Geometric Constructions — Questions and Answers
Welcome to HSLC Guru. This page gives the complete solutions to ASSEB Class 7 New Mathematics Chapter 14, Geometric Constructions — a step-by-step compass-and-straightedge guide to the perpendicular bisector, a right angle, an angle bisector, an equal angle, a parallel line, a regular hexagon and tiling, together with every “Work it Out” box, the hexagon activity and all of Exercise 14, with an inline figure for each construction.
Summary
When a line segment is divided into two equal parts it is said to be bisected; and a line that bisects a segment perpendicularly (making a 90° angle) is called its perpendicular bisector. Every point that is equidistant from the two endpoints of a segment lies on its perpendicular bisector — this single property is behind the leaf-like rangoli shapes and many of the constructions in this chapter.
Using only a compass and a scale we can construct a perpendicular bisector, a right angle at any point on a segment, the bisector of an angle, an angle equal to a given angle (without measuring it), a line parallel to a given line, a 60° angle and a regular hexagon. A 60° angle gives us equilateral triangles and regular hexagons; in fact a regular hexagon is made of 6 congruent equilateral triangles and each of its interior angles measures 120°.
Finally we learn tiling — covering a surface with one or more shapes leaving no gap and no overlap. For a grid to be tileable by 2×1 tiles the number of cells must be even, and, by a checkerboard (chess-board) marking, the number of marked and empty cells must also be equal.
Summary: This ASSEB Class 7 New Mathematics Chapter 14 (Geometric Constructions) guide shows, step by step with compass and straightedge, how to construct a perpendicular bisector, a right angle at any point on a segment, the bisector of an angle, an angle equal to a given angle, a parallel line, a 60° angle and a regular hexagon, and it explains tiling (tessellation) with the checkerboard test. It solves every “Work it Out” box (14.1–14.7), the hexagon activity and all of Exercise 14, with inline figures for each construction.
Methods of Geometric Construction
Construction of a Perpendicular Bisector (14.2)
- Step 1: With P as centre and a radius greater than half of PQ, draw two arcs — one above and one below the segment PQ.
- Step 2: With Q as centre and the same radius, draw two more arcs above and below PQ. Name the points where the arcs intersect above and below as M and N.
- Step 3: Join M and N. The segment MN is the perpendicular bisector of PQ; it cuts PQ at its midpoint O (PO = OQ) and makes a 90° angle there.
A Right Angle (90°) at Any Point on a Segment (14.3)
- Step 1: With the compass, take two points A and B on either side of O such that AO = BO — that is, O is the midpoint of AB.
- Step 2: Since O is the midpoint of AB, the perpendicular bisector of AB passes through O. So from A and B draw one pair of intersecting arcs of equal radius (above or below the line) meeting at P, and join PO. Then PO makes a 90° angle at O.
Bisector of an Angle (14.4)
- Step 1: With the vertex Q of $\angle PQR$ as centre, draw an arc that cuts the arms QP and QR at X and Y. Here QX = QY (same radius).
- Step 2: With X and Y as centres, draw two arcs of equal radius meeting at Z. Join Q and Z; QZ is the bisector of $\angle PQR$, so $\angle PQZ = \angle RQZ$.
An Angle Equal to a Given Angle (14.5)
- Step 1: Draw a segment PQ. With the vertex B of $\angle ABC$ as centre, draw an arc cutting BA and BC at X and Y.
- Step 2: With the same radius and P as centre, draw an arc cutting PQ at M.
- Step 3: With the compass measure the distance XY, and from M draw an arc of that radius cutting the previous arc at N. Join P and N; then $\angle NPQ = \angle ABC$.
Constructing a Parallel Line (14.6)
- Step 1: Draw a transversal q cutting the given line p at X. Take any point Y on q through which the parallel line will pass.
- Step 2: With X and Y as centres, draw arcs of equal radius. The arc at X cuts p and q at A and B; the arc at Y cuts q at C.
