ICSE Class 7 Mathematics · Selina Solutions Concise Mathematics Class 7 ICSE · 77 solved questions
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is a rectangle with a curved cut on its left side. It can be folded into two coincident parts only about the horizontal line joining the mid-points of its left and right sides.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is an isosceles right angled triangle, as its two arms containing the right angle are equal. It can be folded into two coincident parts only about the bisector of the right angle, which is also the perpendicular bisector of its hypotenuse.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is a right angled triangle whose sides are all unequal, i.e., it is a scalene triangle. In whatever manner we fold it, the two parts of the figure do not coincide.
Hence, a line of symmetry is not possible for the given figure.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is the shape of the letter T. It can be folded into two coincident parts only about the vertical line passing through the middle of its stem.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is an arrow pointing to the right. It can be folded into two coincident parts only about the horizontal line passing through the tip of the arrow and the middle of its tail.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is a cone-shaped figure made up of a circle and two equal lines drawn from a point outside it. It can be folded into two coincident parts only about the horizontal line passing through the centre of the circle and the vertex of the figure.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is an angle with equal arms. We know that a given angle with equal arms is symmetrical about the bisector of the angle.
Hence, the given figure has 1 line of symmetry, which is the bisector of the angle, shown dotted in the figure above.
For the figure, given below, draw the line(s) of symmetry, if possible:
The given figure is an arrow-shaped figure. It can be folded into two coincident parts only about the horizontal line passing through the tip of the arrow and the middle of its tail.
Hence, the given figure has 1 line of symmetry, shown dotted in the figure above.
Write capital letters A to Z of English alphabet and in each case, if possible, draw the largest number of lines of symmetry.
The letters A, M, T, U, V, W and Y have only one line of symmetry, which is vertical.
The letters B, C, D, E and K have only one line of symmetry, which is horizontal.
The letters H, I, O and X have two lines of symmetry, one vertical and one horizontal.
The letters F, G, J, L, N, P, Q, R, S and Z cannot be folded into two coincident parts about any line.
Hence, the letters H, I, O and X have the largest number of lines of symmetry, i.e., 2 each; the letters A, B, C, D, E, K, M, T, U, V, W and Y have 1 line of symmetry each, and the letters F, G, J, L, N, P, Q, R, S and Z have no line of symmetry.
Draw a freehand sketch of a scalene triangle and mark its line(s) of symmetry, if any.
A scalene triangle has no line of symmetry. In whatever manner we fold a scalene triangle, the two parts of the figure do not coincide.
Hence, a line of symmetry is not possible in a scalene triangle.
Draw a freehand sketch of an isosceles right angled triangle and mark its line(s) of symmetry, if any.
An isosceles right angled triangle has only one line of symmetry, which is the bisector of its right angle and is also the perpendicular bisector of its hypotenuse.
Hence, an isosceles right angled triangle has 1 line of symmetry, shown dotted in the figure above.
Draw a freehand sketch of a rhombus and mark its line(s) of symmetry, if any.
A rhombus has two lines of symmetry, which are its two diagonals.
Hence, a rhombus has 2 lines of symmetry, shown dotted in the figure above.
Draw a freehand sketch of a kite shaped figure and mark its line(s) of symmetry, if any.
A kite shaped figure has only one line of symmetry, which is the diagonal joining the vertices at which its pairs of equal sides meet.
Hence, a kite shaped figure has 1 line of symmetry, shown dotted in the figure above.
Draw a freehand sketch of a rectangle and mark its line(s) of symmetry, if any.
A rectangle has two lines of symmetry, which are the lines joining the mid-points of its opposite sides.
Hence, a rectangle has 2 lines of symmetry, shown dotted in the figure above.
Draw a freehand sketch of a square and mark its line(s) of symmetry, if any.
A square has four lines of symmetry, which are its two diagonals and the two lines joining the mid-points of its opposite sides.
Hence, a square has 4 lines of symmetry, shown dotted in the figure above.
Draw a freehand sketch of an isosceles triangle and mark its line(s) of symmetry, if any.
An isosceles triangle has only one line of symmetry, which is the bisector of the angle of vertex and is also the perpendicular bisector of its base.
Hence, an isosceles triangle has 1 line of symmetry, shown dotted in the figure above.
Draw a triangle with no line of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the triangle drawn.
