Chapter 6 Constructions and Tilings (Class 7 - Latest Maths NCERT (Ganita Prakash II) Solutions)

To Explore: If different radii can be used for the arcs above and below the line segment $XY$ while constructing a perpendicular bisector.

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Solution:

No, it is not necessary to have the same radius for the arcs above and below the line $XY$.

Justification:

To draw a line, we only need to identify any two distinct points on that line. In the construction of a perpendicular bisector, any point that is equidistant from the two endpoints $X$ and $Y$ will lie on the bisector.

If we use a radius $r_1$ to draw arcs from $X$ and $Y$ that intersect at point $P$ above the line, then $PX = PY = r_1$. Thus, $P$ is on the bisector.

If we use a different radius $r_2$ to draw arcs from $X$ and $Y$ that intersect at point $Q$ below the line, then $QX = QY = r_2$. Thus, $Q$ is also on the bisector.

The line passing through $P$ and $Q$ will be the perpendicular bisector of $XY$, even if $r_1 \neq r_2$.

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To Explore: If both pairs of arcs can be constructed on the same side of the line segment $XY$.

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Solution:

Yes, it is possible to construct both pairs of intersecting arcs on the same side of $XY$.

Justification:

A straight line is uniquely determined by any two points. If we construct two points $P$ and $Q$ such that both are on the same side of the line segment $XY$, and both points are equidistant from $X$ and $Y$, then both points will lie on the perpendicular bisector.

To do this, we draw the first pair of arcs with a radius $r_1$ to find point $P$. Then, we draw a second pair of arcs with a different radius $r_2$ (much larger or smaller) to find point $Q$ on the same side. Joining $P$ and $Q$ and extending the line will give us the perpendicular bisector of $XY$.

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To Explore: If a single pair of arcs must have the same radius.

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Solution:

Yes, it is absolutely necessary to use the same radius for both arcs in a single intersecting pair.

Justification:

Let $P$ be the intersection point of two arcs, one centered at $X$ with radius $r_X$ and another centered at $Y$ with radius $r_Y$.

For point $P$ to lie on the perpendicular bisector, it must be equidistant from $X$ and $Y$, which means $PX = PY$.

Since $PX = r_X$ and $PY = r_Y$, the condition $PX = PY$ can only be satisfied if $r_X = r_Y$.

If we use different radii ($r_X \neq r_Y$), the intersection point will be closer to one endpoint than the other, and therefore, it will not lie on the perpendicular bisector of the segment.

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Construction Required: A four-petal floral design using a ruler and compass.

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Steps of Construction:

1. Draw a line segment and construct its perpendicular bisector to get two perpendicular lines intersecting at center $O$.

2. Draw a circle with center $O$ and any radius $r$. Let it intersect the perpendicular lines at points $A, B$ (horizontal) and $C, D$ (vertical).

3. Keeping the same radius $r$, place the compass pointer at $A$ and draw an arc passing through $O$ to $C$ and $D$.

4. Repeat the process by placing the pointer at $B$, $C$, and $D$ in turn, drawing arcs that pass through the center $O$.

5. The overlapping arcs will form four petals centered at $O$.

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Given:

In the given figure, $XAY$ and $XBY$ are two positions of a rope. Points $A$ and $B$ are the midpoints of the rope.

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To Prove:

$AB$ is the perpendicular bisector of the line segment $XY$.

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Proof:

In $\triangle AXB$ and $\triangle AYB$:

$\therefore \triangle AXB \cong \triangle AYB$ by SSS congruence criterion.

Now, in $\triangle AXO$ and $\triangle AYO$ (where $O$ is the intersection of $AB$ and $XY$):

$\therefore \triangle AXO \cong \triangle AYO$ by SAS congruence criterion.

From the congruence of these triangles, we get:

Since $XY$ is a straight line:

Since $OX = OY$ (bisecting property) and $\angle XOA = 90^\circ$ (perpendicular property), we can conclude that $AB$ is the perpendicular bisector of $XY$.

