---
title: "NCERT Solutions Class 7 Maths Ch 14 Constructions and Tilings"
url: https://www.swavid.com/maths/class/7/chapter/constructions-and-tilings/ncert-solutions
dateModified: 2026-10-07T15:16:53+00:00
---

# NCERT Solutions Class 7 Maths Ch 14 Constructions and Tilings

This chapter contains various in-text questions and activities focused on geometric constructions using a ruler and compass, such as perpendicular bisectors, angle bisection, and parallel lines. It also includes inquiries into tiling patterns, tangram puzzles, and the mathematical properties of shapes like hexagons and equilateral triangles.

Free PDF (28 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-14-constructions-and-tilings-948b6b0f7d.pdf

## 6.1 Geometric Constructions

### Question 1

*2 marks · Very short answer*

Do you recall the ‘Eyes’ construction we did in Grade 6?

**Solution**

1. Yes, the construction involves drawing two arcs from two centres A and B to form symmetrical upper and lower arcs for the eye.
2. The centres A and B are determined such that the upper and lower arcs have the same radius, ensuring symmetry.

**Answer:** The 'Eyes' construction uses two centres A and B to draw symmetrical upper and lower arcs for the eye.

> Common mistake: Confusing the centres A and B with the endpoints of the line segment.

### Question 2

*3 marks · Short answer*

How do we find such A and B?

**Solution**

1. From points X and Y, draw arcs above and below the line segment XY using the same radius.
2. The point where the arcs meet above XY is named A.
3. The point where the arcs meet below XY is named B.

**Answer:** Points A and B are found by drawing arcs of the same radius from X and Y above and below the line segment XY.

> Common mistake: Using different radii for the arcs from X and Y.

### Question 3

*3 marks · Short answer*

In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the angle formed between them?

**Solution**

1. The line AB passes through the midpoint O of the line segment XY.
2. The angle formed between the line AB and the line segment XY is $90^\circ$.
3. Therefore, AB is the perpendicular bisector of XY.

**Answer:** AB intersects XY at its midpoint O, and the angle formed between them is $90^\circ$.

> Common mistake: Assuming the angle is not $90^\circ$ without justification.

### Question 4

*3 marks · Short answer*

Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length?

**Solution**

1. Yes, the line joining the two points of intersection of arcs drawn with the same radius from the endpoints of a line segment is always its perpendicular bisector.
2. This is because any point equidistant from the endpoints of a line segment lies on its perpendicular bisector.
3. Since A and B are equidistant from X and Y, they both lie on the perpendicular bisector of XY.

**Answer:** Yes, the line joining the two points will always be the perpendicular bisector of XY because any point equidistant from the endpoints of a line segment lies on its perpendicular bisector.

> Common mistake: Thinking the length of the radius affects whether it is a perpendicular bisector.

### Question 5

*3 marks · Short answer*

Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)?

**Solution**

1. The triangles $\Delta AOX$ and $\Delta AOY$ should be congruent.
2. If $\Delta AOX \cong \Delta AOY$, then $OX = OY$ and $\angle AOX = \angle AOY$.
3. Since $\angle AOX + \angle AOY = 180^\circ$, we get $\angle AOX = \angle AOY = 90^\circ$, proving AB is the perpendicular bisector.

**Answer:** The triangles $\Delta AOX$ and $\Delta AOY$ should be congruent to prove that O is the midpoint of XY and AB is perpendicular to XY.

> Common mistake: Identifying the wrong pair of triangles for congruence.

### Question 6

*3 marks · Short answer*

How do we get these different shapes? Try!

**Solution**

1. Different shapes are obtained by choosing different pairs of points (like C and D) on the perpendicular bisector of XY.
2. These points must satisfy the condition $CX = CY = DX = DY$.
3. Using these points as centres to draw arcs creates eyes of different shapes.

**Answer:** Different shapes are obtained by choosing different pairs of points on the perpendicular bisector of XY as centres to construct the upper and lower arcs.

> Common mistake: Choosing points not on the perpendicular bisector.

### Question 7

*3 marks · Short answer*

Will C and D lie on the perpendicular bisector AB?

**Solution**

1. The points C and D are at the same distance from both X and Y, meaning CX = CY and DX = DY.
2. Joining any two points that are at equal distances from X and Y gives the perpendicular bisector of XY.
3. Since a line segment XY has only one unique perpendicular bisector (the line AB), the points C and D must lie on the line AB.

**Answer:** Yes, C and D lie on the perpendicular bisector AB.

> Common mistake: Thinking that points equidistant from the endpoints can form a separate line rather than lying on the unique perpendicular bisector.

### Question 8

*4 marks · Proof*

Justify the following statement using the facts that we have established. Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.

**Solution**

1. Given a point P such that PX = PY.
2. Let AB be the perpendicular bisector of XY intersecting XY at O.
3. If P lies on AB, then by definition it is equidistant from X and Y.
4. Conversely, if a point P is equidistant from X and Y (PX = PY), and we join P to the midpoint O of XY, we form two triangles $\triangle POX$ and $\triangle POY$.
5. By SSS congruence condition (using PX = PY, OX = OY, and OP common), $\triangle POX \cong \triangle POY$.
6. Thus, $\angle POX = \angle POY = 90^\circ$, which shows that OP lies along the perpendicular bisector of XY. Hence proved.

**Answer:** Any point equidistant from X and Y lies on the perpendicular bisector of XY.

> Common mistake: Skipping the congruence step when proving the point lies on the perpendicular line.

### Question 9

*3 marks · Short answer*

Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?

**Solution**

1. Taking some fixed radius greater than half of XY, from X and then Y, construct two sufficiently long arcs above XY meeting at A.
2. Using the same radius, from X and then Y, construct two sufficiently long arcs below XY meeting at B.
3. Join points A and B using an unmarked ruler to get the required perpendicular bisector AB.

**Answer:** The line AB obtained by joining the intersection points of arcs above and below XY is the perpendicular bisector.

