CBSE Class 7 Mathematics Constructions and Tilings Worksheet Set 01

Long Answer Questions

Question 1. A line segment PQ of length 8 cm is drawn on paper.
(i) Write the complete step-by-step procedure to construct the perpendicular bisector of PQ using only an unmarked ruler and a compass.
(ii) Let the perpendicular bisector meet PQ at O. Using the congruent triangles formed in the construction, justify that O is the midpoint of PQ and that the bisector is perpendicular to PQ.
(iii) State the general property about points equidistant from P and Q that this construction depends on.
Answer:
(i) Draw PQ = 8 cm. Open the compass to more than half of PQ, say 5 cm.
With centre P draw a long arc above PQ; with the same radius and centre Q draw an arc cutting it at A.
With the same radius, draw arcs below PQ from P and Q, meeting at B.
Join AB; it is the perpendicular bisector of PQ and meets PQ at O.
(ii) By construction AP = AQ and BP = BQ.
In △ABP and △ABQ: AP = AQ, BP = BQ, AB common, so △ABP ≅ △ABQ (SSS) and ∠PAO = ∠QAO.
In △AOP and △AOQ: AP = AQ, ∠PAO = ∠QAO, AO common, so △AOP ≅ △AOQ (SAS).
Hence OP = OQ (O is the midpoint) and ∠AOP = ∠AOQ. These two angles make a straight angle of 180°, so each is 90°, and AB ⟂ PQ.
(iii) Any point at the same distance from P and Q lies on the perpendicular bisector of PQ. The construction finds two such points and joins them.
In simple words: Equal arcs give two points equidistant from P and Q; congruent triangles show the line through them halves PQ at 90°.

Teacher's Note:
1. Use SSS first, then SAS.
2. Two equal angles on a straight line are each 90°.
3. State the property in (iii) in full.

Question 2. An eight-petalled flower design, like the one made from four lines drawn through a single point, needs angles of 45° between adjacent lines.
(i) Explain why the angle between two adjacent supporting lines of the design must be 45°.
(ii) Write the steps to construct a 90° angle at a point O on a line using a ruler and a compass.
(iii) Write the steps to bisect that 90° angle to obtain 45°, and name the congruence condition that justifies the bisection.
Answer:
(i) Four lines through one point make eight angles around it. These add up to 360°, and for the petals to be alike they must be equal: 360° ÷ 8 = 45°.
(ii) Draw a line and mark O on it. With centre O and a convenient radius, cut the line at X and Y, so O is the midpoint of XY.
With a larger radius, the same for both, draw arcs from X and Y meeting above the line at A. Join OA. As O and A are both equidistant from X and Y, OA is the perpendicular bisector of XY, so ∠AOX = 90°.
(iii) With centre O and any radius, cut OA at P and OX at Q, so OP = OQ.
With a large enough radius, the same for both, draw arcs from P and Q meeting at R. Join OR.
△OPR ≅ △OQR by SSS (OP = OQ, PR = QR, OR common), so ∠POR = ∠QOR = 45°. Repeating around O gives all eight 45° angles.
In simple words: 360° ÷ 8 = 45°: make a right angle with the perpendicular bisector idea and then halve it.

Teacher's Note:
1. Bisection of 90° gives 45°.
2. Use the same radius for each pair of arcs.
3. SSS justifies angle bisection.

Question 3. Answer the following about tiling.
(i) Define tiling and state clearly the two conditions that a covering must satisfy.
(ii) Name the three regular polygons that can tile the plane by themselves, and in each case show how their angles add up at a vertex.
(iii) Give one real-life example of each of these three tilings.
Answer:
(i) Covering a region using a set of shapes, without gaps or overlaps, is called tiling. Conditions: no part of the region is left uncovered, and no shape lies on top of another.
(ii) Equilateral triangle (60°): six meet at a point, 6 × 60° = 360°.
Square (90°): four meet at a point, 4 × 90° = 360°.
Regular hexagon (120°): three meet at a point, 3 × 120° = 360°.
In each case the angles fill exactly 360°, so there is no gap and no overlap.
(iii) Triangles: a quilt of triangular patches or a steel truss pattern on a bridge. Squares: square floor tiles, squared graph paper. Hexagons: the cells of a beehive or a wasp nest, hexagonal paver blocks.
In simple words: Tiling means no gaps and no overlaps; triangles, squares and hexagons work because their angles fill 360°.