- Step 3: With the compass take the length AB and, from C, draw an arc cutting the previous arc at D.
- Step 4: Draw the line r through Y and D. Since the corresponding angles are equal, r ∥ p.
Regular Hexagon and 6 Equilateral Triangles (14.8)
Drawing the diagonals of a regular hexagon gives 6 triangles at the centre O. The 6 angles at the centre add up to 360°, so each is $360° ÷ 6 = 60°$. Since O is equidistant from every vertex, each triangle is isosceles with a 60° apex angle — hence each is an equilateral triangle. So a regular hexagon is made of 6 congruent equilateral triangles, and each interior angle is $60° + 60° = 120°$.
Construction: On a segment AX, with A as centre draw an arc meeting AX at B; then with B as centre and the same radius cut the arc at C. As AB = BC = CA, we get $\angle CAB = 60°$. Now take PQ = 3 cm, draw a 60° exterior angle at Q with QR = 3 cm, and repeat a 60° exterior angle at each vertex to complete the regular hexagon.
Tiling (14.10)
A 5×4 grid has 20 unit squares; it can be completely covered by 2×1 tiles (each tile covers 2 unit squares) — for example two horizontal tiles per row give a tiling with 10 tiles.
But even an even number of cells does not always allow tiling. To decide, mark the grid like a chess-board: a 2×1 tile always covers one marked and one empty cell, so tiling is possible only if the marked and empty cells are equal in number. In the 3×5 grid below there are 8 marked but only 7 empty cells — unequal — so it cannot be tiled by 2×1 tiles.
Regular polygons such as regular hexagons, equilateral triangles and squares can also tile a surface without gaps. The figure below shows a honeycomb-like tiling made of regular hexagons.
Textbook Questions and Answers
Work it Out 14.1
1. Can the perpendicular bisector of line segment XY pass through points X and Y? Give reasons.
Answer: No. The perpendicular bisector passes through the midpoint O of XY and meets XY at that one point only. X and Y are the endpoints, not the midpoint. If it also passed through X, it would meet XY at two points (O and X) — but two distinct lines meet at only one point. So it cannot pass through X or Y.
2. Can PX and PY be equal for a point P that does not lie on the perpendicular bisector of XY? Give reasons.
Answer: No. Every point equidistant from X and Y lies on the perpendicular bisector, and conversely only the points on the perpendicular bisector are equidistant from X and Y. So if P is not on it, then $PX \ne PY$.
3. Draw two pairs of intersecting arcs from X and Y with the same radius such that the line joining the two intersection points is not the perpendicular bisector of XY. Is such a construction possible? Give reasons.
Answer: It is not possible. If the arcs from X and Y have equal radius, each intersection point is equidistant from X and Y, so both lie on the perpendicular bisector, and the line joining them is the perpendicular bisector. (Only if the radii were unequal would the joining line fail to be the perpendicular bisector.) Hence, with equal radii, such a construction cannot be made.
4. Instead of one pair of arcs above and one pair below XY, can the perpendicular bisector be obtained by drawing two pairs of intersecting arcs on only one side of XY?
Answer: Yes. Two intersection points on the same side of XY (obtained with two different but within-each-pair equal radii) are both equidistant from X and Y, so both lie on the perpendicular bisector. Since two distinct points determine a line, joining and extending them gives the perpendicular bisector.
5. Is it necessary to draw arcs of the same radius on both the upper and lower sides of XY? What happens if arcs of different radii are drawn? Judge the logic with a figure.
Answer: The radius used for the upper pair need not equal the radius used for the lower pair. What matters is that at a single intersection point the arc from X and the arc from Y have equal radius, so that the point is equidistant from X and Y. If, at one intersection point, the arcs from X and Y have different radii, that point is not equidistant and the joining line is not the perpendicular bisector. A figure shows that only the line through two equidistant points is the perpendicular bisector.