> Common mistake: Changing the compass radius between drawing arcs from X and Y.

### Question 10

*3 marks · Short answer*

Figure it Out: 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.

**Solution**

1. No, it is not necessary to have the same radius for the arcs above XY as the radius used for the arcs below XY.
2. The two points A (above) and B (below) only need to be independently equidistant from X and Y.
3. Thus, the radius used for arcs above XY can be different from the radius used for arcs below XY, and the resulting line AB will still be the unique perpendicular bisector.

**Answer:** It is not necessary; different radii can be used for the top and bottom pairs of arcs as long as each point is equidistant from X and Y.

> Common mistake: Assuming all four arcs must be drawn with the exact same radius value.

### Question 11

*3 marks · Short answer*

Figure it Out: 2. Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.

**Solution**

1. No, we cannot construct both pairs of arcs on the same side of XY to get the perpendicular bisector.
2. Two distinct points are required to uniquely determine a straight line.
3. Constructing arcs on the same side gives only one intersection point, which together with another point does not form the bisector unless two separate pairs of intersecting arcs on opposite sides (or two distinct points equidistant from X and Y) are used.

**Answer:** No, arcs must be on opposite sides (or we need two distinct points) to determine a unique line.

> Common mistake: Trying to draw two intersection points on the same side, which fails to create a line perpendicular and bisecting XY.

### Question 12

*3 marks · Short answer*

Figure it Out: 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.

**Solution**

1. Yes, while constructing a single pair of intersecting arcs from X and Y, we must use the same radius for both arcs.
2. If different radii are used from X and Y, the intersection point will not be at an equal distance from both X and Y.
3. Equal radii ensure that the point of intersection is equidistant from X and Y, which is the fundamental condition for lying on the perpendicular bisector.

**Answer:** Yes, the same radius must be used from both X and Y to ensure the intersection point is equidistant from them.

> Common mistake: Changing the compass width while moving from center X to center Y for the same intersecting arc pair.

### Question 13

*Activity*

Figure it Out: 4. Recreate this design using only a ruler and compass —

**Solution**

1. Draw a line segment and construct its perpendicular bisector using a compass and ruler.
2. Use the intersection points of the arcs as centres to construct the symmetrical flower petal design as shown in the textbook figure.

**Answer:** The design is successfully recreated using only a ruler and a compass.

### Question 14

*3 marks · Short answer*

Can we extend the method of constructing the perpendicular bisector to construct a 90° angle at any point on a line? Draw a line and mark a point O on it. Construct a 90° angle at point O.

**Solution**

1. Draw a straight line and mark a point O on it.
2. Using a compass with any suitable radius and centre O, mark two points X and Y on the line on either side of O such that O is the midpoint of XY.
3. Construct the perpendicular bisector of the line segment XY, which will pass through O and form a $90^\circ$ angle with the given line.

**Answer:** Yes, by marking points X and Y equidistant from O and constructing the perpendicular bisector of XY, we obtain a $90^\circ$ angle at point O.

> Common mistake: Not taking points X and Y at equal distances from O.

### Question 15

*3 marks · Short answer*

Find a segment of this line for which O is the midpoint.

**Solution**

1. Extend the given line on either side of point O.
2. Using a compass, cut off equal lengths on both sides of O along the line.
3. Label the two intersection points as X and Y, making O the exact midpoint of the line segment XY.

**Answer:** The segment is XY, where O is the midpoint such that OX = OY.

> Common mistake: Marking unequal lengths on the two sides of O.

### Question 16

*3 marks · Short answer*

In this case, do we need to draw two pairs of intersecting arcs to get the perpendicular bisector of XY?

**Solution**

1. Observe that when constructing a $90^\circ$ angle at a given point O on a line, point O already lies on the required perpendicular bisector.
2. Since we need only two points to determine a straight line, point O serves as one point.
3. Therefore, we only need to construct one more point above (or below) the line using arcs from X and Y to draw the perpendicular bisector, so two pairs of intersecting arcs are not required.

**Answer:** No, we do not need to draw two pairs of intersecting arcs because point O is already known to lie on the perpendicular bisector.

> Common mistake: Unnecessarily drawing both upper and lower pairs of arcs.

### Question 17

*3 marks · Proof*

Figure it Out: 1. Justify why AB in Fig. 6.4 is the perpendicular bisector.

**Solution**

1. Given a line segment XY with poles at X and Y, a rope of fixed length with its midpoint marked is fastened at X and Y.
2. When the midpoint is pulled fully stretched to position A above XY and position B below XY, we get AX = AY and BX = BY.
3. Since any point that is at equal distance from the two endpoints of a line segment lies on its perpendicular bisector, points A and B both lie on the perpendicular bisector of XY.
4. Hence, the line AB is the required perpendicular bisector of XY. (Hence proved.)

**Answer:** AB is the perpendicular bisector because points A and B are equidistant from both X and Y.

> Common mistake: Failing to state the property that points equidistant from endpoints lie on the perpendicular bisector.

### Question 18

*3 marks · Short answer*

Figure it Out: 2. Can you think of different methods to construct a 90° angle at a given point on a line using a rope?

**Solution**

1. Fix poles at two points X and Y on the line such that the given point O is the midpoint of XY.
2. Take a rope folded to find its midpoint and use the Sulba-Sutra method to find a point A such that AX = AY and BX = BY.
3. Join AB to get the perpendicular bisector of XY, which passes through O and forms a $90^\circ$ angle with the given line.

**Answer:** A $90^\circ$ angle can be constructed by marking points X and Y equidistant from O on the line and constructing the perpendicular bisector using a stretched rope.

> Common mistake: Not ensuring that point O is the exact midpoint of the segment on the line.

### Question 19

*3 marks · Short answer*

How do we construct this figure?

