NCERT Solutions Class 6 Maths Chapter 8 Playing with Constructions

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 6 · M AT H S

NCERT Solutions

Chapter 8: Playing with
Constructions

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

187 – 216 17 66 English

Solutions, notes, sample papers & more at 74 pages

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

CLASS 6 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 8: Playing with
Constructions
Complete, step-by-step NCERT Solutions for Class 6 Maths Chapter 8 Playing with Constructions from the
NCERT textbook Ganita Prakash. Every Figure it Out and every Construct task from pages 188 to 216 is solved
as numbered construction steps — the artwork (A Person, Wavy Wave, Eyes), squares and rectangles with
ruler, compass and set square, breaking rectangles into identical squares, diagonals of rectangles, and the
House built from two arcs.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 6) 187 – 216

SECTIONS QUESTIONS

17 66

MEDIUM

English

In-text Questions — Pages 188 & 189
Section 8.1 Artwork

Q1 Observe the way a compass is made. What can one draw with the compass? Explore!

A compass has two arms joined at a hinge — one arm ends in a sharp metal tip and the other
holds a pencil. Once you tighten the hinge, the gap between the tip and the pencil lead cannot
change while you turn it.

Press the metal tip on the paper and turn the pencil once round — you get a circle.
Turn it only part of the way — you get an arc (a piece of a circle).
Because the opening stays fixed, a compass can also copy a length from one place to
another without a ruler.

Why it happens: the tip stays at one point and the pencil is always the same
distance away from it. So every point the pencil draws is at that same distance from
the tip — and that is exactly what a circle is.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q2 Mark a point ‘P’ in your notebook. Then, mark as many points as possible, in
different directions, that are 4 cm away from P. Think: Imagine marking all the
points of 4 cm distance from the point P. How would they look?

Do it like this:

1. Mark a point and name it P.
2. Place the ruler with its 0 mark on P, turn it to the right, and put a dot at the 4 cm mark.
3. Turn the ruler a little — up, sideways, slanting, downwards — and each time put a dot at 4
cm.
4. Keep going. The dots slowly form a ring around P.

If all such points are marked, they form a circle with centre P and radius 4 cm.

4 cm
P

Every dot is 4 cm from P. Mark enough of them and the dots join up into a circle of radius 4 cm.

Page 2 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Why it happens: “4 cm from P” does not fix a direction — only a distance. Since you
may go in any direction, the points spread evenly all around P, and a curve with every
point at the same distance from a fixed point is a circle.

Q3 You can start by marking a few points of distance 4 cm from P using the compass.
How can this be done?

1. Place the ruler on the table. Put the metal tip of the compass at the 0 mark and open the
compass until the pencil point reaches the 4 cm mark.
2. Tighten the screw so that the opening does not change.
3. Press the metal tip on the point P in your notebook.
4. Let the pencil touch the paper and make a small mark. Lift, turn the compass a little, and
mark again — as many times as you like.
5. Now, keeping the tip fixed at P, turn the pencil all the way round without lifting it. You get the
full circle.

Tip: Hold the compass only by the knob at the top and lean it slightly in the direction
you are turning. Do not press the pencil hard, or the tip will slip and the circle will
come out uneven.

Q4 Take a point on the circle. What will be its distance from P— equal to 4 cm, less than
4 cm or greater than 4 cm? Similarly, what will be the distance between P and
another point on the circle?

Its distance from P is exactly equal to 4 cm — not less, not more. The same is true for any
other point you pick on the circle: every point of the circle is 4 cm from P.

distance from P to any point of the circle = radius = 4 cm

Why it happens: while you were drawing, the metal tip never moved from P and the
compass opening never changed. So the pencil was always 4 cm away from P — and
the curve it traced is made only of such points. P is called the centre and this fixed
distance is called the radius.

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: mark any three points on your circle and measure each one from

m as e
l
P with a ruler. All three readings will be 4 cm.

m .co a g
l a se
a g
Q5 Having explored the use of a compass, go ahead and recreate the images in Fig. 8.1.

co m
ag
Can you make the figures look as good as the figures shown there? Also, has the use

m .
e
of instruments made the construction easier?

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 . a
ase
agl

. c om
m a s em
Fig. 8.1, page 187 — the five figures of the Artwork section, all made from circles.

e m . co agl
g l as
a c
m .
m a s e
co agl
Yes — with a compass the figures of Fig. 8.1 become easy, because every one of them is made

m .
e
only of circles.

g l as
a

co m
m .
m ase
.co


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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

FIGURE IN FIG. HOW TO CONSTRUCT IT
8.1

Two touching Take radius 2 cm. Draw the first circle with centre A. Mark B on the line through A, 4 cm
circles from A, and draw the second circle. They touch at exactly one point.

Circle inside a Keep the tip at the same centre. Draw one circle of radius 1 cm and another of radius 2 cm.
circle

Three Mark three centres on one straight line, 2.5 cm apart, and draw a circle of radius 2 cm at
overlapping each.
circles

The mouse face One big circle of radius 3 cm for the face; two circles of radius 1.5 cm for the ears, with
their centres on the big circle; small circles for the eyes and nose; a straight line for the
mouth.

The crossed Draw two slanting line segments crossing each other, then a small circle of radius 1 cm at
sticks the lower end of each.

Yes, the instruments make it far easier. Freehand circles come out wobbly and of different
sizes; a compass gives a smooth curve, and the same opening gives circles of exactly the same
size — so the picture looks neat and can be repeated.

Try This: keep one compass opening (say 2 cm) and, without changing it, draw six
circles whose centres lie on one circle of the same radius. You get the beautiful six-
petal rangoli flower.

Construct — Pages 190 & 191

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.1 Artwork — A Person, Wavy Wave

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q1 A Person. How will you draw this?

‘A Person’ — a circle joined by a short line to a body whose top edge is a curve.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

The figure has two components — a circle with a short line (the head and neck) and a square-
like body whose top side is replaced by a curve that dips inwards.
Steps of construction

1. Draw the body first. Draw a line segment DE = 4 cm for the base.
2. At D and at E draw perpendiculars of length 4 cm and mark the top corners B and C. (Do not
join B and C by a straight line.)
3. Now the curve. Take a radius of about 2.5 cm in the compass, place the tip above the middle
of BC, and try it out until the pencil passes through both B and C. Draw that arc — it dips
down into the body.
4. Above the arc, draw a short vertical line segment of about 1 cm for the neck, starting from
the lowest point of the arc.
5. With the top of the neck as the point of contact, draw a circle of radius 1.5 cm for the head.

compass tip

B C

D E

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

A Person. The red dot is where the compass tip is placed; the dashed lines show that it is the same
co m
e m.
distance from B and from C. The red curve through B and C is the part that is actually drawn.