Using the 3-4-5 Principle (Pythagorean Triple)

In this construction, we use the principle that if the sides of a triangle are in the ratio $3 : 4 : 5$, then the angle opposite to the longest side ($5$ units) is $90^\circ$. This is a well-known method used since ancient times in India (as seen in the Sulba Sutras) for land measurement and altar construction.

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Construction Required:

1. Draw a line $XY$ and take any point $A$ on it where the $90^\circ$ angle is to be constructed.

2. Fix a small pole at point $A$.

3. Take a rope and mark it at $0$ units, $3$ units, $8$ units, and $12$ units.

4. Attach the $0$ unit mark and $12$ unit mark of the rope together at the pole at point $A$.

5. Stretch the rope along the line $XY$ and fix the $3$ unit mark at a point $B$ with another pole.

6. Now, hold the $8$ unit mark of the rope and pull it away from the line $XY$ until both parts of the rope ($BC$ and $AC$) are tight.

7. Place a pole at this point and label it $C$.

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Solution:

In $\triangle ABC$, the lengths of the sides are calculated as follows:

Here, the longest side is $BC = 5$ units.

According to the 3-4-5 principle:

The angle opposite to the longest side ($BC$) is $\angle BAC$ (or $\angle A$).

$\therefore \angle BAC = 90^\circ$.

Thus, the line $AC$ is perpendicular to the line $XY$ at the point $A$.

To Construct:

Four different angles ($60^\circ$, $90^\circ$, $120^\circ$, and $180^\circ$) and their respective angle bisectors.

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General Steps of Construction:

1. Draw a ray $OA$.

2. Using a protractor or compass, construct the required angle $\angle AOB$.

3. With $O$ as the center and any convenient radius, draw an arc to intersect the rays $OA$ and $OB$ at points $P$ and $Q$ respectively.

4. With $P$ as the center and a radius more than half of $PQ$, draw an arc in the interior of $\angle AOB$.

5. With $Q$ as the center and the same radius, draw another arc to intersect the previous arc at point $C$.

6. Join $OC$. The ray $OC$ is the required angle bisector such that:

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1. Bisecting a $60^\circ$ Angle

When we bisect a $60^\circ$ angle, we obtain two angles of $30^\circ$ each.

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2. Bisecting a $90^\circ$ Angle

When we bisect a $90^\circ$ (Right) angle, we obtain two angles of $45^\circ$ each.

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3. Bisecting a $120^\circ$ Angle

When we bisect a $120^\circ$ (Obtuse) angle, we obtain two angles of $60^\circ$ each.

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4. Bisecting a $180^\circ$ Angle

When we bisect a $180^\circ$ (Straight) angle, we obtain two angles of $90^\circ$ each, which results in a perpendicular line.

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Conclusion: By following the steps of construction, we can successfully bisect any given angle into two equal parts using a compass and a straightedge.

To Construct: An $8$-petalled floral figure as shown in Fig. 6.5.

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Steps of Construction:

1. Draw a circle with center $O$ and a suitable radius $r$.

2. Draw two diameters that are perpendicular to each other. This divides the circle into $4$ equal parts, intersecting the circumference at $4$ points.

3. Bisect each of the four $90^\circ$ angles formed at the center. This will result in $8$ equally spaced points on the circumference of the circle, each at an angle of $45^\circ$ from the next.

4. Let these $8$ points on the circumference be $P_1, P_2, P_3, P_4, P_5, P_6, P_7,$ and $P_8$.

5. Now, use the original radius $r$ of the circle. Place the compass at each of these $8$ points and draw arcs that pass through the center $O$.

6. The overlapping arcs formed by these $8$ centers will create the $8$-petalled design.

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To Justify: Whether drawing arcs on the "other side" (opposite to the interior of the angle) results in a line that is still the angle bisector.

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Solution:

Yes, the line $OC$ will still be the angle bisector.

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Justification:

1. In the figure, points $A$ and $B$ are marked on the two rays such that they are equidistant from the vertex $O$. Hence, $OA = OB$.

2. Arcs of equal radius are drawn from $A$ and $B$, intersecting at point $C$ on the side opposite to the angle. Therefore, $AC = BC$.

