the shape of rectangle which has two lines of symmetry.

Question 12.
The number of lines of symmetry in a divider is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(B)
Divider has one line of symmetry.

Question 13.
The number of lines of symmetry in compasses is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(A)
Compasses has no line of symmetry.

Question 14.
The number of lines of symmetry in a protractor is
(A) 0
(B) 1
(C) 2
(D) more than 2
Solution:
(B)
The protractor is in the shape of letter D which has one line of symmetry.

Question 15.
The number of lines of symmetry in a 45° – 45° – 90° set-square is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(B)
The set-square of measurement 45° – 45° – 90° is in the shape of an isosceles right-angled triangle which has one line of symmetry.

Question 16.
The number of lines of symmetry in a 30° – 60° – 90° set square is
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(A)
The set-square of measurement 30 ° – 60° – 90° is in the shape of a scalene right-angled triangle which has no line of symmetry.

Question 17.
The instrument in the geometry box having the shape of a triangle is called a
(A) Protractor
(B) Compasses
(C) Divider
(D) Set-square
Solution:
(D)
The instrument set-square in the geometry box has the shape of a triangle.

Directions: In questions 18 to 42, fill in the blanks to make the statements true.

Question 18.
The distance of the image of a point (or an object) from the line of symmetry (mirror) is ___ as that of the point (object) from the line (mirror).
Solution:
Same

Question 19.
The number of lines of symmetry in a picture of Taj Mahal is ____.
Solution:
One

Question 20.
The number of lines of symmetry in a rectangle and a rhombus are ____ (equal/unequal).
Solution:
Equal : Since, both rectangle and rhombus have two lines of symmetry.

Question 21.
The number of lines of symmetry in a rectangle and a square are ___ (equal/unequal)
Solution:
Unequal : Since, a rectangle has 2 lines of symmetry and a square has 4 lines of symmetry.

Question 22.
If a line segment of length 5 cm is reflected in a line of symmetry (mirror), then its reflection (image) is a ___ of length ____.
Solution:
Line segment, 5 cm

Question 23.
If an angle of measure 80° is reflected in a line of symmetry, then the reflection is an ___ of measure ___.
Solution:
Angle, 80°

Question 24.
The image of a point lying on a line / with respect to the line of symmetry / lies on ____.
Solution:
l

Question 25.
In figure, if B is the image of the point A with respect to the line l and p is any point lying on l, then the lengths of line segments PA and PB are ____.

Solution:
Equal

Question 26.
The number of lines of symmetry in the given figure is ____.

Solution:
5
Number of lines of symmetry = 5.

Question 27.
The common properties in the two set-squares of a geometry box are that they have a ____ angle and they are of the shape of a ____.
Solution:
Right, triangle

Question 28.
The digits having only two lines of symmetry are ___ and ____.
Solution:
0, 8

Question 29.
The digit having only one line of symmetry is ____.
Solution:
3

Question 30.
The number of digits having no line of symmetry is ____.
Solution:
7
The digits are 1, 2, 4, 5, 6, 7 and 9 having no line of symmetry.

Question 31.
The number of capital letters of the English alphabets having only vertical line of symmetry is ____.
Solution:
7
The letters are A, M, T, U, V, W and Y having only vertical line of symmetry.

Question 32.
The number of capital letters of the English alphabets having only horizontal line of symmetry is ____.
Solution:
5
The letters are B, C, D, E and K having only horizontal line of symmetry.

Question 33.
The number of capital letters of the English alphabets having both horizontal and vertical lines of symmetry is ___.
Solution:
4
The letters are H, I, O and X having both horizontal and vertical line of symmetry.

Question 34.
The number of capital letters of the English alphabets having no line of symmetry is ___.
Solution:
10
The letters are F, G, J, L, N, P, Q, R, S and Z having no line of symmetry.

Question 35.
The line of symmetry of a line segment is the ____ bisector of the line segment.
Solution:
Perpendicular

Question 36.
The number of lines of symmetry in a regular hexagon is ____.
Solution:
6

Question 37.
The number of lines of symmetry in a regular polygon of n sides is ____.
Solution:
n

Question 38.
A protractor has ____ line/lines of symmetry.
Solution:
1

Question 39.
A 30° – 60° – 90° set-square has ___ line/lines of symmetry.
Solution:
No

Question 40.
A 45° – 45° – 90° set-square has ___ line/lines of symmetry.
Solution:
One

Question 41.
A rhombus is symmetrical about ____.
Solution:
Diagonals

Question 42.
A rectangle is symmetrical about the lines joining the ___ of the opposite sides.
Solution:
Mid points

Directions : In questions 43 to 61, state whether the statements are true (T) or false (F).