A scalene triangle has no line of symmetry.
Hence, the required triangle is a scalene triangle.
Draw a triangle with only one line of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the triangle drawn.
An isosceles triangle has only one line of symmetry, which is the bisector of the angle of vertex.
Hence, the required triangle is an isosceles triangle.
Draw a triangle with exactly two lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the triangle drawn.
If a triangle has two lines of symmetry, it must be equilateral, and an equilateral triangle has three lines of symmetry. Therefore a triangle with exactly two lines of symmetry is not possible.
Hence, a triangle with exactly two lines of symmetry is not possible.
Draw a triangle with exactly three lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the triangle drawn.
An equilateral triangle has three lines of symmetry, which are the bisectors of the angles of its vertices.
Hence, the required triangle is an equilateral triangle.
Draw a triangle with more than three lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the triangle drawn.
As the number of sides in a triangle is 3, the largest number of lines of symmetry in it is 3.
Hence, a triangle with more than three lines of symmetry is not possible.
Draw a quadrilateral with no line of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
A quadrilateral with all its sides and angles unequal has no line of symmetry.
Hence, the required quadrilateral is a quadrilateral with all sides unequal.
Draw a quadrilateral with only one line of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
A kite-shaped figure has only one line of symmetry.
Hence, the required quadrilateral is a kite-shaped figure.
Draw a quadrilateral with exactly two lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
A rectangle has two lines of symmetry, which are the lines joining the mid-points of its opposite sides. A rhombus also has two lines of symmetry, which are its diagonals.
Hence, the required quadrilateral is a rectangle or a rhombus.
Draw a quadrilateral with exactly three lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
No quadrilateral can be folded into coincident parts about exactly three lines.
Hence, a quadrilateral with exactly three lines of symmetry is not possible.
Draw a quadrilateral with exactly four lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
A square has four lines of symmetry, which are its two diagonals and the two lines joining the mid-points of its opposite sides.
Hence, the required quadrilateral is a square.
Draw a quadrilateral with more than four lines of symmetry, In each case, if possible, represent the line(s) of symmetry by dotted lines. Also, write the specific name of the quadrilateral drawn.
As the number of sides in a quadrilateral is 4, the largest number of lines of symmetry in it is 4.
Hence, a quadrilateral with more than four lines of symmetry is not possible.
Construct an equilateral triangle with each side 6 cm. In the triangle drawn, draw all possible lines of symmetry.
Steps of construction:
Draw BC = 6 cm.
With B as centre and radius 6 cm, draw an arc.
With C as centre and the same radius, draw another arc which cuts the previous arc at the point A.
Join AB and AC.
Draw the bisectors of ∠A, ∠B and ∠C.
ABC is an equilateral triangle. An equilateral triangle has three lines of symmetry, which are the bisectors of the angles of its vertices and are also the perpendicular bisectors of its sides. They are shown dotted in the figure above.
Hence, ABC is the required triangle and, since it is equilateral, it has 3 lines of symmetry.
Construct a triangle ABC in which AB = AC = 5 cm and BC = 5.6 cm. If possible, draw its line(s) of symmetry.
Steps of construction:
Draw BC = 5.6 cm.
With B as centre and radius 5 cm, draw an arc.
With C as centre and the same radius, draw another arc which cuts the previous arc at the point A.
Join AB and AC.
Draw the bisector of ∠A.
AB = AC ⇒ the triangle ABC is isosceles.
An isosceles triangle has only one line of symmetry, which is the bisector of the angle of vertex A and is also the perpendicular bisector of its base BC. It is shown dotted in the figure above.
Hence, ABC is the required triangle and it has only 1 line of symmetry.
Construct a triangle PQR such that PQ = QR = 5.5 cm and angle PQR = 90°. If possible, draw its lines of symmetry.
Steps of construction:
Draw QR = 5.5 cm.
At Q, construct ∠RQX = 90°.
From QX, cut off QP = 5.5 cm.
Join PR.
Draw the bisector of ∠PQR.
PQ = QR ⇒ △PQR is isosceles.
An isosceles triangle has only one line of symmetry, which is the bisector of the angle of vertex Q and is also the perpendicular bisector of its base PR. It is shown dotted in the figure above.
Hence, PQR is the required triangle and it has only 1 line of symmetry.