Teacher's Note:
1. Show the multiplication for each shape.
2. Real-life examples should be familiar ones.
3. Keep the two conditions separate.

Question 4. A regular hexagon of side 4 cm is to be constructed with a ruler and a compass.
(i) Write the steps to construct a 60° angle, and explain why the method gives exactly 60°.
(ii) Describe how six congruent equilateral triangles of side 4 cm placed around a point O give the regular hexagon.
(iii) Explain why each angle of the hexagon so formed is 120°, and why AOD is a straight line when the hexagon is named ABCDEF.
Answer:
(i) With centre A and any radius, draw an arc cutting the ray \( \overrightarrow{AX} \) at B. With the same radius and centre B, cut this arc at C. Join AC; ∠CAX = 60°. Reason: AB = AC = BC (all equal to the radius), so △ABC is equilateral and each angle is 60°.
(ii) Mark O and draw OA = 4 cm. At O construct 60° and mark B on the new arm with OB = 4 cm; △OAB is equilateral.
Repeat five more times, each time building 60° on the previous arm, to get C, D, E, F with OC = OD = OE = OF = 4 cm.
Join AB, BC, CD, DE, EF, FA: ABCDEF is the regular hexagon of side 4 cm.
(iii) Six 60° angles at O make 360°, so the triangles fit with no gap or overlap. At each hexagon vertex two triangle angles meet: 60° + 60° = 120°.
Also ∠AOB + ∠BOC + ∠COD = 60° + 60° + 60° = 180°, a straight angle, so A, O, D lie on one straight line.
In simple words: Six 4 cm equilateral triangles around O make the hexagon; two 60° angles give each 120° corner and three give a straight line.

Teacher's Note:
1. Every side of every triangle is 4 cm.
2. Three 60° angles make 180°.
3. Check the order of letters around O.

Question 5. A line m is given, and a point B is marked outside it. A line l is drawn through B cutting m at A.
(i) Write the steps of construction for copying the angle that l makes with m at A, onto the point B on l.
(ii) Explain, using triangle congruence, why the copied angle is exactly equal to the original angle.
(iii) Explain why the new line through B is parallel to m, naming the angle property used.
Answer:
(i) With centre A and any radius, draw an arc cutting m at C and l at D.
With the same radius and centre B, draw an arc cutting l at E.
Open the compass to the length CD. With centre E, cut the arc drawn from B at F.
Join BF and extend it both ways to get the line n.
(ii) In △ACD and △BEF: AC = BE and AD = BF (same radius) and CD = EF (length transferred). So △ACD ≅ △BEF by SSS, and the copied angle at B equals the angle at A.
(iii) The line l cuts both m and n, so it is a transversal. The angles at A and at B are equal and lie in matching positions on the same side, so they are corresponding angles. Equal corresponding angles mean n ∥ m.
In simple words: Copy the angle at A onto B with equal arcs and chord; equal corresponding angles make the new line parallel.

Teacher's Note:
1. Copy the angle in the same position, not flipped.
2. SSS proves the copy, corresponding angles prove parallel.
3. Label points as you draw.

