Q2: Construct the 8-petalled figure shown in Fig. 6.5.

To Construct: An 8-petalled flower figure. Steps of Construction: 1. Draw a horizontal line l and a vertical line m intersecting at a point O. This creates four 90^° angles. 2. Bisect each of the four 90^° angles. This results in four new lines passing through O. Now you have eight rays starting…

Q2: Make your own arch designs.

To Construct: A custom arch design, for example, a Trefoil Arch. A trefoil arch has three lobes (foils). Steps of Construction: 1. Define the base: Draw a horizontal line segment AB which represents the width of the arch. 2. Divide the base: Divide the segment AB into two equal halves at point M.…

Q2: Can you think of different methods to construct a 90^° angle at a given point on a line using a rope?

Answer: Yes, here are two methods to construct a 90^° angle at a given point O on a line using a rope. Method 1: Using the Perpendicular Bisector Principle 1. Let the given point be O on a line l drawn on the ground. 2. Use a rope to measure a certain length from O along the line l to a point X.…

Q2: Construct the following figure.

To Construct: A parallelogram with lines drawn through its vertices parallel to its diagonals. Steps of Construction: 1. Construct the parallelogram: a. Draw a line segment AB. b. From point A, draw another line segment AD at any angle to AB. c. Through point B, construct a line parallel to AD. d.…

Constructions and Tilings Class 7 Mathematics NCERT Solutions

Q1Figure it Out (Angle Bisection for a Design)

Construct at least 4 different angles. Draw their bisectors.

Solution

To Construct: Four different angles and their bisectors.

General Steps for Angle Bisection: Let's assume we have an angle

\angle PQR

with vertex at Q.

Place the compass point at the vertex Q and draw an arc that intersects the two arms of the angle, QP and QR. Let the intersection points be A and B, respectively.
Place the compass point at A and draw an arc in the interior of the angle.
Without changing the compass radius, place the compass point at B and draw another arc that intersects the first arc. Let the point of intersection be C.
Use a ruler to draw a ray from the vertex Q passing through point C. This ray QC is the bisector of $\angle PQR$ .

Construction of 4 Angles and their Bisectors:

Angle 1: Acute Angle (e.g., $60^{\circ}$ )
1. Construct a $60^{\circ}$ angle.
2. Follow the general bisection steps above to bisect it into two $30^{\circ}$ angles.
Angle 2: Right Angle ( $90^{\circ}$ )
1. Construct a perpendicular on a line to get a $90^{\circ}$ angle.
2. Follow the general bisection steps to bisect it into two $45^{\circ}$ angles.
Angle 3: Obtuse Angle (e.g., $120^{\circ}$ )
1. Construct a $120^{\circ}$ angle (by constructing two adjacent $60^{\circ}$ angles).
2. Follow the general bisection steps to bisect it into two $60^{\circ}$ angles.
Angle 4: Straight Angle ( $180^{\circ}$ )
1. Draw a straight line. This represents a $180^{\circ}$ angle.
2. Choose a point O on the line. Follow the bisection steps. The bisector will be a line perpendicular to the original line at point O, creating two $90^{\circ}$ angles.

Q2Figure it Out (Angle Bisection for a Design)

Construct the 8-petalled figure shown in Fig. 6.5.

Solution

To Construct: An 8-petalled flower figure.

Steps of Construction:

Draw a horizontal line l and a vertical line m intersecting at a point O. This creates four $90^{\circ}$ angles.
Bisect each of the four $90^{\circ}$ angles. This results in four new lines passing through O. Now you have eight rays starting from O, with an angle of $45^{\circ}$ between any two adjacent rays.
With O as the center, draw a circle of a suitable radius. Let the points where the eight rays intersect the circle be $P_1, P_2, P_3, P_4, P_5, P_6, P_7, P_8$ .
To construct the petals: a. With $P_1$ as the center and a radius equal to the distance from $P_1$ to O, draw an arc inside the circle connecting points $P_8$ and $P_2$ . (This method creates rounded petals). b. A better method for pointed petals as often drawn is: With $P_2$ as the center and radius $P_2P_1$ , draw an arc. With $P_8$ as the center and radius $P_8P_1$ , draw an arc. The intersection of these arcs will be on the line $OP_1$ . This is complex. c. Let's use a simpler, standard method for this design: i. After getting the 8 rays, take any point A on one ray (say $OP_1$ ). ii. With A as the center and radius AO, draw an arc. iii. Take a corresponding point B on the next ray ( $OP_2$ ) such that OB = OA. With B as center and radius BO, draw another arc. iv. The shape bounded by these arcs and the center O is not a petal.

