Chapter 9 Symmetry & Practical Geometry (Class 6 - Maths NCERT Exemplar Solutions)

Correct Option: (C)

Explanation:

A line of symmetry is a line that divides a figure into two identical halves that are mirror images of each other.

1. E has one horizontal line of symmetry.

2. T has one vertical line of symmetry.

3. X has both vertical and horizontal lines of symmetry.

4. N does not have any line of symmetry (it only has rotational symmetry).

Final Answer: The letter N does not have any line of symmetry.

Correct Option: (D)

Explanation:

Using a ruler and compasses, we can construct angles that are multiples of $15^\circ$ (such as $15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ, 90^\circ, 105^\circ, 120^\circ, 135^\circ, 150^\circ,$ etc.) or angles obtained by further bisecting these.

1. $75^\circ$ can be constructed (bisecting the angle between $60^\circ$ and $90^\circ$).

2. $15^\circ$ can be constructed (bisecting $30^\circ$).

3. $135^\circ$ can be constructed (bisecting the angle between $90^\circ$ and $180^\circ$).

4. $85^\circ$ is not a multiple of $15$ and cannot be reached by standard bisection methods.

Final Answer: $85^\circ$ cannot be constructed using ruler and compasses.

Answer: $BD$

Explanation:

Symmetry (reflection in a line) is a transformation that preserves distances between points. This means the distance between any two original points is equal to the distance between their corresponding image points.

Since reflection preserves distance, the length of segment $AC$ must be equal to the length of segment $BD$.

Final Answer: $AC = BD$

Answer: $PR$

Explanation:

From the given figure 9.1, the points $P, R,$ and $Q$ lie on the same line $l$. We can see that the segment $PQ$ is made up of segments $PR$ and $RQ$.

$PQ = PR + RQ$

Subtracting $RQ$ from both sides:

$PR = PQ - RQ$

Given that:

Substituting these values into the equation:

$PR = AB - CD$

Final Answer: $AB - CD = PR$

Answer: two

Explanation:

A standard semicircular protractor used in geometry sets has two scales.

1. The Outer Scale: It reads from $0^\circ$ to $180^\circ$ in a clockwise direction (from left to right).

2. The Inner Scale: It reads from $0^\circ$ to $180^\circ$ in an anti-clockwise direction (from right to left).

These two scales allow us to measure angles starting from either the left side or the right side of the baseline.

Final Answer: The number of scales is two.

Statement: True

Explanation:

We can construct various angles by placing two set squares together. To draw an angle of $75^\circ$, we can use the $30^\circ$ angle from the first set square and the $45^\circ$ angle from the second set square.

By placing these two angles adjacent to each other at a common vertex and along a common arm, the resulting outer arms will form a $75^\circ$ angle.

Final Answer: The statement is True.

Statement: False

Explanation:

A line of symmetry of a circle is any line that passes through its centre (i.e., its diameter). Since we can draw an infinite number of diameters through the centre of a circle, it possesses an infinite number of lines of symmetry.

Final Answer: The statement is False as a circle has infinitely many lines of symmetry.

Given:

The word ALGEBRA.

To Find:

Letters of this word which have no line of symmetry.

Solution:

Let us examine each letter of the word ALGEBRA individually:

1. A: Has one vertical line of symmetry.

2. L: Has no line of symmetry.

3. G: Has no line of symmetry.

4. E: Has one horizontal line of symmetry.

5. B: Has one horizontal line of symmetry.

6. R: Has no line of symmetry.

7. A: Has one vertical line of symmetry.

The letters that do not have any line of symmetry are L, G, and R.

Final Answer: The letters are L, G, and R.

Given:

Two line segments $AB$ and $CD$ as shown in Fig. 9.2.

To Construct:

A line segment whose length is equal to the sum of $AB$ and $CD$.

Construction Required:

1. Draw a line $l$ and mark a point $P$ on it.

2. Open the compasses to measure the length of the segment $AB$.

3. Without changing the settings of the compasses, place the pointer at $P$ and draw an arc to cut the line $l$ at point $Q$. Now, $PQ = AB$.

4. Again, open the compasses to measure the length of the segment $CD$.

5. Place the pointer of the compasses at $Q$ and draw another arc on the line $l$ (on the side opposite to $P$) to cut it at point $R$. Now, $QR = CD$.

6. The resulting line segment $PR$ is the required segment.

Given:

Two angles $\angle DEF$ and $\angle PQR$ are given in Fig. 9.5. By observation, we can see that:

To Construct:

An angle equal to the difference $(\angle DEF - \angle PQR)$.

Construction Required:

1. Draw a ray $OX$.

2. Copy the larger angle: Place the pointer of the compasses at vertex $E$ of $\angle DEF$ and draw an arc cutting the arms $ED$ and $EF$. Without changing the radius, place the pointer at $O$ and draw a similar arc cutting $OX$ at a point $A$.

3. Measure the width of the arc of $\angle DEF$ using the compasses. Place the pointer at $A$ and cut the previous arc at point $B$. Join $OB$ to form ray $OY$.

4. Copy the smaller angle inside: Now, take the width of the arc of the smaller angle $\angle PQR$ using the compasses.

5. Place the pointer of the compasses at $A$ and cut the arc $AB$ at a point $C$ towards the ray $OY$.

6. Join $OC$ to form ray $OZ$. This ray $OZ$ lies inside the angle $\angle XOY$.

7. The remaining angle $\angle ZOY$ is the required difference.

Given:

A partial figure and a line $l$ as the line of symmetry in Fig. 9.7.

Solution:

To complete the figure so that $l$ is the line of symmetry, we must draw the mirror image of the given part on the other side of line $l$.

1. Identify the key vertices (points) of the given figure.

2. For each vertex, find a corresponding point on the opposite side of line $l$ such that both points are at the same perpendicular distance from $l$.

3. Join these new points in the same order as the original figure to complete the symmetric shape.

Correct Option: (B)

Explanation:

1. Figure (i) is a rectangle. It has two lines of symmetry (one horizontal and one vertical).

2. Figure (ii) is a swastika-like shape. If we try to fold it along any line passing through its centre, the two parts will not coincide. Therefore, it has no line of symmetry.

