NCERT Solutions Class 7 Maths Chapter 14 Constructions and Tilings

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 7 · M AT H S

NCERT Solutions

Chapter 14: Constructions and
Tilings

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 137 – 162 25 81 English

Solutions, notes, sample papers & more at 100 pages

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 14: Constructions and
Tilings
Chapter 6 of Ganita Prakash Grade 7 Part II puts an unmarked ruler and a compass in your hands and asks
how much geometry those two tools alone can build. Perpendicular bisectors, 90°, 60°, 45° and 30° angles,
copied angles, parallel lines, arches, regular hexagons and stars — every one of them is justified by a
congruence argument, not by measurement. The second half turns to tiling: covering a region with shapes,
no gaps and no overlaps, and the beautiful black-and-white trick that proves when a covering is impossible.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 137 – 162

SECTIONS QUESTIONS

25 81

MEDIUM

English

In-text Questions — Page 137
Section 6.1 Geometric Constructions — Eyes

Q1 How do we find such A and B?

Draw arcs of the same radius from X and from Y — one pair above the line XY and one pair
below it. The point where the upper arcs cross is A; the point where the lower arcs cross is B.

A is on the arc from X ⇒ AX = radius

A is on the arc from Y ⇒ AY = same radius

So AX = AY

In the same way BX = BY

Both pairs of arcs used the same radius, so

AX = AY = BX = BY

Page 1 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A
arcs from X arcs from Y

X Y

B

Same radius from X and from Y. The arcs cross at A above XY and at B below it.

Why it happens: the compass keeps the radius fixed. Every point on the arc drawn
from X is at that fixed distance from X, and every point on the arc drawn from Y is at
the same fixed distance from Y. A crossing point belongs to both arcs, so it is at that
one distance from X and from Y. That is exactly the condition AX = AY the eye needs.

Tip: make the radius clearly more than half of XY, otherwise the two arcs will not
meet at all.

Page 2 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q2 In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the
angle formed between them?

A

X Y

B

Fig. 6.1, page 137.

AB cuts XY at its midpoint, and the angle between them is 90°.

Let O be the point where AB meets XY.

Then OX = OY — O is the midpoint of XY
and ∠AOX = ∠AOY = 90°

Page 3 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: A and B were both built to be equally far from X and from Y. Fold

m l a se
the picture along AB: X lands exactly on Y. A fold line that carries X onto Y must pass
o g a segment into
.c middle of XY and must stand square to it. A line thatacuts
m
through the
twoas e parts and is perpendicular to it is called the perpendicular bisector.
ag l equal

o m
c ag
Check it yourself: measure OX and OY with a compass, not a scale — open the
compass to OX and swing it across to m .
a s e Y. It should land exactly on Y.

agl

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
em . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q3 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?

A

X Y

B

Fig. 6.1, page 137 — the segment XY with arcs struck from X and from Y, meeting above
at A and below at B.

Yes, always. The only thing we used was AX = AY = BX = BY, and that is true for every length of
XY and every radius that lets the arcs meet.

Page 5 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Given: AX = AY, BX = BY, AB common

Step 1. In ΔABX and ΔABY:

AX = AY, BX = BY, AB = AB
⇒ ΔABX ≅ ΔABY by SSS

⇒ ∠XAB = ∠YAB, i.e. ∠XAO = ∠YAO

Step 2. In ΔAOX and ΔAOY:

AX = AY, ∠XAO = ∠YAO, AO = AO

⇒ ΔAOX ≅ ΔAOY by SAS

⇒ OX = OY and ∠AOX = ∠AOY

Step 3. ∠AOX and ∠AOY together make a straight angle:

∠AOX + ∠AOY = 180°

2 × ∠AOX = 180°

∠AOX = 90°

Why it happens: notice that no number ever entered the argument — not the
length of XY, not the radius of the arcs. Only the four equal distances were used. So
the conclusion holds for every segment and every workable radius. That is what
makes it a construction and not a lucky drawing.

Q4 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)?

ΔAOX and ΔAOY.

Page 6 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

If ΔAOX ≅ ΔAOY, then by corresponding parts

OX = OY ⇒ O is the midpoint of XY

∠AOX = ∠AOY
and since ∠AOX + ∠AOY = 180° (straight angle at O),

∠AOX = ∠AOY = 90° ⇒ AB ⊥ XY

Why it happens: "perpendicular bisector" is really two claims at once — equal
halves, and a right angle. One congruence delivers both, because congruent
triangles hand us equal sides and equal angles in a single step.

Tip: to get ΔAOX ≅ ΔAOY we still need ∠XAO = ∠YAO. That comes from the bigger
pair ΔABX ≅ ΔABY (SSS), which uses AB as the common side.

In-text Questions — Page 139
Eyes of different shapes · Construction of Perpendicular Bisector

Q1 How do we get these different shapes? Try!

The three eye shapes printed on page 138 — a long thin one, a broader one with rings,
and a round one. Each is made only of arcs drawn with a compass.

Keep the same XY, but choose a different pair of centres on the perpendicular bisector. Pick
points C and D with CX = CY = DX = DY, using a bigger or smaller radius than before, and draw
the two arcs from C and D.

Page 7 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Bigger radius ⇒ the centres sit far from XY ⇒ arcs are flatter ⇒ a long, narrow eye

Smaller radius ⇒ the centres sit near XY ⇒ arcs are rounder ⇒ a wide, open eye

Why it happens: the eye is drawn by two arcs that both start at X and end at Y. Their
shape depends only on how curved they are, and the curvature depends on the
radius. A small radius bends the arc a lot; a large radius keeps it almost straight. The
corners X and Y never move, so the eye stays symmetric whatever radius you
choose.

Try This: draw four eyes on the same XY using four different radii. Colour only the
boundaries and rub out the supporting arcs — you get four faces with four different
expressions.

Q2 Will C and D lie on the perpendicular bisector AB?

Yes. C and D must lie on the line AB.

C is at the same distance from X and Y (CX = CY)

D is at the same distance from X and Y (DX = DY)

By the earlier result, the line CD is a perpendicular bisector of XY

But a segment has only one perpendicular bisector, and that is AB

⇒ line CD is the line AB ⇒ C and D lie on AB

Why it happens: we proved that joining any two points equidistant from X and Y
gives the perpendicular bisector. C and D are two such points. Since XY cannot have
two different perpendicular bisectors, the line through C and D has to be the very
same line AB.

Page 8 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
Justify the following statement using the facts that we have established. Any point
e
Q3

m l as
.co
that has the same distance from X and Y lies on the perpendicular bisector of XY.

a g
se m
g l a
a

Let P be any point with PX = PY. We show P is on the perpendicular bisector AB.

