NCERT Solutions Class 7 Mathematics Ganita Prakash 2 Chapter 06 Constructions and Tilings

Question 1. 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.]
Answer: No, it is not necessary to have the same radius for arcs above and below XY. Draw a line segment XY. Choose two different distances, k and k', both slightly larger than half the length of XY. With X and Y as centres, draw arcs of radius k below XY; they will meet at point B. Then, with X and Y as centres, draw arcs of radius k' above XY; they will meet at point A. Join A and B. This line AB intersects XY at O. By the congruence of triangles ABX and ABY (since AX = AY = k, BX = BY = k', and AB is common), we get ∠XAO = ∠YAO. Similarly, triangles AOX and AOY are congruent (AX = AY, ∠XAO = ∠YAO, and AO is common), giving us OX = OY and ∠AOX = ∠AOY. Since ∠AOX + ∠AOY = 180°, we have ∠AOX = 90°. Therefore, AB is the perpendicular bisector of XY. The key insight is that any point equidistant from X and Y must lie on the perpendicular bisector, regardless of whether the arc radii match.
In simple words: You can use different sized arcs above and below the line and still get the perpendicular bisector. What matters is that the two points you find (A and B) are each the same distance from X and from Y.

Exam Tip: Focus on proving that OX = OY and ∠AOX = 90° using triangle congruence - these two properties together define a perpendicular bisector. Examiners reward clear use of SSS or SAS rules.

Question 2. Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.
Answer: Yes, both pairs of arcs can be drawn on the same side of XY, and the perpendicular bisector will still be obtained. Draw a line segment XY. Pick two distances k and k', each slightly more than half of XY. With X and Y as centres, draw arcs of radius k on the same side (say, above) XY. These arcs meet at point A. Then, with X and Y as centres, draw arcs of radius k' also above XY. These arcs meet at point B. Join A and B and extend this line to meet XY at O. Connect A and B to both X and Y. Triangles ABX and ABY are congruent because AX = AY = k, BX = BY = k', and AB is common. Thus ∠XAO = ∠YAO. Triangles AOX and AOY are congruent because AX = AY, ∠XAO = ∠YAO, and OA is common. This gives OX = OY and ∠AOX = ∠AOY. Since these angles sum to 180°, each equals 90°. Therefore, AB is the perpendicular bisector even when both arc pairs are on the same side of the line.
In simple words: You don't need arcs above and below the line. You can put both sets of arcs on just one side, and you will still get the perpendicular bisector.

Exam Tip: Emphasize that the perpendicular bisector depends on equidistance (OX = OY) and the right angle, not on whether construction arcs are above or below the original segment.

Question 3. 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.
Answer: Yes, it is necessary to use the same radius when drawing one pair of arcs for the perpendicular bisector construction. Draw a line segment XY. With X and Y as centres, draw arcs of unequal radii, say k and k'. Let these arcs meet at point A. Let PQ be the perpendicular bisector of XY, and let R be any point on PQ. When we join R to X and Y, triangles ROX and ROY are congruent because OX = OY (O is the midpoint), ∠ROX = ∠ROY (both 90°), and OR is common. Therefore RX = RY. Every point on the perpendicular bisector is equidistant from X and Y. Since we drew the arcs with unequal radii, we have AX ≠ AY, so point A does not lie on the perpendicular bisector. Thus, to correctly construct the perpendicular bisector, we must use the same radius for both arcs in each pair.
In simple words: When you draw arcs from X and Y, they must have the same radius. If they have different radii, the point where they meet won't be the right distance from both X and Y.

Exam Tip: The core principle is equidistance: a point on the perpendicular bisector must be equally far from both endpoints. Use this principle to explain why equal radii are mandatory.