3. Consider $\triangle$OAC and $\triangle$OBC:

4. By the SSS (Side-Side-Side) congruence criterion, $\triangle$OAC $\cong$ $\triangle$OBC.

5. Consequently, $\angle$AOC = $\angle$BOC (by CPCT).

6. Since the angles are equal, the ray $OC$ bisects the "reflex" angle. By extending the line $CO$ through the vertex into the interior of $\angle$XOY, we get the standard angle bisector.

To Find: Other constructible angles and the possibility of constructing a $65.5^\circ$ angle.

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Solution:

Angle bisection allows us to construct any angle that is half of a known constructible angle. By repeatedly bisecting standard angles like $90^\circ$, $60^\circ$, and $120^\circ$, we can obtain a wide variety of angles.

1. Angles derived from $90^\circ$:

$\frac{90^\circ}{2} = 45^\circ$

$\frac{45^\circ}{2} = 22.5^\circ$

$\frac{22.5^\circ}{2} = 11.25^\circ$

2. Angles derived from $60^\circ$:

$\frac{60^\circ}{2} = 30^\circ$

$\frac{30^\circ}{2} = 15^\circ$

$\frac{15^\circ}{2} = 7.5^\circ$

We can also construct angles by adding or subtracting these values (e.g., $60^\circ + 15^\circ = 75^\circ$, or $90^\circ + 45^\circ = 135^\circ$).

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Regarding $65.5^\circ$ Angle:

To construct an angle of $65.5^\circ$ through bisection, we would need to start from an angle of:

$65.5^\circ \times 2 = 131^\circ$

Since $131^\circ$ is not an angle that can be constructed using a standard ruler and compass, we cannot construct a $65.5^\circ$ angle using standard angle bisection methods.

Given:

Let $\angle XOY$ be the given angle that needs to be bisected using a rope.

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Construction Required:

1. Fix a small pole at the vertex point $O$.

2. Take a rope and make a loop at one end. Mark a point at a fixed distance on the rope to represent a constant radius.

3. Fix the loop of the rope at the pole at $O$ and rotate the rope from ray $OX$ to ray $OY$.

4. Mark points $A$ (on $OX$) and $B$ (on $OY$) where the fixed distance mark on the rope meets the rays.

5. Fix small poles at points $A$ and $B$.

6. Take another piece of rope and make loops on both ends. Fix these loops to the poles at $A$ and $B$.

7. Mark the midpoint of this rope and hold the rope at that midpoint. Pull the rope away from $O$ until both segments of the rope are tight.

8. Label this point as $M$.

9. Join $AM$, $BM$, and $OM$.

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Proof / Solution:

In $\triangle OAM$ and $\triangle OBM$:

$\therefore \triangle OAM \cong \triangle OBM$ by SSS (Side-Side-Side) congruence rule.

By Corresponding Parts of Congruent Triangles (C.P.C.T.):

Since $\angle AOM$ is part of $\angle XOM$ and $\angle BOM$ is part of $\angle YOM$, we have:

Thus, $OM$ is the bisector of the given angle $\angle XOY$.

Constructions and Tilings Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 6 Page 144 Q6

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To Construct:

A figure containing four petals of maximum possible size within a square.

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Construction Required:

1. Step 1 (Constructing the Square): Draw a straight line and mark two points $A$ and $B$ on it. The distance $AB$ will be the side of our square.

2. Using a compass, construct perpendicular lines at point $A$ and point $B$. To do this, draw semicircles at $A$ and $B$ to get points $C, D$ and $E, F$. Draw intersecting arcs from these points to find points $G$ and $H$ respectively.

3. Join $AG$ and $BH$. On these perpendicular lines, mark points $I$ and $J$ such that $AI = BJ = AB$.

4. Join $I$ and $J$ to complete the square $ABJI$.

5. Step 2 (Finding Midpoints): Use a ruler and compass to find the perpendicular bisectors of sides $AB, BJ, JI,$ and $IA$ to locate their midpoints. Let these midpoints be $K, L, M,$ and $N$ respectively.