. c om ag
Take a second point Q with QX = QY, on m
s e the other side of XY.
glaPQ is the perpendicular bisector of XY.
By the result already proved, thealine

co m
em.
as
XY has exactly one perpendicular bisector, namely AB.
om= line AB.
So line.cPQ
a g l
emon line PQ ⇒ P lies on AB.
aPslies
agl
m a s
em
.co
Why it happens: a shorter way to see it. Join P to O, the midpoint of XY. In ΔPOX and
agl
a s
agl add to 180°, each is 90°. So PO is perpendicular to XY
ΔPOY we have PX = PY, OX = OY and PO common, so ΔPOX ≅ ΔPOY by SSS. Hence
∠POX = ∠POY, and since they
at its midpoint — that is the perpendicular bisector, and P is on it.

co m
m .
o m l a se
.c g the arch, the hexagon
Did you know? This one sentence is the whole engine of the chapter. Every
m a
ase
construction that follows — the 90° angle, the angle bisector,

agl
— is built by finding points at equal distances from two fixed points.

se m
com g l a
m . a
ase
agl
Q4 Given a line segment XY, how do we draw its perpendicular bisector using only an
unmarked ruler and a compass?

co m
.

se m
m a
Three steps, no measuring at all.
o l
ofgXY. With centre X draw a long
1..cOpen the compass to some fixed radius, more than half a
se m
a
agl
arc above XY; with the same radius and centre Y draw another long arc above XY. Call their
crossing point A.
.c
s e m
m a
2. With the same radius, draw the matching pair of arcs below XY. Call their crossing point B.

m . co
3. Join A and B with the ruler. AB is the perpendicular bisector of XY.
e agl
g l as
a

com
m .
m ase
.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A

O
X Y

B

Step 1 gives A, step 2 gives B, step 3 joins them. O is the midpoint and the angle at O is 90°.

Why it happens: A and B are each at equal distances from X and Y, so both lie on the
perpendicular bisector. Two points fix a line, so the line through A and B is that
perpendicular bisector.

Tip: this is a better way to find the midpoint of a segment than measuring with a
scale. A scale can be read wrong by half a millimetre; the compass method has no
reading in it at all.

Figure it Out — Page 140

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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Construction of Perpendicular Bisector

Q1 When constructing the perpendicular bisector, is it necessary to have the same
radius for the arcs above and below XY? Explore this through construction, and then
justify your answer. [Hint 1: Any point that is of the same distance from X and Y lies
on the perpendicular bisector. Hint 2: We can draw the whole line if any two of its
points are known.]

No, it is not necessary. The upper pair of arcs may use one radius and the lower pair a
completely different radius. You still get the perpendicular bisector.

Upper arcs, radius r₁ ⇒ crossing point A with AX = AY = r₁

Lower arcs, radius r₂ ⇒ crossing point B with BX = BY = r₂

A is equidistant from X and Y ⇒ A is on the perpendicular bisector

B is equidistant from X and Y ⇒ B is on the perpendicular bisector

Two points determine a line ⇒ AB is the perpendicular bisector

Why it happens: the proof never compared r₁ with r₂. Each point only had to be
equally far from X and from Y — its own distance. Once A and B are both on the line,
the ruler does the rest. What you must keep equal is the radius within each pair.

Check it yourself: draw the upper arcs with a small radius and the lower arcs with a
much bigger one. The join AB still passes through the midpoint at a right angle —
the picture just looks lopsided.

Q2 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.

Both pairs may be drawn on the same side — provided the two pairs use different radii, so
that they cross at two different points.

Page 11 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Pair 1, radius r₁, above XY ⇒ point A, with AX = AY

Pair 2, radius r₂ (r₂ ≠ r₁), also above XY ⇒ point C, with CX = CY

A and C both lie on the perpendicular bisector

A ≠ C because r₁ ≠ r₂

⇒ the line AC is the perpendicular bisector of XY, extended down through XY

Why it happens: the bisector is a whole line, not just the bit between A and B. Any
two of its points fix it. Points above XY are just as good as one above and one below
— as long as they are two distinct points. If you used the same radius twice on the
same side you would get the same point back, and one point cannot fix a line.

Tip: in practice A and B on opposite sides are placed far apart, so the ruler line is
more accurate. Two points close together on one side make small drawing errors
grow when you extend the line.

Q3 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.

Yes — within one pair the two radii must be equal. This is the one thing you cannot change.

Arc from X with radius r₁, arc from Y with radius r₂, crossing at P

Then PX = r₁ and PY = r₂

If r₁ ≠ r₂ then PX ≠ PY

⇒ P is not equidistant from X and Y

⇒ P does not lie on the perpendicular bisector

Page 12 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: the whole method rests on one property — the point must be
equally far from X and from Y. Unequal radii destroy exactly that property. The
crossing point then slides towards whichever endpoint used the smaller radius, and
the line you draw tilts off the midpoint.

Check it yourself: draw an arc of 4 cm from X and 6 cm from Y. Join the two crossing
points. Measure where the line meets XY — it will not be the midpoint, and the angle
will not be 90°.

Page 13 of 100

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Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
Recreate this design using only a ruler and compass —
e
Q4

m l as
m .co a g
l a se
a g

co m
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m .
The design to be recreated (page 140) — four petals meeting at one point.
a
l a se
agabove design, you can use a colour pencil with a ruler or
After completing the

m
compass to trace its boundary. This will make the design stand out from the
supporting lines and arcs.
. co
em
com g l as
. a
sem

a
agl
The design is four eyes sharing one centre — one pointing up, one down, one left, one right.
Build the two supporting lines first, then draw an eye on each of the four half-lines.
.c
s e m
m a
co agl
1. Draw a line and mark a point O on it. Construct the perpendicular to it at O (Q4 method,

m .
e
page 139). You now have two perpendicular lines through O.

g l as
a
2. With centre O and a fixed radius, cut the four half-lines at P, Q, R and S. So OP = OQ = OR =
OS.

co m
m .
m ase
.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

3. Take the segment OP. Construct its perpendicular bisector and mark two points C and D on
that bisector, one on each side, at equal distances from the bisector's foot.
4. With centre C draw the arc from O to P; with centre D draw the arc from O to P. These two
arcs make the first petal (an eye with corners O and P).
5. Repeat exactly the same for OQ, OR and OS, using the same compass opening every time.
Four identical petals appear.
6. Trace only the eight arcs with a colour pencil, then rub out the supporting lines.

Why it happens: each petal is symmetric about its own half-line because its two
centres, C and D, are equally far from O and from P — they sit on the perpendicular
bisector of OP. Using the same radius for all four petals makes the four petals
congruent, so the whole design has the same shape whichever way you turn it by
90°.

Tip: keep the compass at one setting for all eight arcs. The moment you change it,
one petal becomes fatter than its neighbour and the four-fold symmetry is lost.

In-text Questions — Page 141
Construction of a 90° Angle at a Given Point

Q1 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.

Yes. Turn the problem around: instead of being given a segment and finding its midpoint, start
from the midpoint and manufacture a segment.

1. Draw the line and mark O on it. Extend the line on both sides of O.
2. Open the compass to any radius. With centre O, cut the line at X on one side and Y on the
other. Now OX = OY, so O is the midpoint of XY.
3. With a larger radius, draw arcs from X and from Y that cross above the line at A.
4. Join A to O. Then ∠AOX = ∠AOY = 90°.

Page 15 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A

X O Y

O is made the midpoint of XY first. The perpendicular bisector of XY then has to pass through O.

Why it happens: AX = AY makes A a point of the perpendicular bisector of XY, and
OX = OY makes O another point of it. So the line AO is that perpendicular bisector —
and a perpendicular bisector meets the segment at a right angle. The 90° lands at O
because we chose O to be the midpoint.

Q2 Find a segment of this line for which O is the midpoint.

Extend the line on both sides of O. Set the compass to any convenient opening, put the point at
O, and cut the line on both sides — at X and at Y.