If possible, draw a rough sketch of a quadrilateral which has exactly two lines of symmetry.
A rectangle has exactly two lines of symmetry, which are the lines joining the mid-points of its opposite sides.
A rhombus has exactly two lines of symmetry, which are its diagonals.
Hence, the required figure is a rhombus or a rectangle.
A quadrilateral ABCD is symmetric about its diagonal AC. Name the sides of this quadrilateral which are equal.
Since the quadrilateral ABCD is symmetric about its diagonal AC, on folding it about AC, the two parts of the figure coincide exactly, i.e., the vertex B coincides with the vertex D.
∴ Side AB coincides with side AD and side CB coincides with side CD.
Hence, AB = AD and CB = CD.
In figure, given below, find the image of the point P in the line AB:
Steps of construction:
From the given point P, draw a perpendicular to AB which meets AB at the point O.
From PO produced, cut OP′ = OP.
Hence, P′ is the required image of the point P in the line AB.
In figure, given below, find the image of the point P in the line AB:
Steps of construction:
From the given point P, draw a perpendicular to AB which meets AB at the point O.
From PO produced, cut OP′ = OP.
Hence, P′ is the required image of the point P in the line AB.
In figure, given below, find the image of the line segment AB in the line PQ:
Steps of construction:
From the point A, draw AO perpendicular to the line PQ and from AO produced cut OA′ such that OA′ = OA.
Similarly from the point B, draw BR perpendicular to the line PQ and from BR produced cut RB′ such that RB′ = RB.
Join A′ and B′.
Hence, A′B′ is the required image of AB in the line PQ.
In figure, given below, find the image of the line segment AB in the line PQ:
Steps of construction:
From the point A, draw AO perpendicular to the line PQ and from AO produced cut OA′ such that OA′ = OA.
Similarly from the point B, draw BR perpendicular to the line PQ and from BR produced cut RB′ such that RB′ = RB.
Join A′ and B′.
Hence, A′B′ is the required image of AB in the line PQ.
Complete the following table:
| Point | Reflection in x-axis | Reflection in y-axis | Reflection in origin |
|---|---|---|---|
| (i) (8, 2) | |||
| (ii) (5, 6) | |||
| (iii) (4, -5) | |||
| (iv) (6, -2) | |||
| (v) (-3, 7) | |||
| (vi) (-4, 5) | |||
| (vii) (-2, -7) | |||
| (viii) (-6, -3) | |||
| (ix) (4, 0) | |||
| (x) (-7, 0) | |||
| (xi) (0, -6) | |||
| (xii) (0, 8) | |||
| (xiii) (0, 0) |
We know that,
Image of P (x, y) in x-axis = P′ (x, -y), i.e., the sign of y changes.
Image of P (x, y) in y-axis = P′ (-x, y), i.e., the sign of x changes.
Image of P (x, y) in origin = P′ (-x, -y), i.e., the signs of both x and y change.
Hence, the completed table is :
| Point | Reflection in x-axis | Reflection in y-axis | Reflection in origin |
|---|---|---|---|
| (i) (8, 2) | (8, -2) | (-8, 2) | (-8, -2) |
| (ii) (5, 6) | (5, -6) | (-5, 6) | (-5, -6) |
| (iii) (4, -5) | (4, 5) | (-4, -5) | (-4, 5) |
| (iv) (6, -2) | (6, 2) | (-6, -2) | (-6, 2) |
| (v) (-3, 7) | (-3, -7) | (3, 7) | (3, -7) |
| (vi) (-4, 5) | (-4, -5) | (4, 5) | (4, -5) |
| (vii) (-2, -7) | (-2, 7) | (2, -7) | (2, 7) |
| (viii) (-6, -3) | (-6, 3) | (6, -3) | (6, 3) |
| (ix) (4, 0) | (4, 0) | (-4, 0) | (-4, 0) |
| (x) (-7, 0) | (-7, 0) | (7, 0) | (7, 0) |
| (xi) (0, -6) | (0, 6) | (0, -6) | (0, 6) |
| (xii) (0, 8) | (0, -8) | (0, 8) | (0, -8) |
| (xiii) (0, 0) | (0, 0) | (0, 0) | (0, 0) |
A point P (7, 3) is reflected in x-axis to point P′. The point P′ is further reflected in y-axis to point P″. Find:
(i) the co-ordinates of P′
(ii) the co-ordinates of P″
(iii) the image of P (7, 3) in origin.