3. Figure (iii) is an isosceles triangle. It has one vertical line of symmetry passing through the vertex between the equal sides.

4. Figure (iv) is a circle. It has infinitely many lines of symmetry (any line passing through its centre).

Final Answer: Figure (ii) is not symmetric with respect to any line.

Correct Option: (A)

Explanation:

A scalene triangle is a triangle where all three sides have different lengths and all three internal angles have different measures. Because there is no equality between any sides or angles, there is no line along which the triangle can be folded to obtain two matching halves.

Final Answer: The number of lines of symmetry in a scalene triangle is $0$.

Correct Option: (D)

Explanation:

A circle is perfectly symmetrical about any line that passes through its centre. Such lines are the diameters of the circle. Since infinitely many diameters can be drawn in a circle, it has infinitely many lines of symmetry. Out of the given options, "more than 4" best describes this infinite property.

Final Answer: The number of lines of symmetry in a circle is more than 4.

Correct Option: (C)

Explanation:

1. M has a vertical line of symmetry.

2. H has both vertical and horizontal lines of symmetry.

3. E has only a horizontal line of symmetry. A vertical line through its centre would result in different left and right halves.

4. V has a vertical line of symmetry.

Final Answer: The letter E does not have a vertical line of symmetry.

Correct Option: (A)

Explanation:

1. X can be folded both vertically and horizontally into identical halves.

2. E has only a horizontal line of symmetry.

3. M has only a vertical line of symmetry.

4. K usually has only a horizontal line of symmetry.

Final Answer: The letter X has both horizontal and vertical lines of symmetry.

Correct Option: (B)

Explanation:

1. M has one line of symmetry (vertical).

2. S does not have any line of symmetry. Even though it looks balanced, if you fold it along any axis, the parts will not match exactly. It only has rotational symmetry.

3. K has one line of symmetry (horizontal).

4. H has two lines of symmetry (vertical and horizontal).

Final Answer: The letter S does not have any line of symmetry.

Correct Option: (D)

Explanation:

Let us analyze the lines of symmetry for each letter:

1. H has two lines of symmetry (one vertical and one horizontal).

2. X has two lines of symmetry (one vertical and one horizontal).

3. Z has no lines of symmetry (it only has rotational symmetry).

4. T has only one line of symmetry, which is a vertical line passing through its center.

Final Answer: The letter T has only one line of symmetry.

Correct Option: (B)

Explanation:

1. Ruler is used to draw straight lines and measure lengths.

2. Protractor is a semi-circular instrument used specifically to measure and draw angles in degrees.

3. Divider is used to compare lengths of line segments.

4. Compasses are used to draw circles and arcs.

Final Answer: The instrument to measure an angle is a Protractor.

Correct Option: (D)

Explanation:

To draw a circle, we need an instrument with a fixed pointer and a moving pencil arm to maintain a constant radius. This instrument is called Compasses.

Final Answer: The instrument to draw a circle is Compasses.

Correct Option: (C)

Explanation:

In a standard Indian geometry box, there are two set squares:

1. One with angles $45^\circ$, $45^\circ$ and $90^\circ$.

2. Another with angles $30^\circ$, $60^\circ$ and $90^\circ$.

Final Answer: The number of set squares is $2$.

Correct Option: (C)

Explanation:

A standard ruler is rectangular in shape. A rectangle has two lines of symmetry:

1. One line passing through the mid-points of the longer sides (vertical symmetry).

2. One line passing through the mid-points of the shorter sides (horizontal symmetry).

Final Answer: The number of lines of symmetry in a ruler is $2$.

Correct Option: (B)

Explanation:

A divider, when its arms are kept equal, is symmetrical about a single line. This line passes through the hinge (vertex) and bisects the angle between the two arms.

Final Answer: The number of lines of symmetry in a divider is $1$.

Correct Option: (A)

Explanation:

Compasses consist of two different arms: one arm has a sharp metal needle at the end, while the other arm is designed to hold a pencil. Since these two arms are not identical in shape or appearance, the instrument is asymmetrical.

Final Answer: The number of lines of symmetry in compasses is 0.

Correct Option: (B)

Explanation:

A protractor is semi-circular in shape. A semi-circle is symmetrical about exactly one line, which is the perpendicular bisector of its diameter. This line passes through the $90^\circ$ mark on the protractor.

Final Answer: The number of lines of symmetry in a protractor is 1.

Correct Option: (B)

Explanation:

A $45^\circ - 45^\circ - 90^\circ$ set-square is an isosceles right-angled triangle because it has two equal angles (and thus two equal sides). Any isosceles triangle has exactly one line of symmetry, which bisects the angle between the equal sides.

Final Answer: The number of lines of symmetry in a $45^\circ - 45^\circ - 90^\circ$ set-square is 1.

Correct Option: (A)

Explanation:

A $30^\circ - 60^\circ - 90^\circ$ set-square is a scalene triangle because all three of its angles are different, which means all three of its sides are also of different lengths. A scalene triangle does not have any lines of symmetry.

Final Answer: The number of lines of symmetry in a $30^\circ - 60^\circ - 90^\circ$ set-square is 0.

Correct Option: (D)

Explanation:

In a standard geometry box, the instruments that are shaped like right-angled triangles are known as Set-squares. They are used to draw parallel and perpendicular lines as well as specific angles.

Final Answer: The triangular instrument is called a Set-square.

Solution:

In reflection symmetry, the line of symmetry acts as a mirror. A fundamental property of reflection is that the object and its image are at the same perpendicular distance from the mirror line.

Final Answer: The distance of the image of a point (or an object) from the line of symmetry (mirror) is same as that of the point (object) from the line (mirror).

Solution:

The Taj Mahal is a world-renowned example of architectural symmetry. A frontal photograph of the Taj Mahal shows a clear bilateral symmetry. If we draw a vertical line through the center of the main dome, the left and right halves of the picture are identical mirror images.