6. Step 3 (Drawing Petals): With point $K$ (midpoint of $AB$) as the center and $AK$ as the radius, draw a semicircle inside the square.

7. Repeat this process by drawing semicircles with centers at $L, M,$ and $N$ inside the square, all with the same radius equal to half the side of the square.

8. Step 4 (Finishing): Erase any extra construction lines and arcs to reveal the four-petal design.

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Observation and Conclusion:

The petals are formed by the overlapping regions of the four semicircles. To ensure the petals are of maximum possible size, we must ensure that:

1. The diameter of each semicircle is equal to the side of the square.

2. The semicircles are drawn using the midpoints of the sides of the square as centers.

Because each semicircle spans the full width or height of the square, the resulting petals occupy the maximum available space within the boundary of the square.

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Final Figure:

Constructions and Tilings Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 6 Page 147 Q1

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To Construct: Four different angles in various orientations and creating their exact copies using a ruler and compass (without numerical measurement).

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1. Copying Angle (i): $\angle ABC$

Step 1: Draw a ray $B'C'$ in the desired direction for the new angle.

Step 2: With vertex $B$ of the original angle as center and any radius, draw an arc intersecting $BA$ at $M$ and $BC$ at $N$.

Step 3: Using the same radius, draw an arc from $B'$ intersecting $B'C'$ at $N'$.

Step 4: Measure the distance $MN$ using the compass. From $N'$, draw an arc of radius $MN$ to cut the previous arc at $M'$. Join $B'M'$.

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2. Copying Angle (ii): $\angle DEF$

Step 1: Draw a ray $E'F'$ along the required orientation. With $E$ as center, draw an arc cutting $ED$ and $EF$ at $O$ and $P$.

Step 2: With $E'$ as center and the same radius, draw an arc cutting $E'F'$ at $P'$. Transfer length $OP$ to the new arc to find $O'$. Join $E'O'$.

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3. Copying Angle (iii): $\angle GHI$

Step 1: At vertex $H$, draw an arc intersecting $HG$ and $HI$ at $Q$ and $R$. This defines the angular spread.

Step 2: Draw ray $H'I'$. Repeat the arc from $H'$ with the same radius to find $R'$. Mark $Q'$ such that $R'Q' = RQ$. Draw ray $H'G'$.

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4. Copying Angle (iv): $\angle JKL$

Step 1: With $K$ as center, draw an arc intersecting $KJ$ and $KL$ at $S$ and $T$.

Step 2: Draw ray $K'L'$. With $K'$ as center and the same radius as before, draw an arc cutting $K'L'$ at $T'$. Mark $S'$ on the arc so $T'S' = TS$. Join $K'S'$.

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Note: In all these constructions, the compass setting used for the first arc must remain unchanged when drawing the second arc at the new vertex. This ensures the radial distance is preserved before transferring the chord length ($MN, OP, QR, ST$).

Given:

A pattern of connected circular sectors as shown in Figure 6.6. Let the vertices of the original figure be labeled as $A, B, C, D, E, F,$ and $G$ for reference during construction.

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To Construct:

An exact replica of the pattern using a ruler and a compass without numerical measurement.

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Construction Procedure:

1. Constructing the first shaded sector:

Draw a straight line and mark a segment $A'B'$ equal in length and direction to the original segment $AB$.

With $B'$ as the center, draw an arc with radius $A'B'$. Measure the distance $AC$ from the original figure with a compass and transfer it to this arc from $A'$ to find point $C'$.

Join $B'C'$ and shade the sector $A'B'C'$ as shown in the original figure.

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2. Constructing the first unshaded part:

With $C'$ as the center, draw an arc of radius $B'C'$. Measure the distance $BD$ from the original figure and transfer it onto this arc from $B'$ to find point $D'$. Join $C'$ and $D'$.

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3. Constructing the second shaded sector:

With $D'$ as the center, draw an arc of radius $C'D'$. Measure the distance $CE$ and transfer it to the arc from $C'$ to get point $E'$. Join $D'E'$ and shade this sector.