Work it Out 14.2
1. Draw 5 angles of any measurement and construct the bisector for each of them.
Answer: For each angle $\angle PQR$ use the same method: with the vertex Q as centre draw an arc marking X and Y on the arms; then from X and Y draw two equal-radius arcs meeting at Z; join QZ, which is the bisector. Repeat for five different angles (e.g. 50°, 70°, 90°, 110° and 140°).
2. In step 2 of the angle bisector we drew equal arcs meeting inside the angle. If these two arcs are drawn outside the angle (as in the adjacent figure), will the line ZQ be the bisector of the angle?
Answer: Yes. Even if the arcs meet at Z outside the angle, Z is still equidistant from X and Y (equal radii) and QX = QY. So $\triangle XQZ \cong \triangle YQZ$ (SSS), giving $\angle XQZ = \angle YQZ$. The line ZQ lies along the same bisector line, so it still bisects $\angle PQR$.
3. Using the angle bisector, transform Figure 14.11(i) into the figure shown (with more petals).
Answer: In the original figure the 4 base lines split the 360° at the centre into four 90° angles. Bisecting each 90° angle gives 45° base lines and creates 4 new base lines. Draw one petal (a leaf-shape made with perpendicular-bisector arcs) along each new 45° base line to get the 8-petal figure.
4. Construct 135° and 67.5° angles using the angle bisector. Try to construct a 22.5° angle.
Answer: First construct a 90° angle on a straight line at a point. Bisecting the 90° angle between the 90° ray and the other half of the straight line gives $90° + 45° = 135°$. Bisecting this 135° angle gives $135° ÷ 2 = 67.5°$. And constructing a 45° angle (bisector of 90°) and bisecting it again gives $45° ÷ 2 = 22.5°$.
5. Try to find a method of constructing an angle bisector with the help of a rope.
Answer: As in the Śulba method, use the rope to lay off equal lengths. From the vertex Q mark equal lengths on both arms (QX = QY) using the rope. Then stretch equal lengths of rope from X and Y to meet at a common point Z (XZ = YZ). Join Q and Z — QZ is the bisector; here the rope does the work of the compass.
6. Construct the adjacent figure. Construct the flower petals so that they cover the maximum area of the square.
Answer: Draw a square and draw its two diagonals together with the two lines joining the midpoints of opposite sides — these 4 lines split the 360° at the centre into eight 45° angles (the central right angles get bisected). Along each 45° base line draw one petal (a leaf-shape using perpendicular-bisector arcs); drawn wide enough, the petals cover the maximum area of the square.
Work it Out 14.3
1. Give the reasoning why $\angle NPQ$ equals $\angle ABC$ in the above construction.
Answer: The arc from B and the arc from P have the same radius, so BX = BY = PM = PN. Also MN was drawn equal to XY, so MN = XY. In $\triangle BXY$ and $\triangle PMN$: BX = PM, BY = PN (equal radii) and XY = MN (transferred). Hence $\triangle BXY \cong \triangle PMN$ (SSS), so the corresponding angles $\angle XBY = \angle MPN$, i.e. $\angle ABC = \angle NPQ$.
2. Using a scale and a compass, construct angles equal to the angles given below.
Answer: For each given angle apply the “equal angle” method (without measuring): draw an arc at the vertex marking X and Y on the arms; on a new segment draw an arc of the same radius giving a point M; then with the compass transfer the distance XY from M to cut the arc at N, and join the endpoint to N. This reproduces each given angle exactly.
Work it Out 14.4
1. Draw parallel lines to each of the given lines (a), (b), (c) using the method of constructing equal angles.
Answer: For each line use the same method: draw a transversal cutting the given line at X, take a point Y on the transversal, and at Y construct an angle equal to the corresponding angle at X. Because equal corresponding angles mean the lines are parallel, the new line through Y is parallel to the given line. Apply this to (a), (b) and (c).