Final Answer: The number of lines of symmetry in a picture of Taj Mahal is one.

Solution:

A rectangle has $2$ lines of symmetry (the lines joining the midpoints of opposite sides).

A rhombus also has $2$ lines of symmetry (its two diagonals).

Since both figures have the same number of lines of symmetry, the counts are equal.

Final Answer: The number of lines of symmetry in a rectangle and a rhombus are equal.

Solution:

A rectangle has $2$ lines of symmetry.

A square has $4$ lines of symmetry (two joining midpoints of opposite sides and two diagonals).

Since $2$ is not equal to $4$, the counts are unequal.

Final Answer: The number of lines of symmetry in a rectangle and a square are unequal.

Solution:

Reflection is a transformation that preserves the shape and size (congruence) of an object. When a line segment is reflected, its image remains a line segment of the exact same length.

Final Answer: If a line segment of length $5\text{ cm}$ is reflected in a line of symmetry (mirror), then its reflection (image) is a line segment of length $5\text{ cm}$.

Solution:

Reflection in a line preserves the magnitude of angles. If an object is an angle of a certain measure, its mirror image will also be an angle of the same measure.

Final Answer: If an angle of measure $80^\circ$ is reflected in a line of symmetry, then the reflection is an angle of measure $80^\circ$.

Explanation:

In reflection symmetry, the line of symmetry $l$ acts as a mirror. If a point is located exactly on the mirror line, its mirror image will be produced at the same location as the point itself. Such points are called invariant points.

Final Answer: The image of a point lying on a line $l$ with respect to the line of symmetry $l$ lies on $l$ itself (or the same line).

Given:

From Fig. 9.10, $B$ is the image of the point $A$ with respect to the line $l$. Point $P$ is any point lying on the line $l$.

Solution:

When point $B$ is the image of point $A$ with respect to line $l$, the line $l$ is the perpendicular bisector of the line segment $AB$. According to the properties of a perpendicular bisector, any point lying on it is equidistant from the two endpoints of the segment.

Final Answer: The lengths of line segments $PA$ and $PB$ are equal.

Solution:

The figure 9.11 represents a circular shape containing a flower with five identical petals. For any regular arrangement of five identical parts around a center, there are five lines of symmetry. Each line passes through the center of the circle and the tip of one of the five petals.

Final Answer: The number of lines of symmetry in Fig. 9.11 is $5$.

Solution:

In a standard geometry box, there are two triangular instruments called set-squares. The first one has angles $30^\circ, 60^\circ, 90^\circ$ and the second one has angles $45^\circ, 45^\circ, 90^\circ$.

Both instruments share a common $90^\circ$ angle and both are triangular in their physical shape.

Final Answer: The common properties in the two set-squares of a geometry box are that they have a $90^\circ$ (or right) angle and they are of the shape of a triangle.

Solution:

Let us examine the digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ for symmetry. In standard block form:

1. The digit $0$ has two lines of symmetry (one horizontal and one vertical).

2. The digit $8$ has two lines of symmetry (one horizontal and one vertical).

Final Answer: The digits having only two lines of symmetry are $0$ and $8$.

Solution:

Reviewing the digits for a single line of symmetry, we find that the digit $3$ is symmetrical about a single horizontal line passing through its middle.

Final Answer: The digit having only one line of symmetry is $3$.

Solution:

The digits that do not have any line of symmetry (neither vertical nor horizontal) are $1, 2, 4, 5, 6, 7,$ and $9$.

Counting these digits, we have a total of $7$ such digits.

Final Answer: The number of digits having no line of symmetry is $7$.

Explanation:

To find the letters having only a vertical line of symmetry, we list all letters with vertical symmetry and exclude those that also have horizontal symmetry. The letters with vertical symmetry are: A, H, I, M, O, T, U, V, W, X, Y (Total 11). Out of these, H, I, O, and X also have horizontal symmetry.

Letters with only vertical symmetry: A, M, T, U, V, W, Y.

$\text{Count} = 7$

Final Answer: The number of capital letters having only vertical line of symmetry is $7$.

Explanation:

The letters with horizontal symmetry are: B, C, D, E, H, I, K, O, X (Total 9). We exclude those that also have vertical symmetry (H, I, O, X).

Letters with only horizontal symmetry: B, C, D, E, K.

$\text{Count} = 5$

Final Answer: The number of capital letters having only horizontal line of symmetry is $5$.

Explanation:

These are the letters that can be folded both vertically and horizontally to yield identical halves. These letters are H, I, O, and X.

Final Answer: The number of capital letters having both horizontal and vertical lines of symmetry is $4$.

Explanation:

We subtract the letters with any symmetry from the total number of alphabets ($26$).

Letters with no line of symmetry: F, G, J, L, N, P, Q, R, S, Z.

$\text{Count} = 26 - 16 = 10$

Final Answer: The number of capital letters having no line of symmetry is $10$.

Explanation:

The line of symmetry of a line segment divides the segment into two equal parts and meets it at a right angle ($90^\circ$). This specific line is known as the perpendicular bisector.

Final Answer: The line of symmetry of a line segment is the perpendicular bisector of the line segment.

Explanation:

For any regular polygon with $n$ sides, the number of lines of symmetry is equal to $n$. A regular hexagon has $6$ equal sides and $6$ equal angles.

The lines of symmetry consist of $3$ lines passing through opposite vertices and $3$ lines passing through the midpoints of opposite sides.

Final Answer: The number of lines of symmetry in a regular hexagon is $6$.

Explanation:

A regular polygon is a polygon that is both equilateral (all sides equal) and equiangular (all angles equal). For any regular polygon with $n$ sides, the number of lines of symmetry is always equal to the number of its sides. For example, an equilateral triangle has $3$ lines of symmetry and a square has $4$ lines of symmetry.

Final Answer: The number of lines of symmetry in a regular polygon of $n$ sides is $n$.

Explanation:

A protractor is semi-circular in shape. A semi-circle has exactly one line of symmetry, which is the perpendicular bisector of its base (diameter). In a protractor, this line passes through the $90^\circ$ mark.