Same compass opening on both sides ⇒ OX = OY

X, O, Y are on one straight line

⇒ O lies between X and Y and splits XY into two equal parts

⇒ O is the midpoint of XY

Page 16 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: we cannot bisect a point, but we can build a segment around it.
Any opening of the compass will do — a big opening gives a long XY, which makes
the arcs in the next step cross more sharply and the drawing more accurate.

Tip: XY does not have to be any particular length. Only OX = OY matters.

Q3 In this case, do we need to draw two pairs of intersecting arcs to get the
perpendicular bisector of XY?

No — one pair is enough.

Point 1 on the perpendicular bisector: O (because OX = OY)

Point 2 on the perpendicular bisector: A from the single pair of arcs

Two known points fix the whole line

⇒ join A to O and the line is drawn

Why it happens: in the original construction we had to hunt for two points, so we
drew two pairs of arcs. Here one of the two points is handed to us — O is on the
bisector the moment we make it the midpoint. So only one more point is needed,
and one pair of arcs supplies it.

Tip: drawing the second pair below the line does no harm, and it gives a longer line
to rule along. It is extra accuracy, not extra logic.

Figure it Out — Page 142

Page 17 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Construction Methods in Śulba-Sūtras

MATH TALK

Q1 Justify why AB in Fig. 6.4 is the perpendicular bisector.

A
A X Y

X Y B

The midpoint pulled above XY — mark The midpoint pulled below XY — mark
it A it B

Fig. 6.4, page 142, redrawn — the folded rope tied to pegs at X and Y, its midpoint pulled
taut first above XY and then below it.

Because the rope construction produces exactly the four equal distances we need.

Page 18 of 100

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Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
e m.
The rope is folded in half, so both halves have the same length, say r.
m l as
.co
One loop is fixed at X, the other at Y.
m a g
l a se
g
aPull the marked midpoint up to A with both halves fully stretched:

com
. ag
AX = r and AY = r ⇒ AX = AY
e m
g l as
a
Pull the same midpoint down to B with both halves stretched:

co m
m.
BX = r and BY = r ⇒ BX = BY

m as e
.co a g l
a s em
So AX = AY = BX = BY

a gl ⇒ A and B are both equidistant from X and Y
m a s
agl
⇒ both lie on the perpendicular bisector of XY
. c o
s e mXY
⇒ AB is the perpendicular bisector of
a
agl
m
Why it happens: a stretched rope of fixed length is a compass. Fixing one end at X
. co
m
and sweeping the other traces a circle of radius r about X — exactly what the

o m l a se
compass leg does on paper. The half-way mark on a taut rope is the same distance

m .c both pegs, which is the one property the whole proofagneeds.
ase
from

agl
Did you know? The Śulba-Sūtras are geometric texts of the Vedic period that
se m
com g l a
.
describe how to lay out fire altars. They belong to the Vedāṅgas — literally the 'limbs
m a
ase
of the Vedas'. This rope method is from the Kātyāyana-Śulbasūtra 1.2.

agl

co m
Q2
m .
Can you think of different methods to construct a 90° angle at a given point on a

as e
com
line using a rope?

. a g l
e m
as
agl

Here are two rope methods. Both need nothing but a rope, pegs and level ground.
.c
s e m
om la
Method 1 — make the point a midpoint (the same idea as on paper).
. c a g
s e mline. Use a short piece of rope as a measure: mark off the
1. Let O be the given point on the
a sides, giving pegs at X and Y. Now OX = OY.
same length from O on lboth
ag

com
m .
m ase
.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

2. Take a longer rope, loop its ends at X and Y, fold it to find its midpoint, and pull that midpoint
taut to one side. Peg that position as A.
3. Stretch the rope from A to O. ∠AOX = 90°.
Method 2 — the 3–4–5 rope (an old surveyor's trick).

1. Knot a rope into a closed loop with 12 equal parts marked on it.
2. Peg the loop into a triangle whose sides are 3 parts, 4 parts and 5 parts, with the 3-part side
and the 4-part side meeting at O along the given line.
3. The angle at O is a right angle, because 3² + 4² = 9 + 16 = 25 = 5².

Why it happens: Method 1 is the perpendicular-bisector property in rope form —
equal distances from two pegs. Method 2 uses the converse of the Pythagoras
relation: a triangle whose sides satisfy a² + b² = c² must be right-angled at the corner
between a and b. The Śulba-Sūtras contain rope constructions of exactly this kind.

Tip: keep the rope fully taut at every step. A slack rope is a compass with a loose
screw — the radius quietly changes and the construction fails.

In-text Questions — Page 143

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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Angle Bisection for a Design

Q1 How do we construct this figure?

Fig. 6.5, page 143.

Draw the eight supporting rays first, then hang one petal on each ray.

1. Draw a line through a point O and construct the perpendicular at O. That gives 4 rays, at 90°
to each other.
2. Bisect each of the four right angles. That gives 4 more rays in between. Now there are 8
rays, at 45° to each other.
3. With centre O and a fixed radius, cut all eight rays at the same distance from O. Call the eight
cut points P₁ … P₈.
4. On the segment OP₁ construct a petal (an eye): find two centres on the perpendicular
bisector of OP₁ and draw the two arcs from O to P₁.
5. Repeat with the same compass opening on the other seven rays. Trace the sixteen arcs and
rub out the rays.

Why it happens: the figure looks even because all eight petals sit on rays that are
equally spread. Equal spread means every gap is the same, and eight equal gaps
around a point must each be 360° ÷ 8 = 45°. A right angle bisected once gives exactly
that.

Page 21 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Tip: do not try to set 45° with a protractor. Bisecting a right angle is both faster and
exact.

Q2 What is the angle between two adjacent lines?

45°.

The 8 rays split the complete angle around O into 8 equal parts
Complete angle around a point = 360°

Angle between adjacent rays = 360° ÷ 8

= 45°

Why it happens: for the design to look the same after every turn, every gap must be
equal. Eight equal gaps must share the full turn of 360° between them, so each gap
is one-eighth of 360°.

Check it yourself: 8 × 45° = 360°. ✓ Also, two opposite rays make 4 × 45° = 180°, a
straight line — which is exactly what the picture shows.

Q3 How do we construct a 45° angle using only a ruler and a compass?

Construct a 90° angle, then bisect it.

Step 1: 90° at a point O on a line (page 141 method)

Step 2: bisect ∠AOX

90° ÷ 2 = 45°

1. At O construct the perpendicular OA, so ∠AOX = 90°.
2. With centre O and any radius, cut OA at P and OX at Q. So OP = OQ.
3. With the same (or any equal) radius, draw arcs from P and from Q crossing at R.

Page 22 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

4. Join OR. Then ∠ROX = 45°.
Why it happens: OP = OQ and RP = RQ, with OR common, so ΔOPR ≅ ΔOQR by SSS.
Corresponding angles give ∠POR = ∠QOR, and together they make the 90°. So each
is half of it.

In-text Questions — Page 144
Steps for Angle Bisection

Q1 How do we construct these congruent triangles, given the angle?

Make the three pairs of sides equal, and SSS does the rest.