**Solution**

1. First, draw a set of supporting lines passing through a common centre O such that the angle between each pair of adjacent lines is $45^\circ$.
2. Mark points on these supporting lines at equal distances from the centre O.
3. Using these points as centres and with a suitable radius, construct upper and lower arcs to form the symmetrical petals of the figure.

**Answer:** The 8-petalled figure is constructed by first drawing supporting lines at $45^\circ$ intervals and then constructing symmetrical arcs using these lines.

> Common mistake: Drawing supporting lines without equal angles between them, leading to asymmetrical petals.

### Question 20

*3 marks · Short answer*

What is the angle between two adjacent lines?

**Solution**

1. A complete angle around a point is $360^\circ$.
2. The figure is divided into 8 equal parts by the adjacent supporting lines.
3. Divide $360^\circ$ by 8 to find the angle between each pair of adjacent lines: $\frac{360^\circ}{8} = 45^\circ$.

**Answer:** $45^\circ$

> Common mistake: Dividing by 4 or 6 instead of the number of parts (8).

### Question 21

*3 marks · Short answer*

How do we construct a 45° angle using only a ruler and a compass?

**Solution**

1. First, construct a $90^\circ$ angle at the given point using a ruler and a compass.
2. Bisect the $90^\circ$ angle into two equal parts using the angle bisection method.
3. Each resulting angle measures $\frac{90^\circ}{2} = 45^\circ$.

**Answer:** A $45^\circ$ angle is constructed by first constructing a $90^\circ$ angle and then bisecting it.

> Common mistake: Trying to construct $45^\circ$ directly with a compass without constructing $90^\circ$ first.

### Question 22

*3 marks · Short answer*

How do we construct these congruent triangles, given the angle?

**Solution**

1. Mark points A and B on the arms of the angle such that $OA = OB$.
2. With A and B as centres, draw arcs of the same sufficiently long radius to intersect at point C.
3. In triangles $\Delta OBC$ and $\Delta OAC$, $OB = OA$, $BC = AC$, and $OC$ is common, so by SSS congruence condition, $\Delta OBC \cong \Delta OAC$, which ensures $\angle BOC = \angle AOC$.

**Answer:** Triangles $\Delta OBC$ and $\Delta OAC$ are proven congruent using the SSS condition to show that $\angle BOC = \angle AOC$.

> Common mistake: Not taking equal radii when cutting arcs from points A and B.

### Question 23

*Activity*

Figure it Out: 1. Construct at least 4 different angles. Draw their bisectors.

**Solution**

1. Construct four different angles using a ruler and compass.
2. Apply the standard angle bisection method to draw the bisector for each angle.

**Answer:** Activity completed by drawing four different angles and their respective bisectors.

### Question 24

*Activity*

Figure it Out: 2. Construct the 8-petalled figure shown in Fig. 6.5.

**Solution**

1. Draw supporting lines radiating from a common centre with $45^\circ$ angles between adjacent lines.
2. Use compasses and suitable radii to draw the upper and lower arcs on each supporting line to form the 8-petalled figure.

**Answer:** Activity completed by constructing the 8-petalled figure using supporting lines and arcs.

### Question 25

*3 marks · Short answer*

Figure it Out: 3. In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.

**Solution**

1. 1. When arcs of equal radius are drawn on the other side intersecting at C, we form triangles similar to the standard angle bisection method.
2. 2. By the SSS congruence condition, the triangles formed by joining O to C and the points on the arms are congruent.
3. 3. Therefore, the line OC still bisects the given angle.

**Answer:** Yes, the line OC will still be an angle bisector.

> Common mistake: Thinking that changing the side of the intersecting arcs changes the bisection property.

### Question 26

*3 marks · Short answer*

Figure it Out: 4. What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle?

**Solution**

1. 1. Repeated bisection of a standard 90▰ or 60▰ angle allows us to construct angles like $45^\circ$, $22.5^\circ$, $30^\circ$, $15^\circ$, and $7.5^\circ$.
2. 2. A $65.5^\circ$ angle cannot be constructed using only ruler and compass because it does not result from standard bisections or additions of $60^\circ$ and $90^\circ$ multiples.
3. 3. Thus, standard angle bisection produces angles whose measures are obtained by dividing fundamental constructible angles by powers of 2.

**Answer:** Angles obtained by successive bisection of $90^\circ$ or $60^\circ$ can be constructed; a $65.5^\circ$ angle cannot be constructed.

> Common mistake: Assuming any decimal angle can be constructed using a compass.

### Question 27

*3 marks · Short answer*

Figure it Out: 5. Come up with a method to construct the angle bisector using a rope.

**Solution**

1. 1. Fix a loop of the rope at the vertex O and mark equal lengths OA and OB along the two arms of the angle.
2. 2. Take a rope of fixed length, tie loops at the ends, and place them at points A and B.
3. 3. Pull the midpoint of the rope tightly to get a point C; the line joining O to C gives the angle bisector.

**Answer:** The angle bisector can be constructed using a rope by marking equidistant points on the arms and pulling the rope taut from those points.

> Common mistake: Not keeping the rope fully stretched while marking the midpoint.

### Question 28

*Activity*

6. Construct the following figure.

**Solution**

1. 1. This activity demonstrates the construction of a symmetric four-petalled design using intersecting arcs.
2. 2. Observation: The design is symmetric about perpendicular axes passing through the center of the enclosing shape.

**Answer:** Completed the four-petalled design using a ruler and compass.

### Question 29

*3 marks · Short answer*

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

**Solution**

1. 1. To construct petals of the maximum possible size within a given square, the centers of the arcs must be chosen at the corners or midpoints of the sides of the square.
2. 2. The radius of the arcs should equal the side length or half the diagonal of the square.
3. 3. This ensures the arcs touch the boundaries of the square symmetrically and occupy the maximum possible area.

**Answer:** Choose the vertices of the square as centers and the side length of the square as the radius to construct maximum-sized petals.

> Common mistake: Choosing incorrect centers or radii, resulting in petals that overflow or underfill the square.