Final Answer: A protractor has one line of symmetry.

Explanation:

A $30^\circ - 60^\circ - 90^\circ$ set-square is a scalene triangle. Since all three angles are different, all three sides are also of different lengths. A scalene triangle does not possess any lines of symmetry.

Final Answer: A $30^\circ - 60^\circ - 90^\circ$ set-square has no (or zero) line of symmetry.

Explanation:

A $45^\circ - 45^\circ - 90^\circ$ set-square is an isosceles right-angled triangle. Since it has two equal angles, the sides opposite to those angles are also equal. Any isosceles triangle has exactly one line of symmetry passing through the vertex of the unequal angle.

Final Answer: A $45^\circ - 45^\circ - 90^\circ$ set-square has one line of symmetry.

Explanation:

A rhombus is a quadrilateral where all sides are equal. The lines of symmetry for a rhombus are the lines containing its diagonals. Folding the rhombus along either diagonal results in two parts that coincide perfectly.

Final Answer: A rhombus is symmetrical about its diagonals.

Explanation:

A rectangle has two lines of symmetry. These lines are formed by connecting the middle points of the opposite parallel sides. Folding a rectangle along these lines divides it into two identical mirror images.

Final Answer: A rectangle is symmetrical about the lines joining the midpoints of the opposite sides.

Statement: True

Explanation:

A right-angled triangle can either be scalene or isosceles.

1. If it is a right-angled scalene triangle, it has zero lines of symmetry.

2. If it is a right-angled isosceles triangle (with angles $45^\circ, 45^\circ$ and $90^\circ$), it has exactly one line of symmetry which bisects the $90^\circ$ angle.

In no case can it have more than one. Therefore, the statement that it can have at most one line of symmetry is correct.

Final Answer: The statement is True.

Explanation:

A kite is a quadrilateral with two pairs of equal adjacent sides. A standard kite (which is not a rhombus) has only one line of symmetry, which is the diagonal passing through the vertices where the unequal sides meet. Folding along the other diagonal does not result in matching halves.

Final Answer: The statement is False.

Statement: True

Explanation:

A general parallelogram has no lines of symmetry. While it has rotational symmetry of order $2$, there is no line along which it can be folded so that the two parts coincide perfectly. Special cases of parallelograms like rectangles and rhombuses do have lines of symmetry, but for a general parallelogram, the count is zero.

Final Answer: The statement is True.

Statement: False

Explanation:

If a triangle has more than one line of symmetry, it must have three lines of symmetry. A triangle with three lines of symmetry is always an equilateral triangle. Therefore, if an isosceles triangle has more than one line of symmetry, it must be an equilateral triangle.

Final Answer: The statement is False.

Statement: True

Explanation:

A standard rectangle has exactly two lines of symmetry (the lines joining the midpoints of opposite sides). A square is a special type of rectangle that has four lines of symmetry (two joining midpoints and two diagonals). If any rectangle is found to have more than two lines of symmetry, it must be a square.

Final Answer: The statement is True.

Statement: True

Explanation:

Using a ruler to draw a line segment and a compass to draw arcs from both endpoints with a radius greater than half the length of the segment, we can always find two points of intersection. Joining these points allows us to bisect the line segment into two equal halves. This is a fundamental construction in geometry.

Final Answer: The statement is True.

Statement: True

Explanation:

For any given line segment, there is only one unique line that is both perpendicular to the segment and passes through its exact midpoint. This uniqueness is a basic property of Euclidean geometry.

Final Answer: The statement is True.

Statement: False

Explanation:

From a point not lying on a given line, exactly one unique perpendicular line can be drawn to that line. This is also referred to as the perpendicular distance from a point to a line.

Final Answer: The statement is False.

Statement: True

Explanation:

A circle is defined as the collection of all points in a plane that are at a fixed distance (called the radius) from a fixed point (called the centre). Since the position of the centre and the length of the radius are uniquely given, only one such set of points can exist in the plane.

Final Answer: The statement is True.

Statement: False

Explanation:

The angles available on the two set-squares are $\{30^\circ, 60^\circ, 90^\circ\}$ and $\{45^\circ, 45^\circ, 90^\circ\}$. By combining these (adding or subtracting), we can draw angles that are multiples of $15^\circ$, such as:

$45^\circ - 30^\circ = 15^\circ$

$45^\circ + 30^\circ = 75^\circ$

$45^\circ + 60^\circ = 105^\circ$

There is no mathematical combination of these standard angles that results in exactly $40^\circ$.

Final Answer: The statement is False.

Statement: True

Explanation:

An angle of $15^\circ$ can be drawn by using the $45^\circ$ angle of one set-square and the $30^\circ$ angle of the other. By placing them such that their vertices and one arm coincide, the difference between the other two arms will form the required angle.

Final Answer: The statement is True.

Statement: True

Explanation:

A triangle can have $0, 1,$ or $3$ lines of symmetry. An isosceles triangle typically has $1$ line of symmetry. If it is specified that it has more than one, the only remaining possibility is $3$ lines of symmetry. A triangle with $3$ lines of symmetry is, by definition, an equilateral triangle.

Final Answer: The statement is True.

Statement: False

Explanation:

A square has $4$ lines of symmetry (two horizontal/vertical lines and two diagonal lines). A rectangle (which is not a square) has only $2$ lines of symmetry (one horizontal and one vertical line passing through the midpoints of opposite sides). Therefore, the numbers are not the same.

Final Answer: The statement is False.

Statement: False

Explanation:

Any line passing through the centre of a circle (a diameter) acts as a line of symmetry. Since we can draw an infinite number of diameters through the centre of a circle, it has infinitely many lines of symmetry.

Final Answer: The statement is False.

Statement: True

Explanation:

1. A $45^\circ - 45^\circ - 90^\circ$ set-square is an isosceles right-angled triangle. It has exactly one line of symmetry, which is the perpendicular bisector of the hypotenuse (the side opposite the $90^\circ$ angle).