Let's follow the most common construction for this type of rosette:

Follow steps 1 and 2 to get the eight rays from O at $45^{\circ}$ angles.
Choose a point A on one of the rays.
With O as the center and radius OA, draw a light circle. This circle intersects all eight rays. Let the intersection points be A, B, C, D, E, F, G, H.
With A as the center and radius AO, draw an arc from H to B.
With B as the center and radius BO, draw an arc from A to C.
With C as the center and radius CO, draw an arc from B to D.
Continue this process for all eight points around the circle.

Result: This will create a symmetric 8-petalled flower design centered at O.

Q3Figure it Out (Angle Bisection for a Design)

In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.

Solution

Answer: Yes, the line joining the vertex to the intersection point of arcs drawn on the other side will also be the angle bisector.

Justification: Let the given angle be

\angle XOY

with vertex O.

First, we mark points A on ray OX and B on ray OY such that $OA = OB$ . This is done by drawing an arc from center O.
Next, from centers A and B, we draw arcs of equal radius that intersect at a point C. The ray OC is the angle bisector.
If we draw the arcs from A and B on the other side of the angle, they will intersect at a different point, let's call it D.

Now, let's justify why OD is also the angle bisector. Consider the triangles

\triangle OAD

and

\triangle OBD

.

$OA = OB$ (by construction).
$AD = BD$ (since the arcs were drawn with equal radii from A and B).
$OD$ is common to both triangles.

By the SSS (Side-Side-Side) congruence criterion,

\triangle OAD \cong \triangle OBD

. Since the triangles are congruent, their corresponding parts are equal. Therefore,

\angle AOD = \angle BOD

. This proves that the ray OD is also the bisector of

\angle XOY

. In fact, the points O, C, and D are collinear, and the entire line CD is the line of symmetry for the angle.

Q4Figure it Out (Angle Bisection for a Design)

What are the other angles that can be constructed using angle bisection? Can you construct $65.5^{\circ}$ angle?

Solution

Angles that can be constructed: Using a ruler and compass, we can construct some basic angles like

90^{\circ}

(perpendicular lines) and

60^{\circ}

(equilateral triangle). By applying the angle bisection method repeatedly, and by adding or subtracting these angles, we can construct many other angles.

From a $90^{\circ}$ angle:
- Bisecting $90^{\circ}$ gives $45^{\circ}$ .
- Bisecting $45^{\circ}$ gives $22.5^{\circ}$ .
- Bisecting $22.5^{\circ}$ gives $11.25^{\circ}$ , and so on. (Any angle of the form $\frac{90^{\circ}}{2^n}$ ).
From a $60^{\circ}$ angle:
- Bisecting $60^{\circ}$ gives $30^{\circ}$ .
- Bisecting $30^{\circ}$ gives $15^{\circ}$ .
- Bisecting $15^{\circ}$ gives $7.5^{\circ}$ , and so on. (Any angle of the form $\frac{60^{\circ}}{2^n}$ ).
By combining angles:
- $75^{\circ} = 60^{\circ} + 15^{\circ}$
- $105^{\circ} = 90^{\circ} + 15^{\circ}$
- $120^{\circ} = 60^{\circ} + 60^{\circ}$
- $135^{\circ} = 90^{\circ} + 45^{\circ}$

Can you construct a $65.5^{\circ}$ angle? No, an angle of

65.5^{\circ}

cannot be constructed using only an unmarked ruler and a compass.

Reason: Constructible angles are those that can be formed from a finite number of additions, subtractions, and bisections of the initial constructible angles (

60^{\circ}

and

90^{\circ}

). The angle

65.5^{\circ}

cannot be expressed in this way. For example,

65.5 = 45 + 20.5

or

60 + 5.5

. The angles

20.5^{\circ}

and

5.5^{\circ}

are not constructible from the basic angles through bisection. Therefore,

65.5^{\circ}

is not a constructible angle.

Q5Figure it Out (Angle Bisection for a Design)

Come up with a method to construct the angle bisector using a rope.

Solution

To Construct: The bisector of a given angle

\angle XOY

using a rope.

Steps of Construction:

The angle $\angle XOY$ is marked on the ground with pegs at the vertex O and at points on the arms OX and OY.
Take a piece of rope. Use it to measure a distance from the vertex O along the arm OX. Mark this point as A and place a peg there.
Without changing the length of the rope segment, measure the same distance from O along the arm OY. Mark this point as B and place a peg there. Now, we have $OA = OB$ .
Take another, longer rope. Fasten its two ends to the pegs at A and B.
Find the exact midpoint of this second rope. Hold the midpoint and pull the rope taut, away from the vertex O. Let the position of the midpoint be C.
Stretch a third rope (or use a ruler) to form a straight line from the vertex O to the point C.

Result: The line OC is the angle bisector of

\angle XOY

.

Justification: In the triangles

\triangle OAC

and

\triangle OBC

:

$OA = OB$ (by construction in steps 2 and 3).
$AC = BC$ (because C is the midpoint of the rope connecting A and B, so it is equidistant from A and B).
$OC$ is a common side to both triangles. By the SSS (Side-Side-Side) congruence criterion, $\triangle OAC \cong \triangle OBC$ . Since the triangles are congruent, their corresponding angles are equal. Therefore, $\angle AOC = \angle BOC$ , which means OC bisects $\angle XOY$ .