### Question 30

*Activity*

Construct the following figure.

**Solution**

1. 1. This activity demonstrates the repetition of geometric units in two different orientations to form a continuous pattern.
2. 2. Observation: Exact copies of the basic triangular or angular unit must be created using equal arm lengths and equal angles.

**Answer:** Completed the construction of the repeating unit figure (Fig. 6.6).

### Question 31

*Activity*

Draw an angle. Create a copy of this angle using only a ruler and compass.

**Solution**

1. Draw an arc from vertex A to form an isosceles triangle ABC.
2. Draw an arc of the same radius from vertex X of the new line segment to get point Z.
3. Measure length BC using a compass and transfer it to the arc from Z to locate point Y.
4. Join XY to form the copied angle, as triangles ABC and XYZ are congruent by SSS.

**Answer:** An exact copy of the given angle is constructed using SSS congruence.

### Question 32

*Activity*

Figure it Out: 1. Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

**Solution**

1. Draw four different angles in various orientations on a page without measuring them with a protractor.
2. Apply the angle copying method using a ruler and compass to reproduce each angle accurately.

**Answer:** Multiple angles are successfully copied using ruler and compass constructions.

### Question 33

*Activity*

Figure it Out: 2. Construct the Fig. 6.6.

**Solution**

1. Analyze the repeating unit in Fig. 6.6 with its specific arm length and angle.
2. Construct multiple exact copies of the unit in two different orientations using the angle copying method to complete Fig. 6.6.

**Answer:** Fig. 6.6 is constructed by repeating the unit with equal arm lengths and congruent angles.

### Question 34

*3 marks · Short answer*

How do we implement this idea using a ruler and a compass?

**Solution**

1. Construct a transversal line l intersecting the given line m at point A.
2. Choose a point B on line l through which the parallel line needs to be drawn.
3. Construct an angle at B corresponding to the angle between m and l by copying it, so that the alternate or corresponding angles are equal, ensuring the lines are parallel.

**Answer:** A line parallel to the given line is constructed by making equal corresponding angles using a ruler and compass.

> Common mistake: Not transferring the exact arc length when copying the corresponding angle.

### Question 35

*Activity*

Figure it Out: 1. Construct 4 pairs of parallel lines in different orientations.

**Solution**

1. Draw a base line in four different orientations on paper.
2. Use the ruler and compass parallel line construction method to draw a parallel line for each orientation.

**Answer:** Four pairs of parallel lines are constructed in different orientations.

### Question 36

*Activity*

Figure it Out: 2. Construct the following figure.

**Solution**

1. Analyze the angles and symmetry required for the complex star-like figure.
2. Use perpendicular bisectors, angle bisection, and circle arcs to construct the complete symmetric design.

**Answer:** The complex star-like figure is constructed step-by-step using fundamental geometric constructions.

### Question 37

*3 marks · Short answer*

How did they make these arches?

**Solution**

1. 1. The first step is to be able to draw them on a plane surface such as paper or stone.
2. 2. Ancient architects used geometric methods and support lines involving symmetry and arcs to construct these arches.
3. 3. By drawing appropriate support lines and using compass arcs of equal radii, these arches were constructed accurately.

**Answer:** They made these arches by drawing them on a plane surface using geometric support lines and arcs.

> Common mistake: Thinking arches were drawn entirely freehand without underlying geometric support lines.

### Question 38

*Activity*

Construct this arch shape on a piece of paper.

**Solution**

1. 1. Take a piece of paper and draw the required base support lines.
2. 2. Use a compass to construct the arcs meeting at points B and C to form the arch shape.

**Answer:** The arch shape is successfully constructed on paper using a ruler and compass.

### Question 39

*3 marks · Short answer*

How would you construct these support lines?

**Solution**

1. 1. For symmetry, we must have $AB = CD$ and $\angle BAD = \angle CDA$.
2. 2. Construct equal angles at endpoints A and D on the base line AD.
3. 3. Mark points B and C on the arms of the angles such that $AB = CD$ to complete the support lines.

**Answer:** Construct equal angles at A and D, and mark points B and C such that AB = CD.

> Common mistake: Failing to ensure that the angles constructed at both ends are equal.

### Question 40

*Activity*

Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

**Solution**

1. 1. Use the constructed support lines to set the centres and radii for the arcs.
2. 2. Draw the arcs connecting the points to form a smooth and aesthetically pleasing arch.

**Answer:** The arch is constructed using the support lines and adjusted arcs.

### Question 41

*3 marks · Short answer*

How do we construct this shape?

**Solution**

1. 1. This shape is constructed using two intersecting arcs meeting at a top vertex.
2. 2. The supporting lines are two line segments of equal length forming an inverted V-shape.
3. 3. Arcs drawn from the endpoints and meeting points create the pointed arch.

**Answer:** The pointed arch is constructed using two equal supporting line segments and intersecting arcs.

> Common mistake: Using unequal line segments for the supporting structure.

### Question 42

*3 marks · Short answer*

What supporting lines will you use to draw this arch?

**Solution**

1. 1. The supporting lines used are just two line segments of equal length.
2. 2. These segments are placed to meet at an angle, similar to the 'Wavy Wave' from the Grade 6 textbook.
3. 3. When their midpoints or endpoints are marked, they serve as the base for drawing the pointed arch.

**Answer:** The supporting lines are two line segments of equal length meeting at a common vertex.

> Common mistake: Choosing lines of different lengths which destroys the symmetry of the pointed arch.

### Question 43

*3 marks · Short answer*

If their midpoints are marked, will you be able to construct a pointed arch?

**Solution**

1. Given the two line segments forming the supporting lines of the pointed arch as shown in Fig. 6.11.
2. Mark the midpoints of both supporting line segments using the perpendicular bisector method.
3. Using these midpoints as centres and the distance to the endpoints as radius, draw arcs that meet at the top to form the pointed arch.