Question 4. Recreate this design using only a ruler and compass
Answer: Let ABCD be a square. Draw the perpendicular bisectors of sides AB and BC. These bisectors intersect the square at points P, Q, R, and S (P on AB, Q on BC, R on CD, S on AD). With P, Q, R, and S as centres, draw semicircles inside the square, each with radius equal to AP (or equivalently, BQ, CR, and DS, since these are all equal in a square). The arcs from all four centres together form the flower-like pattern shown in the figure. To emphasize the design, colour the boundary of the pattern using a coloured pencil. This makes the curved design stand out clearly from the construction lines beneath it.
In simple words: Find the middle of each side of the square using perpendicular bisectors. Draw curved lines (semicircles) from each midpoint, all the same size. Together they make a four-petalled flower shape.

Exam Tip: Ensure that all four semicircles have equal radius (the distance from one vertex to the midpoint of the adjacent side). Proper use of the perpendicular bisector to locate P, Q, R, S is essential for symmetry.

Question 1. Justify why AB in the figure given below is the perpendicular bisector of the line XY.
Answer: In the figure, XAY and XBY represent two different positions of a rope with fixed total length. Points A and B mark the midpoints of the rope in each position. This means AX = AY and BX = BY. Since AB is common to both, triangles AXB and AYB are congruent (by SSS: AX = AY, BX = BY, AB = AB). Therefore ∠XAO = ∠YAO, where O is the intersection of AB and XY. Triangles AXO and AYO are also congruent (AX = AY, ∠XAO = ∠YAO, AO common), which gives OX = OY and ∠XOA = ∠YOA. Since ∠XOA + ∠YOA = 180°, we get ∠XOA = 90°. By definition, a perpendicular bisector is a line that is perpendicular to a segment and passes through its midpoint. Since AB is perpendicular to XY (∠XOA = 90°) and passes through the midpoint O (OX = OY), AB is the perpendicular bisector of line XY.
In simple words: A and B are at the midpoint of the rope in both positions. This makes A and B equally far from both X and Y. A line through two such points must be the perpendicular bisector.

Exam Tip: Clearly show that OX = OY (bisects the segment) and ∠XOA = 90° (is perpendicular). Both properties together define the perpendicular bisector; don't omit either one.

Question 2. Can you think of different methods to construct a 90° angle at a given point on a line using a rope?
Answer: One effective method uses the 3 - 4 - 5 principle: if a triangle has sides in the ratio 3:4:5, then the angle opposite the longest side (the hypotenuse) is always 90°. To construct a right angle at point A on line XY: Fix a small pole at A. Mark a rope at 0 units, 3 units, 8 units (= 3 + 5), and 12 units (= 3 + 4 + 5). Attach both the 0 and 12 unit marks to the pole at A. Attach the 3 unit mark at point B on line XY using a second pole. Hold the rope taut at the 8 unit mark and pull it perpendicular to XY, securing it at point C with a third pole. The rope now forms triangle ABC with sides AB = 3 units, BC = 4 units (from 8 to 12 units), and CA = 5 units (from 0 to 5 units). By the 3 - 4 - 5 principle, the angle at B (the angle between AC and XY) equals 90°. Thus AC is perpendicular to line XY at point A.
In simple words: Mark a rope at distances 0, 3, 8, and 12. Attach the ends at point A, pin the 3-mark to make a triangle with sides 3, 4, and 5. The angle at A will be 90 degrees.

Exam Tip: The 3 - 4 - 5 right triangle is a classical construction method. Show clearly how the rope marks create the three side lengths and why this guarantees a 90° angle opposite the side of length 5.

Question 1. Construct at least 4 different angles. Draw their bisectors.
Answer: Draw four distinct angles: ∠ABC, ∠DEF, ∠GHI, and ∠JKL, each in a different position and with a different measure. To bisect each angle: With the angle's vertex (B, E, H, or K) as centre, draw an arc that cuts both arms of the angle, marking the intersection points. For ∠ABC, the arc meets the arms at A' and C'; for ∠DEF at D' and F'; for ∠GHI at G' and I'; and for ∠JKL at J' and L'. With A' and C' as centres, draw arcs of equal radius that intersect at a point; call this point B'. Join B and B' to get the bisector BB' of ∠ABC. Repeat this process for the other three angles: join E and E' for ∠DEF, join H and H' for ∠GHI, and join K and K' for ∠JKL. Each resulting line (BB', EE', HH', KK') divides its respective angle into two equal parts.
In simple words: For each angle, draw an arc from the vertex that touches both sides. Then draw two smaller arcs from those touch-points that meet in the middle. The line from the vertex through that middle point divides the angle in half.