Question 6. Arches in old buildings are drawn with support lines before the arcs are made.
(i) For a trefoil-type arch on the base AD, the support lines are AB and DC. Explain why the design needs AB = DC and ∠BAD = ∠CDA, and describe how these support lines are constructed with a ruler and a compass.
(ii) For a pointed arch, the support lines are two line segments of equal length whose midpoints are marked. Describe how the arcs of the pointed arch are then drawn.
(iii) What happens to the look of the arch if the two arcs are drawn with unequal radii?
Answer:
(i) The arch must look the same on both sides, that is, be symmetric about the perpendicular bisector of the base AD. This needs the two supports to be equally long and equally slanted: AB = DC and ∠BAD = ∠CDA.
Construction: draw the base AD; construct the chosen angle at A; copy the same angle at D on the same side using the copy-an-angle construction; then with the compass set to the support length, cut off AB and DC. The arcs of the trefoil are drawn with centres on these supports.
(ii) Draw the two equal support segments and mark their midpoints M and N by perpendicular bisectors. With centre M and radius equal to the distance from M to the far endpoint, draw an arc; with centre N and the same radius draw a second arc. The two arcs cross above the base, giving the pointed top.
(iii) With unequal radii the two sides no longer match; the apex shifts away from the line of symmetry and the arch looks lopsided.
In simple words: Equal supports and equal angles keep the arch symmetric; equal radii make both sides of a pointed arch match.

Teacher's Note:
1. Symmetry is the reason behind both conditions.
2. Copying an angle gives the equal slant.
3. Different radii change the shape.

Question 7. A rectangular grid has m rows and n columns of unit squares and is to be tiled with 2 × 1 tiles that may be turned.
(i) Give a general strategy when both m and n are even.
(ii) Give a general strategy when one of m and n is even and the other is odd.
(iii) Explain, with a reason based on counting, why the grid can never be tiled when both m and n are odd. Illustrate with the 5 × 7 grid.
Answer:
(i) Work column by column. Each column has m squares, an even number, so it takes \( \frac{m}{2} \) vertical tiles. Doing this for all n columns tiles the grid with \( \frac{mn}{2} \) tiles.
(ii) Use the even side. If m is even, fill each column with \( \frac{m}{2} \) vertical tiles; if n is even, fill each row with \( \frac{n}{2} \) horizontal tiles. Either way \( \frac{mn}{2} \) tiles are used; for example a 4 × 7 grid takes 14 tiles.
(iii) If m and n are both odd, mn is odd. Every tile covers 2 squares, so the squares covered always add up to an even number, which can never equal an odd number. At least one square is always left bare.
For the 5 × 7 grid: 35 squares; 17 tiles cover 34 squares and one square is always left over, so it can never be tiled.
In simple words: Tile along an even side; if both sides are odd the total is odd and one square is always left.

Teacher's Note:
1. One even side is enough.
2. Odd × odd = odd.
3. Show the 5 × 7 numbers.

Question 8. A six-pointed star is constructed by drawing an equilateral triangle outwards on each side of a regular hexagon GHIJKL.
(i) Describe the steps of construction, starting from the regular hexagon.
(ii) Explain why every one of the six point-triangles is equilateral.
(iii) At a vertex of the hexagon, find the total of the hexagon’s interior angle and the two triangle angles meeting there, and say what this tells you about the notch of the star.
Answer:
(i) Construct a regular hexagon GHIJKL, say of side 4 cm, using 60° angles at its centre.
On side GH: with the compass opened to 4 cm, draw arcs from G and H meeting at a point A outside the hexagon. Join GA and HA.
Repeat on the other five sides to get the outer points B, C, D, E, F. Lightly rub out the hexagon's sides; the outline left is the six-pointed star.
(ii) Both new sides of each point-triangle equal the compass opening, which is the hexagon's side, and the third side is a side of the hexagon. All three sides are equal, so each point-triangle is equilateral with 60° angles.
(iii) At a vertex such as G: 120° + 60° + 60° = 240°. This is more than 180°, so the star has an inward (reflex) corner, a notch, there; the angle outside the star at that point is 360° − 240° = 120°. The same happens at all six vertices, so the star has rotational symmetry.
In simple words: Equilateral triangles on all six sides make the star; at each hexagon corner the angles total 240°, leaving a 120° notch.