**Answer:** Yes, by marking the midpoints of the supporting line segments, we can use them as centres to construct the upper arcs of the pointed arch.

> Common mistake: Choosing incorrect centres that do not lie on the supporting lines.

### Question 44

*Activity*

Figure it Out: 1. Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.

**Solution**

1. Draw the two supporting line segments as given in Fig. 6.11.
2. Construct arcs using different radii from the centres to observe how the shape of the pointed arch changes.
3. Observation: Changing the radius alters the curvature and height of the pointed arch.

**Answer:** Constructed different pointed arches by varying the radius of the arcs.

### Question 45

*Activity*

Figure it Out: 2. Make your own arch designs.

**Solution**

1. Use ruler and compass to draw custom supporting lines.
2. Apply arcs and geometric constructions to create unique arch designs.
3. Observation: Symmetrical arch designs are produced by using equal radii and matching centres.

**Answer:** Created personal arch designs using geometric constructions.

### Question 46

*Activity*

How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths.

**Solution**

1. Try constructing a pentagon and hexagon with equal side lengths using a ruler and compass.
2. Observation: A regular hexagon can be easily constructed using 60° angles, whereas a regular pentagon requires more advanced methods.
3. Conclusion: Constructing a regular hexagon is achievable with standard compass and ruler tools.

**Answer:** Explored constructions of pentagons and hexagons with equal side lengths.

### Question 47

*3 marks · Short answer*

Can we break a regular hexagon into smaller pieces that can be constructed?

**Solution**

1. A regular hexagon can be broken down into smaller, simpler geometric shapes that are easier to construct.
2. By joining the opposite vertices of a regular hexagon, it can be divided into six smaller triangles.
3. These smaller pieces are congruent equilateral triangles which can be easily constructed using a compass and ruler.

**Answer:** Yes, a regular hexagon can be broken down into six congruent equilateral triangles.

> Common mistake: Assuming the triangles are not equilateral without checking the angles.

### Question 48

*3 marks · Short answer*

What happens when we join the ‘opposite’ points of a regular hexagon? Since a regular hexagon has equal sides and angles, can we expect a figure like this?

**Solution**

1. When the opposite points of a regular hexagon are joined as shown in Fig. 6.12, they intersect at the centre O.
2. This division creates six triangles meeting at the centre point with angles around the point adding up to $360^\circ$.
3. Since the hexagon is regular, all triangles formed are congruent equilateral triangles with each angle measuring $60^\circ$.

**Answer:** Yes, joining the opposite points of a regular hexagon results in a figure composed of six congruent equilateral triangles meeting at the centre.

> Common mistake: Failing to verify that all angles around the centre sum up to $360^\circ$.

### Question 49

*3 marks · Short answer*

Will all the triangles in the figure be equilateral triangles?

**Solution**

1. In a regular hexagon, all sides are equal and all interior angles are $120^\circ$.
2. When opposite vertices are joined, each of the six triangles formed at the centre has two sides equal to the circumradius (which equals the side length of the hexagon).
3. Thus, all the six triangles formed in Fig. 6.12 are equilateral triangles.

**Answer:** Yes, all the six triangles in Fig. 6.12 are equilateral triangles.

> Common mistake: Assuming triangles inside any polygon are equilateral without checking side lengths.

### Question 50

*3 marks · Short answer*

Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon?

**Solution**

1. Six congruent equilateral triangles each have an angle of $60^\circ$ at the vertex meeting at the centre.
2. The sum of these six angles around the point is $6 \times 60^\circ = 360^\circ$, which completely covers the region without gaps or overlaps.
3. Therefore, placing them together results in a regular hexagon.

**Answer:** Yes, six congruent equilateral triangles can be placed together to form a regular hexagon.

> Common mistake: Forgetting that angles around a point must sum to $360^\circ$.

### Question 51

*3 marks · Short answer*

Consider this figure. Will the 70° angle fit into the gap? What is the gap angle ∠AOI?

**Solution**

1. The sum of all angles around a point is equal to $360^\circ$.
2. Given the angles are $40^\circ, 60^\circ, 50^\circ, 30^\circ, 40^\circ$, and $90^\circ$, their sum is $40^\circ + 60^\circ + 50^\circ + 30^\circ + 40^\circ + 90^\circ = 310^\circ$.
3. The gap angle $\angle AOI = 360^\circ - 310^\circ = 50^\circ$.

**Answer:** The gap angle $\angle AOI$ is $50^\circ$, so the $70^\circ$ angle will not fit.

> Common mistake: Calculation errors when summing angles around a point.

### Question 52

*3 marks · Short answer*

Use this to determine whether the 70°angle fits the gap.

**Solution**

1. The calculated gap angle is $50^\circ$.
2. For an angle to fit into a gap without overlapping or leaving extra space, its measure must be equal to the gap angle.
3. Since $70^\circ > 50^\circ$, the $70^\circ$ angle is larger than the gap and will overlap, so it does not fit.

**Answer:** No, the $70^\circ$ angle does not fit the gap because the gap is only $50^\circ$.

> Common mistake: Comparing angles without checking the exact remaining gap measure.

### Question 53

*3 marks · Short answer*

In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines?

**Solution**

1. Each triangle in Fig. 6.12 is an equilateral triangle with all interior angles equal to $60^\circ$.
2. At vertex O, the angles forming the line segment AOD consist of two adjacent angles of the equilateral triangles meeting on a straight path, each measuring $60^\circ + 60^\circ = 120^\circ$ combined with adjacent angles to form $180^\circ$.
3. Since the sum of adjacent angles along segments AOD, BOE, and COF is $180^\circ$, they form straight lines.

**Answer:** AOD, BOE, and COF are straight lines because the sum of adjacent angles along each segment is $180^\circ$.

> Common mistake: Assuming lines are straight visually without verifying that the angle sum is $180^\circ$.

### Question 54

*Activity*

Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.