Question 43.
A right triangle can have at most one line of symmetry.
Solution:
True
A right triangle can have at most one line of symmetry if its remaining two angles are equal.

Question 44.
A kite has two lines of symmetry.
Solution:
False
Because a kite has only one line of symmetry.

Question 45.
A parallelogram has no line of symmetry.
Solution:
True

Question 46.
If an isosceles triangle has more than one line of symmetry, then it need not be an equilateral triangle.
Solution:
False
Since, isosceles triangle is symmetrical about the bisector of the angle included between the
equal sides, i.e., only one line of symmetry.
∴ If it has more than one line of symmetry then it must be an equilateral triangle.

Question 47.
If a rectangle has more than two lines of symmetry, then it must be a square.
Solution:
True

Question 48.
With ruler and compasses, we can bisect any given line segment.
Solution:
True

Question 49.
Only one perpendicular bisector can be drawn to a given line segment.
Solution:
True

Question 50.
Two perpendiculars can be drawn to a given line from a point not lying on it.
Solution:
False
Since, only one perpendicular can be drawn to a given line from a point not lying on it.

Question 51.
With a given centre and a given radius, only one circle can be drawn.
Solution:
True

Question 52.
Using only the two set-squares of the geometry box, an angle of 40° can be drawn.
Solution:
False

Question 53.
Using only the two set-squares of the geometry box, an angle of 15° can be drawn.
Solution:
True
Using 45° – 45° – 90° set square we can draw an angle of 45°. Now using 30° – 60° – 90° set square we can draw an angle of 30° inside the 45° angle taking one arm common in both. So, the angle between 45° and 30° is 15°.

Question 54.
If an isosceles triangle has more than one line of symmetry, then it must be an equilateral triangle.
Solution:
True

Question 55.
A square and a rectangle have the same number of lines of symmetry.
Solution:
False
A square has 4 lines of symmetry whereas rectangle has 2 lines of symmetry.

Question 56.
A circle has only 16 lines of symmetry.
Solution:
False
Since, a circle is symmetrical about each of its diameters. So it has infinite lines of symmetry.

Question 57.
A 45° – 45° – 90° set-square and a protractor have the same number of lines of symmetry. Solution:
True
Since, both 45° – 45° – 90° set-square and protractor have only one line of symmetry.

Question 58.
It is possible to draw two bisectors of a given angle.
Solution:
False
Since, we can draw only one bisector of a given angle.

Question 59.
A regular octagon has 10 lines of symmetry.
Solution:
False
Since, a regular octagon has 8 sides.
∴ It must have 8 lines of symmetry.

Question 60.
Infinitely many perpendiculars can be drawn to a given ray.
Solution:
True .

Question 61.
Infinitely many perpendicular bisectors can be drawn to a given ray.
Solution:
False

Question 62.
Is there any line of symmetry in the given figure? If yes, draw all the lines of symmetry.

Solution:
Yes, the given figure has one line of symmetry.

Question 63.
In the given figure PQRS is a rectangle. State the lines of symmetry of the rectangle.

Solution:
As we know that a rectangle has two lines of symmetry, which are obtained by joining the midpoints of opposite sides.
∴ The given figure shows that AC and BD are the lines of symmetry.

Question 64.
Write all the capital letters of the English alphabets which have more than one lines of symmetry.
Solution:
The capital letters of English alphabets which have more than one lines of symmetry are as follows:

Question 65.
Write the letters of the word ‘MATHEMATICS’ which have no line of symmetry.
Solution:
The letter S of the given word have no line of symmetry.

Question 66.
Write the number of lines of symmetry in each letter of the word ‘SYMMETRY’.
Solution:
S and R has no line of symmetry whereas all the remaining letters has only one line of symmetry.

Question 67.
Match the following.

Solution:
(i) ➝ (f); (ii) ➝ (c); (iii) ➝ (f); (iv) ➝ (d); (v) ➝ (e); (vi) ➝ (a); (vii) ➝ (g)

Question 68.
Open your geometry box. There are some drawing tools. Observe them and complete the following table:

Solution:
(i) The ruler has 2 lines of symmetry.
(ii) The divider has 1 line of symmetry.
(iii) The compasses has no line of symmetry.
(iv) The protactor has 1 line of symmetry.
(v) Triangular piece with two equal sides has 1 line of symmetry.
(vi) Triangular piece with unequal sides has no line of symmetry.