(i) When a point is reflected in x-axis, the sign of its ordinate changes.
∴ Image of P (7, 3) in x-axis = P′ (7, -3)
Hence, the co-ordinates of P′ are (7, -3).
(ii) When a point is reflected in y-axis, the sign of its abscissa changes.
∴ Image of P′ (7, -3) in y-axis = P″ (-7, -3)
Hence, the co-ordinates of P″ are (-7, -3).
(iii) When a point is reflected in origin, the signs of both of its components change.
∴ Image of P (7, 3) in origin = (-7, -3)
Hence, the image of P (7, 3) in origin is (-7, -3).
A point A (-5, 4) is reflected in y-axis to point B. The point B is further reflected in origin to point C. Find:
(i) the co-ordinates of B
(ii) the co-ordinates of C
(iii) the image of A (-5, 4) in x-axis.
(i) When a point is reflected in y-axis, the sign of its abscissa changes.
∴ Image of A (-5, 4) in y-axis = B (5, 4)
Hence, the co-ordinates of B are (5, 4).
(ii) When a point is reflected in origin, the signs of both of its components change.
∴ Image of B (5, 4) in origin = C (-5, -4)
Hence, the co-ordinates of C are (-5, -4).
(iii) When a point is reflected in x-axis, the sign of its ordinate changes.
∴ Image of A (-5, 4) in x-axis = (-5, -4)
Hence, the image of A (-5, 4) in x-axis is (-5, -4).
The point P (3, -8) is reflected in origin to point Q. The point Q is further reflected in x-axis to point R. Find:
(i) the co-ordinates of Q
(ii) the co-ordinates of R
(iii) the image of P (3, -8) in y-axis.
(i) When a point is reflected in origin, the signs of both of its components change.
∴ Image of P (3, -8) in origin = Q (-3, 8)
Hence, the co-ordinates of Q are (-3, 8).
(ii) When a point is reflected in x-axis, the sign of its ordinate changes.
∴ Image of Q (-3, 8) in x-axis = R (-3, -8)
Hence, the co-ordinates of R are (-3, -8).
(iii) When a point is reflected in y-axis, the sign of its abscissa changes.
∴ Image of P (3, -8) in y-axis = (-3, -8)
Hence, the image of P (3, -8) in y-axis is (-3, -8).
Each of the points A (3, 0), B (7, 0), C (-8, 0), D (-7, 0) and E (0, 0) is reflected in x-axis to points A′, B′, C′, D′ and E′ respectively. Write the co-ordinates of each of the image points A′, B′, C′, D′ and E′.
We know that,
Image of P (x, y) in x-axis = P′ (x, -y)
∴ Image of A (3, 0) in x-axis = A′ (3, 0)
Image of B (7, 0) in x-axis = B′ (7, 0)
Image of C (-8, 0) in x-axis = C′ (-8, 0)
Image of D (-7, 0) in x-axis = D′ (-7, 0)
Image of E (0, 0) in x-axis = E′ (0, 0)
Hence, A′ = (3, 0), B′ = (7, 0), C′ = (-8, 0), D′ = (-7, 0) and E′ = (0, 0).
Each of the points A (0, 4), B (0, 10), C (0, -4), D (0, -6) and E (0, 0) is reflected in y-axis to points A′, B′, C′, D′ and E′ respectively. Write the co-ordinates of each of the image points A′, B′, C′, D′ and E′.
We know that,
Image of P (x, y) in y-axis = P′ (-x, y)
∴ Image of A (0, 4) in y-axis = A′ (0, 4)
Image of B (0, 10) in y-axis = B′ (0, 10)
Image of C (0, -4) in y-axis = C′ (0, -4)
Image of D (0, -6) in y-axis = D′ (0, -6)
Image of E (0, 0) in y-axis = E′ (0, 0)
Hence, A′ = (0, 4), B′ = (0, 10), C′ = (0, -4), D′ = (0, -6) and E′ = (0, 0).
Each of the points A (0, 7), B (8, 0), C (0, -5), D (-7, 0) and E (0, 0) is reflected in origin to points A′, B′, C′, D′ and E′ respectively. Write the co-ordinates of each of the image points A′, B′, C′, D′ and E′.