Exam Tip: Ensure that the two arcs used to find the bisector point have equal radius - this is the key to accuracy. Label all intersection points clearly to show your construction method.

Question 2. Construct the 8-petaled figure shown below.
Answer: Draw a line AB. With A and B as centres and equal radius, draw arcs both above and below line AB so they intersect at points C (above) and D (below). Join C and D; let it intersect AB at O. Now draw the bisectors of the four angles formed at O: ∠BOC, ∠COA, ∠AOD, and ∠DOB. Each bisector line passes through O and extends in both directions. You now have eight rays emanating from O (the original lines AB and CD, plus their four angle bisectors). Draw a circle with centre O and radius equal to the length of one petal shown in the reference figure. This circle intersects each of the eight rays at a point. Mark these eight intersection points. Using O as the centre and the marked points as endpoints, draw eight petals (each petal is a curved segment from one radius to another). Erase the extra construction lines and arcs to reveal the final 8-petalled flower design.
In simple words: Draw two perpendicular lines through the centre point O. Then bisect all four angles they make. You now have eight directions. Draw a circle and place petals along each of these eight lines.

Exam Tip: The eight petals must be evenly spaced around O, separated by 45° each. Verify this by checking that your angle bisectors are accurate, as any error in bisection will create asymmetry in the final figure.

Question 3. In the process 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.
Answer: Yes, line OC (and its extension OD) will still be an angle bisector. Let ∠XOY be the given angle. With O as centre, draw an arc intersecting OX at A and OY at B. With A and B as centres, draw arcs of equal radius intersecting on one side at C. Join O and C, and extend this line to D on the opposite side. In triangle OAC and triangle OBC, we have OA = OB (both radii from O), AC = BC (equal arc radii), and OC is common. By SSS congruence, ∠AOC = ∠BOC. Therefore OC bisects ∠XOY. Extending CO to D, we get ∠AOD + ∠BOD = 360° - (∠AOC + ∠BOC) = 360° - 2∠AOC. Since ∠AOC + ∠BOC = ∠AOB and ∠AOD = ∠BOD (by the same congruence), line OD also bisects the reflex angle ∠XOY. Thus, whether the construction arcs are drawn on one side of the original angle or split above and below, the resulting line is still a valid angle bisector.
In simple words: It doesn't matter which side of the angle you draw the small arcs on. As long as those two arcs are the same size and meet at a point, the line from the angle's vertex through that point will always split the angle in half.

Exam Tip: Use SSS triangle congruence to show ∠AOC = ∠BOC. This equality is the defining property of an angle bisector, so make this step explicit.

Question 4. What are the other angles that can be constructed using angle bisection? Can you construct a 65.5° angle?
Answer: By using angle bisection, you can construct certain special angles by combining a basic right angle (90°) and successive bisections. Bisecting 90° gives 45°. Bisecting 45° gives 22.5°. You can also combine these angles: 90° + 45° = 135°, 90° + 22.5° = 112.5°, and 45° + 22.5° = 67.5°. These angles - 45°, 22.5°, 135°, 112.5°, and 67.5° - can all be constructed using only a ruler and compass through the perpendicular bisector method and angle bisection. However, 65.5° cannot be built from any combination of 90°, 45°, 22.5°, or their sums and differences, because 65.5° does not fit the pattern of angles obtainable by repeated bisection of 90°. Therefore, using angle bisection alone, an angle of 65.5° cannot be constructed.
In simple words: By halving a 90 degree angle over and over, and by adding up the pieces you get, you can make 45, 22.5, 135, 112.5, and 67.5 degrees. But 65.5 degrees doesn't fit this pattern, so you can't make it.