Q6Figure it Out (Angle Bisection for a Design)

Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square?

Solution

To Construct: A four-petaled figure within a square, where the petals are of maximum size.

Steps of Construction:

Construct a square ABCD using a ruler and compass.
Find the midpoints of the four sides of the square. Let M be the midpoint of AB, N be the midpoint of BC, P be the midpoint of CD, and Q be the midpoint of DA.
To construct petals of the maximum possible size, the circles that form them should be centered on the midpoints of the sides and have a radius equal to half the side length of the square.
With M as the center and radius $MA$ (or $MB$ ), draw a circle.
With N as the center and radius $NB$ (or $NC$ ), draw a circle.
With P as the center and radius $PC$ (or $PD$ ), draw a circle.
With Q as the center and radius $QD$ (or $QA$ ), draw a circle.

Result: The four circles will overlap inside the square. The regions where the circles intersect form a four-petaled flower shape in the center. The petals are rounded and touch each other at the center of the square. This construction ensures the petals are as large as possible because the circles they are part of are tangent to the sides of the square at its vertices.

Q1Figure it Out (Arch Designs)

Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.

Solution

To Construct: A pointed arch.

Steps for a Basic Pointed Arch (Equilateral Arch):

Draw a horizontal line segment XY. This will be the base or opening of the arch.
Place the compass point at X.
Set the compass radius to be equal to the length of the base, XY.
Draw a long arc above the line segment XY.
Now, place the compass point at Y.
Keeping the same radius (equal to YX), draw another arc that intersects the first arc. Let the point of intersection be Z.
The shape formed by the arc XZ and the arc YZ is a pointed arch.

Making Different Arches by Changing the Radius: Let the length of the base be

d

.

Taller/Steeper Arch (Lancet Arch):
- Choose a radius $r$ that is greater than the base length $d$ (i.e., $r > d$ ).
- With X and Y as centers and radius $r$ , draw two intersecting arcs above XY. The resulting arch will be taller and more pointed than the equilateral arch.
Shorter/Wider Arch:
- Choose a radius $r$ that is less than the base length $d$ , but still greater than half the base length (i.e., $d/2 < r < d$ ).
- With X and Y as centers and radius $r$ , draw two intersecting arcs above XY. The resulting arch will be shorter and wider.

By systematically changing the radius

r

used for the construction, you can create a variety of pointed arches with different heights and sharpness.

Q2Figure it Out (Arch Designs)

Make your own arch designs.

Solution

To Construct: A custom arch design, for example, a Trefoil Arch.

A trefoil arch has three lobes (foils).

Steps of Construction:

Define the base: Draw a horizontal line segment AB which represents the width of the arch.
Divide the base: Divide the segment AB into two equal halves at point M. So, $AM = MB$ .
Construct guiding triangles: a. On the segment AM as a side, construct an equilateral triangle $\triangle AMN$ pointing upwards. b. On the segment MB as a side, construct an equilateral triangle $\triangle MBP$ pointing upwards.
Draw the side lobes: a. With N as the center and radius $NA$ (which is equal to AM), draw an arc from A to M. b. With P as the center and radius $PB$ (which is equal to MB), draw an arc from M to B.
Draw the top lobe: a. The points N and P form the base of the top lobe. The center of the top lobe's arc will be the third vertex of an equilateral triangle built on NP. b. Construct an equilateral triangle $\triangle NPQ$ on the segment NP, with Q pointing upwards. c. With Q as the center and radius $QN$ (or $QP$ ), draw an arc connecting point N to point P.

Result: The combination of the three arcs (from A to M, M to B, and N to P) forms a symmetric trefoil arch. One can create other designs by using different shapes for the guiding lines (e.g., squares instead of triangles) or by changing the radii and centers of the arcs.

Q1Figure it Out (Construction Methods in Śulba-Sūtras)

Justify why AB in Fig. 6.4 is the perpendicular bisector.

Solution

To Justify: The line AB constructed using the rope method is the perpendicular bisector of the line segment XY.

Justification: The construction involves a rope of a fixed length, say

L

, with its ends fastened at points X and Y.

Finding point A: The midpoint of the rope is pulled taut to a point A above the line segment XY. Since A is the midpoint of the stretched rope that runs from X to Y, the length of the rope from X to A is equal to the length of the rope from A to Y. Therefore, the distance $AX$ is equal to the distance $AY$ . Any point that is equidistant from the endpoints of a line segment lies on its perpendicular bisector. Thus, point A lies on the perpendicular bisector of XY.
Finding point B: Similarly, the midpoint of the rope is pulled taut to a point B below the line segment XY. For the same reason, the distance $BX$ is equal to the distance $BY$ . Therefore, point B also lies on the perpendicular bisector of XY.
Conclusion: Since both A and B are two distinct points on the perpendicular bisector of XY, the straight line that passes through A and B must be the perpendicular bisector of XY.