We know that,
Image of P (x, y) in origin = P′ (-x, -y)
∴ Image of A (0, 7) in origin = A′ (0, -7)
Image of B (8, 0) in origin = B′ (-8, 0)
Image of C (0, -5) in origin = C′ (0, 5)
Image of D (-7, 0) in origin = D′ (7, 0)
Image of E (0, 0) in origin = E′ (0, 0)
Hence, A′ = (0, -7), B′ = (-8, 0), C′ = (0, 5), D′ = (7, 0) and E′ = (0, 0).
Mark points A (4, 5) and B (-5, 4) on a graph paper. Find A′, the image of A in x-axis and B′, the image of B in x-axis. Mark A′ and B′ also on the same graph paper. Join AB and A′B′ and find if AB = A′B′?
We know that,
Image of P (x, y) in x-axis = P′ (x, -y)
∴ Image of A (4, 5) in x-axis = A′ (4, -5)
Image of B (-5, 4) in x-axis = B′ (-5, -4)
The points A, B, A′ and B′ are marked on the graph paper and AB and A′B′ are joined as shown below :
On measuring, we find that AB = A′B′.
Hence, A′ = (4, -5), B′ = (-5, -4) and yes, AB = A′B′.
Mark points A (6, 4) and B (4, -6) on a graph paper. Find A′, the image of A in y-axis and B′, the image of B in y-axis. Mark A′ and B′ also on the same graph paper.
We know that,
Image of P (x, y) in y-axis = P′ (-x, y)
∴ Image of A (6, 4) in y-axis = A′ (-6, 4)
Image of B (4, -6) in y-axis = B′ (-4, -6)
The points A, B, A′ and B′ are marked on the graph paper as shown below :
Hence, A′ = (-6, 4) and B′ = (-4, -6).
Mark points A (-6, 5) and B (-4, -6) on a graph paper. Find A′, the image of A in origin and B′, the image of B in origin. Mark A′ and B′ also on the same graph paper. Join AB and A′B′ and find if AB = A′B′?
We know that,
Image of P (x, y) in origin = P′ (-x, -y)
∴ Image of A (-6, 5) in origin = A′ (6, -5)
Image of B (-4, -6) in origin = B′ (4, 6)
The points A, B, A′ and B′ are marked on the graph paper and AB and A′B′ are joined as shown below :
On measuring, we find that AB = A′B′.
Hence, A′ = (6, -5), B′ = (4, 6) and yes, AB = A′B′.
How many lines of symmetry does a rhombus have?
A rhombus can be folded into two coincident parts about each of its two diagonals only.
Hence, a rhombus has 2 lines of symmetry.
What is the order of rotational symmetry of a rhombus?
On rotating a rhombus about the point of intersection of its diagonals, it fits onto itself when rotated through 180° and 360°.
So, during one complete rotation, the rhombus fits onto itself two times.
Hence, the order of rotational symmetry of a rhombus is 2.
Show that the following figure has two lines of symmetry and a rotational symmetry of order 2.
The letter H can be folded into two coincident parts about the vertical line through its middle and about the horizontal line through its middle. So, it has two lines of symmetry.
On rotating it about its centre, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation. So, it has a rotational symmetry of order 2.
Hence, the letter H has two lines of symmetry and a rotational symmetry of order 2.
Show that the following figure has two lines of symmetry and a rotational symmetry of order 2.
The letter I can be folded into two coincident parts about the vertical line through its middle and about the horizontal line through its middle. So, it has two lines of symmetry.
On rotating it about its centre, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation. So, it has a rotational symmetry of order 2.
Hence, the letter I has two lines of symmetry and a rotational symmetry of order 2.
Show that the following figure has two lines of symmetry and a rotational symmetry of order 2.
The letter O can be folded into two coincident parts about the vertical line through its middle and about the horizontal line through its middle. So, it has two lines of symmetry.
On rotating it about its centre, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation. So, it has a rotational symmetry of order 2.
Hence, the letter O has two lines of symmetry and a rotational symmetry of order 2.
Show that the following figure has two lines of symmetry and a rotational symmetry of order 2.
The letter X can be folded into two coincident parts about the vertical line through its centre and about the horizontal line through its centre. So, it has two lines of symmetry.