Work it Out 14.5
1. Draw some designs of your favourite arches.
Answer: This is a drawing activity. The trefoil-arch method (draw an equilateral triangle, take the midpoint of the base, draw arcs and finally a semicircle on top) can be adapted; using arcs of various radii, semicircles and the equilateral-triangle base, draw a few arch designs of your own choice.
2. Try to draw the arches seen in Diwan-i-Aam, Red Fort, etc.
Answer: This is also a drawing activity. Most arches of the Diwan-i-Aam and the Red Fort are cusped/multifoil (scalloped) arches. To draw them, use an equilateral-triangle base and draw several circular arcs from different centres, joining them up — the underlying idea is the same: equal-radius arcs and the 60° angle of the equilateral triangle give symmetric shapes.
Activity (Regular Hexagon)
Draw the regular hexagon of Figure 14.19, cut out the 6 triangles formed by the diagonals and place them one on top of the other. What did you see? Did the triangles match completely? Is $\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 5 = \angle 6$?
Answer: Placed one on the other, the 6 triangles match completely — they are congruent. The 6 angles at the centre add up to $360°$, so $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 = 360°$, i.e. $6\angle 1 = 360°$, giving $\angle 1 = 60°$. So $\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 5 = \angle 6 = 60°$. Since the centre is equidistant from all vertices (radii of one circle), each triangle is isosceles with a 60° apex, hence equilateral. Therefore a regular hexagon is made of 6 congruent equilateral triangles, and each interior angle is $60° + 60° = 120°$.
Work it Out 14.6
1. Draw the following geometric figures — (a) Kite, (b) an equilateral triangle inscribed in a circle, (c) a hexagon inscribed in a circle, (d) a square formed by joining the centres of four touching circles.
Answer:
- (a) Kite: Draw a vertical diagonal; with the top end as centre and one radius mark two points on either side, and with the bottom end as centre and a different radius mark the same two points. Joining the four points gives a kite with two pairs of adjacent equal sides.
- (b) Equilateral triangle in a circle: Draw a circle; with the radius as the arc length, mark 6 points around the circle, then join alternate points to get the inscribed equilateral triangle.
- (c) Hexagon in a circle: Join all 6 of those points in order. Since the hexagon’s side equals the radius, this gives a regular hexagon inscribed in the circle.
- (d) Square from four touching circles: Draw four equal circles of radius r, each touching two neighbours; joining the four centres gives a square of side $2r$.
2. Draw a star as shown. It has 6 congruent triangles of one size and 2 congruent triangles of another size. Can you identify the triangles?
Answer: The six-pointed star is formed by two congruent large equilateral triangles (one pointing up, one pointing down) overlapped — these are the ‘2 congruent triangles’. The 6 points of the star are 6 congruent small equilateral triangles — these are the ‘6 congruent triangles’.
3. Draw a line ℓ and take a point P outside the line. Draw a line through P parallel to ℓ.
Answer: Draw a transversal from P cutting ℓ at X. At P construct an angle equal to the angle the transversal makes with ℓ at X (equal corresponding angle, using the equal-angle method). Since the corresponding angles are equal, the new line through P is parallel to ℓ.
Work it Out 14.7
1. Check whether the following grids can be tiled with 2×1 tiles. (a), (b), (c), (d)
Answer: Two conditions — (i) the total number of cells must be even, and (ii) on a chess-board marking the number of marked and empty cells must be equal.
- (a) A staircase with rows of 1, 2, 3, 4 cells, total $1+2+3+4 = 10$ (even). But on a chess-board marking there are 6 marked and 4 empty cells — unequal. So even though the count is even, tiling is not possible.
- (b) Total cells $2+2+4+5 = 13$ — odd. 13 cannot be split evenly into 2-cell tiles, so tiling is not possible.
- (c) Total cells $1+2+3+3+2+1 = 12$ — even; and the chess-board marking gives 6 marked and 6 empty — equal. So tiling is possible (each vertically adjacent pair can be covered by an upright tile).