**Solution**

1. Draw a line segment AX of length $4\text{ cm}$.
2. Construct a $60^\circ$ angle at point A using a compass and ruler.
3. Mark the remaining sides of length $4\text{ cm}$ successively by drawing arcs of radius $4\text{ cm}$ to complete the six vertices of the regular hexagon.

**Answer:** A regular hexagon of side length $4\text{ cm}$ is constructed using a ruler and compass.

> Common mistake: Not maintaining a consistent radius of $4\text{ cm}$ for all sides.

### Question 55

*3 marks · Short answer*

How do we do it?

**Solution**

1. Draw a line segment AX and construct a $60^\circ$ angle at point A using a compass.
2. Extend the ray of the $60^\circ$ angle to form a straight line or construct two consecutive $60^\circ$ arcs from point B along the arc.
3. The angle adjacent to the $60^\circ$ angle on the straight line measures $180^\circ - 60^\circ = 120^\circ$.

**Answer:** A $120^\circ$ angle is constructed by making two successive $60^\circ$ arcs along the compass arc.

> Common mistake: Measuring with a protractor instead of using a compass and ruler as required.

### Question 56

*3 marks · Short answer*

How do we construct a 60° angle?

**Solution**

1. Draw a line segment AX and take point A as the centre. Construct an arc with any convenient radius that intersects AX at a point B.
2. With the same radius and centre B, cut another arc that intersects the first arc at a point C.
3. Join AC and extend it. The angle formed $\angle CAX = 60^\circ$.

**Answer:** A $60^\circ$ angle is constructed by drawing an equilateral triangle using equal radii for the arcs.

> Common mistake: Changing the radius of the compass between marking the first and second arcs.

### Question 57

*3 marks · Short answer*

Why is ∠CAX = 60°? Is there an equilateral triangle here?

**Solution**

1. Join points A, B, and C by line segments AB, BC, and CA.
2. Since we used the same radius for all arcs, we have $\text{AB} = \text{BC} = \text{CA}$, which means triangle ABC is an equilateral triangle.
3. Each interior angle of an equilateral triangle is $60^\circ$, therefore $\angle CAX = 60^\circ$.

**Answer:** $\angle CAX = 60^\circ$ because triangle ABC is an equilateral triangle formed by equal radii.

> Common mistake: Stating the angle is $60^\circ$ without mentioning that triangle ABC is equilateral.

### Question 58

*3 marks · Short answer*

Construct a regular hexagon of sidelength 5 cm.

**Solution**

1. Draw a circle or a sequence of arcs with a radius of $5\text{ cm}$.
2. Step-by-step mark six arcs of radius $5\text{ cm}$ along the circumference.
3. Join the consecutive points of intersection to obtain a regular hexagon of sidelength $5\text{ cm}$.

**Answer:** A regular hexagon of sidelength $5\text{ cm}$ is constructed using compass arcs of radius $5\text{ cm}$ connected consecutively.

> Common mistake: Not keeping the compass radius strictly equal to $5\text{ cm}$ throughout.

### Question 59

*3 marks · Short answer*

How will you construct 30° and 15° angles?

**Solution**

1. First construct a $60^\circ$ angle using a ruler and compass.
2. Bisect the $60^\circ$ angle to obtain a $30^\circ$ angle.
3. Bisect the $30^\circ$ angle to obtain a $15^\circ$ angle.

**Answer:** $30^\circ$ is obtained by bisecting a $60^\circ$ angle, and $15^\circ$ is obtained by bisecting a $30^\circ$ angle.

> Common mistake: Bisecting the wrong angle or confusing angle bisector steps.

### Question 60

*3 marks · Short answer*

Construct the following 6-pointed star. Note that it has a rotational symmetry.

**Solution**

1. Construct a regular hexagon with six equilateral triangles meeting at a central point.
2. Extend the alternating sides of the hexagon to form the points of the 6-pointed star.
3. Erase the unnecessary construction lines to leave the final 6-pointed star.

**Answer:** The 6-pointed star is constructed by extending the sides of an underlying regular hexagon or intersecting two equilateral triangles.

> Common mistake: Incorrectly positioning the intersecting triangles so the star loses rotational symmetry.

### Question 61

*3 marks · Short answer*

Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG — equilateral? Why?

**Solution**

1. Yes, the six triangles forming the points of the star are equilateral triangles.
2. Each point of the star is formed by an isosceles triangle whose base angles are equal to the interior angles of a regular hexagon, which are each $120^\circ$ supplementary or calculated via the central angles.
3. Alternatively, since the angles around the center of each triangular point add up such that each triangle has angles of $60^\circ, 60^\circ, 60^\circ$, all six triangles are equilateral.

**Answer:** Yes, the six triangles are equilateral because all their interior angles measure $60^\circ$.

> Common mistake: Assuming the triangles are only isosceles without checking that their vertex and base angles are all equal to $60^\circ$.

### Question 62

*Activity*

Figure it Out: 1. Construct the following figures: (a) An Inflexed Arc (b) The fun part about this figure is that it can also be constructed using only a compass! Can you do it?

**Solution**

1. This activity demonstrates how to construct complex curved geometric patterns like an inflexed arc and a flower-like rosette using only a compass and ruler.
2. Observation: By adjusting the radius and keeping the center points on supporting lines or intersecting arcs, smooth inflexed curves can be successfully drawn.

**Answer:** Constructed the inflexed arc and rosette designs successfully using a compass.

### Question 63

*Activity*

Figure it Out: 2. Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

**Solution**

1. This activity involves drawing specific lines and shapes that trick the human eye into perceiving depth, distortion, or movement due to contrasting background angles.
2. Observation: Recreating the figure in the notebook using exact parallel and intersecting lines reproduces the optical illusion effect.

**Answer:** Recreated the optical illusion figure successfully in the notebook.

### Question 64

*Activity*

Figure it Out: 3. Construct this figure.

**Solution**

1. This activity requires analyzing the given figure to identify its underlying angles and supporting lines.
2. Observation: Using angle bisection and circle arcs of fixed radii, the required geometric figure is successfully constructed.