m l as
m .co a g
l a se
Why the tip goes above the body: an arc always bends away from the compass tip.
g
aTo make the top of the body sink downwards, the centre must be placed above it.
Move the tip higher and the curve becomes flatter; bring it lower and the curve
co m
becomes deeper.
e m . ag
g l as
a
Q2 Wavy Wave. Construct this. (As the length of the central line is not specified, we can
co m
em.
take it to be of any length. Let us take AB to be the central line such that the length

m l as
.co
of AB is 8 cm. Here, the first wave is drawn as a half circle.)
a g
se m
g l a
a Steps of construction

m a s
.co agl
1. Draw a line segment AB = 8 cm. This is the central line.

a s em
2. Mark its mid-point X, so that AX = XB = 4 cm.
3. Mark the mid-point of AX; it isg2lcm from A. Place the compass tip there, open it to 2 cm and
a
draw a half circle above AB from A to X.
4. Without changing the opening, place the tip at the mid-point of XB (2 cm from X) and draw a
co m
m .
as e
half circle below AB from X to B.
m l
.co a g
5. Rub out nothing — the two half circles together make one smooth wave.

se m
g l a
a
se m
m l a
.co
radius e2mcm a g
g l as
a

co m
m .
m as e
.co A X a g l B
se m
g l a
a c
radius 2 cm m .
a s e
. om
cAB agl
em = 8 cm
g l as
a

co m
m .
m ase
.co


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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

The wavy wave on AB = 8 cm. The two red dots are where the compass tip is placed; both half circles
have radius 2 cm.

Why the radius is 2 cm: each half circle sits on half of AB, that is on a diameter of 4
cm. Radius is half of the diameter, so radius = 4 ÷ 2 = 2 cm.

Figure it Out — Page 191
Section 8.1 Artwork — Wavy Wave

TRY THIS

Q1 What radius should be taken in the compass to get this half circle? What should be
the length of AX?

The two waves must be identical, so the central line AB = 8 cm is shared equally between them.

AX = XB = 8 ÷ 2 = 4 cm

AX is the diameter of the first half circle

radius = 4 ÷ 2 = 2 cm

Answer: radius = 2 cm and AX = 4 cm. The tip of the compass is placed at the mid-point of AX,
that is 2 cm from A.

Why: a half circle stands on its diameter. Here the half circle starts at A and ends at
X, so AX itself is the diameter, and the radius is always half the diameter.

Q2 Take a central line of a different length and try to draw the wave on it.

The method never changes — only the numbers change. Take, for example, AB = 10 cm.

1. Draw AB = 10 cm and mark the mid-point X, so AX = XB = 5 cm.
2. Radius needed = 5 ÷ 2 = 2.5 cm.
3. Place the tip 2.5 cm from A and draw the half circle above AB, from A to X.
4. Place the tip 2.5 cm from X and draw the half circle below AB, from X to B.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

CENTRAL LINE AB AX = XB RADIUS OF EACH HALF CIRCLE

8 cm 4 cm 2 cm

10 cm 5 cm 2.5 cm

12 cm 6 cm 3 cm

6 cm 3 cm 1.5 cm

Tip: the radius is always one quarter of the central line — AB ÷ 4. Check: 8 ÷ 4 = 2,
10 ÷ 4 = 2.5, 12 ÷ 4 = 3. ✔

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q3 Try to recreate the figure where the waves are smaller than a half circle (as
appearing in the neck of the figure, ‘A Person’). The challenge here is to get both the
waves to be identical. This may be tricky!

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

‘A Person’, page 190 — the wave at the neck that this question refers to.

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
e m.
as
m l
m .co a g
A wave smaller than a half circle means the arc is flatter. For that the compass tip is not put on

l a se
the line AB but away from it, and a bigger radius is used.

a g
Steps of construction

m
1. Draw AB = 8 cm and mark the mid-point X (AX = XB = 4 cm).
. co ag
se m
2. Take a radius bigger than 2 cm — say 2.5 cm — in the compass.

l a
g A and X. Draw the arc above AB.
3. Place the tip below the line, on the perpendicular through the mid-point of AX, and adjust it
until the pencil passes through a
both
4. Do not change the compass opening. Now place the tip above the line, on the

co m
m.
perpendicular through the mid-point of XB, until the pencil passes through X and B. Draw

m
that arc below AB.
as e
.co a g l
se m
g l a
a
tip above
m a s
m .co agl
l a se
a g

co m
m .
mA X as e B
.co a g l
se m
g l a
a
se m
tip below com g l a
m . a
ase
agl
Flatter waves. The same compass opening is used for both arcs, and the two tips are placed at equal

m
distances on opposite sides of AB — that is what makes the waves identical.

. co
em
m l as
.co a g
m same radius and the same chord. Here both chords are 4 cm (AX and XB) and the
Why both waves come out identical: two arcs are identical when they have the

l a se
ag
c
compass opening is never changed, so the two arcs must match exactly. The trick is
m .
only to remember to put the tip on the opposite side of the line the second time.
m a s e
em . co agl
g l as
Construct — Page 192 a

co m
m .
m ase
.co


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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.1 Artwork — Eyes

Q1 Eyes. How do you draw these eyes with a compass? (For a hint, go to the end of the
chapter.)

Each eye is made of two arcs of the same radius, one bulging up and one bulging down,
meeting at the two corners of the eye. A small filled circle in the middle is the pupil.
Steps of construction (one eye)

1. Draw a light horizontal helper line. Mark two points on it 4 cm apart — these are the two
corners of the eye.
2. Take a radius of about 2.5 cm in the compass. Place the tip at a point B below the helper
line, on the perpendicular through the middle, so that the pencil passes through both
corners. Draw the arc — it bulges upwards and becomes the upper eyelid.
3. Keeping the same opening, place the tip at the point A above the line, exactly as far above as
B was below. Draw the arc — it bulges downwards and becomes the lower eyelid.
4. Mark the mid-point of the eye and draw a small circle of radius 0.7 cm there. Fill it in — that
is the pupil.
5. Repeat everything for the second eye, using the same helper line and the same compass
openings, so that both eyes look alike.

A

B

Page 15 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

One eye. The tip at B (below) draws the upper curve; the tip at A (above) draws the lower curve. A and
B are equally far from the helper line, so the eye comes out symmetrical.

Why A is above and B is below: an arc bends away from the compass tip. To make a
curve that arches upwards, the tip must be below it; to make one that sags
downwards, the tip must be above. Keeping A and B at equal distances from the
helper line makes the upper and lower curves mirror images — a symmetrical eye.

Tip: this may need a few trials. Draw the helper lines very lightly with a hard pencil
so that you can rub them out at the end.

Q2 Make other artwork of your choice with a ruler and a compass.

Here are some designs that use only a ruler and a compass. Try them in your notebook and
colour them.

Six-petal flower (rangoli): draw a circle. Without changing the compass opening, put the tip
anywhere on the circle and cut an arc; move the tip to the cut point and repeat. Six arcs go all
the way round and make a flower.
Chain of circles: mark points on a straight line at equal distances and draw a circle of the
same radius at each. Overlapping circles give a lovely chain.
Crescent moon: draw a circle, then draw a second circle of the same radius with its centre a
little to one side. The part between the two curves is a crescent.
Leaf (vesica): draw two circles of the same radius, each passing through the other's centre.
The common part is a leaf shape.
Rings of a target: keep the tip at one point and draw circles of radius 1 cm, 2 cm, 3 cm, 4
cm. Colour alternate rings.