2. A protractor is in the shape of a semi-circle. A semi-circle also has exactly one line of symmetry, which passes through its centre and is perpendicular to its diameter (the $90^\circ$ line).

Since both instruments have exactly one line of symmetry, the statement is true.

Final Answer: The statement is True.

Statement: False

Explanation:

The bisector of an angle is a unique ray that originates from the vertex and divides the angle into two equal parts. For any given angle, there is only one unique internal bisector that can be drawn.

Final Answer: The statement is False.

Statement: False

Explanation:

In geometry, a regular polygon with $n$ sides has exactly $n$ lines of symmetry. An octagon is a polygon with $8$ sides. Therefore, a regular octagon has $8$ lines of symmetry, not $10$.

Final Answer: The statement is False.

Statement: True

Explanation:

A ray is a collection of an infinite number of points extending in one direction. At each and every point on that ray, a unique line can be drawn that is perpendicular to the ray. Because there are infinitely many points on the ray, infinitely many perpendiculars can be drawn to it.

Final Answer: The statement is True.

Statement: False

Explanation:

A perpendicular bisector is a line that is perpendicular to a line segment and passes through its midpoint. A ray has one endpoint and extends infinitely in the other direction; it does not have a finite length or a midpoint. Therefore, the concept of a "perpendicular bisector" cannot be applied to a ray.

Final Answer: The statement is False.

Given:

Figure 9.12 shows a kite $ABCD$ where:

To Find:

Whether the figure has lines of symmetry, and if so, to identify them.

Solution:

Yes, there is one line of symmetry in the given figure.

A kite is a quadrilateral that is symmetrical about the diagonal connecting the vertices where the pairs of equal sides meet. In Fig. 9.12, the line passing through the points A and C (diagonal $AC$) is the line of symmetry.

If the figure is folded along the line $AC$, vertex $B$ will coincide with vertex $D$. Thus, the two halves are mirror images of each other.

Final Answer: Yes, there is one line of symmetry, which is the line passing through A and C.

Given:

A rectangle $PQRS$ and several dashed lines ($AC$, $BD$, $PR$, $QS$) passing through it as shown in Fig. 9.13.

To Find:

The lines of symmetry of the rectangle.

Solution:

A rectangle has exactly two lines of symmetry. These are the lines that join the midpoints of the opposite sides of the rectangle.

Based on Fig. 9.13, we can identify the following:

1. The line AC joins the midpoints of the sides $PQ$ and $SR$. Folding the rectangle along this vertical line makes the left and right halves coincide.

2. The line BD joins the midpoints of the sides $PS$ and $QR$. Folding the rectangle along this horizontal line makes the top and bottom halves coincide.

Note regarding diagonals:

The lines PR and QS are the diagonals of the rectangle. While they divide the rectangle into two triangles of equal area, they are not lines of symmetry for a rectangle (unless it is a square), because the parts do not coincide perfectly when folded along a diagonal.

Final Answer: The lines of symmetry of the rectangle are Line AC and Line BD.

Analysis:

A capital letter of the English alphabet has more than one line of symmetry if it possesses both vertical and horizontal lines of symmetry. Let us examine the letters:

H: Has one vertical and one horizontal line of symmetry.

I: Has one vertical and one horizontal line of symmetry.

O: Has vertical, horizontal, and many other lines of symmetry (depending on the font, usually treated as two).

X: Has one vertical and one horizontal line of symmetry.

Final Answer: The letters are H, I, O, and X.

Given:

The word ‘MATHEMATICS’.

Solution:

Let us check the lines of symmetry for each unique letter in the word:

M: $1$ line (vertical)

A: $1$ line (vertical)

T: $1$ line (vertical)

H: $2$ lines (vertical and horizontal)

E: $1$ line (horizontal)

I: $2$ lines (vertical and horizontal)

C: $1$ line (horizontal)

S: $0$ lines

Final Answer: In the word ‘MATHEMATICS’, only the letter S has no line of symmetry.

Given:

The word ‘SYMMETRY’.

Solution:

The number of lines of symmetry for each letter is as follows:

1. S: $0$

2. Y: $1$ (vertical)

3. M: $1$ (vertical)

4. M: $1$ (vertical)

5. E: $1$ (horizontal)

6. T: $1$ (vertical)

7. R: $0$

8. Y: $1$ (vertical)

Final Answer: The sequence of the number of lines of symmetry is $0, 1, 1, 1, 1, 1, 0, 1$.

Matching of Shapes with their Number of Lines of Symmetry:

Based on the geometric properties of regular and irregular polygons, the correct matches are as follows:

Note: Option (b) 5 is not used in this match as no regular pentagon is listed.

Analysis of Geometry Box Drawing Tools:

Observing the tools found in a standard Indian geometry box, we find the following lines of symmetry for each instrument:

Given:

Points $A$, $B$ and a line $l$ as shown in Fig. 9.14.

To Construct and Find:

1. Draw the images of $A$ and $B$ in line $l$ and name them $A'$ and $B'$.

2. Measure $AB$ and $A'B'$ and check if they are equal.

Construction Required:

1. From point $A$, draw a line perpendicular to line $l$ using a set-square or ruler. Let it intersect line $l$ at point $M$.

2. Extend this perpendicular line to the other side of $l$ and mark a point $A'$ such that the distance $AM$ is equal to the distance $MA'$.

3. Similarly, from point $B$, draw a perpendicular to line $l$ intersecting it at point $N$.

4. Extend the line and mark point $B'$ such that $BN$ is equal to $NB'$.

5. Join $A'$ and $B'$ to form the line segment $A'B'$.

Given:

In Fig. 9.15, point $C$ is the image of point $A$ in the line of symmetry $l$. The line segment $BC$ intersects the line $l$ at point $P$.

Solution:

(a) Is the image of $P$ in line $l$ the point $P$ itself?

Yes. Any point that lies on the line of symmetry (the mirror line) is its own image. Since point $P$ lies on line $l$, its image is the point $P$ itself.

(b) Is $PA = PC$?