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4. Constructing the second unshaded part:

With $E'$ as the center, draw an arc of radius $D'E'$. Measure distance $DF$ and transfer it from $D'$ to find point $F'$. Join $E'F'$.

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5. Constructing the third shaded sector:

With $F'$ as the center, draw an arc of radius $E'F'$. Measure the distance $EG$ and transfer it from $E'$ to find point $G'$. Join $F'G'$ and shade this final sector.

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Final Step:

Erase any extra construction arcs to obtain the required figure. The final construction will be a series of three shaded sectors and two unshaded connecting parts, perfectly mimicking the proportions of the original Figure 6.6.

To Construct: Four pairs of parallel lines in various orientations (horizontal, vertical, and slanted).

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Construction Method (Using Corresponding Angles):

To construct a line parallel to a given line $l$ passing through a point $P$ not on it:

1. Draw a line $l$ and a point $P$ outside it.

2. Take any point $Q$ on line $l$ and join $PQ$.

3. With $Q$ as the center and any radius, draw an arc intersecting $l$ at $X$ and $PQ$ at $Y$.

4. With $P$ as the center and the same radius, draw an arc intersecting the ray $PQ$ at $R$.

5. Adjust the compass to the length of the arc $XY$.

6. With $R$ as the center, draw an arc intersecting the previous arc at $S$.

7. Draw a line $m$ passing through $P$ and $S$. The line $m$ is parallel to line $l$.

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Orientations:

By repeating this process while rotating the paper or the initial line $l$, we can create $4$ pairs:

1. Horizontal Pair: Lines running left to right.

2. Vertical Pair: Lines running up and down.

3. Slanted (Left-to-Right): Lines at an angle of approximately $45^\circ$.

4. Slanted (Right-to-Left): Lines at an angle of approximately $135^\circ$.

Constructions and Tilings Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 6 Page 148 Q2

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To Construct:

A geometric figure consisting of 8 identical rhombuses arranged around a central vertex to form a star shape.

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To Find:

The measure of the acute angle of each rhombus at the center.

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Construction Required:

1. Step 1: Draw a horizontal line and mark a point $A$ on it. This point will be the central vertex of the star.

2. Step 2: With $A$ as the center, draw a semicircle. From the endpoints $B$ and $C$ of the diameter on the line, draw arcs of equal radius to intersect at point $D$. Join $AD$ to construct a $90^\circ$ angle ($\angle CAD = 90^\circ$).

3. Step 3: Construct the bisector of $\angle CAD$. With centers at $C$ and the point where $AD$ meets the semicircle (let's call it $E$), draw arcs of equal radius. Let them intersect at $F$. Join $AF$. Now, $\angle CAF = 45^\circ$.

4. Step 4: With $A$ as the center and a fixed radius, draw an arc intersecting $AC$ and $AF$ at points $G$ and $H$ respectively.

5. Step 5: Keeping the same radius, draw two arcs—one with center $G$ and one with center $H$. Let these arcs intersect at point $I$. Join $AI$. The figure $AGIH$ is a rhombus with a $45^\circ$ base angle.

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Solution:

1. Erase the construction arcs. Shade or color the upper triangle $AHI$ of the rhombus as per the given design.

2. Using tracing paper or by repeating the construction of $45^\circ$ angles, create 8 identical replicas of this rhombus.

3. Place these 8 units side-by-side around the central point $A$. Since each has an angle of $45^\circ$, they will perfectly complete a full circle ($8 \times 45^\circ = 360^\circ$).

4. The final arrangement forms the required star-shaped figure shown in the question.

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Final Figure:

Given:

A pair of support lines meeting at an apex as shown in Fig. $6.11$. Points are marked on the support lines to guide the construction of the pointed arch.

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To Construct:

A pointed arch using the provided support lines and creating different variations by changing the radius of the arcs.

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Steps of Construction:

Step 1: Draw a vertical line $B'D'$ to act as the central axis of the arch symmetry.

Step 2: With centres at $B$ and $B'$, draw arcs of equal radius to establish the base curvature.

Step 3: Measure the length $RS$ using a compass from the reference figure.