Try This: the six-petal flower works because the radius of a circle can be stepped
around it exactly six times. Count the arcs — you will always get 6, never 5 or 7.

In-text Questions — Page 193

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.2 Squares and Rectangles

Q1 So, can a rectangle be named using any combination of the labels around its
corners? For example, it cannot be named ABDC or ACBD. Can you see what names
are allowed and what names are not?

No, any combination is not allowed. A name is valid only if the letters follow the corners in
the order in which you travel around the boundary, starting from any corner and going
either clockwise or anticlockwise.
For the rectangle of Fig. 8.4 (A top-left, B top-right, C bottom-right, D bottom-left):

ALLOWED NAMES (8 IN ALL) NOT ALLOWED

ABCD, BCDA, CDAB, DABC ABDC

ADCB, DCBA, CBAD, BADC ACBD, ACDB, BDAC …

Why: the name has to tell you which corners are joined. In ABCD the sides are AB,
BC, CD and DA — all four are real sides. In ABDC the letters ask you to join B to D,
which is a diagonal, not a side. So the name describes a different figure altogether.

Quick test: read the letters in a circle. If every neighbouring pair (including the last
with the first) is a side of the figure, the name is valid.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q2 Which of the following is not a name for this square? 1. PQSR 2. SPQR 3. RSPQ 4.
QRSP

S P

R Q

The square of page 193, with its corners marked S, P, Q and R.

In the given square, going round the boundary the corners come in the order S → P → Q → R →
S (S top-left, P top-right, Q bottom-right, R bottom-left).

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
m.
NAME SIDES IT ASKS FOR VALID?

mPQ, QS, SR, RP as e
1. PQSR
.co a g l
No — QS and RP are diagonals

se m
g l a
2. SPQR Yes
a
SP, PQ, QR, RS

3. RSPQ RS, SP, PQ, QR Yes

co m Yes
m . ag
se
4. QRSP QR, RS, SP, PQ

l a
g for this square.
Answer: option 1 — PQSR is not aaname

co m
m.
Why it happens: options 2, 3 and 4 are the same round trip started from different

o m l a se
corners. PQSR jumps from Q straight to S, and Q and S are opposite corners — that
.c a g
se m
is a diagonal, so the letters no longer travel around the boundary.

g l a
a
m a s
. c o agl
Q3 Here is a square piece of paper having all its sides equal in length and all angles

s e m in the figure. Is it still a square?
equal to 90°. It is rotated as shown
a
agl

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
a s em A square piece of paper, and the same piece after it has been turned.

agl
.c
s e m
m a
. co agl

em
l as
Yes, it is still a square. Check the two properties of a square on the rotated paper:

a g
S1 — Are all the sides still equal? Yes. Turning the paper does not stretch or shrink any

m
side.

. co
e m
m l as
.co a g
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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

S2 — Are all the angles still 90°? Yes. Turning the paper does not open or close any corner.

→

a square
still a square
The same piece of paper, turned. Sides and angles are untouched, so both properties S1 and S2 still
hold.

Why it happens: a shape is a square because of its lengths and angles, not because
of the direction it faces. Rotation moves the figure but changes no measurement. By
the same reasoning, a rotated rectangle is still a rectangle.

Figure it Out — Page 194

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.2 Squares and Rectangles

Q1 Draw the rectangle and four squares configuration (shown in Fig. 8.3) on a dot
paper. What did you do to recreate this figure so that the four squares are placed
symmetrically around the rectangle? Discuss with your classmates.

Fig. 8.3, page 192 — a rectangle with four equal squares placed around it.

Steps on dot paper

1. First draw the rectangle in the middle — say 6 dots long and 4 dots high, with its corners on
dots.
2. Now go out diagonally from each corner of the rectangle by the same number of dots (leave
one dot gap in both directions).
3. From that dot, draw a small square of side 3 dots so that its inner corner points at the
corner of the rectangle.
4. Repeat at all four corners, always leaving the same gap and using the same size of square.

What makes it symmetric: equal treatment of all four corners — same gap, same square size,
same diagonal direction. Counting dots does this automatically, without any measuring.

Page 21 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Check it yourself: fold the drawing along the vertical middle line and then along the
horizontal middle line. In a correct figure the four squares fall exactly on one
another.

Q2 Identify if there are any squares in this collection. Use measurements if needed.

Only A is a square. Read the corners off the dot grid — that is quicker and safer than
measuring.

FIGURE HALF-DIAGONALS (IN DOT SIDES ANGLES WHAT IT IS
STEPS)

A 4 across, 4 down all four equal all 90° a square (a rotated
one)

B 3 across, 4 down all four equal not 90° a rhombus, not a
square

C 5 across, 4 down all four equal not 90° a rhombus, not a
square

D sides 1 right 2 down, and 2 right 1 up two long, two all 90° a rectangle, not a
short square

D✘
A✔ B✘ C✘
square rhombus rhombus rect.
Only A has equal sides and right angles, so only A is a square. B and C have equal sides but slanted
corners; D has right angles but unequal sides.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Why B and C fail: in A the two diagonals are equally long (4 + 4 across and 4 + 4
down), so the four sides meet at 90°. In B the diagonals are 6 and 8 dot steps, in C
they are 10 and 8 — the corners get squeezed and the angles are no longer right
angles, even though all four sides are still of equal length.

Think Think: Is it possible to reason out if the sides are equal or not, and if the angles
are right or not without using any measuring instruments in the above figure?
Can we do this by only looking at the position of corners in the dot grid?

A B C D

The collection on page 194 — four 4-sided figures A, B, C and D drawn on a dot grid,
every corner sitting on a dot.

Yes — the dots alone are enough. No ruler, no protractor.

For equal sides: describe each side as a move on the grid, e.g. “4 right and 4 down”. Two
sides are equal if they need the same pair of steps (in some order). In A every side is a “4 and
4” move, so all four sides are equal.
For right angles: a side that goes “4 right, 4 down” meets a side that goes “4 right, 4 up” —
turning one move sideways gives exactly the other. Whenever one move is the other one
turned by a quarter turn, the corner is a right angle.
In D the moves are “2 right, 1 up” and “1 right, 2 down” — one is the other turned by a
quarter turn, so D does have right angles, but the moves are of different sizes, so its sides are
unequal.

Why this works: the dots are equally spaced in both directions, so counting steps is
measuring — with the dot spacing as the unit. Equal step-pairs mean equal lengths,
and a quarter-turn of a step-pair means a quarter-turn of the line, that is 90°.