- (d) A diagonal staircase band with rows of 1, 2, 3, 4, 5, 4, 3, 2, 1 cells, total $1+2+3+4+5+4+3+2+1 = 25$ — odd. So tiling is not possible.
Exercise 14
1. Construct two line segments of 6 cm such that one is the perpendicular bisector of the other. From this, construct a square.
Answer: Draw a 6 cm segment AC and construct its perpendicular bisector BD, cutting AC at right angles at the midpoint O (take BO = OD = 3 cm so that BD is also 6 cm). Join A, B, C, D. Since the diagonals are equal (6 = 6) and bisect each other at right angles, ABCD is a square.
2. Construct two line segments of 5 cm and 4 cm such that one is the perpendicular bisector of the other. From this, construct a rhombus.
Answer: Draw a 5 cm segment AC and, as its perpendicular bisector, a 4 cm segment BD cutting AC at right angles at the midpoint O (O is the midpoint of both). Join A, B, C, D. The diagonals are unequal (5 ≠ 4) but bisect each other at right angles, so all four sides are equal — hence ABCD is a rhombus.
3. Construct right angles at any two points on a line segment of 8 cm.
Answer: Draw an 8 cm segment PQ. Take any two points R and S on it. At each point construct a right angle (by the method of 14.3): mark two points equidistant on either side of the point, draw equal-radius arcs from them to meet at a point, and join. Then both R and S carry a 90° angle on PQ.
4. Bisect the following angles using a compass and a scale: (A) 40° (B) 70° (C) 120° (D) 150°.
Answer: Draw each angle (using a protractor), and with the vertex as centre draw an arc cutting both arms at X and Y; from X and Y draw equal-radius arcs meeting at Z, and join the vertex to Z to get the bisector. The half-angles obtained are — (A) $40° ÷ 2 = 20°$, (B) $70° ÷ 2 = 35°$, (C) $120° ÷ 2 = 60°$, (D) $150° ÷ 2 = 75°$.
5. Construct angles equal to the given angles (about 55°, 65°, 80° and 130°) using a compass and a scale.
Answer: For each given angle apply the equal-angle method (without measuring): (i) draw an arc at the vertex marking X and Y on the arms; (ii) on a new segment draw an arc of the same radius giving a point M; (iii) with the compass transfer the distance XY from M to cut the arc at N, and join the endpoint to N. This reproduces the 55°, 65°, 80° and 130° angles exactly.
6. Construct two parallel lines AB and CD using a compass and a scale such that the distance between them is 5 cm.
Answer: Draw line AB. At a point on AB erect a perpendicular (method of 14.3) and, along this perpendicular, mark a point 5 cm away from AB. Through that point draw a line CD parallel to AB (using the equal-corresponding-angle method, or by erecting a second perpendicular of the same length). Then AB ∥ CD with the perpendicular distance between them equal to 5 cm.
7. Determine whether True or False.
Answer: (i) A line segment can have two perpendicular bisectors — False (it has only one). (ii) Each interior angle of a regular hexagon measures 60° — False (it is 120°). (iii) A regular hexagon contains 6 equilateral triangles — True. (iv) An angle equal to a given angle cannot be drawn using only a compass and a scale — False (it can). (v) Points equidistant from the two endpoints of a segment lie on its perpendicular bisector — True.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The only two instruments used in geometric construction are — (a) protractor and eraser (b) compass and scale (c) scale and eraser (d) pencil and protractor
Answer: (b) compass and scale.
2. How many perpendicular bisectors can be drawn for a line segment? (a) 1 (b) 2 (c) 3 (d) infinitely many
Answer: (a) 1.
3. A perpendicular bisector cuts the segment at its midpoint at an angle of — (a) 45° (b) 60° (c) 90° (d) 120°
Answer: (c) 90°.
4. Each angle of an equilateral triangle measures — (a) 45° (b) 60° (c) 90° (d) 120°
Answer: (b) 60°.
5. Each interior angle of a regular hexagon measures — (a) 60° (b) 90° (c) 108° (d) 120°
Answer: (d) 120°.