**Answer:** Constructed the geometric figure using standard ruler and compass techniques.

### Question 65

*Activity*

Figure it Out: 4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P.

**Solution**

1. Draw a line $l$ and mark a point $P$ outside it.
2. Using $P$ as center, draw an arc that intersects line $l$ at two points, say $A$ and $B$.
3. Construct the perpendicular bisector of line segment $AB$, which will pass through point $P$ and form a perpendicular to line $l$.

**Answer:** Constructed a perpendicular to the given line $l$ passing through the external point $P$.

### Question 66

*Activity*

Figure it Out: How can the tangram pieces be rearranged to form each of the following figures?

**Solution**

1. Take the 7 standard tangram pieces obtained by dividing a square.
2. Rotate and translate the pieces without overlapping to match the silhouettes of the given target figures.

**Answer:** Rearranged the 7 tangram pieces to form the given shapes successfully.

### Question 67

*3 marks · Short answer*

Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles?

**Solution**

1. State that a $4 \times 6$ grid contains 24 unit squares and each $2 \times 1$ tile covers 2 unit squares.
2. Use vertical or horizontal tiles to cover each column or row completely.
3. Conclude that a $4 \times 6$ grid can be tiled using multiple copies of $2 \times 1$ tiles.

**Answer:** Yes, a $4 \times 6$ grid can be tiled using $2 \times 1$ tiles.

> Common mistake: Thinking that only one specific orientation of tiles is allowed.

### Question 68

*3 marks · Short answer*

Can a 4 × 7 grid be tiled using 2 × 1 tiles?

**Solution**

1. State that a $4 \times 7$ grid has a total of 28 unit squares.
2. Observe that 28 is an even number, which is a multiple of the tile area (2 unit squares).
3. Conclude that the grid can be tiled using $2 \times 1$ tiles by placing vertical tiles in columns and horizontal tiles as needed.

**Answer:** Yes, a $4 \times 7$ grid can be tiled using $2 \times 1$ tiles.

> Common mistake: Assuming odd dimensions cannot be tiled at all.

### Question 69

*3 marks · Short answer*

What about a 5 × 7 grid?

**Solution**

1. State that a $5 \times 7$ grid has 35 unit squares.
2. Observe that each $2 \times 1$ tile covers exactly 2 unit squares.
3. Conclude that since 35 is an odd number and cannot be divided evenly by 2, a $5 \times 7$ grid cannot be tiled using $2 \times 1$ tiles.

**Answer:** No, a $5 \times 7$ grid cannot be tiled using $2 \times 1$ tiles.

> Common mistake: Forgetting that total area must be a multiple of the tile area.

### Question 70

*3 marks · Short answer*

Complete the justification.

**Solution**

1. State that a region to be tiled must have a total area equal to the sum of the areas of the tiles used.
2. Note that each $2 \times 1$ tile has an area of 2 square units.
3. Conclude that any tileable grid must have an even total number of unit squares.

**Answer:** Any tileable grid with $2 \times 1$ tiles must have an even number of unit squares.

> Common mistake: Ignoring the area requirement of individual tiles.

### Question 71

*3 marks · Short answer*

Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it.

**Solution**

1. State that if both $m$ and $n$ are even, the total number of unit squares is even.
2. Use the general strategy to cover each column completely with vertical $2 \times 1$ tiles since the number of rows is even.
3. Conclude that an $m \times n$ grid is always tileable when both dimensions are even.

**Answer:** Yes, an $m \times n$ grid is tileable when both $m$ and $n$ are even, by covering each column with vertical tiles.

> Common mistake: Assuming horizontal tiles are strictly necessary.

### Question 72

*3 marks · Short answer*

Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it.

**Solution**

1. State that if one dimension is even and the other is odd, the total number of unit squares in the $m \times n$ grid is the product of an even and an odd number, which is always even.
2. Apply the strategy of using vertical or horizontal strips to tile the grid completely.
3. Conclude that the grid is tileable with $2 \times 1$ tiles.

**Answer:** Yes, an $m \times n$ grid is tileable when one of the dimensions is even and the other is odd.

> Common mistake: Assuming that an odd dimension makes tiling impossible even if the total area is even.

### Question 73

*3 marks · Short answer*

Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons.

**Solution**

1. An $m \times n$ grid has a total of $m \times n$ unit squares.
2. If both $m$ and $n$ are odd, their product $m \times n$ is an odd number of unit squares.
3. Since each $2 \times 1$ tile covers 2 unit squares, an odd number of squares cannot be completely covered without a leftover square. Therefore, the grid is not tileable.

**Answer:** No, an $m \times n$ grid with both $m$ and $n$ odd cannot be tiled with $2 \times 1$ tiles because the total number of squares is odd.

> Common mistake: Thinking that having an even total number of squares is the only condition required for tiling.

### Question 74

*3 marks · Short answer*

Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?

**Solution**

1. A $5 \times 3$ grid has a total of 15 unit squares.
2. When one unit square is removed, the remaining grid has $15 - 1 = 14$ unit squares, which is an even number.
3. However, whether it is tileable depends on the position of the removed square using the black-and-white coloring method, and not just the total count.

**Answer:** Having an even number of squares does not automatically make it tileable; it depends on the position of the removed square.

> Common mistake: Assuming all regions with an even number of squares are tileable.

### Question 75

*3 marks · Short answer*

Is the following region tileable with 2 × 1 tiles?

**Solution**

1. Color the region using an alternating black-and-white grid pattern as taught in the chapter.
2. Count the number of black squares and white squares covered by the region.
3. If the number of black squares and white squares is unequal, the region cannot be tiled with $2 \times 1$ tiles.

**Answer:** A region is tileable only if the number of black squares equals the number of white squares in its black-and-white coloring.

> Common mistake: Failing to check the color parity of the squares.