Teacher's Note:
1. Keep the compass at the hexagon's side.
2. Reflex means more than 180°.
3. Add 120° + 60° + 60° carefully.

Question 9. A 5 × 3 grid of unit squares has the single middle square cut away, as shown.
(i) Count the unit squares left and explain why counting alone cannot settle whether the region can be tiled with 2 × 1 tiles.
(ii) Colour the squares of the full 5 × 3 grid black and white so that neighbouring squares always get different colours. Count the squares of each colour in the region that is left, and use this to decide whether the region is tileable.
(iii) Name a different single square whose removal does leave a tileable region, and give a tiling strategy for it.
Answer:
(i) The full grid has 5 × 3 = 15 squares; after removing the middle one, 14 remain. 14 is even, so counting only says 7 tiles would be needed; it does not tell us whether they can actually be placed.
(ii) Colour like a chessboard starting with white at a corner: the full grid has 8 white and 7 black squares. The middle square is black, so the region left has 8 white and 6 black.
Each 2 × 1 tile covers one white and one black square, so 7 tiles would cover 7 of each. Since 8 ≠ 6, the region cannot be tiled.
(iii) Remove a corner square instead (a white one): 7 white and 7 black remain.
Tiling: the row that lost its corner has 2 squares left, covered by one horizontal tile. The other 4 rows form a 4 × 3 block; fill each of its 3 columns with 2 vertical tiles. In all 1 + 6 = 7 tiles cover the 14 squares.
In simple words: Removing the black middle square leaves 8 white and 6 black, so no tiling; removing a corner keeps the colours balanced.

Teacher's Note:
1. Corners and the centre have different colours here.
2. Show an actual tiling in (iii).
3. Even count is necessary but not enough.

Question 10. The interior angle of a regular polygon with n sides is \( \frac{(n-2)\times 180^\circ}{n} \).
(i) Use this to find the interior angle of a regular polygon with n = 3, 4, 5, 6, 8 and 12 sides.
(ii) Explain the rule a regular polygon must satisfy if its copies alone are to tile the plane, and apply the rule to each of the six polygons in part (i).
(iii) Explain why no regular polygon with more than six sides can tile the plane by itself.
Answer:
(i) n = 3: \( \frac{1\times 180^\circ}{3} \) = 60°; n = 4: \( \frac{2\times 180^\circ}{4} \) = 90°; n = 5: \( \frac{3\times 180^\circ}{5} \) = 108°.
n = 6: \( \frac{4\times 180^\circ}{6} \) = 120°; n = 8: \( \frac{6\times 180^\circ}{8} \) = 135°; n = 12: \( \frac{10\times 180^\circ}{12} \) = 150°.
(ii) At each point of a one-shape tiling, several corners meet and their angles must total exactly 360°. So the interior angle must divide 360° a whole number of times.
360 ÷ 60 = 6, 360 ÷ 90 = 4, 360 ÷ 120 = 3: whole numbers, so the triangle, square and hexagon tile.
360 ÷ 108 = 3 \( \frac{1}{3} \), 360 ÷ 135 = 2 \( \frac{2}{3} \), 360 ÷ 150 = 2 \( \frac{2}{5} \): not whole numbers, so the pentagon, octagon and 12-gon do not tile alone.
(iii) For n > 6 the interior angle is more than 120° but less than 180°. So 360° divided by it lies strictly between 2 and 3 and is never a whole number: two tiles leave a gap and three overlap. Hence no regular polygon with more than six sides tiles the plane by itself.
In simple words: Only angles that divide 360° exactly work, which leaves just 60°, 90° and 120°.

Teacher's Note:
1. Substitute n carefully in the formula.
2. Whole-number test decides tiling.
3. Between 2 and 3 means never a whole number.