Yes. By the property of reflection symmetry, the line of symmetry $l$ is the perpendicular bisector of the line segment joining a point and its image ($AC$). Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

(c) Is $PA + PB = PC + PB$?

Yes. We can verify this using the equality found in part (b).

From equation (i), we have $PA = PC$.

Adding $PB$ to both sides of the equation:

(d) Is $P$ that point on line $l$ from which the sum of the distances of points $A$ and $B$ is minimum?

Yes. The sum of the distances from $A$ and $B$ to point $P$ is $PA + PB$. From part (c), we know $PA + PB = PC + PB$.

In the figure, points $B, P,$ and $C$ lie on a straight line (since $P$ is the intersection of $BC$ and $l$). Therefore, $PC + PB$ is equal to the length of the straight line segment $BC$.

The shortest distance between any two points ($B$ and $C$) is a straight line. For any other point $P'$ on line $l$, the sum $P'C + P'B$ would form a triangle $BP'C$, and by the triangle inequality, $P'C + P'B > BC$. Since $P'A = P'C$, it follows that $P'A + P'B > PA + PB$.

Final Answer: All statements (a), (b), (c), and (d) are Yes/True.

Solution:

To complete the figure so that line $l$ becomes the line of symmetry, we must perform a reflection of the existing upper portion across the line $l$.

1. Identify the vertices of the given shape above line $l$.

2. For each vertex, plot a corresponding point below line $l$ at the same perpendicular distance.

3. Join these reflected points with straight lines to create a mirror image of the top part.

The resulting figure will look like a symmetric block-style shape centered vertically on line $l$.

Given:

A triangle $ABC$ and a line $m$ passing through it as shown in Fig. 9.17.

To Find:

1. The images of points $A, B,$ and $C$ in line $m$ (named $A', B', C'$).

2. To verify if the lengths of the corresponding segments are equal ($AB = A'B'$, $BC = B'C'$, and $CA = C'A'$).

Construction Required:

1. From point $A$, drop a perpendicular to line $m$. Extend it to the other side of the line and mark point $A'$ such that the distance of $A$ from $m$ is equal to the distance of $A'$ from $m$.

2. Repeat the same process for points $B$ and $C$ to find their images $B'$ and $C'$ respectively.

3. Join $A'$ to $B'$, $B'$ to $C'$, and $C'$ to $A'$ to form $\triangle A'B'C'$.

Solution:

Reflection is a transformation that preserves distances between points. This property is known as isometry.

After measuring the sides of the original $\triangle ABC$ and the reflected $\triangle A'B'C'$ using a ruler, we find:

Final Answer: Yes, the lengths of the corresponding sides are equal, i.e., $AB = A'B'$, $BC = B'C'$, and $CA = C'A'$.

Given:

An angle $\angle PQR$ and a line $n$ as the line of symmetry in Fig. 9.18.

To Find:

1. The images $P', Q',$ and $R'$ of points $P, Q,$ and $R$ in line $n$.

2. To verify if $\angle PQR = \angle P'Q'R'$.

Construction Required:

1. Draw perpendiculars from points $P, Q,$ and $R$ to line $n$.

2. Extend these perpendiculars to the opposite side of line $n$ to mark points $P', Q',$ and $R'$ at the same distance from $n$ as their original points.

3. Join $P'$ to $Q'$ and $Q'$ to $R'$ to form the angle $\angle P'Q'R'$.

Solution:

Reflection in a line preserves the magnitude of angles. This means that the shape and size of an object remain identical in its mirror image.

On measuring the original angle $\angle PQR$ and the image angle $\angle P'Q'R'$ using a protractor, we observe:

Final Answer: Yes, the two angles are equal.

Given:

A partial figure of an arrow and a line $l$ acting as the line of symmetry in Fig. 9.19.

To Construct:

Complete the figure using $l$ as the axis of symmetry.

Solution:

To complete the figure, we must draw the exact mirror image of the given arrow part on the other side of line $l$.

1. The original figure shows the tail of an arrow touching the line $l$ and the head pointing to the left.

2. To create the symmetric part, we draw an identical arrow pointing to the right, starting from the line $l$.

3. The completed figure will look like a double-headed arrow that is bisected vertically by the line $l$.

Given:

A line segment $AB = 7\text{ cm}$.

To Construct:

Perpendicular bisector of $AB$.

Construction Required:

1. Draw a line segment $AB$ of length $7\text{ cm}$ using a ruler.

2. With $A$ as centre and radius more than half of $AB$ (i.e., more than $3.5\text{ cm}$), draw two arcs, one above and one below the line segment $AB$.

3. With $B$ as centre and the same radius, draw two more arcs intersecting the previous arcs at points $P$ and $Q$.

4. Join $P$ and $Q$. Let $PQ$ intersect $AB$ at point $M$.

5. $PQ$ is the required perpendicular bisector of $AB$.

Final Answer: The line $PQ$ bisects $AB$ at $90^\circ$ such that $AM = MB$.

Given:

A line segment $XY = 6.5\text{ cm}$.

To Construct:

Divide $XY$ into four equal parts.

Construction Required:

1. Draw $XY = 6.5\text{ cm}$.

2. Draw the perpendicular bisector of $XY$ which meets $XY$ at point $M$. Now $XY$ is divided into two equal parts $XM$ and $MY$.

3. Draw the perpendicular bisector of $XM$ which meets $XM$ at point $P$.

4. Draw the perpendicular bisector of $MY$ which meets $MY$ at point $Q$.

5. Now, the points $P, M,$ and $Q$ divide $XY$ into four equal parts: $XP, PM, MQ,$ and $QY$.

Calculation:

Length of each part:

$\frac{6.5}{4} = 1.625\text{ cm}$

Final Answer: The segments $XP = PM = MQ = QY$ are the four equal parts of the line segment.

Given:

An angle of $140^\circ$.

To Construct:

The bisector of the angle.

Construction Required:

1. Draw an angle $\angle ABC = 140^\circ$ using a protractor.

2. With $B$ as centre and any convenient radius, draw an arc intersecting arms $BA$ and $BC$ at points $E$ and $F$ respectively.