On the two arms of ∠XOY mark A and B with OA = OB (one compass opening from O)

From A and from B cut arcs of one equal radius, meeting at C ⇒ AC = BC

OC is common to both triangles

In ΔOAC and ΔOBC:

OA = OB, AC = BC, OC = OC

⇒ ΔOAC ≅ ΔOBC by SSS

⇒ ∠AOC = ∠BOC ⇒ OC bisects ∠AOB

Page 23 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
e m.
m X l as
m .co a g
l a se
a g

co m
e m . ag
B g l as C
a

co m
e m.
m l as
m .co a g
l a se
a g O Y
A
m a s
m .co agl
l a se
OA = OB from one arc; AC = BC from two equal arcs. SSS makes ΔOAC ≅ ΔOBC, so OC bisects the

a g angle.

co m
Why it happens: the compass can only do two things — carry a length and draw
m .
o m l a se
g forces the two angles
.c from O, and AC = BC is one radius used twice. SSS then
equal distances. Both congruent sides we need are of that kind: OA = OB is one

m a
aseat O to be equal, without any measuring.
sweep

agl
se m
com g l a
. a
Tip: the radius used for the arcs from A and B need not equal OA. It only has to be
m
ase
big enough for the two arcs to cross.

agl

co m
Figure it Out — Pages 144 – 145
m .
m as e
.co
Angle Bisection for a Design
a g l
se m
l a
MATH TALK

ag
.c
s e m
m a
Construct at least 4 different angles. Draw their bisectors.

co agl
Q1

m .
as e
a g l
Use the same three steps for every angle — the method does not care how big the angle is.

com
m .
m ase
.co


a g l Page 24 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

1. Draw ∠AOB of any size (try one acute, one right, one obtuse, one reflex-looking wide angle).
2. With centre O and one radius, cut both arms at P and Q, so OP = OQ.
3. With one equal radius, cut arcs from P and from Q meeting at R.
4. Join OR — it is the bisector.

ANGLE YOU DRAW EACH HALF CHECK

70° 35° 35 + 35 = 70 ✓

90° 45° 45 + 45 = 90 ✓

120° 60° 60 + 60 = 120 ✓

150° 75° 75 + 75 = 150 ✓

Why it happens: the construction proves ΔOPR ≅ ΔOQR by SSS every single time, so
the two halves are equal every single time. Measure with a protractor afterwards
only to check your drawing, never to make it.

Check it yourself: after bisecting, put the compass point on R and check that RP =
RQ. If they differ, one arc slipped.

Page 25 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q2 Construct the 8-petalled figure shown in Fig. 6.5.

Fig. 6.5, page 143.

Two right angles bisected twice give the eight rays; then eight identical petals.

1. Mark O on a line and construct the perpendicular at O. Four rays, 90° apart.
2. Bisect each right angle. Eight rays, 45° apart.
3. With centre O and one fixed radius, cut all eight rays at P₁ … P₈, so OP₁ = OP₂ = … = OP₈.
4. For OP₁: construct its perpendicular bisector, mark two centres C and D on it at equal
distances from OP₁, and draw the arcs C→(O to P₁) and D→(O to P₁). That is one petal.
5. Repeat with the same compass settings for the other seven rays.
6. Ink the sixteen arcs; rub out the rays and the supporting bisectors.

Number of petals = 8

Angle between neighbouring petals = 360° ÷ 8 = 45°

Check: 8 × 45° = 360° ✓

Why it happens: equal rays plus equal petals give the figure its 8-fold rotational
symmetry — turn it by 45° and it looks unchanged. The symmetry is a consequence
of the construction, not of careful eyeballing.

Page 26 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q3 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.

X

B

C O A Y

The construction shown on page 144 — the equal arcs from A and B are drawn on the far
side of O and meet at C.

Yes. The line OC is still the bisector — but it now bisects the angle on the other side of O, and the
bisecting ray of ∠AOB is the opposite ray of OC.

Page 27 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A and B are still marked with OA = OB

C is now on the far side, with CA = CB (equal radii again)

In ΔOAC and ΔOBC: OA = OB, AC = BC, OC common

⇒ ΔOAC ≅ ΔOBC by SSS

⇒ ∠AOC = ∠BOC

So the line OC is a line of symmetry for the two arms.

∠AOC = ∠BOC are the two angles outside ∠AOB,

and the ray opposite to OC divides ∠AOB into two equal parts.

Why it happens: the congruence only knows that O and C are both equidistant from
A and B — so OC is the perpendicular bisector of the chord AB. A perpendicular
bisector is a full line, and a full line through O splits the plane into two pairs of equal
angles. Whichever side C falls on, the same line appears; only the ray you name
changes.

Check it yourself: extend OC backwards through O with a ruler. The extension falls
exactly inside ∠AOB and cuts it in half.

Q4 What are the other angles that can be constructed using angle bisection? Can you
construct 65.5° angle?

From 90° and 60°, repeated bisection (and adding or subtracting the pieces) gives a long list of
angles — but 65.5° is not one of them.

START FROM BISECT ONCE TWICE THREE TIMES

90° 45° 22.5° 11.25°

60° 30° 15° 7.5°

180° 90° 45° 22.5°

Page 28 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
e m.
Adding and subtracting these gives many more:
m l as
.co
60° + 15° = 75° · 90° + 15° = 105° · 45° + 22.5° = 67.5°
m a g
l a se
g
60° − 7.5° = 52.5° · 30° + 7.5° = 37.5° · 90° + 60° = 150°
a

co m
. ag
65.5°? Every angle built this way is a whole number of steps of 15°, or 7.5°, or 3.75°, …
e m
(each step being 15° halved again and again).
g l as
a
65.5 ÷ 15 = 4.366… 65.5 ÷ 7.5 = 8.733… 65.5 ÷ 3.75 = 17.466…

co m
m.
None of these is a whole number, and halving further never fixes it.

m as e
.co l
⇒ 65.5° cannot be constructed by bisection.
a g
se m
g l a
a Why it happens: bisection can only halve. Starting from 90° and 60°, everything you

s
can reach is made of pieces of size 15° cut in half some number of times. 65.5° is
m a
.co agl
65½°, and a half-degree simply never appears among those pieces. The nearest

se m
a
constructible angle is 67.5° (= 45° + 22.5°), which is 2° away.

a g l

m
Tip: if you need 65.5° for a drawing, use a protractor. Ruler-and-compass work is
. co
m
exact, but it cannot reach every angle.

m as e
.co a g l
a s em
a gl Q5 Come up with a method to construct the angle bisector using a rope.
sem
com g l a
m . a
ase
agl
Copy the compass steps, using the rope as both compass and ruler.

1. Let the angle be ∠AOB with a peg at the corner O and the two arms marked on the ground.

co
2. Take a rope and knot it at a fixed length r. Hold one end at O and stretch it along one arm —
m
m .
se
peg the far end as P. Stretch the same rope along the other arm — peg the far end as Q.

o m
Now OP = OQ = r.
l a
gswing the other end to sweep
3..cTake a second rope of some length s. Hold one end at P and
a
m
se an arc. Do the same from Q. Peg the crossing point as R. Now PR = QR = s.
l a
ag 4. Stretch a rope from O to R and mark the line. OR bisects ∠AOB.
m .c
o m l a se
.c a g
se m
l a
ag

co m
m .
m ase
.co


a g l Page 29 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

In ΔOPR and ΔOQR:

OP = OQ (= r), PR = QR (= s), OR common

⇒ ΔOPR ≅ ΔOQR by SSS
⇒ ∠POR = ∠QOR

Why it happens: a taut rope of fixed length does exactly what a compass does — it
marks all points at one distance from a fixed peg. Since the proof of the bisector
uses only equal lengths, it works just as well with rope on the ground as with a
compass on paper. This is how large layouts, such as fire altars, were set out.