Question 69.
Draw the images of points A and B in line l of the given figure and name them as A’ and B’ respectively. Measure AB and A’B’. Are they equal?

Solution:

Yes, the measurement of AB and A’B’ will be equal.

Question 70.
In the given figure, the point C is the image of point A in line l and line segment BC intersects the line l at P.

(a) Is the image of P in line l the point P itself?
(b) Is PA = PO
(c) Is PA + PB = PC + PB?
(d) Is P that point on line l from which the sum of the distances of points A and B is minimum?
Solution:
(a) Yes (b) Yes (c) Yes
Since, PA = PC
⇒ PA + PB = PC + PB
[By adding PB on both sides]
(d) Yes, since, PA + PB = CP + PB
Now, CP + PB will be minimum when C, P and B lie on a line.

Question 71.
Complete the figure so that line l becomes the line of symmetry of the whole figure.

Solution:

Question 72.
Draw the images of the points A, B and C in the line m (see figure). Name them as A’, B’ and C’, respectively and join them in pairs. Measure AB, BC, CA, A’B’, B’C’ and CA’. Is AB = A’B’, BC = B’C’ and CA = C’A’?

Solution:
Yes, AB = A’B’, BC = B’C’ and CA = CA’

Question 73.
Draw the images P’, Q’ and R’ of the points P, Q and R, respectively in the line n (see figure). Join P’Q’ and Q’R’ to form an angle P’Q’R’. Measure ∠PQR and ∠P’Q’R’. Are the two angles equal?

Solution:

Yes, ∠PQR and ∠P’Q’R’

Question 74.
Complete the figure by taking l as the line of symmetry of the whole figure.

Solution:

Question 75.
Draw a line segment of length 7 cm. Draw its perpendicular bisector, using ruler and compasses.
Solution:

Steps of construction:
(i) Draw a line segment AB = 7 cm.
(ii) With A as centre and radius more than half of AB, draw arcs, one on each side of AB.
(iii) With B as centre and same radius as before, draw arcs, cutting the previously drawn arcs at X and Y respectively.
(iv) Now join XY, meeting AB at O.
Thus, XY is the perpendicular bisector of AB.

Question 76.
Draw a line segment of length 6.5 cm and divide it into four equal parts, using ruler and compasses.
Solution:

Steps of construction:
(i) Draw a line segment AB = 6.5 cm.
(ii) With A and B as centres and radius more than \(\frac{1}{2}(A B)\), draw arcs one on each side of AB, these arcs cuts each other at E and F.
(iii) Join EF, which cuts AB at Q.
(iv) With centres A and Q and radius more than \(\frac{1}{2} A Q\) draw arcs one on each side of AB, these arcs cuts each other at G and H.
(v) Join GH, which cuts AB at P.
(vi) Similarly, with centres B and Q and radius more than \(\frac{1}{2}(B Q)\), draw arcs one on each side of AB, these arcs cuts each other at I and J.
(vii) Join IJ, which cuts AB at R.
(viii) Now, point P, Q and R divide the line segment AB in four equal parts.
Thus AP = PQ = QR = RB.

Question 77.
Draw an angle of 140° with the help of a protractor and bisect it using ruler and compasses.
Solution:

Steps of construction:
(i) Draw ∠BAC = 140° with the help of protractor.
(ii) With A as centre and any convenient radius, draw an arc, cutting AC and AB at Q and P respectively.
(iii) With centre P and radius more than \(\frac{1}{2}(\widehat{P Q})\), draw an arc.
(iv) With centre Q and the same radius as before, draw another arc, cutting the previously drawn arc at a point S.
(v) Join AS and produce it to any point X. Then, ray AX bisects ∠CAB.

Question 78.
Draw an angle of 65° and draw an angle equal to this angle, using ruler and compasses.
Solution:

Steps of construction:
(i) Draw ∠AOB = 65° with the help of protractor.
(ii) Draw a ray O’X.
(iii) With O as centre and any radius, draw an arc cutting OA and OB at C and D respectively.
(iv) With O’ as centre and with the same radius, draw an arc, cutting O’X at P.
(v) With P as centre and radius equal to CD, cut the arc through P at Q.
(vi) Join O’Q and produce it to Y.
Then ∠YO’X is the required angle equal to ∠AOB.