On rotating it about its centre, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation. So, it has a rotational symmetry of order 2.
Hence, the letter X has two lines of symmetry and a rotational symmetry of order 2.
Name a figure that has a line of symmetry but does not have any rotational symmetry.
A kite shaped figure has one line of symmetry, which is the diagonal joining the vertices at which its pairs of equal sides meet.
On giving it one complete rotation, it fits onto itself only once, i.e., on being rotated through 360°. So, it does not have any rotational symmetry.
Hence, a kite shaped figure has a line of symmetry but does not have any rotational symmetry.
In the following figure, draw all possible lines of symmetry and also write the order of rotational symmetry:
The given figure is made up of three identical triangular blades meeting at a point. There is no line about which it can be folded so that its two parts coincide. So, it has no line of symmetry.
On rotating it about the common point, it fits onto itself when rotated through 120°, 240° and 360°, i.e., three times during one complete rotation.
Hence, the given figure has no line of symmetry and its order of rotational symmetry is 3.
In the following figure, draw all possible lines of symmetry and also write the order of rotational symmetry:
The given figure is a rhombus with equal marks on its sides. It can be folded into two coincident parts about each of its two diagonals.
On rotating it about the point of intersection of its diagonals, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation.
Hence, the given figure has 2 lines of symmetry, shown dotted in the figure above, and its order of rotational symmetry is 2.
In the following figure, draw all possible lines of symmetry and also write the order of rotational symmetry:
The given figure is a rectangle with identical notches cut in the middle of its upper and lower sides. It can be folded into two coincident parts about the vertical line through the notches and about the horizontal line joining the mid-points of its left and right sides.
On rotating it about its centre, it fits onto itself when rotated through 180° and 360°, i.e., two times during one complete rotation.
Hence, the given figure has 2 lines of symmetry, shown dotted in the figure above, and its order of rotational symmetry is 2.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
In the following figures, line of symmetry is given by dotted line. Complete the figure so that the completed figure is symmetric about the given dotted line.
Hence, the completed figure is symmetric about the given dotted line.
The number of lines of symmetry of a regular polygon with n sides is:
If a polygon has n sides, the largest number of lines of symmetry it can have is also n. A regular polygon, having all sides and all angles equal, has this largest number of lines of symmetry.
So, a regular polygon with n sides has n lines of symmetry.
Hence, option 3 is the correct option.
The number of lines of symmetry in the given figure, is:
![The number of lines of symmetry in the given figure, is: [Given: Circles have equal radii]. Symmetry, Mathematics Solutions ICSE Class 7.](/assets/img/a38e639d90a58c4430.png)
[Given: Circles have equal radii]
The given figure is made up of three circles of equal radii whose centres are the vertices of an equilateral triangle.
The figure can be folded into two coincident parts about each of the three lines drawn through a vertex of the triangle (centre of a circle) and the mid-point of the opposite side, i.e., about the three lines of symmetry of the equilateral triangle.
![The number of lines of symmetry in the given figure, is: [Given: Circles have equal radii]. Symmetry, Mathematics Solutions ICSE Class 7.](/assets/img/d3dc98bc222c18b0d8.png)
So, the given figure has 3 lines of symmetry.
Hence, option 2 is the correct option.
The number of lines of symmetry, which the given figure has:
The given figure is a circle with four identical small circles placed at equal distances inside it, at the top, bottom, left and right.
The figure can be folded into two coincident parts about the vertical line and the horizontal line through the centre, and also about the two lines through the centre which bisect the angles between them.
So, the given figure has 4 lines of symmetry.
Hence, option 2 is the correct option.
The maximum lines of symmetry, which a quadrilateral can have:
As the number of sides in a quadrilateral is 4, the largest number of lines of symmetry in it is 4. A square has 4 lines of symmetry.
Hence, option 4 is the correct option.
A triangle has three lines of symmetry, the triangle is:
A scalene triangle has no line of symmetry and an isosceles triangle has only one line of symmetry.
An equilateral triangle has three lines of symmetry. An equilateral triangle has all its angles equal, i.e., it is equiangular.
Hence, option 3 is the correct option.
The point (5, 3) is reflected in y-axis to point (-a, b); then:
We know that,
Image of P (x, y) in y-axis = P′ (-x, y)
∴ Image of (5, 3) in y-axis = (-5, 3)
Given, image = (-a, b)
∴ (-a, b) = (-5, 3)
⇒ a = 5 and b = 3.