Page 30 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q6 Construct the following figure. How do we construct the petals so that they are of
the maximum possible size within a given square?

The four-petal design, page 145.

The four petals are largest when each arc is a semicircle drawn on a side of the square as
diameter.

1. Construct the square ABCD (four right angles, four equal sides).
2. Construct the perpendicular bisector of each side. This marks the midpoints P, Q, R, S of the
four sides — and the four bisectors meet at the centre O of the square.
3. With centre P (midpoint of AB) and radius PA, draw the semicircle inside the square. It runs
from A through O to B.

Page 31 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

4. Repeat from Q, R and S with the same radius. The four semicircles cut each other and four
petals appear.

P

O
S Q

R

Four semicircles on the four sides as diameters. Each passes through two corners and through the
centre O, and the overlaps are the four petals.

Let the side of the square be s.

Radius of each arc = s ÷ 2

Distance from P (midpoint of AB) to A = s ÷ 2 ✓

Distance from P to the centre O = s ÷ 2 ✓

So the semicircle from P passes through A, O and B.

Page 32 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: the arc must stay inside the square. Its centre is on a side and it
has to reach the two nearest corners, which are s ÷ 2 away. Any larger radius spills
outside the square; any smaller radius leaves the arc short of the corners and the
petals shrink. So radius = s ÷ 2 is the largest that fits — the petals are then as big as
they can be, with their four tips exactly at the four corners.

In-text Questions — Page 145
Repeating Units and Repeating Angles

MATH TALK

Q1 Construct the following figure.

Fig. 6.6, page 145.

The whole figure is one unit, repeated. The unit is a fan shape: two straight arms of equal
length meeting at a point, closed off by an arc. Copies are laid down alternately point-up and
point-down.

1. Draw the unit once: mark a vertex V, draw two arms VA and VB of equal length with some
angle between them, and join A to B with an arc of centre V.
2. To repeat it, you must copy both the arm length and the angle. The arm length is easy —
carry it with the compass.
3. Copy the angle by the arc-and-chord method: draw an arc from V cutting the arms at A and
B; draw the same arc at the new vertex; then transfer the chord AB with the compass.
4. Place the copies alternately: point down, point up, point down, … so that neighbouring units
share an endpoint.

Page 33 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: two fans are exact copies only if their arms match and the angle

m l a se
between the arms matches. Equal arms alone are not enough — a wide fan and a
o g now needs a way
.c can have the same arm length. That is why the chapter
a
m
narrow fan

l a se an angle.
ag
to copy

o m
c ag
Tip: mark the shared endpoints on a straight line first. Then every unit starts and
.
ends where its neighbours do, and themwave stays level.
l a se
ag

co m
m.
Q2 Draw an angle. Create a copy of this angle using only a ruler and compass.

o m l a se
ANSWER .c a g
m
sethe angle into a triangle, then rebuild the same triangle somewhere else. SSS congruence
l a
ag
Turn
carries the angle across.

om a s
. c agl
1. Let the given angle be at A. Draw a ray from the new point X.

s
2. With centre A and any radius, draw an
e marc cutting the two arms of the angle at B and C. Now
ΔABC is isosceles with AB = AC. la
ag
3. With the same radius and centre X, draw an arc cutting the new ray at Z.
4. Open the compass to the length BC. With centre Z, cut the new arc at Y, so YZ = BC.
com
∠YXZ = ∠BAC. m .
e
5. Join X to Y. Then
m l as
m .co a g
l a se AB = XZ (same radius) — call it r
ag AC = XY (same radius) — also r

se m
BC = YZ (chord transferred with the compass)
com g l a
m . a
ase
⇒ ΔABC ≅ ΔXYZ by SSS agl
∠A = ∠X m
co
⇒

m .
o m l a se
m .cWhy it happens: a compass cannot measure an angle,agbut it can measure a length.
l a se
g
So we convert the angle into a length — the chord BC across the arc. With the two
a c
radii fixed, the chord decides the angle completely: a wider angle gives a longer
m .
chord. Copy the chord and the angle comes with it.
m a s e
em . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 34 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Check it yourself: after copying, place the arc-and-chord of the copy over the
original by tracing on thin paper. They should fall exactly on each other.

Figure it Out — Page 147
Steps of Construction to Copy an Angle

Q1 Construct at least 4 different angles in different orientations without taking any
measurement. Make a copy of all these angles.

Draw four angles freely — one narrow, one nearly a right angle, one wide, one opening
downwards — and copy each by the arc-and-chord method.

1. Draw the angle at A. Draw a fresh ray from X, pointing any way you like.
2. Arc of radius r from A, cutting the arms at B and C.
3. Same radius r from X, cutting the new ray at Z.
4. Compass opened to BC; from Z cut the arc at Y.
5. Join XY. Then ∠YXZ = ∠BAC.

Every copy rests on the same congruence:

AB = AC = XZ = XY = r and BC = YZ

⇒ ΔABC ≅ ΔXYZ (SSS) ⇒ ∠A = ∠X

Why it happens: the orientation of the copy does not matter at all. The construction
fixes only the opening between the arms, never the direction they point. That is why
the same three steps work for an angle opening left, right, up or down.

Tip: use the same radius r for the original and the copy. If you change it, the chord
BC no longer matches the new arc and the copy will be wrong.

Page 35 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q2 Construct the Fig. 6.6.

Fig. 6.6, page 145.

Build one unit, then copy its angle again and again.

1. Draw a horizontal base line and mark equally spaced points on it with the compass: V₁, V₂, V₃,
… (all gaps equal).
2. Draw the first unit: from V₁ draw two arms of equal length l with the chosen angle between
them, and close it with an arc of centre V₁ and radius l.
3. Copy that angle at V₂ using the arc-and-chord method, but place it upside down — the copy
points the other way.
4. Copy again at V₃ in the first orientation, and so on, alternating.
5. Because all arms are of length l and all angles are copies of one angle, the units are
congruent and the ends meet neatly.

Why it happens: a repeating pattern needs its unit to repeat exactly. Two
measurements decide the unit — arm length and angle. The compass fixes the first;
the copy-an-angle construction fixes the second. Alternating the orientation is just a
flip; the unit itself is unchanged.

In-text Questions — Page 147
Construction of a Line Parallel to the Given Line

Q1 How do we implement this idea using a ruler and a compass?

Draw a transversal, then copy the corresponding angle at the new point. Equal corresponding
angles force the lines to be parallel.

1. Let m be the given line. Draw a line l crossing it at A. Line l is the transversal.

Page 36 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

2. Choose the point B on l through which the parallel must pass.
3. With one radius, draw an arc from A cutting m at C and cutting l at D. With the same radius
draw an arc from B, cutting l at E.
4. Open the compass to CD and, from E, cut the arc at F.
5. Join B to F and extend — this line n satisfies m ∥ n.

Angle a = angle between m and l at A

The copy at B is the corresponding angle for the transversal l

Corresponding angles equal ⇒ m ∥n

Why it happens: two lines cut by a transversal are parallel exactly when a pair of
corresponding angles is equal. We cannot check "never meeting" by drawing, since
we can only draw a finite bit of a line. But we can guarantee the angle, and the angle
guarantees the parallelism.