Question 79.
Draw an angle of 80° using a protractor and divide it into four equal parts, using ruler and compasses. Check your construction by measurement.
Solution:

Steps of construction:
(i) Draw ∠AOB = 80° with the help of protractor.
(ii) With centre O and any radius, draw an arc, meeting \(\overrightarrow{O B}\) and \(\overrightarrow{O B}\) at P and Q respectively.
(iii) With centres P and Q and radius more than \(\frac{1}{2}(\widehat{P Q})\) draw arcs which cuts each other at R.
(iv) Join OR which bisects PQ at S.
(v) With centres P and S and radius more than \(\frac{1}{2}(\widehat{P S})\), draw arcs which cuts each other at U.
(vi) Join OU.
(vii) With centres Q and S and radius more than \(\frac{1}{2}(\widehat{Q S})\) draw arcs which cuts each other at V
(viii) Join OV. Thus, ∠AOV, ∠VOR, ∠ROU and ∠UOB divide ∠BOA in four equal parts and ∠AOV = ∠VOR = ∠ROU = 20°

Question 80.
Copy figure on your notebook and draw a perpendicular to l through P, using (i) set squares (ii) Protractor (iii) ruler and compasses. How many such perpendiculars are you able to draw?

Solution:
Do it yourself.

Question 81.
Copy figure on your notebook and draw a perpendicular from P to line m, using (i) set squares (ii) Protractor (iii) ruler and compasses. How many such perpendiculars are you able to draw?

Solution:
Do it yourself.

Question 82.
Draw a circle of radius 6 cm using ruler and compasses. Draw one of its diameters. Draw the perpendicular bisector of this diameter. Does this perpendicular bisector contain another diameter of the circle?
Solution:

Steps of construction:
(i) Draw a circle with centre O of radius 6 cm.
(ii) Now, draw a diameter AB of the circle.
(iii) With the centre B and the radius more than \(\frac{1}{2}(\widehat{A B})\), draw the arcs on each side of AB.
(iv) With the centre A and the same radius, draw the arcs on each side of AB which cuts the previously drawn arc at X and Y.
(v) Now join OX and OY and produce them to any points P and Q.
Thus PQ bisects the diameter of a circle. Yes, the perpendicular PQ contains another diameter of this circle.

Question 83.
Bisect ∠XYZ of figure.

Solution:

Steps of construction:
(i) With centre Y and any radius draw an arc which cuts ZY at P and XY at Q.
(ii) With centre Q and radius more than \(\frac{1}{2}(\widehat{P Q})\) draw an arc.
(iii) With centre P and same radius draw an arc which cuts the previously drawn arc at R.
(iv) Join YR and produce it to any point S. Thus \(\overrightarrow{Y S}\) is the bisector of ∠XYZ.

Question 84.
Draw at angle of 60° using ruler and compasses and divide it into four equal parts. Measure each part.
Solution:

Steps of construction:
(i) Draw a ray \(\overrightarrow{O A}\).
(ii) With O as centre and any convenient radius draw an arc meeting the rav OA at X.
(iii) With X as centre and the same radius draw an arc which cuts the previously drawn arc at U.
(iv) Join OU and produce it to any point B. So ∠BOA = 60°.
(v) Now with centre X and U and radius more than \(\frac{1}{2}(\widehat{U X})\) draw the arcs on the same side of ∠BOA which cuts each other at V.
(vi) Join OV and produce it any point C as well as it cuts the arc \(\widehat{U X}\) at W.
(vii) Again with the centre X and W and radius more than \(\frac{1}{2}(\widehat{W X})\) draw the arcs on the same side of ∠BOA which cuts each other at Z.
(viii) Join OZ and produce it to any point D.
(ix) With the centre U and W and radius more than \(\frac{1}{2}(\widehat{U W})\) draw the arcs on the same side of ∠BOA with cuts each other at the point P.
(x) Join OP and produce it to any point E. Now with the help of protractor we observe that each angle i.e, ∠BOE = ∠EOC = ∠COD = ∠DOA = 15°.