Exam Tip: List all constructible angles explicitly (90°, 45°, 22.5°, 135°, 112.5°, 67.5°) and explain that 65.5° is not among them. Showing why a particular angle is impossible is just as important as showing how to construct the ones that are possible.

Question 5. Come up with a method to construct the angle bisector using a rope.
Answer: Let ∠XOY be the given angle. Fix a pole at point O. Take a rope, make a loop at one end, and mark a point at a fixed distance along the rope. Fix the looped end at the pole at O, then rotate the rope from ray OX to ray OY, keeping it taut. Mark points A and B on the rope at the fixed distance mark (so OA = OB). Fix poles at both A and B. Take a second rope, make loops at both ends, and fix these loops to the poles at A and B. Find and mark the midpoint M of this second rope. Hold the rope at M, pull it taut away from the angle, and mark point M. Now A, M, and B form a triangle with AM = MB (since M is the midpoint) and OA = OB (by construction). Join A to M, B to M, and O to M. In triangles OAM and OBM, OA = OB, AM = BM, and OM is common. By SSS congruence, ∠AOM = ∠BOM. Therefore OM is the bisector of angle ∠XOY.
In simple words: Mark two points A and B equally far from O on the two sides of the angle. Attach a rope between A and B, find its midpoint M, pull it tight, and mark that spot. The line from O through M bisects the angle.

Exam Tip: Emphasize that the rope method relies on two key equalities: OA = OB and AM = BM. These guarantee that triangles OAM and OBM are congruent by SSS, which proves the angle bisection.

Question 6. Construct the following figure: How do we construct the petals so that they are of the maximum possible size within a given square?
Answer: To create four petals of maximum size within a square: Draw a line and place points A and B on it. With A and B as centres and equal radius, draw semicircles that intersect the line at points C, D, E, and F. With C and D as centres (on one side), draw arcs of equal radius intersecting at G. With E and F as centres (on the other side), draw arcs intersecting at H. Join A and G, and B and H. Mark points I on AG and J on BH such that AI = BJ = AB. Join I and J. The quadrilateral ABJC becomes a square (since all sides equal AB and all angles are 90°). Using a ruler and compass, construct the perpendicular bisectors of sides AB and BJ of this square. These bisectors intersect the square at points K, L, M, and N. With K, L, M, and N as centres, draw semicircles inside the square, each with radius equal to AK (the distance from a corner to the midpoint of an adjacent side). Erase all construction lines except the four semicircles. These four petals are now the largest possible size that fit inside the square without overlapping.
In simple words: Build a square using perpendicular bisectors and right-angle constructions. Find the midpoints of each side. Draw curved petals from each midpoint with the largest radius that still stays inside the square.

Exam Tip: The maximum petal size is achieved when the radius equals half the side length of the square. Show clearly how the perpendicular bisector locates this optimal radius point.

Question 1. Construct at least 4 different angles in different orientations without taking any measurements. Make a copy of all these angles.
Answer: Draw four angles of different sizes and orientations - let's call them ∠ABC, ∠DEF, ∠GHI, and ∠JKL. To copy angle (i) ∠ABC: Draw a reference line A'B' in the new location. With B as centre, draw an arc cutting BA at M and BC at N. With B' as centre and the same radius, draw an arc cutting A'B' at M'. Using a compass, measure the arc distance MN from the original angle. Transfer this distance on the new arc from M' to mark N'. Draw a ray from B' through N' to complete the copied angle ∠A'B'C'. Repeat this three-step process (draw a reference line, create an arc at the new vertex, transfer the arc-chord distance) for each of the other three angles: angle (ii) becomes ∠D'E'F', angle (iii) becomes ∠G'H'I', and angle (iv) becomes ∠J'K'L'. Each copied angle is congruent to its original.
In simple words: For each angle, draw an arc from the vertex. Measure how far apart the two points where the arc touches the sides are. In the new location, draw the same arc and measure off the same distance on it. Draw a line through that point to complete the copied angle.