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
m.
Draw at least 3 rotated squares and rectangles on a dot grid. Draw them such that
e
Q3

om their respective properties.
their corners are on the dots. Verify if the squares and rectangles that you have
csatisfy g l as
. a
em
drawn

a s
a gl

m
Use the “step” trick: pick a move, then turn it by a quarter turn for the next side.
. co ag
e m
FIGURE
l as
STEPS FOR THE SIDES (REPEAT THE PATTERN
g
CHECK
ROUND) a
✔ .com
Rotated square 3 right 1 down, then 1 left 3 down, then 3 left 1 up, then 1 right 3 all sides equal, all angles

em
1 up 90°

a s
. com a gl all sides equal, all angles
2 em
Rotated square 2 right 2 down, 2 left 2 down, 2 left 2 up, 2 right 2 up

as
✔
l
90°

a g Rotated square 4 right 1 down, 1 left 4 down, 4 left 1 up, 1 right 4 up all sides equal, all angles
✔
a s
com
3

agl
90°

.
long sides (4 right 2la sem
Rotated 2 right 1 down, then 1 left 2 down — repeat twice as long on the opposite sides equal, all

ag
rectangle 1 down) angles 90° ✔
Rotated
m✔
3 right 1 down for the long side, 1 left 3 down for the short side opposite sides equal, all
rectangle 2
. co
angles 90°

em
m l as
.co a
How to verify: measure the four sides with a ruler (equal for a square; equal in pairs for a
g
a s em
rectangle) and check each corner with the corner of a set square or a piece of paper — every

agl corner must fit it exactly.
se m
com
Tip: to turn a move by a quarter turn, swap the two numbers and change one
g l a
m . a
e
as closes by itself.
direction: “3 right 1 down” becomes “1 left 3 down”. Do this four times and you come

a g l
back to the start — the figure

co m
m .
m
In-text Questions — Pages 195 & 196
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 as e
.co


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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.3 Constructing Squares and Rectangles

Q1 Now, let us start constructing squares and rectangles. How would you construct a
square with a side of 6 cm?

Here is the full construction of the square PQRS of side 6 cm, exactly as in the six steps of the
book.

1. Step 1. Draw a line segment PQ = 6 cm with a ruler.
2. Step 2. At P, draw a perpendicular to PQ. (Place the centre of the protractor on P, mark the
90° point, and join; or use the right-angled corner of a set square.)
3. Step 3. On this perpendicular mark S so that PS = 6 cm. Method 1: measure 6 cm with a ruler.
Method 2: open the compass to 6 cm, put the tip at P and cut the perpendicular — the cut is
S.
4. Step 4. At Q, draw a perpendicular to PQ in the same way.
5. Step 5. With the compass still open to 6 cm, put the tip at Q and cut this second
perpendicular. Call the cut R.
6. Step 6. Join RS. PQRS is the required square.

S 6 cm R

6 cm 6 cm

P 6 cm Q

Page 25 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

The square PQRS of side 6 cm. The two grey arcs are the compass, opened to 6 cm, cutting the
perpendiculars at S and at R.

Tip: the compass method (Method 2) is more accurate than measuring twice with a
ruler, because the same opening is used for both sides — so PS and QR are certainly
equal.

Q2 Can you see why PS should be 6 cm long?

Yes — because of the very definition of a square.

Property S1: all the sides of a square are equal

PQ = 6 cm (already drawn)

so PS = QR = RS = 6 cm

Why it happens: PQ and PS are two sides of the same square meeting at the corner
P. If PS were shorter or longer than PQ, the figure would still have four right angles
but unequal sides — it would be a rectangle, not a square. Taking PS = 6 cm is what
forces the figure to close as a square.

Q3 How long is the side RS and what are the measures of ∠R and ∠S?

RS = 6 cm

∠R = 90° ∠S = 90°

Measure them in your figure — the ruler shows 6 cm and the protractor shows 90° at both
corners.

Page 26 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Why it comes out this way without being asked for: we fixed PQ = 6 cm, made the
two side lines perpendicular to PQ and cut off 6 cm on each. So S and R are both 6
cm above the base and the two perpendiculars are 6 cm apart — the segment RS
joining them is therefore parallel to PQ, of the same length 6 cm, and it meets the
two upright sides at right angles. The last two properties come free: you only need
four correct steps to force a square.

Construct — Page 197
Section 8.3 Constructing Squares and Rectangles

TRY THIS

Q1 Draw a rectangle with sides of length 4 cm and 6 cm. After drawing, check if it
satisfies both the rectangle properties.

Steps of construction

1. Draw AB = 6 cm.
2. At A draw a perpendicular to AB; at B draw another perpendicular to AB.
3. Open the compass to 4 cm. With the tip at A cut the first perpendicular at D; with the tip at B
cut the second perpendicular at C.
4. Join DC. ABCD is the required rectangle.

Page 27 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

D C

4 cm

A 6 cm B
Rectangle ABCD with AB = 6 cm and AD = 4 cm. All four corners are right angles.

Checking the two properties

R1 AB = DC = 6 cm, AD = BC = 4 cm → opposite sides equal ✔

R2 ∠A = ∠B = ∠C = ∠D = 90° ✔

So the figure drawn is indeed a rectangle.

Q2 Draw a rectangle of sides 2 cm and 10 cm. After drawing, check if it satisfies both
the rectangle properties.

The steps are exactly the same, only the numbers change.

1. Draw PQ = 10 cm.
2. Draw perpendiculars to PQ at P and at Q.
3. Open the compass to 2 cm and cut both perpendiculars, at S (from P) and R (from Q).
4. Join SR. PQRS is the rectangle.

Page 28 of 74

as e
Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
S sem .co a g
R
a
agl
m 2 cm
.co ag
a sem
P agl 10 cm Q
. c
A long, thin rectangle: 10 cm by 2 cm. It is still a rectangle — the shape looks unusualom
but both

m
properties hold.
a s em
. co a gl
m
ase
agl R1 PQ = SR = 10 cm, PS = QR = 2 cm ✔
R2 ∠P = ∠Q = ∠R = ∠S = 90° ✔
m a s
em
.co agl
a s
agl draw the 10 cm base along the long side of your
Tip: for such a long rectangle,
notebook page and press the set square firmly, otherwise the two short sides come

co m
.
out slanting.

e m
m l as
.co a g
a s em Is it possible to construct a 4-sided figure in which — all the angles are equal to 90º
a gl Q3
but opposite sides are not equal?

se m
com g l a
m . a
e

l s are right angles, the opposite sides are forced to be
aangles
g
No, it is not possible. If all four
equal — you cannot escapeait.
Try it and see what happens:
c o m
.
m both make 90° with
s e
1. Draw AB, and draw perpendiculars to AB at A and at B. These two lines

. omso they run in the same direction — they are parallel aand
cAB, g lastay the same distance apart
a s em for ever.
agl 2. Mark D on the first line and C on the second. For ∠D and ∠C to be 90° as well, DC must be
c
perpendicular to both lines.
m .
m a s e
co agl
3. But the distance between two parallel lines is the same everywhere, so DC = AB, and D and C

m .
e
must be at the same height, so AD = BC.

g l as
a

co m
m .
m ase
.co


a g l Page 29 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

All four angles 90° → opposite sides parallel → opposite sides equal

Why it happens: a four-sided figure has only four corners to play with. Once you fix
all four angles at 90°, the only freedom left is the two lengths — and those lengths
repeat on the opposite sides. So every 4-sided figure with four right angles is a
rectangle (and if the two lengths are also equal, a square).