3. With $E$ as centre and radius more than half of $EF$, draw an arc in the interior of the angle.

4. With $F$ as centre and the same radius, draw another arc intersecting the previous arc at point $D$.

5. Join $BD$. Ray $BD$ is the bisector of $\angle ABC$.

Final Answer: Ray $BD$ bisects the $140^\circ$ angle into two $70^\circ$ angles.

Given:

An angle $\angle ABC = 65^\circ$.

To Construct:

An angle equal to $\angle ABC$ without using a protractor for the second angle.

Construction Required:

1. Draw $\angle ABC = 65^\circ$ using a protractor.

2. Draw a ray $OX$.

3. With $B$ as centre, draw an arc of any radius intersecting arms $BA$ and $BC$ at $E$ and $F$.

4. With $O$ as centre and the same radius, draw an arc cutting ray $OX$ at point $P$.

5. Measure the distance $EF$ using compasses. With $P$ as centre and radius equal to $EF$, draw an arc intersecting the previous arc at point $Q$.

6. Join $OQ$ and extend it to $Y$.

Final Answer: Angle $\angle YOX$ is constructed equal to $65^\circ$.

Given:

An angle of $80^\circ$.

To Construct:

Divide the $80^\circ$ angle into four equal parts.

Construction Required:

1. Draw an angle $\angle POQ = 80^\circ$ using a protractor.

2. Construct the bisector $OR$ of $\angle POQ$. Now, $\angle POR = \angle ROQ = 40^\circ$.

3. Construct the bisector $OS$ of $\angle POR$.

4. Construct the bisector $OT$ of $\angle ROQ$.

5. The rays $OS, OR,$ and $OT$ divide the $80^\circ$ angle into four equal parts.

Verification:

On measuring with a protractor, each of the four angles $\angle POS, \angle SOR, \angle ROT,$ and $\angle TOQ$ should be exactly $20^\circ$.

Final Answer: The $80^\circ$ angle is successfully divided into four equal parts of $20^\circ$ each.

Given:

A line $l$ and a point $P$ lying on it.

Construction Required:

(i) Using Set-squares:

1. Align the edge of a ruler with the line $l$.

2. Place a set-square such that one of its edges forming the right angle is against the ruler.

3. Slide it until the vertical edge reaches point $P$.

4. Draw a line through $P$ along this vertical edge.

(ii) Using Protractor:

1. Place the protractor center at point $P$ and align the $0^\circ-180^\circ$ line with line $l$.

2. Mark a point at the $90^\circ$ graduation.

3. Join this point with $P$ to get the perpendicular.

(iii) Using Ruler and Compasses:

1. With $P$ as centre, draw an arc cutting line $l$ at $X$ and $Y$.

2. With $X$ and $Y$ as centres and radius $> PX$, draw arcs intersecting at $Z$.

3. Join $PZ$ to get the perpendicular.

Solution:

Through a point $P$ on a line $l$, we can draw only one unique perpendicular line.

Final Answer: We are able to draw only one such perpendicular.

Given:

A line $m$ and a point $P$ not on it.

Construction Required:

(i) Using Set-squares:

1. Place a ruler along line $m$.

2. Place a set-square on the ruler and slide it until the edge perpendicular to the ruler passes through $P$.

3. Draw a line segment from $P$ to the line $m$.

(ii) Using Protractor:

1. Align the protractor baseline with line $m$.

2. Slide it horizontally until the $90^\circ$ mark is vertically aligned with point $P$.

3. Drop a perpendicular line from $P$ to the centre of the protractor on line $m$.

(iii) Using Ruler and Compasses:

1. With $P$ as centre, draw an arc cutting line $m$ at $A$ and $B$.

2. With $A$ and $B$ as centres, draw arcs on the opposite side of $P$ intersecting at $Q$.

3. Join $PQ$. The line $PQ$ is perpendicular to $m$.

Solution:

From a point $P$ outside a line $m$, only one unique perpendicular can be dropped to the line.

Final Answer: We are able to draw only one such perpendicular.

Given:

Radius of the circle ($r$) = $6\text{ cm}$

To Construct:

1. A circle with radius $6\text{ cm}$.

2. A diameter of the circle.

3. The perpendicular bisector of this diameter.

Construction Required:

1. Mark a point $O$ as the centre of the circle.

2. Using a ruler, open the compasses to a radius of $6\text{ cm}$.

3. Placing the pointer at $O$, draw a complete circle.

4. Draw a straight line passing through the centre $O$ that intersects the circle at two points, $A$ and $B$. $AB$ is the diameter of the circle.

5. With $A$ as centre and a radius more than half of $AB$ (i.e., $> 6\text{ cm}$), draw two arcs, one above and one below the diameter $AB$.

6. With $B$ as centre and the same radius, draw two more arcs intersecting the previous arcs at points $P$ and $Q$.

7. Join $P$ and $Q$ using a ruler. Let this line intersect the circle at points $C$ and $D$.

Solution:

The perpendicular bisector of a diameter always passes through the centre of the circle because the centre is the midpoint of any diameter.

Since the line $PQ$ (the perpendicular bisector) passes through the centre $O$ and connects two points $C$ and $D$ on the boundary of the circle, the segment $CD$ is also a diameter of the circle.

Final Answer: Yes, the perpendicular bisector contains another diameter of the circle.

Given:

An angle $\angle XYZ$ as shown in Fig. 9.22.

To Construct:

The bisector of $\angle XYZ$.

Construction Required:

1. With $Y$ as the centre and any convenient radius, draw an arc that intersects the arms $YX$ and $YZ$ at two points, say $P$ and $Q$ respectively.

2. With $P$ as the centre and a radius more than half of the distance $PQ$, draw an arc in the interior of $\angle XYZ$.

3. With $Q$ as the centre and the same radius, draw another arc that intersects the previous arc at a point, say $S$.

4. Join the vertex $Y$ and the point $S$ with a straight line and extend it to form a ray.

5. The ray $YS$ is the required bisector of $\angle XYZ$.

Final Answer: The ray $YS$ bisects $\angle XYZ$ into two equal parts.