Exam Tip: The key step is accurately transferring the chord length MN (the distance between the arc's intersection points on the angle's arms) to the new arc. Precision in this measurement ensures the copied angle is congruent to the original.

Question 2. Construct the following figure:
Answer: Label the vertices of the target figure as A, B, C, D, E, F, G in order. Draw a line A'B' equal in length to segment AB and positioned along the same direction as AB. With B' as centre, draw an arc of radius A'B'. Measure the distance AC from the original figure using a compass. Mark point C' on the arc so that A'C' = AC. Join B' and C' and shade this sector. With C' as centre, draw an arc of radius B'C'. Measure distance BD and mark D' such that B'D' = BD. Join C' and A', and shade the new sector. Continue this process: with D' as centre and radius C'D', mark E' so that C'E' = CE. With E' as centre and radius D'E', mark F' so that D'F' = DF. Finally, with F' as centre and radius E'F', mark G' so that E'G' = EG. Join F' and G' and shade this final sector. Erase the extra construction arcs to reveal the finished spiral-like figure made of shaded sectors chained together in sequence.
In simple words: Start with a line segment AB. Build a shaded slice (sector) with B' and C' as endpoints. From C', build the next slice with endpoints C' and D', and keep chaining sectors together, each one built from the endpoint of the previous one.

Exam Tip: Maintain consistency in measuring and transferring distances at each step - small errors accumulate. Clearly label all vertices and arcs to show the logical sequence of the construction.

Question 1. Construct 4 pairs of parallel lines in different orientations.
Answer: Start with a line called AB and select any line CD that crosses it. Pick a point P on line CD. Draw two arcs with the same radius, one centered at E (where CD meets AB) and another centered at P. Measure the space between two marked points on the first arc using a compass. Copy this measurement onto the arc near P, marking point I such that the measured length equals the original. Join point P to point I and extend the line in both directions to create line JK. Since the corresponding angles ∠PEB and ∠CPK are equal, lines AB and JK are parallel. Repeat this construction three more times in different orientations to create a total of 4 pairs of parallel lines.
In simple words: Use a compass and straight edge to copy angles at the place where a line crosses another line. Equal angles mean the lines will be parallel to each other.

Exam Tip: Show all arc markings and angle measurements clearly - examiners need to see you used equal angles to prove the lines are parallel.

Question 2. Construct the following figure:
Answer: The figure is made of 8 identical rhombuses. Since 360° ÷ 8 = 45°, each rhombus has an acute angle of 45°. Start by drawing a line and marking point A on it. From A, draw a semicircle. Mark two more points B and C, then draw equal arcs centered at these points until they meet at point D. Join A to D. Next, find the bisector of angle ∠CAD: draw equal arcs from C and E to find point F, then join A to F. From A, draw an arc crossing both AC and AF at points G and H. With equal radius AG, draw arcs centered at G and H that meet at point I. Join A to I. This creates rhombus AGIH with a 45° base angle. Erase extra construction lines and arcs. Using tracing paper, make 8 identical copies of this rhombus and arrange them together without gaps to form the final pattern.
In simple words: Create one small rhombus with 45-degree angles, then make 8 copies of it and fit them together like puzzle pieces to fill the whole design.

Exam Tip: Keep the rhombus shape exact by using a compass for all arc work - any mistake in angle measurement will prevent the pieces from fitting together properly.

Question 3. Use support lines in the given figure to construct a pointed arch. Make different arches by changing the radius of the arcs.
Answer: Draw a vertical line B'D' as the center line of the arch. With equal radii, draw arcs centered at B and at B', and measure the distance RS using a compass. Transfer this measurement onto the arc from S', making R'S' = RS. Join B' to R' and extend it. Set B'A' equal to BA along this line. Again using a compass, measure ST and copy this onto the arc from S', making S'T' = ST. Join B' to T' and extend it. Set B'C' equal to BC along this line. Find the midpoints M and N of segments A'B' and B'C'. Draw similar arcs along the lines A'M, MB', B'N, and NC' as shown in reference diagrams. Erase all extra letters, construction lines, and arcs to reveal the final pointed arch with the given support lines. To create different arch styles, vary the radius of the arcs you draw in each step.
In simple words: Use a compass to measure and copy curved arc shapes onto your support lines. Different sized arcs make different shaped arches.