Hence Justified.

Q2Figure it Out (Construction Methods in Śulba-Sūtras)

Can you think of different methods to construct a $90^{\circ}$ angle at a given point on a line using a rope?

Solution

Answer: Yes, here are two methods to construct a

90^{\circ}

angle at a given point O on a line using a rope.

Method 1: Using the Perpendicular Bisector Principle

Let the given point be O on a line l drawn on the ground.
Use a rope to measure a certain length from O along the line l to a point X. Place a peg at X.
Measure the same length from O in the opposite direction along l to a point Y. Place a peg at Y. Now, O is the midpoint of the segment XY.
Take a longer rope and fasten its ends to the pegs at X and Y.
Find the midpoint of this second rope and pull it taut to a point A. The line AO will be the perpendicular bisector of XY.
Since the line AO passes through the midpoint O of XY and is perpendicular to XY, the angle $\angle AOX$ (or $\angle AOY$ ) is $90^{\circ}$ .

Method 2: Using the 3-4-5 Pythagorean Triple

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides ( $a^2 + b^2 = c^2$ ). A triangle with side lengths 3, 4, and 5 units satisfies this condition ( $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ).
Take a long rope and mark it into 12 equal sections (since $3+4+5=12$ ).
Let the given point where the $90^{\circ}$ angle is needed be O on a line l.
From O, measure 3 units along the line l to a point P. Place a peg at P.
Hold the rope such that the start (0 mark) and end (12th mark) are at point P.
Hold the 3rd mark of the rope at point O.
Take the 7th mark (which is 4 units from the 3rd mark) and stretch the rope taut. Let this point be Q.
The rope now forms a triangle OPQ with side lengths OP = 3 units, OQ = 4 units, and PQ = 5 units.
Since this is a 3-4-5 triangle, it is a right-angled triangle with the right angle at vertex O.
Therefore, $\angle QOP = 90^{\circ}$ .

Q1Figure it Out (Construction of a Line Parallel to the Given Line)

Construct 4 pairs of parallel lines in different orientations.

Solution

To Construct: Four pairs of parallel lines.

General Method (using corresponding angles):

Draw a line and a point: Draw a line m and choose a point P that is not on m.
Draw a transversal: Draw a straight line t that passes through point P and intersects line m at a point Q.
Copy the angle: The transversal t forms an angle with line m at point Q. We need to copy this angle at point P. Use the standard angle copying procedure to construct an angle at P, on the same side of the transversal t, that is equal to the angle at Q.
Draw the parallel line: Draw a line n through point P along the arm of the newly constructed angle. Since the corresponding angles are equal, line n is parallel to line m.

Task: Repeat this construction four times with lines in different orientations:

A pair of horizontal parallel lines.
A pair of vertical parallel lines.
A pair of parallel lines slanting upwards to the right.
A pair of parallel lines slanting downwards to the right.

Q2Figure it Out (Construction of a Line Parallel to the Given Line)

Construct the following figure.

Solution

To Construct: A parallelogram with lines drawn through its vertices parallel to its diagonals.

Steps of Construction:

Construct the parallelogram: a. Draw a line segment AB. b. From point A, draw another line segment AD at any angle to AB. c. Through point B, construct a line parallel to AD. d. Through point D, construct a line parallel to AB. e. The intersection of these two new lines is the fourth vertex, C. ABCD is the required parallelogram.
Draw the diagonals: a. Use a ruler to join A and C to form the diagonal AC. b. Join B and D to form the diagonal BD.
Construct the outer parallel lines: a. Through vertex A, construct a line parallel to the diagonal BD. b. Through vertex C, construct a line parallel to the diagonal BD. c. Through vertex B, construct a line parallel to the diagonal AC. d. Through vertex D, construct a line parallel to the diagonal AC.

Result: The four newly constructed lines will form a larger parallelogram that encloses the original parallelogram ABCD.

Q1Figure it Out (Construction of Perpendicular Bisector)

When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer. [Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector. Hint 2: We can draw the whole line if any two of its points are known.]

Solution

Answer: No, it is not necessary to use the same radius for the arcs above and below the line segment XY.

Justification: The perpendicular bisector of a line segment XY is the line containing all points that are equidistant from X and Y. To define a line, we only need two distinct points.