Step 4: Transfer this length on the arc from $S'$ to get the point $R'$.

Step 5: Join $B'$ and $R'$ and extend this line further.

Step 6: Mark a point $A'$ on this extended line such that the length $B'A'$ is equal to $BA$.

Step 7: Now, measure the length $ST$ using a compass.

Step 8: Transfer this length on the arc from $S'$ to locate point $T'$, ensuring $S'T' = ST$.

Step 9: Join $B'$ and $T'$ and extend this line.

Step 10: Mark a point $C'$ on this line such that $B'C'$ is equal to $BC$.

Step 11: Using a compass, find the midpoints $M$ and $N$ of the segments $A'B'$ and $B'C'$ respectively.

Step 12: On the lines $A'M$, $MB'$, $B'N$, and $NC'$, draw similar arcs as shown in the construction figure to form the curves of the pointed arch.

Step 13: Carefully erase the extra construction letters, lines, and overlapping arcs to reveal the final required pointed arch supported by the given lines.

Solution:

In this construction, we will design a custom arch, often referred to as a Cusped Arch or a Multifoil Arch, which is frequently seen in Indian historical monuments like the palaces of Rajasthan and Mughal architecture.

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Given:

A central vertical axis and base points to begin the geometric layout of a decorative arch.

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To Construct:

A unique arch design using intersecting arcs, midpoints, and semicircular components.

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Construction Steps:

Step 1: Draw a vertical line $AB$ to serve as the line of symmetry for the arch design.

Step 2: With the centre at $A$, draw a large arc that will define the upper boundary of the arch area.

Step 3: Choose a point $C$ (base point) and with $C$ as the centre, draw another arc such that it intersects the first arc at two distinct points, $D$ and $E$.

Step 4: Join the points $AD$ and $AE$ using a straightedge and extend these lines outwards.

Step 5: Mark a point $F$ on the line $AD$ and a point $G$ on the line $AE$. Ensure that the distance from the apex $A$ to these points is equal.

Step 6: Find the midpoints $M$ and $N$ of the segments $AF$ and $AG$ respectively using a compass (perpendicular bisector method).

Step 7: Using a compass set to the radius of half the segment length (e.g., $AM$), draw decorative semicircles on the segments $MF$, $AM$, $AN$, and $NG$.

Step 8: Finally, erase the construction lines, extra letters, and auxiliary arcs to reveal the pointed cusped arch.

Solution for Figure (a): An Inflexed Arc

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Given: A horizontal segment and a requirement to build an inflexed (ogee-style) arch.

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To Construct: An Inflexed Arc using a ruler and a compass.

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Steps of Construction:

1. Draw a horizontal line and mark two points $A$ and $B$ on it.

2. Draw perpendicular lines at points $A$ and $B$ extending downwards using a ruler and a compass.

3. Mark points $C$ and $D$ on these perpendiculars such that the segments $AC$ and $BD$ are equal.

4. Find the midpoint $M$ of the line segment $AB$ by using the perpendicular bisector method.

5. With $A$ as the centre and radius equal to $AM$, draw an arc starting from the vertical support.

6. Similarly, with $B$ as the centre and the same radius $AM$, draw another arc to meet at the apex.

7. Erase the construction lines and extra arcs to obtain the clean Inflexed Arc design, which is commonly seen in Indian architecture and decorative doorways.

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Solution for Figure (b): The ‘Flower of Life’

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To Construct: A six-petalled floral pattern known as the ‘Flower of Life’, often used in Indian Rangoli and Mandalas.

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Steps of Construction:

1. Draw a central circle with a fixed radius $r$ and centre $O$.

2. Select any point $A$ on the circumference of this circle.

3. With $A$ as the centre and the same radius $r$, draw an arc that touches the circumference at two points, say $B$ and $C$.

4. Move the compass needle to point $B$. With the same radius $r$, draw another arc.

5. Continue this process around the circle, using each new intersection point as the centre for the next arc.

6. After completing six such arcs, a perfect six-petalled flower pattern is formed within and around the central circle.

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Solution for Figure (c): Regular Hexagon in a Circle

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To Construct: A regular hexagon inscribed inside a circle.