Exam Tip: The key is measuring distances accurately with a compass and transferring them precisely - any measurement error will distort the arch shape.

Question 4. Make your own arch design.
Answer: Start by drawing a vertical line AB. From point A, draw an arc. From point C, draw another arc that crosses the first arc at two points D and E. Join A to D and A to E, then extend both lines outward. Select points F on line AD and G on line AE such that AF = AG. Find the midpoints of AF and AG, calling them M and N. Using a compass, draw semicircles on the four segments MF, AM, AN, and NG. Erase all extra letters, construction lines, and arcs to produce the final arch design. This creates a decorative arch with a series of alternating semicircular curves.
In simple words: Draw two lines spreading outward from a point, mark equal distances on each line, find the middle of those distances, then draw semicircles to make a pretty curved arch.

Exam Tip: Equal spacing and accurate midpoint location are critical - measure carefully and use your compass to mark the center of each segment before drawing the semicircles.

Question 1. Construct the following figures: (a) An Inflexed Arc
Answer: Draw a line and mark two points A and B on it. Draw perpendiculars at both A and B below the line using a ruler and compass. Draw two equal lines AC and BD perpendicular to the original line. Find the midpoint M of segment AB. With A as center and radius AM, draw an arc below. With B as center and radius BM (which equals AM), draw another arc below. Erase the extra lines, arcs, and letters to show the final inflexed arc figure.
In simple words: Find the middle point of a line segment, then draw two equal curved arcs from the endpoints that meet in a curved shape.

Exam Tip: Ensure the two perpendiculars AC and BD are exactly equal in length - this guarantees the arcs are symmetric and creates the correct inflexed arc shape.

Question 1. Construct the following figures: (b) Flower of Life
Answer: Draw a circle - this becomes the central circle. Pick any point A on the circumference. From A, draw an arc using the same radius as the original circle; this arc touches the circle at two new points B and C and extends outside the circle. Move to point B and draw another arc of the same radius, again touching the circle at its circumference. Repeat this pattern by moving around the circle, always drawing arcs of the same radius that touch the circle at adjacent points. Continue until you have completed the full circular pattern. Erase all the extra arc lines and labels, leaving only the overlapped circular regions that form the "Flower of Life" pattern.
In simple words: Draw one circle, then draw more circles of the same size around it, each touching the middle circle and its neighbors, until you have a beautiful flower-like shape.

Exam Tip: Keep all circle radii identical and ensure each new circle touches the previous one at exactly one point - this creates the characteristic overlapping petal pattern.

Question 1. Construct the following figures: (c) Regular Hexagon
Answer: Draw a circle and select any point A on it. Using the circle's radius, draw an arc centered at A that crosses the circle at point B. From B, draw another arc of the same radius that crosses the circle at point C. Repeat this process around the circle, marking points D, E, and F. Connect these six points in order: AB, BC, CD, DE, EF, and FA. Erase the arcs and all labels to reveal the regular hexagon inscribed in the circle.
In simple words: Starting at a point on a circle, step around the circle using the radius as your measuring tool. Connect all six step marks to make a six-sided shape.

Exam Tip: The radius of each arc must equal the circle's radius for the six points to be evenly spaced - this is what creates a regular hexagon with all sides and angles equal.

Question 1. Construct the following figures: (d) Six Circles Pattern
Answer: Draw a circle and mark any point A on it. Using the circle's radius, draw an arc centered at A that intersects the circle at point B. From B, draw an arc of the same radius that intersects the circle at point C. Continue this process to locate points D, E, and F around the circle. Join O (the center) to point A and find the midpoint of segment OA; call this midpoint M. The radius OM is the key measurement. From each of the six points A, B, C, D, E, and F, draw circles with radius equal to OM. Erase all construction lines, arcs, and labels to produce the final figure showing six equal circles arranged around a central point.
In simple words: Find six evenly-spaced points on a circle, measure halfway from the center to one of those points, then draw small circles at each of the six points using this half-distance as the radius.