Case Study Based Questions

Question 11. A bathroom floor is being relaid with square tiles.
Riya’s family is getting the bathroom floor relaid. The floor is a rectangle of length 240 cm and breadth 180 cm. The mason has brought square tiles of side 20 cm and lays them in neat rows and columns so that the whole floor is covered, with no cement showing between the tiles and no tile resting on top of another. Riya notices that at every corner point inside the floor, the corners of a few tiles meet neatly around that point. The mason also has a box of rectangular tiles of size 20 cm × 10 cm left over from an earlier job.
(i) Using the chapter’s definition, explain what is meant by saying that the tiles tile the floor.
(ii) How many square tiles meet at a corner point inside the floor? Support your answer with the angles.
(iii) Find the number of square tiles of side 20 cm needed for the whole floor.
(iv) Can the rectangular 20 cm × 10 cm tiles also cover this floor completely? Give a reason.
Answer:
(i) The tiles cover the whole floor with no part left uncovered (no cement showing) and no tile on top of another. Covering without gaps or overlaps is tiling.
(ii) Four tiles. Each corner of a square is 90°, and 4 × 90° = 360°, the complete angle around the point.
(iii) Along the length: 240 ÷ 20 = 12; along the breadth: 180 ÷ 20 = 9. Total = 12 × 9 = 108 tiles.
(iv) Yes. Two 20 cm × 10 cm tiles side by side make a 20 cm square, so they cover the floor exactly; 12 × 18 = 216 tiles are needed.
In simple words: Square tiles meet four at a point, 108 are needed, and pairs of half-tiles can replace each square.

Teacher's Note:
1. Divide each side by the tile side.
2. Two half-tiles make one square tile.
3. Angles at a point total 360°.

Question 12. A honeycomb in the school garden.
The children of Class 7 are shown a photograph of the front face of a beehive. The cells are all regular hexagons of the same size, packed side by side with no gaps between them and no cell overlapping another. Their teacher tells them that wasps also build nests of this kind, and that the cells are used to keep eggs, larvae and food. She points out that because the hexagons fit perfectly, no space inside the comb is wasted. The children measure one cell on an enlarged photograph and find that each angle of the cell is 120°.
(i) How many hexagonal cells meet at a point in the comb?
(ii) Show by addition that the angles at that point leave no gap.
(iii) Explain why the bees waste no space in the comb.
(iv) Name one other regular polygon whose copies could have been used to build cells with no space wasted, and give its angle.
Answer:
(i) Three cells meet at each point.
(ii) 120° + 120° + 120° = 360°, the whole turn around the point, so no gap is left.
(iii) The cells form a tiling: they cover the comb with no gaps and no overlaps, so every bit of space is used exactly once.
(iv) The square, angle 90° (4 × 90° = 360°). The equilateral triangle, angle 60°, would also work.
In simple words: Three 120° angles make 360°, so hexagon cells fit with no wasted space.

Teacher's Note:
1. Addition, as asked, not multiplication.
2. Any one correct polygon is enough in (iv).
3. Hexagons also use less wax for the same space.

Question 13. A tangram in the craft period.
In the craft period the students are given a square sheet of card. Following the picture in their book, they cut the square into the seven tangram pieces. Each child then rearranges all seven pieces to make a new figure. Aarav makes an arrow, Meera makes a running cat, and Zoya turns one piece before placing it. The teacher reminds them that every piece must be used, that the pieces must not be placed one on top of another, and that no empty space should be left inside the figure they make.
(i) Write the meaning of tiling in your own words.
(ii) Do the seven pieces tile the original square? Give a reason.
(iii) Aarav says his arrow covers more area than the square did. Is he right? Give a reason.
(iv) Zoya rotated a piece before placing it. Does her figure still count as a tiling by the seven pieces? Give a reason.
Answer:
(i) Tiling means covering a region completely with shapes so that no empty space is left and no shape sits on another.
(ii) Yes. The pieces were cut from that square, so put back they fill it exactly, with no gaps or overlaps.
(iii) No. The same seven pieces are used, so the area equals that of the square; only the outline changes.
(iv) Yes. Turning a piece is allowed, just as a 2 × 1 tile may be placed either way. All pieces are used with no gaps or overlaps.
In simple words: Rearranging the same pieces keeps the area, and turning a piece is allowed in a tiling.