Try This: draw AB = 5 cm, perpendiculars at A and B, then mark AD = 3 cm and BC =
4 cm and join DC. Measure ∠D and ∠C — they will not be 90°, one is more and the
other less.

In-text Questions — Pages 197 – 199
Section 8.4 An Exploration in Rectangles

MATH TALK

Q1 Construct a rectangle ABCD with AB = 7 cm and BC = 4 cm. Imagine X to be a point
that can be moved anywhere along the side AD. Similarly, imagine Y to be a point
that can be moved anywhere along the side BC.

Steps of construction

1. Draw AB = 7 cm as the top side.
2. At A and at B draw perpendiculars to AB, going downwards.
3. Open the compass to 4 cm; from A cut the first perpendicular at D, from B cut the second at
C.
4. Join DC — ABCD is the rectangle. (AD = BC = 4 cm, DC = AB = 7 cm.)
5. Mark any point X on AD and any point Y on BC, and join XY with a dashed line.

Page 30 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

A 7 cm B

X
4 cm

Y

D C

Rectangle ABCD (7 cm × 4 cm). X slides along AD, Y slides along BC, and we watch the length XY.

Tip: draw the rectangle four or five times on one page, so that you can try different
positions of X and Y without rubbing out.

Q2 At which positions will the points X and Y be at their closest? When do you think
they will be the farthest? What does your intuition say? Discuss with your
classmates.

Closest: when X and Y are at the same distance from A and from B — that is, when XY is
parallel to AB. Then

XY = AB = 7 cm (the smallest possible value)

Farthest: when X and Y are at opposite ends of their sides — X at A with Y at C, or X at D with Y
at B. Then XY is a diagonal of the rectangle.

XY = AC = BD = about 8.1 cm (the largest possible value)

Page 31 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

POSITION OF X POSITION OF Y LENGTH XY

X at A Y at B 7 cm (smallest)

1 cm from A 1 cm from B 7 cm (smallest)

X at D Y at C 7 cm (smallest)

X at A Y at C 8.1 cm (largest)

X at D Y at B 8.1 cm (largest)

Why it happens: going from X to Y you must cross the full width of the rectangle, 7
cm. If X and Y are level with each other you cross straight, and 7 cm is all you travel.
If one of them is higher than the other you must also climb, and a slanting path is
always longer than the straight crossing. The biggest climb possible is the full 4 cm
— that gives the diagonal.

Q3 How does the minimum distance between the points X and Y compare to the length
of AB?

They are exactly equal.

minimum XY = AB = 7 cm

The smallest XY can ever be is the width of the rectangle, and that width is AB itself.

Why: AD and BC are opposite sides of a rectangle, so they are parallel and always 7
cm apart. XY joins a point of one to a point of the other, so XY can never be shorter
than 7 cm — and it becomes exactly 7 cm when XY is drawn straight across, parallel
to AB.

Page 32 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q4 How will you keep track of the lengths XY for different positions of X and Y? Is there
a shorthand way of writing it down? (When X is 5 mm away from A and Y is 3 cm
away from B, XY = ___ cm ___ mm; when X is 1 cm away from A and Y is 1 cm away
from B, XY = ___ cm ___ mm; when X is 2 cm away from A and Y is 4 cm away from B,
XY = ___ cm ___ mm.)

Yes — instead of writing a whole sentence each time, make a table with one column for each
thing that changes.

DISTANCE OF X FROM A DISTANCE OF Y FROM B LENGTH OF XY

5 mm 3 cm 7 cm 4 mm

1 cm 1 cm 7 cm 0 mm

2 cm 4 cm 7 cm 3 mm

Measure each XY in your own figure — you should get these readings (correct to the nearest
millimetre).

Why the middle row is exactly 7 cm: X and Y are both 1 cm below the top, so XY
runs straight across, parallel to AB, and equals AB = 7 cm. In the other two rows
there is a difference in height (2.5 cm in the first row, 2 cm in the third), so the line
slants and comes out a few millimetres longer.

Tip: a table like this is a mathematician's shorthand. Only three things change, so
three columns are enough — the rest of the sentence is written once, in the
headings.

Page 33 of 74

as e
Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
m.
Have you checked what happens to the length XY when X and Y are placed at the
se
Q5

o m l a
same distance away from A and B, respectively? (5 mm and 5 mm; 1 cm and 1 cm; 1
g the length XY
m .c and 1 cm 5 mm.) In each of these cases, observe 1. how
cm 5 mm a
l a se
compares to that of AB and 2. the shape of the 4-sided figure ABYX.
a g

com

.
emY FROM B ag
DISTANCE OF X FROM A
a
DISTANCE OF
s LENGTH OF XY SHAPE OF ABYX

5 mm 5 mma
gl 7 cm a rectangle

co m
m.
1 cm 1 cm 7 cm a rectangle

m as e a rectangle
1 cm 5 mm
.co
1 cm 5 mm 7 cm
a g l
se m
g l a
a 2. The 4-sided figure ABYX is always a rectangle — a small one at the top of the big rectangle.
1. In every such case XY = AB = 7 cm.

m a s
e
. c o
m parts of two equal, parallel sides, so X and Y agl
s
Why it happens: AX and BY are equal
a ∠B are right angles because they are corners of
stay level with each other. ∠Aland
g
a
the big rectangle, and XY crosses the two upright sides at right angles too. Four right
angles and equal opposite sides — that is a rectangle, and its side XY must equal its
co m
m .
e
opposite side AB.
m l as
.co a g
m it yourself: as X slides down from A to D with Y always level with it, the small
a s eCheck
a gl rectangle ABYX grows taller and taller, but its width never changes. When X reaches
D and Y reaches C, ABYX becomes the whole rectangle ABCD.
se m
com g l a
m . a
ase
Q6
agl
How does the farthest distance between X and Y compare with the length of AC?

m
BD?

. co
se m
o m l a
.c farthest distance is exactly equal to AC, and also to BDa—gthe two diagonals of the

se m
The
a
agl
rectangle.

.c
s e m
m a
. co agl
farthest XY = AC = BD ≈ 8.1 cm

em
l as
(check: 7 cm across and 4 cm down gives about 8.1 cm)
a g

co m
m .
m as e
.co


a g l Page 34 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Why it happens: X can go no further than the ends A and D of its side, and Y no
further than the ends B and C of its side. The longest join is from one end to the
opposite end — X at A with Y at C gives the line AC, and X at D with Y at B gives DB.
Those are the diagonals, and in a rectangle the two diagonals are always equal in
length.

Check it yourself: draw both diagonals of your rectangle and measure them. Both
come out about 8.1 cm, and they cut each other exactly in half.

Construct: Breaking Rectangles — Pages 199 – 201
Section 8.4 An Exploration in Rectangles

Q1 Breaking Rectangles. Construct a rectangle that can be divided into 3 identical
squares as shown in the figure.

A rectangle divided into 3 identical squares.