Exam Tip: The six outer circles must have identical radii and be centered exactly at the marked points for the pattern to look balanced and symmetric.

Question 1. Construct the following figures: (e) Complex Geometric Pattern
Answer: Draw a circle and pick point A on it. Using the circle's radius, draw an arc from A that crosses the circle at B. From B, draw an arc of the same radius crossing at C. Repeat to find points D, E, and F around the circle. Next, draw a second circle centered at O with radius equal to twice OA (the original radius doubled). Extend line OA through the circle to meet the outer circle at A'. Draw similar lines from O through the other five points and extend them to the outer circle at B', C', D', E', and F'. Connect the inner and outer points with straight lines as shown in the reference pattern. This creates triangles between the two circles. Count the pattern: rows contain 5, 7, 7, and 5 triangles respectively. Join the vertices of each triangle to the center of that triangle. Erase all construction lines, arcs, and labels to obtain the final intricate figure.
In simple words: Mark six points on a circle, then make a bigger circle around it. Draw straight lines from the center through each point to the outer circle. Connect these to make triangles that form a beautiful star-like pattern.

Exam Tip: The inner and outer circles must be exactly concentric (same center) and the radius ratio must be exactly 1:2 for the triangle count to match the specified pattern of 5, 7, 7, 5.

Question 2. Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.
Answer: Draw a circle and select point A at the top. Using the circle's radius, draw an arc centered at A that intersects the circle at B. From B, draw an arc of equal radius intersecting at C. Continue around the circle marking points D, E, and F. Connect the six points: join AC, CE, EA, BD, DF, and FB. This forms two interlocking triangles. Draw three equal circles at positions A, C, and E, but remove a wedge-shaped section from each circle. The removed wedges face inward toward the center. Erase the extra arcs, the main circle, and all letters to reveal the final figure - the Kanizsa triangle. This optical illusion works because the brain perceives a white equilateral triangle pointing upward in the center, even though no triangle is actually drawn. The three circles with wedges removed and the three V-shaped gaps are arranged so cleverly that our visual system fills in the missing triangle automatically.
In simple words: Draw three circles with bites taken out of them, and space them around a center point. Your brain sees a triangle in the middle that is not really there.

Exam Tip: The three circles must be identical and positioned so their removed wedges point toward a common center - this positioning is what triggers the optical illusion in the viewer's mind.

Question 3. Construct this figure. [Hint: Find the angles in this figure.]
Answer: Draw a circle and mark point A at the top. Using the circle's radius as your arc radius, draw an arc centered at A that intersects the circle at point B. From B, draw an arc of the same radius intersecting the circle at C. Repeat this process around the full circle, marking all six intersection points D, E, and F. Connect all the points in the pattern shown: join AF to FB, FB to BD, BD to DC, DC to CE, CE to EA, and EA to AF, creating interlocking triangles. Once the lines are drawn, connect all the vertices of each triangle to its own center point, creating smaller internal divisions. Erase the extra arcs, the circle outline, and all construction labels to show the final six-pointed star or similar geometric figure with internal detail.
In simple words: Mark six evenly-spaced points on a circle, connect them to make triangles, then draw lines from each triangle's corner to its center point.

Exam Tip: All angles in this construction are determined by the equal radius arcs - the 60-degree angles form naturally from the six evenly-spaced points on the circle.

Question 4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P. [Hint: Find a line segment on l whose perpendicular bisector passes through P.]
Answer: Draw a line l and select point P outside of it. Using P as center, draw an arc large enough to cross line l at two points; call these points A and B. You now need the perpendicular bisector of segment AB. Set your compass to more than half the distance AB. With A as center, draw an arc above and below line l. With B as center and the same radius, draw arcs that intersect the previous arcs at points C and D. Draw a line through C and D - this is the perpendicular bisector of AB. If P does not already lie on line CD, extend line CD until it passes through P. The line CD is perpendicular to line l and passes through point P, completing the construction.
In simple words: Draw an arc from point P that hits the line at two spots. Find the midpoint between those two spots by making equal arcs from each spot. The line through these arc intersections is perpendicular to the original line and goes through P.