6. Bisecting a 90° angle gives an angle of — (a) 30° (b) 45° (c) 60° (d) 75°
Answer: (b) 45°.
7. How many congruent equilateral triangles make up a regular hexagon? (a) 3 (b) 4 (c) 6 (d) 8
Answer: (c) 6.
8. Covering a surface with shapes leaving no gap and no overlap is called — (a) bisection (b) tiling (c) inscribing (d) symmetry
Answer: (b) tiling.
9. A single 2×1 tile always covers — (a) 1 cell (b) 2 cells (c) 3 cells (d) 4 cells
Answer: (b) 2 unit cells.
10. The sum of the two angles on a straight line at a point (a straight angle) is — (a) 90° (b) 180° (c) 270° (d) 360°
Answer: (b) 180°.
Fill in the Blanks
1. Dividing a line segment into two equal parts is called ______.
Answer: bisection.
2. While constructing an angle bisector, an ______ is first drawn with the vertex as centre.
Answer: arc.
3. From a 60° angle we can construct ______ triangles and regular hexagons.
Answer: equilateral.
4. Each angle formed at the centre of a regular hexagon is ______.
Answer: 60°.
5. Covering a surface with the same or different shapes leaving no gap is called ______.
Answer: tiling.
True or False
1. A perpendicular bisector cuts a segment at its midpoint.
Answer: True.
2. Using a compass and a scale, an angle equal to a given angle can be drawn without measuring it.
Answer: True.
3. A 3×5 grid (15 cells) can be completely tiled with 2×1 tiles.
Answer: False (15 is odd, so it is not possible).
4. Each side of a regular hexagon equals the radius of its circumscribed circle.
Answer: True.
5. The bisector arcs must meet inside the angle; meeting outside does not give the bisector.
Answer: False (whether the arcs meet inside or outside, the same bisector line is obtained).
Short Answer Questions
1. Why does a line segment have only one perpendicular bisector?
Answer: A perpendicular bisector must pass through the midpoint of the segment and make a 90° angle there. A segment has only one midpoint, and only one line can be drawn through that midpoint making a 90° angle; hence there is only one perpendicular bisector.
2. Show briefly why each interior angle of a regular hexagon is 120°.
Answer: A regular hexagon is made of 6 equilateral triangles. At each vertex two 60° angles of these triangles meet, so the interior angle $= 60° + 60° = 120°$.
3. Show, with an example, that an even number of cells does not guarantee that tiling is possible.
Answer: An even count alone is not enough. On a chess-board marking the marked and empty cells must also be equal, because each 2×1 tile covers one marked and one empty cell. For example, the staircase with rows 1, 2, 3, 4 has 10 cells (even) but 6 marked and 4 empty — unequal — so it cannot be tiled.
4. How do you draw a line through a point outside a given line, parallel to it?
Answer: Draw a transversal from the point cutting the given line, and at the point construct an angle equal to the angle the transversal makes with the line (equal corresponding angle, by the equal-angle method). As the corresponding angles are equal, the new line is parallel to the given line.
Key Terms
| Term | Meaning |
|---|---|
| Bisection | Dividing a line segment or an angle into two equal parts |
| Perpendicular bisector | A line that bisects a segment at its midpoint at a 90° angle |
| Angle bisector | A line that divides an angle into two equal parts |
| Arc | A part of the circumference of a circle |
| Radius | The distance from the centre to the circle |
| Midpoint | The point that divides a segment into two equal parts |
| Parallel lines | Lines that never meet and stay the same distance apart |
| Regular hexagon | A polygon with six equal sides and equal angles |
| Equilateral triangle | A triangle with all three sides and angles equal |
| Interior angle | The inside angle at a vertex of a polygon |
| Tiling / Tessellation | Covering a surface with no gap and no overlap |
| Śulba-Sūtra | The ancient method of drawing geometric shapes with rope and stakes |