Tip: when you used a ruler and set square in Grade 6, you were doing the same
thing — the set square slid along, carrying a fixed angle. Here the compass carries
the angle instead.

Figure it Out — Page 148
Construction of a Line Parallel to the Given Line

Q1 Construct 4 pairs of parallel lines in different orientations.

Repeat the copy-the-corresponding-angle construction four times, choosing a different slant for
the given line each time.

PAIR GIVEN LINE M TRANSVERSAL L WHAT YOU COPY

1 Horizontal Slanting up-right The angle at A onto B

2 Vertical Slanting down-right The angle at A onto B

3 Slanting up-right Horizontal The angle at A onto B

4 Slanting down-right Vertical The angle at A onto B

Page 37 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: the construction never refers to "horizontal" or "vertical". It only
says: equal corresponding angles with respect to a transversal. That statement is
true whatever the page is turned to, so the same steps work in every orientation.

Check it yourself: measure the perpendicular distance between the two lines at two
widely separated places. For a true pair of parallels the two distances are equal.

Q2 Construct the following figure.

S T

A

H B
Z U

G C

Y V
F D

E

X W

The eight-pointed star, page 148.

The figure is eight congruent rhombi arranged around one centre. Each rhombus has one
vertex at the centre and the opposite vertex at a star tip. Opposite sides of a rhombus are
parallel — that is where the parallel-line construction is used.

Page 38 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
e m.
Eight rhombi meet at the centre with no gap and no overlap
m l as
.co
Angle of each rhombus at the centre = 360° ÷ 8 = 45°
m a g
l a se
g
Check: 8 × 45° = 360° ✓
a

co m
. ag
In a rhombus, adjacent angles add to 180°, so the angle at a star tip
e m
= 180° − 45° = 135°
g l as
a
m
1. Mark the centre O. Draw a line through O and its perpendicular, then bisect each right angle.
co
m.
That gives 8 rays, 45° apart.

o m l a se
2. With one compass opening, cut the eight rays at S, T, U, V, W, X, Y, Z — all the same distance
.c a g
m
se two neighbouring rays OS and OT. Through T construct the line parallel to OS, and
from O.
3.aTake
l
ag through S the line parallel to OT. The two lines meet at a point — call it A. Then O, S, A, T are
s
the four corners of the first rhombus.
m a
.co
4. Repeat for each of the eight neighbouring pairs of rays. The eight inner points A, B, C, … H
em agl
a s
agl alternate rhombi if you want the star effect of the book.
appear.
5. Ink the rhombus outlines; shade

. c om
Why it happens: a rhombus is a parallelogram with all four sides equal. Because all

s e
eight rays are cut at the same distance, every rhombus has the same m side length;
. com the rays are 45° apart, every rhombus has the same
because
a glacorner angle at O. So
s m eight are congruent, and the design has 8-fold rotational symmetry.
eall
a
agl
se m
com g l a
In-text Questions — Page 149m. a
ase
agl
Arch Designs — Trefoil Arch

How did they make these arches?
co m
.
Q1

em
m l as
.co a g
emThe stonework starts as a drawing. The mason first sets the arch out full size on a flat surface —

a s
agl paper, board or the stone itself — using nothing but a straight edge and a compass (or a peg
.c
and cord on a large job). The stones are then cut to follow those drawn arcs.
s e m
m a
em . co
1. Draw the supporting lines that fix the width and the springing points of the arch. agl
g l as
2. Find the centres of the arcs on those supporting lines.
a
3. Sweep the arcs with a compass or a cord.

m
4. Cut the stones to the drawn curves and build.

. co
e m
m l as
.co a g
Page 39 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: an arch has to be symmetric or it will look wrong and stand badly.
Symmetry cannot be judged by eye across several metres of stone. But it can be
constructed — equal lengths and equal angles on the two sides — and that is exactly
what a compass guarantees.

Did you know? The pictures in the book are of the Diwan-i-Aam at the Red Fort and
an arch in Central Park, New York City — the same geometry, centuries and
continents apart.

Q2 Construct this arch shape on a piece of paper.

B C

A D

The trefoil arch of page 149, with its four corner points A, B, C and D marked.

Draw the four support points first — A and D on the base line, B and C above them — and then
hang the arcs on that frame.

1. Draw the base segment AD.
2. Construct equal angles at A and at D, opening inwards (copy one angle to the other end).
This makes ∠BAD = ∠CDA.
3. With one compass opening, mark B on the arm from A and C on the arm from D, so AB = CD.
4. Join B to C. The frame ABCD is now symmetric about the perpendicular bisector of AD.
5. Draw the three arcs of the trefoil on this frame — one small arc over AB, one small arc over
CD, and the large arc over BC — adjusting the radii until the curve looks right.

Page 40 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Why it happens: AB = CD and ∠BAD = ∠CDA make the left half of the frame a
mirror image of the right half. Once the frame is symmetric, arcs drawn with
matching radii on the two sides are symmetric too, and the finished arch balances.

Q3 For symmetry, we should have AB = CD, and ∠BAD = ∠CDA. How would you
construct these support lines?

One length carried by the compass, one angle carried by the copy-an-angle construction.

WHAT MUST MATCH TOOL HOW

∠BAD = ∠CDA Compass Draw the angle at A, then copy it at D using arc-and-chord (SSS)

AB = CD Compass Set the compass to AB and cut the arm at D to get C

Copying the angle: arc of radius r from A cuts AD at P and AB at Q

Same radius r from D cuts DA at P′

Transfer the chord PQ from P′ to get Q′

⇒ ΔAPQ ≅ ΔDP′Q′ (SSS) ⇒ ∠BAD = ∠CDA

Why it happens: symmetry about the perpendicular bisector of AD means the left
half must fold exactly onto the right half. Folding carries A to D, so it must carry B to
C. For that, the angle at A must equal the angle at D and the arm lengths must be
equal — precisely the two conditions given.

Tip: construct the perpendicular bisector of AD lightly first. It is the fold line, and it
tells you at a glance whether your frame is really symmetric.

Q4 Use these support lines to construct an arch. If required, adjust the radii of the arcs
to make the arch look more aesthetically pleasing.

With the frame ABCD ready, draw three arcs — but always change the two side arcs together.

Page 41 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

1. Left lobe: choose a centre on the perpendicular bisector of AB and draw an arc from A to B.
2. Right lobe: use the same compass opening on the perpendicular bisector of CD, and draw
the arc from D to C.
3. Top lobe: choose a centre on the perpendicular bisector of BC (which is also the axis of the
whole figure) and draw the arc from B to C.
4. Step back and look. A smaller radius makes the lobes rounder and the arch chubbier; a
larger radius flattens them and the arch looks taller and calmer. Redraw until it pleases you.

Why it happens: the frame fixes where the arcs begin and end; the radius decides
only how much they bulge. So you can change the mood of the arch freely without
ever losing its symmetry — as long as the left and right radii stay equal.

Try This: draw the same frame three times and use three different radii. Put the
three arches side by side and decide which one you would build.

In-text Questions — Page 150

Page 42 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A Pointed Arch

Q1 How do we construct this shape?

The pointed arch shape of Fig. 6.11, page 150.