Question 85.
Bisect a straight angle, using ruler and compasses. Measure each part.
Solution:

Steps of construction:
(i) Draw a line AB of any length.
(ii) Draw an arc of 180° which meets the ray \(\overrightarrow{O A} \) at P and ray \(\overrightarrow{O B} \) at Q.
(iii) Taking P and Q as centres and radius more than \(\frac{1}{2}(\widehat{P Q})\), draw two arcs which intersect each other at R.
(iv) Join OR and produce it to any point C.
(v) Thus, \(\overrightarrow{O C}\) bisect the straight angle. Now, on measuring each angle we get, ∠BOC = ∠AOC = 90°.

Question 86.
Bisect a right angle, using ruler and compasses. Measure each part. Bisect each of these parts. What will be the measure of each of these parts?
Solution:

Steps of construction:
(i) Draw ∠AOB = 90° with the help of protractor.
(ii) Draw an arc with centre O and any radius, which cuts the ray OB and OA at U and Y.
(iii) With centres U and Y draw the arcs with radius more than \(\frac{1}{2}(\widehat{U Y})\), which cuts each other at Q.
(iv) Join OQ and produce it to any point D as well as it cuts the arc \(\widehat{Y U} \) at W
(v) With centres U and W and radius more than \(\frac{1}{2}(\widehat{W U})\) draw the arcs which cuts each other at P.
(vi) Join OP and produce it to any point E.
(vii) Now with centres Y and W and radius more than \(\frac{1}{2}(\widehat{Y W})\) draw the arcs which cuts each other at R.
(viii) Join OR and produce it to any point C. On measuring, we have ∠BOD = ∠AOD = 45° and ∠BOE = ∠EOD = ∠DOC = ∠COA \(=22 \frac{1}{2}^{\circ}\)

Question 87.
Draw an angle ABC of measure 45°, using ruler and compasses. Now draw an angle DBA of measure 30°, using ruler and compasses as shown in figure. What is the measure of ∠DBC ?

Solution:

Steps of construction:
(i) Draw a ray \(\overrightarrow{B C}\)
(ii) With B as centre and any radius draw an arc which cuts BC at P.
(iii) With centre P and same radius, draw an arc which cuts the previously drawn arc at Q.
(iv) Similarly, with Q as centre and same radius draw an another arc which cuts the previously drawn arc at R.
(v) With centre R and Q and same radius draw the arcs which cuts each other at S.
(vi) Join BS which cuts the arc RP at T.
(vii) With centre P and T and radius more than \(\frac{1}{2}(\widehat{P T})\) draw the arcs which cuts each other at U.
(viii) Join BU and produce it to a point A. Thus ∠ABC = 45°.
(ix) Now, with B as centre draw another arc which cuts the line AB at X.
(x) With centre X and same radius draw an arc which cut the arc drawn in step (ix) at Y.
(xi) With centre X and Y and radius more than \(\frac{1}{2}(\widehat{X Y})\) draw two arcs which cuts each other at Z.
(xii) Join BZ and produced it to a point D. Thus ∠ABD = 30°.
On measuring, we get ∠DBC = 75°

Question 88.
Draw a line segment of length 6 cm. Construct its perpendicular bisector. Measure the two parts of the line segment.
Solution:

Steps of construction:
(i) Draw a line segment AB = 6 cm
(ii) With centre A and radius more than \(\frac{1}{2}(A B)\) draw the arcs on each side of AB.
(iii) With centre B and same radius draw the arcs on each side of AB, which cuts the previously drawn arcs at X and Y.
(iv) Join XY which cuts the line segment AB at O.
Thus XY is the perpendicular bisector of AB.
On measuring the two parts, we get AO = OB = 3 cm.

Question 89.
Draw a line segment of length 10 cm. Divide it into four equal parts. Measure each of these parts.
Solution:

Steps of construction:
(i) Draw a line segment AB = 10 cm.
(ii) With A and B as centres and radius more than \(\frac{1}{2}(A B)\), draw arcs on each side
AB, these arcs cuts each other at R at S.
(iii) Join RS, which cuts AB at a point O.
(iv) Now, with centres A and O and radius more than \(\frac{1}{2}(A O)\), draw arcs on each side of AB, these arcs cuts each other at P and Q.
(v) Join PQ, which cuts AB at M.
(vi) Similarly, with centres B and O and radius more than \(\frac{1}{2}(B O)\), draw arcs on
each side of AB, these arcs cuts each other at T and U.
(vii) Join TU, which cuts AB at N. Thus the measurement of each i.e., AM, MO, ON and NB is equal to 2.5 cm.

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