Hence, option 1 is the correct option.
The point (6, -8) is first reflected in x-axis followed by reflection in origin. The co-ordinates of final image are:
We know that,
Image of P (x, y) in x-axis = P′ (x, -y)
∴ Image of (6, -8) in x-axis = (6, 8)
Image of P (x, y) in origin = P′ (-x, -y)
∴ Image of (6, 8) in origin = (-6, -8)
Hence, option 3 is the correct option.
The point (0, 13) is first reflected in origin followed by reflection in y-axis. The co-ordinates of final image are:
We know that,
Image of P (x, y) in origin = P′ (-x, -y)
∴ Image of (0, 13) in origin = (0, -13)
Image of P (x, y) in y-axis = P′ (-x, y)
∴ Image of (0, -13) in y-axis = (0, -13)
Hence, option 3 is the correct option.
Which of the following quadrilaterals has rotational symmetry of order 4?
A parallelogram and a rectangle fit onto themselves only when rotated through 180° and 360°, so they have rotational symmetry of order 2. A trapezium fits onto itself only once in a complete rotation.
A square fits onto itself when rotated through 90°, 180°, 270° and 360°, so it has a rotational symmetry of order 4.
Hence, option 4 is the correct option.
Which of the following quadrilaterals has four lines of symmetry and rotational symmetry of order 4?
A square has 4 lines of symmetry — the diagonals and the lines through the mid-points of opposite sides, and it has a rotational symmetry of order 4 when rotated through 90°, 180°, 270° and 360°.
Hence, option 4 is the correct option.
Statement 1: A circle has a point symmetry and infinite many lines symmetry.
Statement 2: A hexagon has 6 lines symmetry (lines joining the mid-points of opposite sides and the diagonals passing through the centre).
Which of the following options is correct?
A circle looks exactly the same when rotated through 180° about its centre, so it has a point symmetry. Also, every line through its centre is a line of symmetry, so it has infinite lines of symmetry.
So, Statement 1 is true.
If a polygon has n sides, the largest number of lines of symmetry it can have is n. So, a hexagon can have at most 6 lines of symmetry, and it has 6 lines of symmetry only when it is a regular hexagon. Every hexagon does not have 6 lines of symmetry.
So, Statement 2 is false.
∴ Statement 1 is true, and Statement 2 is false.
Hence, option 3 is the correct option.
Assertion (A): Square is a quadrilateral having four lines of symmetry.
Reason (R): A geometrical figure is said to be symmetrical about a line, if the line divides the geometrical figure into two congruent halves.
Refer the following square :
A square has four lines of symmetry. Two of them are its diagonals and the other two are the right-bisectors of both pairs of its opposite sides.
So, Assertion (A) is true.
Refer the following parallelogram :
In the given figure, the diagonal divides the parallelogram into two congruent triangles, but it is not a line of symmetry of the figure, as on folding about it the two parts do not coincide.
So, Reason (R) is false.
∴ A is true, R is false.
Hence, option 1 is the correct option.
Assertion (A): Equilateral triangle has rotational symmetry of order 3.
Reason (R): If a figure is given one complete rotation and during rotation the number of times the figure fits onto itself is called the order of the rotational symmetry.
An equilateral triangle fits onto itself when rotated through 120°, 240° and 360°, i.e., three times during one complete rotation. So, it has a rotational symmetry of order 3.
So, Assertion (A) is true.
The number of times a figure fits onto itself during one complete rotation is called the order of rotational symmetry.
So, Reason (R) is true.
∴ Both A and R are true.
Hence, option 3 is the correct option.
Assertion (A): A square has 4 lines symmetry and rotational symmetry of order 4.
Reason (R): A parallelogram has a line symmetry.
A square has 4 lines of symmetry — its two diagonals and the two lines through the mid-points of its opposite sides. It fits onto itself when rotated through 90°, 180°, 270° and 360°, so it has a rotational symmetry of order 4.
So, Assertion (A) is true.
In whatever manner we fold a parallelogram, the two parts do not coincide. So, a parallelogram has no line of symmetry.
So, Reason (R) is false.
∴ A is true, R is false.
Hence, option 1 is the correct option.
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