Finding the first point (A): We can find a point A above XY that is equidistant from X and Y. To do this, we draw an arc with center X and a certain radius, let's say $r_1$ . Then we draw another arc with center Y and the same radius $r_1$ . The intersection of these arcs gives us point A, such that $AX = AY = r_1$ . Since A is equidistant from X and Y, it lies on the perpendicular bisector.
Finding the second point (B): We can find a second point B below XY that is also equidistant from X and Y. We can choose a different radius, say $r_2$ , for this construction. We draw an arc with center X and radius $r_2$ , and another arc with center Y and radius $r_2$ . The intersection of these arcs gives us point B, such that $BX = BY = r_2$ . Since B is also equidistant from X and Y, it also lies on the perpendicular bisector.
Constructing the line: Since both points A and B lie on the perpendicular bisector, the straight line passing through A and B is the perpendicular bisector of XY.

Therefore, we can use different radii (

r_1

and

r_2

) to find the two points, as long as for each point, the two arcs used to find it are of the same radius.

Q2Figure it Out (Construction of Perpendicular Bisector)

Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.

Solution

Answer: No, it is not necessary to construct the arcs on opposite sides of XY. Both pairs of arcs can be constructed on the same side.

Justification: To construct the perpendicular bisector, we need two distinct points that are equidistant from X and Y.

Finding the first point (A): Choose a radius $r_1$ (which must be greater than half the length of XY). With X and Y as centers, draw two arcs of radius $r_1$ on the same side of XY (e.g., above XY). Let their intersection be point A. By construction, $AX = AY = r_1$ , so A is on the perpendicular bisector.
Finding the second point (C): Choose a different radius $r_2$ (also greater than half the length of XY, and $r_2 \neq r_1$ ). With X and Y as centers, draw two arcs of radius $r_2$ on the same side of XY. Let their intersection be point C. By construction, $CX = CY = r_2$ , so C is also on the perpendicular bisector.
Constructing the line: Since we have two distinct points, A and C, that both lie on the perpendicular bisector, the line passing through A and C is the perpendicular bisector of XY.

Therefore, it is possible to construct the perpendicular bisector by finding two points on the same side of the line segment.

Q3Figure it Out (Construction of Perpendicular Bisector)

While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them ? Explore this through construction, and then justify your answer.

Solution

Answer: Yes, it is absolutely necessary to use the same radius for the two arcs that intersect to find a single point on the perpendicular bisector.

Justification: A point lies on the perpendicular bisector of a segment XY if and only if it is equidistant from both X and Y. Let's say we are trying to find a point A.

If we use a radius $r_1$ from center X, any point on that arc is at a distance $r_1$ from X. So, $AX = r_1$ .
If we use a different radius $r_2$ from center Y, any point on that arc is at a distance $r_2$ from Y. So, $AY = r_2$ .

If the radii are different (

r_1 \neq r_2

), then at the point of intersection A, we would have

AX \neq AY

. A point that is not equidistant from X and Y does not lie on the perpendicular bisector.

To ensure that the intersection point A is equidistant from X and Y (i.e.,

AX = AY

), we must use the same radius for both arcs. This ensures that

AX = AY = r

, satisfying the condition for the point to be on the perpendicular bisector.

Q4Figure it Out (Construction of Perpendicular Bisector)

Recreate this design using only a ruler and compass -

Solution

To Construct: The described geometric design, which is a four-petal flower shape inside a square.

Steps of Construction:

Use a ruler to draw a horizontal line segment. Let's call its endpoints P and Q.
Construct the perpendicular bisector of the segment PQ. Let the bisector be line l, and let it intersect PQ at point O.
With O as the center, and a suitable radius (e.g., OP), draw a circle. This circle will pass through P and Q.
Let the circle intersect the perpendicular bisector l at points R and S. The points P, R, Q, and S form the vertices of a square inscribed in the circle.
To create the four petals: a. With P as the center and radius PO, draw an arc that connects S and R. b. With R as the center and radius RO, draw an arc that connects P and Q. c. With Q as the center and radius QO, draw an arc that connects R and S. d. With S as the center and radius SO, draw an arc that connects Q and P.

Result: The resulting figure is a four-petaled flower-like design, where each petal is formed by two arcs, and the entire design is symmetric within a square.

Q1Figure it Out (Related Constructions)

Construct the following figures:

(a)

A 6-pointed star

(b)

A curved, six-petaled flower

(c)

A pattern of seven intersecting circles

(d)

An interwoven pattern based on a hexagon

(e) A square with quarter circles

Solution

(a) Construction of a 6-pointed star:

Draw a circle with center O and any radius.
Pick a point A on the circumference. With the same radius, step off six points around the circle: A, B, C, D, E, F.
Join the points to form a regular hexagon ABCDEF.
Join alternate vertices to form an equilateral triangle $\triangle ACE$ .
Join the other set of alternate vertices to form another equilateral triangle $\triangle BDF$ .
The resulting figure is a 6-pointed star (Star of David).

(b) Construction of a six-petaled flower:

Construct a regular hexagon ABCDEF with center O.
With vertex A as the center and the side length AB as the radius, draw an arc connecting the adjacent vertices F and B.
With vertex B as the center and the same radius, draw an arc connecting A and C.
Repeat this for all six vertices (C, D, E, F). The six arcs will meet at the center O and form the flower.