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Steps of Construction:

1. Draw a circle with centre $O$ and any radius $r$.

2. Mark a point $A$ anywhere on the circumference.

3. With $A$ as the centre and the same radius $r$, draw an arc to intersect the circle at point $B$.

4. With $B$ as the centre and radius $r$, mark point $C$ on the circumference. Repeat this until points $D, E,$ and $F$ are found.

5. Join the points in sequence: $AB, BC, CD, DE, EF,$ and $FA$.

6. The resulting figure is a Regular Hexagon where each side equals the radius of the circumscribing circle.

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Solution for Figure (d): Ring of Six Circles

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To Construct: A pattern of six smaller circles touching each other around a central point.

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Steps of Construction:

1. Draw a base circle with centre $O$ and radius $OA$.

2. Use the radius $OA$ to mark six equidistant points $A, B, C, D, E,$ and $F$ on the circumference (as done for the hexagon).

3. Find the midpoint $M$ of the radius $OA$.

4. Using a compass, set the radius to $OM$.

5. With centres at $A, B, C, D, E,$ and $F$, draw six individual circles.

6. These six circles will perfectly touch each other and the central point $O$, creating the required symmetric pattern.

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Solution for Figure (e): Complex Geometric Tiling

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To Construct: A star-hexagonal tiling pattern.

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Steps of Construction:

1. Follow the steps in Figure (c) to create a central hexagon with vertices $A, B, C, D, E, F$ and centre $O$.

2. Draw a larger concentric circle with centre $O$ and radius equal to $2 \times OA$.

3. Extend the radii $OA, OB, \dots$ to meet the outer circle at points $A', B', \dots$.

4. Create a grid by joining these points to form a series of equilateral triangles. The pattern consists of rows of triangles in a $5, 7, 7, 5$ sequence.

5. Inside each small triangle, find the centroid and join it to the three vertices to create the 3D-effect "Y" shape within each cell.

6. This creates a Tessellation or Jaali pattern common in Indian Mughal monuments.

Given:

A set of three circles with wedges removed (often called 'Pac-Man' shapes) and three V-shaped angles.

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To Construct:

Recreate the Kanizsa Triangle using a ruler and a compass to understand the geometric alignment that creates the illusion.

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Steps of Construction:

Step 1: Draw a circle with centre $O$ and any suitable radius $r$.

Step 2: Mark a point $A$ at the top of the circle. Using the same radius $r$, mark six equidistant points $A, B, C, D, E,$ and $F$ around the circumference.

Step 3: Join points $A, C,$ and $E$ to form an equilateral triangle $\triangle ACE$.

Step 4: Join points $B, D,$ and $F$ to form another equilateral triangle $\triangle BDF$ pointing in the opposite direction.

Step 5: Using $A, C,$ and $E$ as centres, draw three small identical circles. Remove a wedge (60-degree sector) from each circle such that the open mouth of the "Pac-Man" faces the centre of the figure.

Step 6: At the points $B, D,$ and $F$, highlight the V-shaped angles formed by the overlapping triangle $\triangle BDF$.

Step 7: Erase the central circle, the construction letters, and the lines of triangle $\triangle ACE$.

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Observation and Explanation:

The resulting figure contains only three "Pac-Man" shapes and three V-shapes. However, the brain perceives a bright white equilateral triangle pointing upwards. This happens because:

1. Subjective Contours: Our visual system tries to make sense of the gaps by "filling in" the edges of a shape that would logically block the circles.

2. Brightness Illusion: The perceived central triangle appears "whiter" or brighter than the surrounding white background, even though the colour is identical.

Given:

A circle and a requirement to construct a pattern where the vertices are equidistant, suggesting the use of $60^\circ$ angles.

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To Construct:

A geometric star figure by dividing a circle into six equal parts.

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Steps of Construction:

Step 1: Draw a circle with centre $O$ and any radius $r$.

Step 2: Take any point $A$ on the top of the circumference. Using $A$ as the centre and the same radius $r$, draw an arc that intersects the circle at point $B$.