What supporting lines will you use to draw this arch?

The supporting lines are just two line segments of equal length meeting at the top — the
same idea as 'Wavy Wave' from Grade 6.

1. Draw the base and mark the two springing points, P and Q.
2. Draw two segments PT and QT of equal length meeting at a point T above the base. They
form the skeleton of the pointed arch.
3. Mark the midpoint of each segment with the perpendicular-bisector construction.
4. Draw an arc from P to T, and a matching arc from Q to T, using centres chosen on the
perpendicular bisectors of PT and QT.
5. The two arcs meet at T in a point — that is the pointed top.

Why it happens: equal supporting segments make the two halves mirror images.
Because both arcs end at the same top point T and start at the same height, the arch
closes cleanly at the apex instead of rounding off. Curving the arcs one way gives an
ogee (S-shaped) profile; curving them the other way gives the plain pointed arch of
the photograph.

Page 43 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
If their midpoints are marked, will you be able to construct a pointed arch?
e
Q2

m l as
.co a g
a
s em
aglThe midpoint is exactly what the arc needs.
Yes.

com
. ag
Let M be the midpoint of the supporting segment PT.
e m
Every centre that gives an arc through both P and T
g l as
must be equidistant from P and T
a

co m
m.
⇒ it lies on the perpendicular bisector of PT, which passes through M.

m as e
.co a g l
se m
a
Choose any point on that perpendicular bisector as the centre;

a g l set the compass to its distance from P; sweep from P to T.

om a s
e
. c
m property of the chapter again — a point is agl
s
Why it happens: this is the very first

a gla when it lies on the perpendicular bisector of PT.
equally far from P and T exactly
Marking the midpoint gives you a point of that bisector, so you can draw the bisector
and pick your centre anywhere along it. Moving the centre further out flattens the
co m
m .
arc; bringing it closer to M makes the arc bulge.
m as e
.co a g l
s m use the same distance-from-the-midpoint on both sides, otherwise the left half
eTip:
gl a
a of the arch will bulge more than the right half.

se m
com g l a
m . a
gl ase
Figure it Out — Page 151
a

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
em . co agl
g l as
a

co m
m .
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.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A Pointed Arch

Q1 Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by
changing the radius of the arcs.

Fig. 6.11, page 150 — the two support lines, with their midpoints marked.

Keep the two supporting segments fixed and change only where the arc centres sit on their
perpendicular bisectors.

WHERE THE CENTRE IS PLACED RADIUS ARCH YOU GET

Far out along the perpendicular bisector large Almost straight sides — a sharp, tall arch

Moderately out medium A gentle, classic pointed arch

Close to the midpoint small Strongly bulging sides — a plump arch

1. Draw the two equal supporting segments PT and QT.

Page 45 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

2. Construct the perpendicular bisector of PT and of QT.
3. Pick a centre on the bisector of PT; open the compass from that centre to P; sweep the arc
from P up to T.
4. Use the same distance from the midpoint on the other side, and sweep the arc from Q to
T.
5. Change the distance and repeat for a new arch.

Why it happens: the two endpoints of each arc are locked by the supporting
segments, so the arc can only vary in how much it curves. Curvature is decided by
the radius alone — small radius, sharp curve; large radius, gentle curve. Everything
else stays put.

Q2 Make your own arch designs.

Start from a symmetric frame, then decide how many lobes and which way each arc curves.

1. Frame first. Draw the base AD, construct equal angles at A and D, and cut equal lengths.
Draw the axis (the perpendicular bisector of AD) — every arc must be mirrored across it.
2. Choose the lobes. One big arc gives a plain round arch. Three arcs give a trefoil. Five give a
cinquefoil.
3. Choose the curvature. Arcs bulging outwards give a full, rounded look; arcs bulging
inwards (inflexed arcs) give the elegant ogee of Mughal architecture.
4. Mirror everything. For every arc on the left, draw its partner on the right with the same
radius.
5. Ink the outline in colour and rub out all supporting lines and centres.

Why it happens: every good arch design is the same two decisions repeated —
where the arc starts and ends (fixed by the frame) and how much it bends (fixed by
the radius). Mirroring across the axis is what keeps the design from looking
accidental.

Try This: look at a doorway, a temple gopuram or a mosque arch near your home.
Sketch its outline and try to guess where the mason put the centres of the arcs.

In-text Questions — Page 151

Page 46 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Regular Hexagons · Regular Hexagon and Equilateral Triangles

TRY THIS

Q1 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.

Making all the sides equal is easy with a compass. Making all the angles equal as well is the
hard part — and that is what "regular" demands.

REGULAR POLYGON SIDES EACH ANGLE RULER AND COMPASS?

Equilateral triangle 3 60° Yes — done in Grade 6

Square 4 90° Yes — done in Grade 6

Regular pentagon 5 108° Needs more theory — later years

Regular hexagon 6 120° Yes — from 60°, in this chapter

With a compass you can quickly draw a five-sided or six-sided closed figure with equal sides:
mark off one length, turn the ruler a little, mark it again, and keep going until the shape closes.
But when you measure the angles you will find they are not equal, so the figure is not regular.

Why it happens: equal sides do not force equal angles. A rhombus has four equal
sides but is not a square. For a regular polygon we need to control the angle at each
corner, and to do that with only a ruler and compass we must be able to construct
that angle. 120° comes from 60°, which comes from an equilateral triangle — so the
hexagon is within reach. 108° cannot be reached from 60° and 90° by bisecting, so
the pentagon must wait.

Q2 Can we break a regular hexagon into smaller pieces that can be constructed?

Yes — into six equilateral triangles.

Page 47 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Join the centre O to each of the six vertices A, B, C, D, E, F.

The hexagon splits into 6 triangles: ΔOAB, ΔOBC, ΔOCD, ΔODE, ΔOEF, ΔOFA.

Angle of each triangle at O = 360° ÷ 6 = 60°
OA = OB = … = OF (all radii of the same circle), so each triangle is isosceles

An isosceles triangle with a 60° apex has base angles (180° − 60°) ÷ 2 = 60°

⇒ each triangle is equilateral

Why it happens: equilateral triangles are the one polygon we already know how to
construct exactly, straight from the compass. If a hexagon is nothing but six of them
stuck together, then constructing the hexagon reduces to constructing an equilateral
triangle six times — a problem we have already solved.

Page 48 of 100

as e
Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
m.
What happens when we join the ‘opposite’ points of a regular hexagon? Since a
e
Q3

com in the figure be equilateral triangles?
regular hexagon has equal sides and angles, can we expect a figure like this? Will all
g l as
. a
em
the triangles

a s
agl

. comA ag
e m
g l as
a

co m
em.
m l as
.co F a g B
a s em
ag l

m a s
m .co agl
l a se
a g
O co m
m .
m as e
.co E a g l C
a sem
agl
se m
com g l a
m . a
ase
agl D
co m
m .
m as e
.co a g l
se m
a
Fig. 6.12, page 151.

ag l
.c
s e m
m a
. co agl

em
l as
The three long diagonals AD, BE and CF all pass through one point O, and they cut the hexagon
g
a
into six equilateral triangles.

co m
m .
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.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

A

F B
60°

O

E C

D
The three main diagonals of a regular hexagon meet at O and cut it into six equilateral triangles, each
with a 60° angle at O.