First plan with a rough diagram. If the small squares have side s, the rectangle is s tall and 3s
long. So the long side must be three times the short side.
Take the short side as 3 cm, so the long side is 9 cm.

1. Draw AF = 3 cm (the short side, drawn vertically).
2. At A draw a perpendicular to AF, long enough to hold 9 cm.

Page 35 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

3. Open the compass to AF (3 cm) and, starting from A, step it along the perpendicular three
times to get B, then the next point, then C. Now AC = 9 cm.
4. At C draw a perpendicular to AC and cut off 3 cm to get D. Join FD.
5. Join the marks on AC to the matching marks on FD. The rectangle ACDF is now cut into three
identical squares of side 3 cm.

A 9 cm = 3 × 3 cm C

3 cm

F D
A 9 cm × 3 cm rectangle falls into three identical 3 cm squares.

Why the long side must be 3 × the short side: the three squares sit side by side
and each of them is as tall as the rectangle. So each square has side = the short side,
and three of them laid in a row make a length of three times that.

Q2 Explore: What about constructing a rectangle that can be divided into two identical
squares? Can you try it? It is wise to first plan and then construct. But how do we
plan? Can you think of a way?

The way to plan is to draw a rough diagram of the finished figure first and mark on it
everything that must be equal.
Rough diagram: rectangle ACDF, cut by the segment BE into the squares ABEF and BCDE.

Page 36 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Since the two squares are identical: AB = BC and FE = ED

Since ABEF is a square: AF = AB = BE = FE

Since BCDE is a square: BE = BC = CD = ED

So all the short segments are equal — and AC = 2 × AF

Steps of construction (taking the short side 3 cm):

1. Draw AF = 3 cm.
2. Draw a perpendicular to AF at A.
3. Open the compass to AF and step it twice along that perpendicular: first step gives B, second
gives C. So AC = 6 cm.
4. At C draw a perpendicular, cut off 3 cm to get D, and join FD.
5. Join B to E (the matching point on FD). ABEF and BCDE are the two identical squares.

Tip: put a small tick mark ‘|’ on each segment that is equal to the others. That single
habit turns a rough sketch into a construction plan.

Q3 To draw the rectangle ACDF, one could assign any length to AF. For example, if we
assign AF = 4 cm, then what must the length of AC be? Explore: Can the rectangle
now be completed?

AF = 4 cm

AB = BC = AF = 4 cm (all the short segments are equal)

AC = AB + BC = 4 + 4 = 8 cm

Yes, the rectangle can now be completed, because both of its sides are known: 8 cm and 4
cm.

1. Draw AF = 4 cm; draw the perpendicular to AF at A.
2. Mark B at 4 cm and C at 8 cm along it.
3. Draw the perpendicular to AC at C and mark D with CD = 4 cm.
4. Join FD, and join BE. The rectangle 8 cm × 4 cm splits into two 4 cm squares.

Page 37 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Did you know? For three identical squares the same rule gives AC = 3 × 4 = 12 cm;
for four squares, 16 cm. The rectangle 8 × 4 is exactly the one used in the next task,
“A Square within a Rectangle”.

Q4 In fact, one could proceed by drawing AF without even measuring its length using a
ruler. As, AB = AF, we need to somehow transfer the length of AF to get the point B.
How do we do it without a ruler? Can it be done using a compass?

Yes — this is exactly what a compass is best at. A compass can carry a length from one place
to another without ever telling you what the length is.

1. Draw AF of any length and draw the perpendicular to AF at A.
2. Put the metal tip on A and open the compass until the pencil reaches F. Now the opening is
the length AF.
3. Without changing the opening, keep the tip at A and cut the perpendicular. The cut is B, and
AB = AF.
4. Shift the tip to B and cut again — that gives C, with BC = AB. (One more step gives a third
square.)
5. Finish the rectangle with perpendiculars at C and the segment FD.

Why it works: the compass opening does not change when you move it, so every
arc you draw has the same radius. Stepping it along a line therefore lays out equal
lengths one after another — like using a pair of dividers instead of a ruler.

Check it yourself: measure AF, AB and BC at the end with a ruler. All three come out
the same, even though you never measured while constructing.

Q5 Give the lengths of the sides of a rectangle that cannot be divided into — two
identical squares; three identical squares.

A rectangle splits into n identical squares exactly when its long side is n times its short side. So
choose lengths that break this rule.

Page 38 of 74

as e
Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

co m
m.
RECTANGLE IS LONG = 2 × IS LONG = 3 × CONCLUSION

as e
com
SHORT? SHORT?

. a g l
cm as
em
4 cm × 2.5 No (2 × 2.5 = 5) No (3 × 2.5 = 7.5) cannot be divided into 2 or 3

agl
identical squares

7 cm × 2 cm No (2 × 2 = 4) No (3 × 2 = 6) cannot be divided into 2 or 3
m
.co ag
identical squares

a s e(3m× 3 = 9)
gl
5 cm × 3 cm No (2 × 3 = 6) No cannot be divided into 2 or 3
a identical squares

6 cm × 3 cm Yes No
co m
splits into 2 squares, but not into 3

em.
m l as
.co
9 cm × 3 cm No Yes splits into 3 squares, but not into 2

a g
se m
g l a
Answers: a rectangle of 4 cm × 2.5 cm (or 5 cm × 3 cm) cannot be divided into two identical
a squares; a rectangle of 7 cm × 2 cm (or 5 cm × 3 cm) cannot be divided into three identical
s
squares. Many more are possible — try your own.
m a
em
.co agl
a s
Why: if n identical squares fill the rectangle in a row, each square must be as tall as

agl the short side; the total length is then n times the
the rectangle, so its side equals
short side. If the long side is not that exact multiple, no such division exists.

co m
m .
m as e
. c o a g l
e m
Construct
s
— Pages 201 – 203
a
agl
se m
com g l a
m . a
ase
agl

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

co m
m .
m ase
.co


a g l Page 39 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Section 8.4 — squares inside squares and rectangles

TRY THIS

Q1 A Square within a Rectangle. Construct a rectangle of sides 8 cm and 4 cm. How will
you construct a square inside, as shown in the figure, such that the centre of the
square is the same as the centre of the rectangle? (Hint: Draw a rough figure. What
will be the sidelength of the square? What will be the distance between the corners
of the square and the outer rectangle?)

8 cm
4 cm

A rectangle of sides 8 cm and 4 cm with a square drawn inside it.

Plan first. In the figure the square touches the top and the bottom sides of the rectangle. So its
side must equal the height of the rectangle.

side of the square = height of rectangle = 4 cm

space left over along the length = 8 − 4 = 4 cm

this is shared equally on the two sides: 4 ÷ 2 = 2 cm on the left and 2 cm on the right

Steps of construction

1. Construct the rectangle ABCD with AB = 8 cm and BC = 4 cm (base, two perpendiculars, then
the top side).

Page 40 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

2. On the base AB mark a point 2 cm from A and another point 2 cm from B.
3. At each of these two points draw a perpendicular to AB, right across to the top side.
4. The strip between these two perpendiculars is the required square, 4 cm by 4 cm.