(c) Construction of seven intersecting circles (Seed of Life):

Draw a circle with center O and radius r.
Choose any point P on its circumference.
With P as the center and the same radius r, draw a second circle. It will pass through O.
This second circle intersects the first one at two new points. Choose one of these, say Q.
With Q as the center and radius r, draw a third circle.
Continue this process, always choosing a new intersection point on the first circle as the center for the next, until six circles have been drawn around the central one.

(d) Construction of an interwoven hexagonal pattern:

Construct a regular hexagon ABCDEF.
For each side of the hexagon (AB, BC, etc.), draw a semicircle inside the hexagon using the side as its diameter.
This will create a flower-like shape with six rounded petals.
To create the interwoven (Celtic knot) effect, you must carefully erase small segments of the semicircles where they cross to make it appear as if the bands are weaving over and under each other.

(e) Construction of a square with quarter circles:

Construct a square ABCD.
Find the midpoints of the four sides: M (of AB), N (of BC), P (of CD), and Q (of DA).
Join MP and NQ. They intersect at the center O and divide the large square into four smaller congruent squares: AMOQ, MBNO, ONCP, and QOPD.
In the top-left small square (AMOQ), place the compass at vertex A and with radius AM, draw a quarter-circle arc from M to Q.
In the top-right square (MBNO), place the compass at vertex B and with radius BM, draw a quarter-circle arc from M to N.
In the bottom-right square (ONCP), place the compass at vertex C and with radius CN, draw a quarter-circle arc from N to P.
In the bottom-left square (QOPD), place the compass at vertex D and with radius DQ, draw a quarter-circle arc from Q to P.

Q2Figure it Out (Related Constructions)

Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

Solution

Observation: The figure, known as the Café Wall Illusion, appears distorted. Although it is constructed from a grid of perfectly straight horizontal and vertical lines, the horizontal lines appear to be tilted or sloped, and the squares can look like they are bulging.

How it happens: This optical illusion is caused by the interaction of several visual cues:

High Contrast: The alternating black and white squares create strong visual contrast.
Staggered Rows: The rows are offset from each other by half a square's width.
Mortar Lines: The thin lines between the squares (the 'mortar') are crucial. When this line is a neutral color (like grey) or has a specific brightness relative to the black and white squares, our brain's edge-detection mechanisms get confused. The brain interprets the segments of the horizontal lines differently based on whether they are bordered by black or white squares, leading to the perception of a tilt.

Steps to Recreate the Illusion:

Use a ruler to draw a series of parallel horizontal lines.
Draw a series of parallel vertical lines to create a grid of squares.
In the first row, color the squares alternately black and white.
In the second row, shift the pattern by half a square. Start with a white square below the first row's black square, and so on. Continue this staggered coloring for all rows.
It is important that the grid lines (the mortar) are visible. Using a pencil for the grid and then coloring in the squares will naturally leave the pencil lines as the mortar, creating the illusion.

Q3Figure it Out (Related Constructions)

Construct this figure. [Hint: Find the angles in this figure.]

Solution

To Construct: A five-pointed star (pentagram) inside a regular pentagon.

Note: The exact construction of a regular pentagon with only a ruler and compass is an advanced geometric procedure not typically covered at this level. The textbook itself mentions that this topic will be discussed in later years. An approximate construction can be done using a protractor.

Construction using a Protractor (for approximation):

Draw a circle with center O.
The central angle of a regular pentagon is $360^{\circ} \div 5 = 72^{\circ}$ .
Draw a radius OA.
Using a protractor, measure $72^{\circ}$ from OA and draw another radius OB. Continue this process to get five radii OA, OB, OC, OD, OE, each separated by $72^{\circ}$ .
Join the points A, B, C, D, and E on the circumference to form the regular pentagon ABCDE.

Construction of the Star:

Once the regular pentagon ABCDE is constructed, draw its diagonals.
Join vertex A to the non-adjacent vertices C and D.
Join vertex B to the non-adjacent vertices D and E.
Join vertex C to the non-adjacent vertices E and A (already drawn).
The five diagonals (AC, BD, CE, DA, EB) will intersect to form the five-pointed star inside the pentagon.

Q4Figure it Out (Related Constructions)

Draw a line $l$ and mark a point P anywhere outside the line. Construct a perpendicular to the given line $l$ through P.

Solution

To Construct: A perpendicular to a line l from an external point P.

Steps of Construction:

Given a line l and a point P not on l.
Place the compass point at P.
Open the compass to a radius that is large enough to cross the line l at two distinct points. Draw an arc that intersects line l at two points. Let's call these points A and B.
Now, the segment AB is on the line l, and the point P is equidistant from A and B (since $PA = PB$ , both being radii of the same arc).
We can now construct the perpendicular bisector of the segment AB, which we know will pass through P.
Place the compass point at A. Set its radius to be more than half the distance of AB. Draw an arc on the side of line l opposite to P.
Keeping the same radius, place the compass point at B and draw another arc that intersects the arc from step 6. Let the intersection point be Q.
Use a ruler to draw a straight line passing through points P and Q.