Teacher's Note:
1. Area does not change when pieces are moved.
2. Rotation is allowed in tiling.
3. All seven pieces must be used.

Question 14. A carpenter marks the middle of a wooden batten.
A carpenter has a straight wooden batten AB and must fix one screw exactly at its middle point. His measuring tape is worn out at the edges, so he decides to use a compass and a straight edge instead. He opens the compass to a little more than half the length of the batten, and with this opening he draws arcs from A and from B on one side of the batten and then on the other side. He joins the two points where the arcs cross and marks the point where this joining line meets the batten. He also wants a line at right angles to the batten at that point.
(i) Which construction of the chapter is the carpenter using?
(ii) Write the first two steps of this construction in order.
(iii) Why is this method more accurate than reading a marked scale?
(iv) The carpenter marks another point P on his bench with PA = PB. On which line must P lie?
Answer:
(i) The perpendicular bisector of AB; it meets the batten at its midpoint.
(ii) Step 1: with a radius more than half of AB, draw arcs from A and B on one side and mark where they cross. Step 2: with the same radius, draw arcs from A and B on the other side and mark where they cross.
(iii) It is geometrically exact; it does not depend on worn markings or on halving a measured length, so there is no measuring or rounding error.
(iv) P lies on the perpendicular bisector of AB, since any point equidistant from A and B lies on it.
In simple words: Equal arcs from both ends find the exact middle and the right-angled line through it.

Teacher's Note:
1. The same line gives the midpoint and the right angle.
2. Equidistant points lie on the bisector.
3. Exact constructions avoid reading errors.

Question 15. A paver-block footpath.
A school is laying a footpath with rectangular paver blocks. The ground staff first chalk the footpath into a grid of equal unit squares. One stretch of the path is 5 squares wide and 7 squares long. Each paver block covers exactly 2 unit squares, and a block may be laid lengthwise or crosswise, but it must sit exactly on the chalked squares. The staff want the stretch covered completely, with no square left bare and no block overlapping another. After trying for a long time they find that one square is always left over.
(i) How many unit squares does this stretch of the path contain?
(ii) Explain, by counting, why the stretch can never be covered completely by such blocks.
(iii) The staff reduce the width to 4 squares, keeping the length 7. Can the new stretch be covered? Describe a strategy.
(iv) How many blocks are needed for a stretch that is 6 squares wide and 9 squares long?
Answer:
(i) 5 × 7 = 35 unit squares.
(ii) Each block covers 2 squares, so blocks always cover an even number of squares. 35 is odd, so at least one square is always left bare.
(iii) Yes. 4 × 7 = 28 squares. The width 4 is even, so each of the 7 rows across the path takes 2 blocks laid across: 2 × 7 = 14 blocks.
(iv) 6 × 9 = 54 squares; 54 ÷ 2 = 27 blocks. Since 6 is even, this works with 3 blocks across in each of the 9 rows.
In simple words: An odd number of squares can never be covered by 2-square blocks; with an even side it always can.

Teacher's Note:
1. Odd total means failure.
2. Lay blocks along the even side.
3. Divide the squares by 2 for the number of blocks.