Step 3: With $B$ as the centre and the same radius, draw an arc intersecting the circle at $C$.

Step 4: Repeat this process sequentially to find points $D, E,$ and $F$. Since the radius is used to mark these points, the central angle between any two consecutive points is:

Step 5: Join the points to form two large overlapping equilateral triangles:

$\bullet$ Join $A \to C \to E \to A$ to form the first triangle.

$\bullet$ Join $B \to D \to F \to B$ to form the second triangle.

Step 6: Join the vertices to the centre $O$ or join the intersection points as shown in the specific figure provided in the question.

Step 7: Erase the extra construction arcs and the outer circle to get the final required star design.

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Mathematical Analysis of Angles:

The hint asks to find the angles. Since the circle is divided into 6 equal parts:

Part	Angle Calculation	Value
Central Angle	$360^\circ / 6$	$60^\circ$
Interior Angle of Hexagon	$(6-2) \times 180^\circ / 6$	$120^\circ$
Angle at Star Point	Equilateral Triangle vertex	$60^\circ$

Solution:

To construct a perpendicular to a line from a point outside it, we essentially create an isosceles triangle where the apex is the given point $P$, and then find the perpendicular bisector of its base.

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Given:

A straight line $l$ and a point $P$ located at a distance from the line.

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To Construct:

A line segment or line passing through point $P$ that is perpendicular ($\perp$) to line $l$.

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Steps of Construction:

Step 1: Draw the given line $l$ using a ruler and mark point $P$ anywhere outside it.

Step 2: Place the compass needle at point $P$. Adjust the width of the compass so it is greater than the distance to line $l$. Draw an arc that cuts line $l$ at two distinct points, label them $A$ and $B$.

Step 3: Now, using $A$ and $B$ as centres and with a radius greater than half of $AB$, draw two arcs on the opposite side of the line $l$ (away from $P$). Let these arcs intersect at point $Q$.

Step 4: Join point $P$ and point $Q$ using a ruler. Let this line intersect line $l$ at point $M$.

Step 5: The line $PQ$ (or the segment $PM$) is the required perpendicular to line $l$ passing through $P$.

Given:

A standard 7-piece tangram set used for geometric puzzles and spatial reasoning.

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Visual Breakdown of Tans:

Instead of a tabular representation, the following figure illustrates the 7 individual pieces (tans) that constitute a standard tangram set:

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Reconstruction Steps:

Below are the visual solutions showing the internal boundaries of the pieces:

Solution for Question 1:

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Given:

1. A region (often called a Polyomino) consisting of $12$ unit squares.

2. A tile shaped like an 'L' consisting of $3$ unit squares (a tromino).

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To Find:

Whether the given region can be perfectly tiled using the provided L-shaped tiles.

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Solution:

First, let us calculate the total area of the region and the area of a single tile to see if the division is possible.

To find the number of tiles required, we divide the total area by the area of one tile:

Since the area is divisible, we check for the geometric arrangement. By rotating and flipping the L-shaped tiles, we can perfectly fit $4$ tiles into the region without any gaps or overlaps.

Conclusion: Yes, the tiling is possible. The following image shows the arrangement of the four tiles:

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Solution for Question 2:

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Given:

1. A grid region which is an $8 \times 8$ square with two opposite corners removed.

2. A $2 \times 1$ rectangular tile (a domino).

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To Find:

Whether the given region can be tiled using the $2 \times 1$ tiles.

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Solution:

This is a classic problem in Combinatorial Geometry. To solve this, we use the method of Coloring (similar to a Chessboard).

In a standard $8 \times 8$ chessboard, there are $64$ squares in total, divided equally into $32$ black squares and $32$ white squares.

In the given figure, two opposite corners are removed. On a chessboard, opposite corners are always the same colour. Let us assume two white corners are removed:

Every $2 \times 1$ tile (domino) placed on the board must cover exactly one black square and one white square.

Since the number of black squares ($32$) is not equal to the number of white squares ($30$), we will eventually be left with $2$ black squares that cannot be covered by a single domino.

Conclusion: The tiling is not possible.