8 cm

4 cm

2 cm square 4 cm 2 cm

The 4 cm square sits in the middle of the 8 cm × 4 cm rectangle, leaving 2 cm on the left and 2 cm on
the right. The green dot is the common centre.

Why 2 cm on each side: “same centre” means the picture must look the same from
the left and from the right. The leftover length 8 − 4 = 4 cm therefore has to be split
into two equal halves of 2 cm.

Q2 Falling Squares. Construct three squares of side 4 cm arranged as shown (make sure
that the squares are aligned the way they are shown). Now, try this with a square of
side 3 cm, a square of side 5 cm and a square of side 7 cm.

Look carefully at how the squares are joined: the bottom-left corner of one square is the top-
right corner of the next. They only touch, corner to corner, going down to the left.
Steps of construction (three squares of side 4 cm)

1. Construct the first square ABCD of side 4 cm at the top right (base, perpendiculars, top).

Page 41 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

2. Take its bottom-left corner. Starting there, construct the second square of side 4 cm going
down and to the left, so that this corner becomes the second square's top-right corner.
3. Repeat once more from the bottom-left corner of the second square to get the third square.
4. Keep every side either horizontal or vertical, so that the squares stay aligned like steps of a
staircase.

4 cm

4 cm
shared corner

4 cm

Falling squares. Each red dot is a corner shared by two squares — the lower square hangs from the
bottom-left corner of the one above it.

For the second figure the same rule is used, but the squares get bigger as they fall: side 3 cm
at the top, then 5 cm, then 7 cm. Start with the 3 cm square; from its bottom-left corner build
the 5 cm square (that corner is the 5 cm square's top-right corner); from the bottom-left corner
of the 5 cm square build the 7 cm square.

Page 42 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Tip: draw one long horizontal helper line for each row of squares. If a square slips
even a little, the “falling” look is lost.

Q3 Shadings. Construct this. Choose measurements of your choice. Note that the
larger 4-sided figure is a square and so are the smaller ones.

Shadings, page 202 — a large square cut into smaller squares, with half of many of them
shaded.

Steps of construction (taking a big square of side 8 cm)

1. Construct a square of side 8 cm.

Page 43 of 74

as e
Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

2. Mark points every 2 cm along all four sides and join opposite marks. The square is now a
co m
grid of 16 small squares, each 2 cm × 2 cm.
se m.
o m g l a
.c corner and shade the triangle above it with slanting lines.
3. In the small squares you want shaded, draw the diagonal from the top-left corner to the
m a
sea 4 cm square (four small squares together) unshaded at the bottom-left, and one
bottom-right
l a
ag
4. Leave
more 2 cm square unshaded next to it, exactly as in the picture.

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
. a
left plain
a s em
l ag

co m
m .
o m l a se
m agupper triangle is filled in. A 4 cm
.cThe shading pattern: each shaded cell is cut by a diagonal and the
l a se
g
square and one 2 cm square at the bottom left are left plain.

a c
m .
s e
. c om
Tip: all the diagonals slant the same way, so the shaded triangles line up and the
a g la
s e m rain. Shade lightly with a pencil, using strokes parallel
whole design looks like falling
a
agl
to the diagonal.

co m
m .
m ase
.co


a g l Page 44 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q4 Square with a Hole. Observe that the circular hole is the same as the centre of the
square. (Hint: Think where the centre of the circle should be.)

The centre of the circle must be the centre of the square, and the centre of a square is the
point where its two diagonals cross.
Steps of construction (square of side 6 cm, hole of radius 1.5 cm)

1. Construct a square ABCD of side 6 cm.
2. Join the diagonals AC and BD with light pencil lines. Call their meeting point O.
3. Place the compass tip at O, open it to 1.5 cm and draw the circle.
4. Rub out the two diagonals — they were only helpers.

O

The diagonals of the square meet at O, its centre. The hole is the circle drawn with O as centre.

Page 45 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Why the diagonals give the centre: each diagonal cuts the square into two equal
halves, so the mid-point of a diagonal is equally far from all four sides. Both
diagonals pass through that one point, so their crossing is the centre — no
measuring needed.

Q5 Square with more Holes. Construct this.

This is the previous figure done four times. Divide the big square into four equal squares and
put one hole in each.
Steps of construction (big square of side 8 cm)

1. Construct a square of side 8 cm.
2. Mark the mid-points of all four sides. Join the mid-point of the top side to the mid-point of
the bottom side, and the mid-point of the left side to the mid-point of the right side. The big
square is now four small squares of side 4 cm.
3. In each small square draw its two diagonals lightly to find its centre.
4. With each centre as the tip position, draw a circle of radius 1.5 cm. Use the same opening
for all four, so that the holes are identical.
5. Rub out the light diagonals.

Page 46 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Four identical holes, one at the centre of each of the four small squares.

Tip: never change the compass opening between the four circles. If you re-set it
each time, the holes come out slightly different and the design loses its symmetry.

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q6 Square with Curves. This is a square with 8 cm sidelengths. (Hint: Think where the
tip of the compass can be placed to get all the 4 arcs to bulge uniformly from each
of the sides. Try it out!)

Each side of the square is replaced by an arc that bulges inwards. An arc always bends away
from the compass tip, so for each side the tip must be placed outside the square, on the
perpendicular bisector of that side.
A neat choice is a radius of 5 cm with the tip 3 cm outside the middle of the side, because 3, 4
and 5 fit together perfectly:

half of a side = 8 ÷ 2 = 4 cm

tip is 3 cm outside the middle of the side

distance from the tip to each corner = 5 cm (a 3–4–5 right triangle)
so an arc of radius 5 cm passes exactly through both corners

Steps of construction

1. Construct a square of side 8 cm.
2. Mark the mid-point of each side and draw a short perpendicular there, going outwards.
3. On each of these, mark the point 3 cm outside the square.
4. Open the compass to 5 cm. Put the tip on one of these four points and draw the arc joining
the two corners of that side.
5. Repeat for the other three sides without changing the opening. All four arcs then bulge in
by the same amount (2 cm).

Page 48 of 74

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

In-text Questions — Pages 203 & 204
Section 8.5 Exploring Diagonals of Rectangles and Squares

MATH TALK

Q1 Consider a rectangle PQRS. Join PR and QS. Compare the lengths of the diagonals.
First predict the answer. Then construct a rectangle marking the points as shown
and measure the diagonals.

P Q

S R

Rectangle PQRS with both diagonals PR and QS joined.

Prediction and result agree: the two diagonals of a rectangle are equal in length.
Construct a rectangle, say PQRS with PQ = 7 cm and QR = 4 cm, join PR and QS and measure:

PR = QS ≈ 8.1 cm

Page 50 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

RECTANGLE DIAGONAL 1 DIAGONAL 2 EQUAL?