Question 16. A patchwork quilt.
A tailor is making a patchwork quilt from cloth patches. Every patch is either a square or an equilateral triangle, and all the patches have the same side length, so that any side of a patch can be stitched to any side of another. The tailor wants the quilt to have no gaps between patches and no patch folded over another. She studies the corner points of her design, where the corners of several patches meet, and checks each time whether the angles there fill the space around the point completely.
(i) Write the angle at a corner of each kind of patch.
(ii) At one corner point three triangular patches meet along with some square patches. How many square patches meet there?
(iii) Can two triangular patches and two square patches meet at one corner point? Show your working.
(iv) Describe a corner point of the quilt where only triangular patches meet.
Answer:
(i) Square patch: 90°; equilateral triangular patch: 60°.
(ii) 3 × 60° = 180°; the squares supply 360° − 180° = 180°, so 180° ÷ 90° = 2 squares.
(iii) No. 2 × 60° + 2 × 90° = 120° + 180° = 300° < 360°, leaving a 60° gap.
(iv) Six triangular patches meet there, since 6 × 60° = 360°.
In simple words: The angles at each corner must add to 360°: three triangles need two squares, and six triangles alone work.

Teacher's Note:
1. Find what is left of 360°.
2. A gap of 60° could take one more triangle.
3. Equal sides let any patches join.

Question 17. A jaali screen for the school verandah.
An artisan is cutting a stone jaali screen for the school verandah. The main motif is an eight-petalled rosette. To set it out he first marks a centre point and draws four straight supporting lines through it, so that eight equal angles are formed around the centre. The petals are then drawn as arcs on these supporting lines. The artisan works only with a straight edge and a pair of compasses, because the stone cannot be marked with a protractor. He later plans a second screen with a twelve-petalled rosette made the same way.
(i) Find the angle between two adjacent supporting lines of the eight-petalled rosette.
(ii) Describe how this angle is constructed with a ruler and a compass only.
(iii) What angle is needed between adjacent supporting lines of the twelve-petalled rosette, and how can it be constructed?
(iv) A visitor asks for a nine-petalled rosette, which would need an angle of 40° between adjacent arms. Explain why the artisan cannot set this out by these methods.
Answer:
(i) 360° ÷ 8 = 45°.
(ii) First construct 90° at the centre: cut the base line at two points equally far from the centre, draw equal arcs from them and join their meeting point to the centre. Then bisect the 90° angle to get 45°.
(iii) 360° ÷ 12 = 30°. Construct 60° with an equilateral triangle and bisect it.
(iv) From 60° and 90°, bisecting and adding only give 15° multiplied by a whole number or by a half, a quarter, an eighth, and so on (7.5°, 15°, 22.5°, 30°, 45°, …).
40° ÷ 15° = 2 \( \frac{2}{3} \), and a third is never obtained by halving. So 40° cannot be set out by these methods.
In simple words: 45° and 30° come from halving 90° and 60°, but 40° can never be reached by halving and adding.

Teacher's Note:
1. Divide 360° by the number of petals.
2. Bisection only halves angles.
3. A third is the stumbling block.

Question 18. Laying out an altar with a rope.
The Sulba-Sutras are ancient Indian texts that describe how fire altars were to be laid out on the ground. Instead of a compass, the worker used a rope. A line segment XY is drawn on the ground and a peg is fixed at X and at Y. A long rope with a loop at each end is folded in half and its midpoint is marked. The two loops are then slipped over the pegs at X and Y. The worker pulls the marked midpoint of the rope to one side of XY until both halves are fully stretched, and marks that position as A. He repeats this on the other side and marks the position as B. Finally he joins A and B.
(i) Which construction of this chapter is being carried out?
(ii) Why is AX = AY, and why is BX = BY?
(iii) Justify that AB is the perpendicular bisector of XY.
(iv) Using the same rope, how would the worker obtain a right angle at a given point O of the line?
Answer:
(i) The perpendicular bisector of XY, made with a rope instead of a compass.
(ii) The rope was folded in half, so the two parts on either side of the mark are equal. Fully stretched from the pegs, the mark is equally far from both: AX = AY, and in the same way BX = BY.
(iii) A and B are both equidistant from X and Y, so both lie on the perpendicular bisector of XY. Two points fix a line, so AB is the perpendicular bisector.
(iv) Use the rope to mark X and Y on the line at equal distances from O, so O is the midpoint. Find one point A as above and join AO; AO is at right angles to the line at O.
In simple words: The folded rope gives points equally far from X and Y, which lie on the perpendicular bisector.