Result: The line PQ is the required line. It passes through the external point P and is perpendicular to the given line l.

Q1Figure it Out (Repeating Units and Repeating Angles)

Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

Solution

To Construct: Four arbitrary angles and their copies.

General Steps for Copying an Angle: Let the given angle be

\angle PQR

and we want to copy it to a new vertex S on a ray ST.

Draw the original angle: Use a ruler to draw two rays from a vertex Q to form an angle $\angle PQR$ .
Start the copy: Draw a ray ST, which will be one arm of the new angle.
Draw first arc: Place the compass point at vertex Q of the original angle and draw an arc that intersects both arms QP and QR. Let the intersection points be A and B.
Draw second arc: Without changing the compass radius, place the compass point at the new vertex S and draw a similar arc that intersects the ray ST at a point, say U.
Measure the opening: Set the compass width to the distance between points A and B on the original angle.
Mark the opening: Place the compass point at U and draw an arc that intersects the arc drawn in step 4. Let the intersection point be V.
Complete the angle: Draw a ray from S passing through V. The resulting angle $\angle VST$ is an exact copy of $\angle PQR$ .

Task: Perform the above steps for four different angles drawn freely with a ruler:

An acute angle pointing upwards.
An obtuse angle pointing to the right.
A reflex angle (an angle greater than $180^{\circ}$ ) pointing downwards.
An acute angle pointing to the left.

Q2Figure it Out (Repeating Units and Repeating Angles)

Construct the Fig. 6.6.

Solution

To Construct: The repeating V-shape pattern shown in the figure.

Steps of Construction:

Draw a baseline: Use a ruler to draw a long horizontal line l.
Create the first unit: a. Choose a point A on the line l. b. At point A, draw a ray AP that makes an acute angle with the line l (e.g., construct a $60^{\circ}$ angle or just draw an arbitrary one). c. Choose a fixed length for the arm of the V-shape, say 5 cm. With A as the center and a radius of 5 cm, draw an arc to cut the ray AP at point Q. d. With Q as the center and the same radius (5 cm), draw an arc to cut the line l at a point B. Join AQ and QB. The shape AQB is the first V-unit. e. A simpler way to define the unit is just the angle $\angle PAB$ . The line segment AB will be the base of our repeating unit.
Copy the unit: a. Now, at vertex B on the line l, we need to place the next V-shape. We do this by copying the angle $\angle PAB$ . b. At B, construct an angle $\angle RBC$ equal to $\angle PAB$ , where R is a point such that the angle opens in the same direction as the first one. c. The pattern is formed by continuing this process: at point C, copy the angle again, and so on.
Construct the inverted pattern: a. At point A, construct the angle vertically opposite to $\angle PAB$ . Let this be $\angle SAT$ . b. Copy this angle at points B, C, and so on, below the line l.

Result: This process creates a continuous, repeating pattern of V-shapes both above and below the baseline l.

Q1Figure it Out (Tangrams)

How can the tangram pieces be rearranged to form each of the following figures?

Solution

Answer: This is a set of spatial reasoning puzzles. The goal is to use all seven tangram pieces (two large triangles, one medium triangle, two small triangles, one square, and one parallelogram) without any overlap to form the given silhouettes. Since providing a solution for each image is not feasible without seeing them, here is a general strategy and a descriptive example for one common shape, a cat.

General Strategy:

Identify large parts: Look for large, simple geometric shapes within the silhouette, such as a triangular body or a square head. Try to fit the larger tangram pieces (the two large triangles, the square) there first.
Use corners and edges: Pay attention to the corners and straight edges of the silhouette. The right-angled corners of the triangles and the square are useful for forming these features.
Fill the gaps: Use the smaller and more unusually shaped pieces (parallelogram, small triangles) to fill in the remaining gaps or to form more complex parts like limbs or tails.
Trial and Error: Tangram puzzles often require experimentation. Don't be afraid to try different arrangements until the pieces fit correctly.

Example Solution for a Stylized Cat Figure:

Body: The two large triangles can be placed together to form the main torso and hind legs.
Head: The square piece often works well as the head.
Ears/Paws: The two small triangles are versatile and can be used for features like ears or front paws.
Neck/Shoulders: The medium triangle can connect the head to the body.
Tail: The parallelogram is uniquely shaped and is perfect for forming a tail at an angle.

Q1Figure it Out (Tiling)

Are the following tilings possible? An 8x8 chessboard with two opposite corners removed, to be tiled by 2x1 dominoes.

Solution

Answer: No, this tiling is not possible.