Exam Tip: The arc radius when finding the perpendicular bisector must be more than half of AB - if too small, the arcs won't intersect and you cannot find the bisector.

Question 1. How can the tangram pieces be rearranged to form each of the following figures?
Answer: A tangram consists of 7 pieces cut from a square: two large triangles, one medium triangle, two small triangles, one square, and one parallelogram (often numbered 1-7). To form different shapes, these pieces must be repositioned and rotated while keeping them flat in the same plane. The solution diagrams show piece placements for each target figure. For example, one arrangement shows pieces 1, 3, 6, 7, 2, 4, 5 positioned to form an "L" shape. Another arrangement rearranges the same pieces to create a butterfly or bird silhouette. A third shows how to form a boat, and additional arrangements create different animal shapes and geometric patterns. In each case, no pieces overlap, and all pieces must be used exactly once. The key skill is rotating and flipping each piece to fit into the target outline while respecting the boundaries of the design.
In simple words: The seven tangram pieces can be moved and turned around in many different ways. By shifting pieces, you can make them fit together into many different shapes.

Exam Tip: Always use all 7 pieces in every solution and ensure no gaps or overlaps - if your arrangement uses fewer pieces or leaves spaces, the solution is incorrect.

Question 1. Is the following tiling possible?
Answer: Examine the region to be tiled and the shape of the available tile. Study the dimensions and angles of both shapes. The given tile can be used to tile the specified region. Arrange the tiles so they fit together without overlaps or gaps, filling the entire region completely. For this particular problem, exactly 4 tiles of the given shape are needed to cover the entire region. Position the tiles carefully, rotating and flacing them as needed, so they interlock perfectly to match the boundary of the region.
In simple words: Yes, the tiling is possible. You need exactly 4 of the tile shapes to cover the whole region. Arrange and turn them until they fit together perfectly with no spaces left.

Exam Tip: Count the area units carefully - the total area of all tiles must equal the total area of the region being tiled.

Question 2. Is the following tiling possible?
Answer: Analyze the region to be tiled by coloring it in a checkerboard pattern of alternating black and white squares. Count how many black squares and white squares are in this pattern. Examine the tile shape - determine how many black and white squares it contains when placed on the checkerboard pattern. If each tile always covers an equal number of black and white squares, or a specific ratio, calculate whether the total number of tiles needed can cover the region's black and white balance. In this case, the checkerboard analysis reveals that the region contains an unequal number of black and white squares, while the tile shape covers a specific ratio that cannot match this imbalance. Therefore, it is impossible to tile this region using the given tile shape, regardless of how the tiles are arranged or rotated.
In simple words: Color the region like a checkerboard. The tiles cannot cover equal black and white amounts the way the region needs, so the tiling is impossible.

Exam Tip: Use checkerboard coloring to prove tiling is impossible - show that tiles always cover a fixed ratio of colors that cannot match the region's color balance.

In the given region, there are 30 black squares and 32 white squares. Since these numbers are not equal, the given region cannot be tiled by using the tiles of the given shape.
Explanation: When you try to cover a checkerboard pattern with tiles, each tile must cover the same number of black and white squares (or a specific fixed ratio). Because the region has 30 black squares but 32 white squares - a mismatch of 2 - no arrangement of identical tiles can fill the entire space without leaving gaps or overlaps. This is a powerful impossibility argument that works regardless of the tile's exact shape, as long as each tile maintains a balanced color coverage.
In simple words: If a region has more white squares than black squares, you cannot cover it completely with tiles that always cover equal numbers of each colour.

Exam Tip: This is a classic parity/colouring argument in combinatorics - remember that unequal counts of two alternating types make complete tiling impossible, and use this to rule out infeasible configurations quickly.