Six equal angles fill the turn at O:

each = 360° ÷ 6 = 60°

All six segments OA, OB, …, OF are equal ⇒ each triangle is isosceles

Isosceles + 60° apex ⇒ other two angles = (180° − 60°) ÷ 2 = 60° each

⇒ all six triangles are equilateral

Also, the hexagon's angle at each vertex = 60° + 60° = 120° ✓

Why it happens: a regular hexagon is as symmetric as a shape can be — turn it by
60° and nothing changes. That turning symmetry forces the six segments from the
centre to be equal and the six angles at the centre to be equal. Equal radii plus a 60°
angle is exactly the recipe for an equilateral triangle.

Page 50 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Check it yourself: the book asks us to argue the other way round, which is safer —
start with six equilateral triangles, show they fit around a point, and see that a
regular hexagon comes out. That is done in the next questions.

Page 51 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q4 Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will
it result in a regular hexagon?

A

F B

O
E C

D

Fig. 6.12, page 151.

Yes to both. Six 60° angles exactly fill the turn around a point, and the outer boundary is a
regular hexagon.

Page 52 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Angle of each equilateral triangle = 60°

Six of them at the centre: 60° × 6 = 360°

360° is the complete angle around a point

⇒ no gap is left and no two triangles overlap ✓

Sides of the resulting figure: each is a side of one triangle,

and the triangles are congruent ⇒ all six sides are equal ✓

Angle at each outer vertex = 60° + 60° = 120°, the same at all six ✓

Equal sides + equal angles ⇒ regular hexagon

Why it happens: a degree was defined by taking the full turn around a point to be
360°. So a set of angles can be packed around a point without gap or overlap exactly
when they add up to 360°. Six 60° angles do; that is the whole reason a hexagon
works and, for example, a regular pentagon does not (5 × 108° = 540°, and 108°
does not divide 360° evenly either).

Tip: this "angles around a point add to 360°" test is the same test used later in the
chapter to decide which regular polygons can tile the plane.

In-text Questions — Pages 152 – 153

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Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

Regular Hexagon and Equilateral Triangles · Construction of a 60° angle
co m
em.
m l as
.co
Consider this figure. Will the 70° angle fit into the gap? What is the gap angle
a g ∠AOI?
emhave, 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°. Use this to determine
Q1

a s
We

a gl whether the 70° angle fits the gap.

co m
e m . ag
aDs
E
a g l

co m
em.
m l as
m .co a g G

l a se F
a g
s
50° 30°
m a
.co 60° agl
C

B
a s em 40°
a gl 40°
O 90° H
co m
m .
m as e
.co a g l
se m
g l a A
a ?
se m
com g l a
.
I
m a
ase
agl

co m
70°
m .
m ase
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

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m .
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.co


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Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

The figure on page 152 — six angles placed edge to edge around the point O, with the
70° angle waiting below.

The gap angle is 50°, so the 70° angle does not fit — it is 20° too wide.

Angles already placed around O:

40° + 60° + 50° + 30° + 40° + 90°

= 100° + 50° + 30° + 40° + 90°

= 150° + 30° + 40° + 90°
= 180° + 40° + 90°

= 220° + 90°

= 310°

Complete angle around a point = 360°

Gap angle ∠AOI = 360° − 310° = 50°

70° > 50°, so the 70° angle overlaps the neighbouring pieces.

It is 70° − 50° = 20° too big.

Page 55 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

60°
50°

30° 40°

O
40°
50°

90°

dashed piece = the gap ∠AOI

Six angles use up 310° of the turn around O. Only 50° is left, so a 70° piece cannot be dropped in.

Why it happens: the pieces sit around a single point O, and one full turn about a
point is 360°. Whatever is not used up by the six pieces is the gap. Since the gap is
smaller than the piece we want to insert, the piece would have to ride over a
neighbour — that is an overlap, not a tiling.

Check it yourself: 40 + 60 + 50 + 30 + 40 + 90 + 50 = 360 ✓. The right piece for this
gap is a 50° angle.

Page 56 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Q2 In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines?

A

F B

O
E C

D

Fig. 6.12, page 151.

Because the two angles at O on each side of the line add up to a straight angle, 180°.

Page 57 of 100

Class 7 Maths Chapter 14 Constructions and Tilings AglaSem · NCERT Solutions

Take AOD. At O, the angles on one side of it are

∠AOB + ∠BOC + ∠COD = 60° + 60° + 60° = 180°

Angles that add to 180° at a point on one side of a ray form a straight angle
⇒ OA and OD are opposite rays ⇒ A, O, D are collinear

Same for BOE: ∠BOC + ∠COD + ∠DOE = 60° + 60° + 60° = 180° ✓

Same for COF: ∠COD + ∠DOE + ∠EOF = 60° + 60° + 60° = 180° ✓

Why it happens: six equal 60° angles fill the 360° around O. Skipping three of them
takes you exactly half way round — and half of a full turn is a straight angle. So each
vertex has its "opposite" vertex directly across O, and the segment joining them
passes straight through the centre.

Tip: this is why the three long diagonals of a regular hexagon are concurrent — they
are not three separate diagonals but three straight lines through the centre.

Q3 Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.

Six equilateral triangles of side 4 cm, packed around one point.

1. Mark a point O. With the compass set at 4 cm, draw a circle with centre O.
2. Mark any point A on the circle.
3. Without changing the compass opening, put the point at A and cut the circle at B. From B
cut at C, from C at D, from D at E, from E at F.
4. F should land exactly 4 cm from A. Join A–B–C–D–E–F–A with the ruler.

Page 58 of 100

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Class 7 Maths Chapter 14 Constructions and Tilings
a g l AglaSem · NCERT Solutions

co m
e m.
Every step used the same radius, so
m l as
.co
OA = OB = OC = OD = OE = OF = 4 cm (radii)
m a g
l a se
g
AB = BC = CD = DE = EF = FA = 4 cm (the compass steps)
a

com
. ag
So ΔOAB has all three sides 4 cm ⇒ equilateral ⇒ ∠AOB = 60°
e m
l
Six such triangles: 6 × 60° = 360°, the complete turn ✓
g as
a
Angle of the hexagon at each vertex = 60° + 60° = 120°

co m
se m.
o m l a
Why it happens: the radius of a circle steps exactly six times around its own
g equilateral triangle
m .c That is not a coincidence — each step makes a
se the centre, each such triangle contributes 60° at the centre, and 6 × 60° = 360°
circumference. an

g l a
with
a closes the circle perfectly. The construction needs no protractor at all.

om a s
Check it yourself: if the sixth stepm
e
. c agl
s
misses A, the compass opening slipped. Re-set it
a
agl
to 4 cm and start again from A.

. com
How do we do it? (constructing a 120° angle using a ruler andeam
m a s
gl
compass)
co
Q4

. a
a s em
l
ag Construct a 60° angle at a point on a line — then the angle on the other side of it is

se m
automatically 120°.
com g l a
m . a
ase
agl
Let the line be XAY with A between X and Y.

Construct ∠CAX = 60° on one side.

co m
∠CAX and ∠CAY are angles on a straight line:
m .
m as e
.co
∠CAX + ∠CAY = 180°
a g l
se m 60° + ∠CAY = 180°
g l a
a c
∠CAY = 120°
m .
m a s e
em . co agl
g l as
a

com
m .
m ase
.co


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