Reasoning (Coloring Argument):

A standard 8x8 chessboard has 64 squares.
These squares are colored in an alternating pattern of black and white. There are 32 black squares and 32 white squares.
The two opposite corners of a chessboard are always of the same color. For example, both might be white.
When we remove two opposite corners (both white), the remaining board has $64 - 2 = 62$ squares.
The number of squares of each color is now unequal. We are left with 32 black squares and $32 - 2 = 30$ white squares.
A single 2x1 domino, regardless of whether it is placed horizontally or vertically, must always cover exactly one black square and one white square.
To tile the entire 62-square region, we would need $62 \div 2 = 31$ dominoes.
These 31 dominoes would cover 31 black squares and 31 white squares.
However, our modified board has 32 black squares and 30 white squares. Since the number of black and white squares is not equal, it is impossible to cover the board with dominoes that always cover one of each color.

Conclusion: The tiling is impossible.

Q2Figure it Out (Tiling)

Are the following tilings possible? A region to be tiled by T-trominoes.

Solution

Answer: Whether the tiling is possible depends on the specific shape of the region. We can use general principles to determine possibility for common shapes.

A T-tromino is a shape made of 4 squares in the form of a 'T'.

Necessary Conditions for Tiling:

Area: The total area of the region (number of unit squares) must be a multiple of 4, since each T-tromino covers 4 squares.
Coloring: A more advanced coloring argument can often prove impossibility.

Analysis for a common example, a 10x10 grid:

Area: A 10x10 grid has 100 squares. Since $100$ is divisible by 4, the tiling is not ruled out by the area condition. We would need $100 \div 4 = 25$ T-trominoes.
Coloring Argument:
1. Color the 10x10 grid like a standard chessboard.
2. A 10x10 grid has 50 black squares and 50 white squares.
3. Consider a T-tromino. When placed on the board, it can be in one of two configurations regarding color:
  - It can cover 3 squares of one color and 1 of the other (e.g., 3 black, 1 white).
  - It can cover 1 square of one color and 3 of the other (e.g., 1 black, 3 white).
4. Let $n_1$ be the number of trominoes covering (3 black, 1 white) and $n_2$ be the number covering (1 black, 3 white). The total number of trominoes is 25, so $n_1 + n_2 = 25$ .
5. The total number of black squares covered must be 50: $3n_1 + 1n_2 = 50$ .
6. The total number of white squares covered must be 50: $1n_1 + 3n_2 = 50$ .
7. From the last two equations, we can see that $3n_1 + n_2 = n_1 + 3n_2$ , which simplifies to $2n_1 = 2n_2$ , or $n_1 = n_2$ .
8. If we substitute $n_1 = n_2$ into the first equation ( $n_1 + n_2 = 25$ ), we get $2n_1 = 25$ , or $n_1 = 12.5$ .
9. The number of tiles, $n_1$ , must be a whole number. Since we arrived at a non-integer value, our initial assumption that a tiling exists must be false.

Conclusion: It is impossible to tile a 10x10 grid with T-trominoes. For any other shape, a similar analysis of its area and coloring properties would be required.

Constructions and TilingsClass 7 Mathematics NCERT Solutions

Construct at least 4 different angles. Draw their bisectors.

Solution

Construct the 8-petalled figure shown in Fig. 6.5.

Solution

In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.

Solution

What are the other angles that can be constructed using angle bisection? Can you construct 65.5∘65.5^{\circ}65.5∘ angle?

Solution

Come up with a method to construct the angle bisector using a rope.

Solution

Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square?

Solution

Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.

Solution

Make your own arch designs.

Solution

Justify why AB in Fig. 6.4 is the perpendicular bisector.

Solution

Can you think of different methods to construct a 90∘90^{\circ}90∘ angle at a given point on a line using a rope?

Solution

Construct 4 pairs of parallel lines in different orientations.

Solution

Construct the following figure.

Solution

Solution

Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.

Solution

While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them ? Explore this through construction, and then justify your answer.

Solution

Recreate this design using only a ruler and compass -

Solution

Construct the following figures:

Solution

Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

Solution

Construct this figure. [Hint: Find the angles in this figure.]

Solution

Draw a line lll and mark a point P anywhere outside the line. Construct a perpendicular to the given line lll through P.

Solution

Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

Solution

Construct the Fig. 6.6.

Solution

How can the tangram pieces be rearranged to form each of the following figures?

Solution

Are the following tilings possible? An 8x8 chessboard with two opposite corners removed, to be tiled by 2x1 dominoes.

Solution

Are the following tilings possible? A region to be tiled by T-trominoes.

Solution

What are the other angles that can be constructed using angle bisection? Can you construct $65.5^{\circ}$ angle?

Can you think of different methods to construct a $90^{\circ}$ angle at a given point on a line using a rope?

Draw a line $l$ and mark a point P anywhere outside the line. Construct a perpendicular to the given line $l$ through P.