### Question 76

*3 marks · Short answer*

What about this one?

**Solution**

1. Examine the given region in Fig. 6.13 and color it with alternating black and white squares.
2. Count the black and white squares in the colored region.
3. Since the number of black and white squares is unequal, the region cannot be tiled.

**Answer:** No, the region in Fig. 6.13 is not tileable with $2 \times 1$ tiles.

> Common mistake: Trying to fit tiles without checking square counts.

### Question 77

*3 marks · Short answer*

Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?

**Solution**

1. We know the region is not tileable because coloring the $5 \times 3$ grid like a chessboard results in an unequal number of black and white squares.
2. Any $2 \times 1$ tile always covers one black square and one white square.
3. Removing a corner square of the $5 \times 3$ grid leaves an unequal number of black and white squares, making it non-tileable.

**Answer:** Removing any square of the same color as the majority color in the chessboard coloring of the $5 \times 3$ grid makes it non-tileable.

> Common mistake: Assuming any square removal from an odd grid behaves the same way.

### Question 78

*3 marks · Short answer*

If the plain grid is tileable, is the black-and-white-grid tileable?

**Solution**

1. Yes, if the plain grid is tileable, we can place the $2 \times 1$ tiles on it.
2. Each tile placed on the plain grid will cover one black square and one white square of the corresponding black-and-white grid.
3. Thus, a valid tiling of the plain grid directly gives a valid tiling of the black-and-white grid.

**Answer:** Yes, if the plain grid is tileable, the black-and-white grid is also tileable.

> Common mistake: Confusing grid tiling with color matching rules.

### Question 79

*3 marks · Short answer*

If the black-and-white grid is tileable, is the plain grid tileable?

**Solution**

1. Yes, if the black-and-white grid is tileable, the plain grid is also tileable.
2. Each tile covers one black square and one white square of the grid.
3. Removing the colours leaves the same region, which can therefore be tiled using plain $2 \times 1$ tiles.

**Answer:** Yes, the plain grid is tileable.

> Common mistake: Assuming colouring changes the geometric tileability of the grid.

### Question 80

*3 marks · Short answer*

Is the black-and-white region in Fig. 6.14 tileable?

**Solution**

1. No, the black-and-white region in Fig. 6.14 is not tileable.
2. Each $2 \times 1$ black-and-white tile covers exactly one black square and one white square.
3. The region has 8 white squares and 6 black squares, so the number of black and white squares is unequal, making tiling impossible.

**Answer:** No, it is not tileable because the number of black and white squares is unequal.

> Common mistake: Forgetting that each domino must cover one square of each colour.

### Question 81

*3 marks · Short answer*

Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?

**Solution**

1. A $5 \times 3$ grid has a total of 15 unit squares.
2. If we colour the grid like a chessboard, one colour will have 8 squares and the other will have 7 squares.
3. Removing any square of the majority colour leaves 7 squares of one colour and 7 squares of the other, but depending on the position (such as a corner square), it may leave unequal counts of black and white squares adjacent or total, making it non-tileable.

**Answer:** Removing any corner square of the $5 \times 3$ grid makes it non-tileable.

> Common mistake: Choosing a square that preserves equal counts of both colours.

### Question 82

*3 marks · Short answer*

Figure it Out: Are the following tilings possible? 1. Region to be tiled 2. Region to be tiled

**Solution**

1. Count the total number of unit squares in the given region.
2. Colour the squares in an alternating black-and-white pattern.
3. If the number of black squares and white squares is different, the region cannot be tiled by $2 \times 1$ tiles.

**Answer:** Check by colouring; if black and white squares are unequal, the tiling is not possible.

> Common mistake: Only checking total area without checking colour parity.

### Question 83

*3 marks · Short answer*

Can you think of a shape whose copies can tile the entire plane?

**Solution**

1. A shape whose copies can cover the entire plane without gaps or overlaps is called a tiling shape.
2. Squares are a primary example of such shapes.
3. Equilateral triangles and regular hexagons can also tile the entire plane.

**Answer:** Squares, equilateral triangles, and regular hexagons can tile the entire plane.

> Common mistake: Listing shapes that leave gaps, like regular pentagons.

### Question 84

*3 marks · Short answer*

Are there other regular polygons that can tile the plane? What about equilateral triangles?

**Solution**

1. Yes, other regular polygons can tile the plane.
2. Equilateral triangles can tile the plane because six equilateral triangles meet at a point with angles adding up to $360^\circ$.
3. Regular hexagons can also tile the plane for the same reason.

**Answer:** Yes, equilateral triangles and regular hexagons can tile the plane.

> Common mistake: Thinking only squares can tile a plane.

## Frequently asked questions

### How many questions are there in Class 7 Maths Chapter 14 Constructions and Tilings of the new NCERT book?

This chapter in the new NCERT book for the 2026-27 session contains a total of 84 questions under geometric constructions. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page.

### What topics do the questions cover in the Class 7 Maths Constructions and Tilings chapter?

The questions cover various concepts such as angle bisection, angle bisection with rope, angles around a point, arch construction, and the construction of figures like the 6-pointed star and the 8-petalled figure. SwaVid's free PDF on this page provides detailed explanations for all these topics.

### Which question types are included in Class 7 Maths Chapter 14 and how should I approach them?

The chapter includes activity-based questions, proofs, short answer questions, and very short answer questions. To score well, you should carefully read the geometric instructions and follow a logical sequence for every construction step.

### How can I write answers to get full marks in Class 7 Maths Chapter 14 Constructions and Tilings?

To get full marks, you need to write clear step-by-step construction procedures alongside neat geometric diagrams. You can refer to the solutions provided in SwaVid's free PDF on this page to understand the correct presentation format.

### Is the free PDF for Class 7 Maths Chapter 14 available for download?

Yes, SwaVid provides a free PDF with comprehensive step-by-step solutions for this chapter. You can access and download this material directly from this page to help with your exam preparation for the 2026-27 session.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student