Teacher's Note:
1. The rope halves act like equal compass arcs.
2. Two equidistant points fix the bisector.
3. For (iv), make O the midpoint first.

Question 19. Pentagon tiles for a hotel lobby.
An interior designer orders floor tiles shaped like regular pentagons for a hotel lobby, because she likes their look. The tiles are all of the same size and the same shape. When the masons begin laying them, they find that after fitting a few tiles around one point, a thin wedge-shaped gap is always left, which has to be filled with cement. The designer is told that this will happen at every such point, and that no tile of the same shape can ever close the gap. She is asked to choose a different regular shape so that the lobby can be tiled with no cement showing at all.
(i) Find the measure of each angle of a regular pentagon.
(ii) How many pentagon tiles can be fitted around a point without overlapping, and what is the size of the gap left?
(iii) Explain why no regular polygon tile can exactly fill that gap.
(iv) Suggest a regular shape for the new order and justify your choice by an angle calculation.
Answer:
(i) Each angle = \( \frac{(5-2)\times 180^\circ}{5} = \frac{540^\circ}{5} \) = 108°.
(ii) Three tiles: 3 × 108° = 324° < 360°, while 4 × 108° = 432° > 360° would overlap. Gap = 360° − 324° = 36°.
(iii) The smallest angle of any regular polygon is 60° (the equilateral triangle), and angles grow as sides increase. 36° is smaller than 60°, so no regular polygon has a 36° angle. (Solving \( \frac{(n-2)\times 180}{n} \) = 36 gives n = 2.5, not a whole number.)
(iv) Regular hexagons: each angle is 120° and 3 × 120° = 360°, so three tiles close up exactly at every point. (Squares, 4 × 90° = 360°, would also work.)
In simple words: Pentagons leave a 36° gap that no regular tile fits; hexagons or squares fill 360° exactly.

Teacher's Note:
1. Find the gap by subtracting from 360°.
2. No regular polygon has an angle below 60°.
3. Justify the new shape with a sum.

Question 20. A new tile made from a square.
In an art class the students make a tile of their own. Each starts with a square piece of card. A piece is cut away from the left edge of the square, and the very same piece is then joined on to the right edge, at the same height and without turning it. The new shape has a notch on its left and a matching bump on its right, as shown. The students then trace many copies of this new shape, sliding each copy sideways and upwards, and colour them to make a wall pattern.
(i) Is the area of the new tile the same as the area of the square it was made from? Give a reason.
(ii) Explain why copies of this new tile still cover the plane with no gaps and no overlaps.
(iii) If the cut-off piece is thrown away instead of being joined to the right edge, can copies of the shape that is left tile the plane? Give a reason.
(iv) Describe one more change the students can make to a square so that copies of the new shape still tile the plane.
Answer:
(i) Yes. The piece cut from the left is not thrown away but joined to the right. Nothing is lost and nothing added, so the area is unchanged.
(ii) The notch on the left and the bump on the right are the same shape at the same height. When the next copy is slid across by the width of the square, its bump fits exactly into the notch of its neighbour. The top and bottom edges are still straight, so rows stack like squares. Hence no gaps and no overlaps.
(iii) No. The right edge is now straight, so nothing fills the notch of the neighbouring tile; a hole the shape of the cut-off piece is left at every join, so there are gaps.
(iv) Cut a piece from the top edge and join the same piece to the bottom edge, at the same place and without turning it. The bump at the bottom of one tile fits the notch at the top of the tile below. Both changes can be made on the same square.
In simple words: Moving a piece from one side to the opposite side keeps the area and lets copies fit together like squares.

Teacher's Note:
1. Bump and notch must match exactly.
2. Slide copies; do not turn them.
3. This is the idea behind Escher-style tiles.