7 cm × 4 cm 8.1 cm 8.1 cm Yes

6 cm × 4 cm 7.2 cm 7.2 cm Yes

5 cm × 5 cm (a square) 7.1 cm 7.1 cm Yes

Why it happens: each diagonal joins two opposite corners, and both diagonals span
the same width and the same height of the rectangle — one goes “7 across and 4
down”, the other “7 across and 4 up”. Same journey, only mirrored, so the two
lengths must be the same.

Check it yourself: the diagonals also cut each other exactly in half. Measure the four
halves — all four are equal.

Page 51 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q2 Observe that a diagonal divides each of the pair of opposite angles into two smaller
angles. The diagonal PR divides angle R into g and h, and angle P into c and d. Are g
and h equal? Are c and d equal? First predict the answers, and then measure the
angles. What do you observe? Identify pairs of angles that are equal.

P Q
d e
c f

b g
a h
S R

Rectangle PQRS, page 203 — the diagonals split the angle at each corner into two parts,
named a, b, c, d, e, f, g and h.

Take the rectangle 7 cm long and 4 cm high and measure the eight angles with a protractor:

c ≈ 60°, d ≈ 30°, e ≈ 30°, f ≈ 60°, g ≈ 60°, h ≈ 30°, a ≈ 30°, b ≈ 60°

each corner: 60° + 30° = 90° ✔

So g and h are NOT equal, and c and d are NOT equal — the diagonal does not cut the right
angle in half (unless the rectangle happens to be a square).
The equal pairs are:

a = d = e = h (the four “flat” angles, ≈ 30°)

b = c = f = g (the four “steep” angles, ≈ 60°)

Page 52 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

P Q
d e
c f

b g
a h

S R

The two diagonals of rectangle PQRS make eight angles. The four flat ones (a, d, e, h) are equal, and
so are the four steep ones (b, c, f, g).

Why: the diagonal PR is a slanting line that rises the same way at both of its ends, so
the angle it makes with the long side at P equals the angle it makes with the long
side at R — that is d = h. Because the long side is 7 cm and the short side only 4 cm,
the diagonal leans much closer to the long side, so the “flat” angle is small and the
“steep” angle is large. They can only be equal when the two sides are equal.

Q3 Explore: How should the rectangle be constructed so that the diagonal divides the
opposite angles into equal parts?

The rectangle must be constructed with its two adjacent sides equal — that is, it must be a
square.

In a square: each corner = 90°

diagonal divides it into 45° + 45° = two equal parts

Page 53 of 74

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

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Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

3. A diagonal splits each of the two opposite right angles into two parts, and the four “flat”
parts are equal to each other, as are the four “steep” parts.
4. The two parts of a corner are equal (45° each) only when the rectangle is a square.
5. In a square the diagonals also meet at right angles.

How can we be sure? Measuring can never prove a law — a ruler and a protractor can only tell
us that it holds for the few figures we drew, and always with a small error. To be sure we must
give a reason that does not depend on the figure. For example:

A reason for law 1: fold or turn the rectangle by half a turn about the point where
the diagonals cross. The rectangle falls exactly on itself, and the diagonal PR falls on
QS. Two lines that fall on each other must be of the same length. Since this
argument uses only the properties R1 and R2, it works for every rectangle, big or
small — that is what makes it a law.

Math Talk: Discuss in class — “measuring ten rectangles” and “giving one reason”
are very different things. In mathematics only the second one settles the matter.

Construct — Pages 205 – 210
Section 8.5 — rectangles from an angle or a diagonal

Q1 Construct a rectangle in which one of the diagonals divides the opposite angles into
60° and 30°.

Start with a rough diagram of the rectangle ABCD, with the diagonal AC making 60° with AB
and 30° with AD. Then decide the order of the steps: the length of AB is not given, so we may
take it as we like.
Steps of construction

1. Step 1. Draw AB of any convenient length, say 4 cm. At B draw a perpendicular to AB.
2. Step 2. At A, draw a ray making an angle of 60° with AB. Where it cuts the perpendicular at B,
mark the point C.
3. Step 3. At A draw a line perpendicular to AB. The point D must lie on this line.
4. Step 4, Method 1. At C draw a perpendicular to BC; it cuts the line through A at D.
Method 2. Or open the compass to BC, put the tip at A and cut the perpendicular at A — that
point is D, since AD = BC.
5. Join CD. ABCD is the required rectangle.

Page 55 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

D C

30°

60°
A B

The diagonal AC splits the right angle at A into 60° and 30°. With AB = 4 cm, BC works out to about 6.9
cm.

Why the shape is fixed even though no length was given: the two angles fix how
steep the diagonal is, and that fixes the ratio of the sides. Choosing AB = 4 cm simply
decides the size — every such rectangle has the same shape, only bigger or smaller.

Page 56 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

Q2 Now ∠A is divided into two angles. One measures 60°. Check what the other angle
is.

∠A = 90° (all angles of a rectangle are right angles)

one part = 60°

other part = 90 − 60 = 30° ✔

This matches the question, which asked for parts of 60° and 30°.

Why it must be 30°: the diagonal lies inside the corner, so the two parts together
make up the whole corner. Since the whole corner is 90°, fixing one part at 60°
automatically fixes the other at 30°. This is why the problem gives you two numbers
that add up to 90 — if they did not, no rectangle could exist.

Check it yourself: measure the two parts of the opposite angle C as well. They are
also 30° and 60°, but the other way round.

Q3 Construct a rectangle where one of its sides is 5 cm and the length of a diagonal is 7
cm.

Rough diagram first: rectangle ABCD with DC = 5 cm and the diagonal DB = 7 cm. The base can
be drawn at once; the trouble is to find B.
Steps of construction

1. Step 1. Draw the base DC = 5 cm.
2. Step 2. At C draw a perpendicular to DC and call this line l. The point B lies somewhere on l.
3. Step 3. Open the compass to 7 cm, place the tip at D and draw an arc cutting the line l. The
cut is the point B (it is 7 cm from D and lies on l — both conditions at once).
4. Step 4. Draw a perpendicular to DC at D and a perpendicular to BC at B. They meet at A. (Or
simply take the compass opening CB and cut from D.)
5. ABCD is the required rectangle. Check it against R1 and R2.

Page 57 of 74

Class 6 Maths Chapter 8 Playing with Constructions AglaSem · NCERT Solutions

l
A B

7 cm
≈ 4.9 cm

D 5 cm C
Side 5 cm and diagonal 7 cm. The orange arc of radius 7 cm, drawn from D, cuts the perpendicular l
exactly at B. The second side comes out about 4.9 cm.

Why the arc finds B without trial and error: the arc contains all the points that are
7 cm from D, and the line l contains all the points that make a right angle at C. B has
to satisfy both, so it must be where the two meet.

Q4 Consider the point at which the circle and the line intersect. What is its distance
from point D? If needed, check your figure. What do you observe?

Its distance from D is exactly 7 cm — the radius of the circle.

B is on the circle → DB = radius = 7 cm

B is on the line l → ∠BCD = 90°

Page 58 of 74

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Class 6 Maths Chapter 8 Playing with Constructions
a g l AglaSem · NCERT Solutions

What we observe: the crossing point satisfies both conditions at the same time, so it is the
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