Given:

An angle of $60^\circ$ to be constructed and divided into four equal parts.

Construction Required:

1. Draw a ray $OA$.

2. With $O$ as centre and any radius, draw an arc cutting $OA$ at a point $B$.

3. With $B$ as centre and the same radius, draw an arc cutting the first arc at $C$.

4. Join $OC$. $\angle COA$ is the $60^\circ$ angle.

5. Draw the bisector $OD$ of $\angle COA$. Now $\angle COA$ is divided into two $30^\circ$ parts ($\angle COD$ and $\angle DOA$).

6. Draw the bisector $OE$ of $\angle DOA$.

7. Draw the bisector $OF$ of $\angle COD$.

Solution:

Since the total angle is $60^\circ$ and it has been divided into four equal parts by successive bisection:

$\text{Measure of each part} = \frac{60^\circ}{4}$

$\text{Measure of each part} = 15^\circ$

Final Answer: Upon measurement with a protractor, each of the four parts measures $15^\circ$.

Given:

A straight angle ($180^\circ$).

To Construct:

The bisector of the straight angle.

Construction Required:

1. Draw a line $AB$ and mark a point $O$ on it. $\angle AOB$ is a straight angle ($180^\circ$).

2. With $O$ as centre and any radius, draw a semi-circle cutting $AB$ at $X$ and $Y$.

3. With $X$ as centre and radius more than $OX$, draw an arc above the line.

4. With $Y$ as centre and the same radius, draw another arc intersecting the previous arc at point $C$.

5. Join $OC$. Ray $OC$ is the bisector of the straight angle.

Solution:

A straight angle measures $180^\circ$. Bisection divides the angle into two equal parts.

$\text{Measure of each part} = 90^\circ$

Final Answer: On measuring with a protractor, each part measures $90^\circ$ (a right angle).

Given:

A right angle ($90^\circ$).

Construction Required:

1. Draw a right angle $\angle ABC = 90^\circ$ (using compasses by constructing $60^\circ$ and $120^\circ$ and bisecting their difference).

2. Draw the bisector $BD$ of $\angle ABC$. This divides the $90^\circ$ angle into two equal parts.

3. Now, bisect $\angle ABD$ and $\angle DBC$ using the standard angle bisection method.

Solution:

Step 1: Bisecting the right angle

$\text{Measure of each part} = \frac{90^\circ}{2} = 45^\circ$

Step 2: Bisecting each of these $45^\circ$ parts

$\text{Measure of each resulting part} = \frac{45^\circ}{2}$

$\text{Measure} = 22.5^\circ$

Final Answer: The measure of each part after the first bisection is $45^\circ$. After bisecting those parts again, each final part measures $22.5^\circ$.

Given:

According to Fig. 9.23:

Measure of $\angle ABC = 45^\circ$

Measure of $\angle DBA = 30^\circ$

To Find:

The measure of $\angle DBC$.

Construction Required:

1. Draw a ray $BC$.

2. Construct 45°: At point $B$, construct a $90^\circ$ angle using ruler and compasses by bisecting the straight angle. Now, bisect this $90^\circ$ angle to get ray $BA$. Thus, $\angle ABC = 45^\circ$.

3. Construct 30°: With $BA$ as the initial arm, construct a $60^\circ$ angle using a compass. Bisect this $60^\circ$ angle to get ray $BD$. Thus, $\angle DBA = 30^\circ$.

4. Resulting Angle: Ray $BD$ and ray $BC$ now form $\angle DBC$.

Solution:

From the construction and the given figure, we see that $\angle DBA$ and $\angle ABC$ are adjacent angles sharing the common arm $BA$.

The total angle $\angle DBC$ is the sum of these two angles:

Substituting the values of the angles:

$\angle DBC = 30^\circ + 45^\circ$

$\angle DBC = 75^\circ$

Final Answer: The measure of $\angle DBC$ is $75^\circ$.

Given:

A line segment of length $6\text{ cm}$.

Construction Required:

1. Draw a line segment $AB = 6\text{ cm}$ using a ruler.

2. With $A$ as the centre and radius more than half of $AB$ (i.e., $> 3\text{ cm}$), draw two arcs, one above and one below $AB$.

3. With $B$ as the centre and the same radius, draw two more arcs intersecting the previous arcs at points $P$ and $Q$.

4. Join $P$ and $Q$. Let $PQ$ intersect $AB$ at point $M$.

5. $PQ$ is the perpendicular bisector of $AB$.

Solution:

Since $PQ$ is the perpendicular bisector, it divides the segment $AB$ into two equal parts at point $M$.

$\text{Length of each part} = \frac{\text{Total Length}}{2}$

$\text{Length} = \frac{6}{2} = 3\text{ cm}$

Verification:

On measuring with a ruler, we find:

$AM = 3\text{ cm}$

$MB = 3\text{ cm}$

Final Answer: The two parts of the line segment measure $3\text{ cm}$ each.

Given:

A line segment of length $10\text{ cm}$.

Construction Required:

1. Draw a line segment $PQ = 10\text{ cm}$.

2. Construct the perpendicular bisector of $PQ$ which intersects $PQ$ at point $M$. Now, $PQ$ is divided into two equal parts: $PM$ and $MQ$.

3. Construct the perpendicular bisector of the segment $PM$. Let it intersect $PM$ at point $A$.

4. Construct the perpendicular bisector of the segment $MQ$. Let it intersect $MQ$ at point $B$.

5. The points $A, M,$ and $B$ divide $PQ$ into four equal parts: $PA, AM, MB,$ and $BQ$.

Solution:

The total length of $10\text{ cm}$ is divided into $4$ equal segments.

$\text{Length of each part} = \frac{10\text{ cm}}{4}$

$\text{Length of each part} = 2.5\text{ cm}$

Final Answer: Upon measurement, each of the four equal parts measures $2.5\text{ cm}$.