NCERT Solutions Class 7 Maths Chapter 9 Geometric Twins

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 9: Geometric Twins

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 1 – 22 10 56 English

Solutions, notes, sample papers & more at 55 pages

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 9: Geometric Twins
Chapter 1 of Ganita Prakash Grade 7 Part II asks a simple question — when are two figures exact copies of
each other? Starting from a signboard symbol that has to be redrawn, the chapter builds the idea of
congruence, and then finds the shortest lists of measurements that force two triangles to be congruent: SSS,
SAS, ASA, AAS and RHS.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 1 – 22

SECTIONS QUESTIONS

10 56

MEDIUM

English

In-text Questions — Pages 1–3
Section 1.1 Geometric Twins

Q1 The symbol on this signboard needs to be recreated on another board. How do we
do it?

One way is to trace the outline of the symbol on tracing paper and copy that outline on to the
new board.
This works for a small symbol. For a big signboard it is not practical — you would need a huge
sheet of tracing paper and a way to hold it flat against the board.

Why it happens: tracing copies every point of the figure, so nothing can go wrong.
But it needs the original and the copy to be placed one over the other. When that is
impossible, we must carry the information as numbers instead — a few
measurements written down on paper.

Tip: A photograph will not do. A photo can be enlarged or reduced, so it fixes the
shape but not the size.

Page 1 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q2 Can we take some measurements that would allow us to exactly recreate this
figure? If yes, what measurements should we take?

The symbol on the signboard, page 1 — two straight arms meeting at a point.

Yes. Name the corner points of the symbol A, B and C as the book does. The symbol is just two
straight arms joined at B.
Take these three measurements:

the length of the arm AB,
the length of the arm BC,

Page 2 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

the angle ∠ABC between the two arms.

Why it happens: once BC is drawn, the arm BA can only be swung about B like the
hand of a clock. The angle ∠ABC tells you exactly where to stop swinging, and the
length AB tells you where to stop drawing. Nothing is left free, so the copy must be
identical.

Q3 Are the arm lengths AB and BC sufficient to exactly recreate this figure?

C
A

B

The same symbol with its corner points named A, B and C (page 1).

No. Two arm lengths are not enough.
Take AB = 4 cm and BC = 8 cm. Keeping both lengths fixed, the arms can still be opened out
wide or closed up narrow. Every opening gives a different symbol.

Page 3 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
.co g
Same arms AB = 4 cm and BC = 8 cm, three different angles at B

m a
l a se
ag
A
m A
.co ag
A

asem
a gl
B C B C B C

co m
e m.
The arm lengths are the same in all three, but the figures are clearly not copies of each other.

m l as
.co a g
a s emit happens: the two lengths pin down how far A and C are from B, but they say
gl
Why
a nothing about how far A is from C. The joint at B is still free to turn, like the hinge of
a pair of scissors.
m a s
m .co agl
l a se
a g
Q4 To get the exact replica, would it help to take any other measurement?

co m
m .
as e
om the angle ∠ABC between the two arms.

Yes —.cmeasure
a g l
a s em
a gl
m
AB = 4 cm

a se
. com g l
BC = 8 cm
m a
ase
agl
∠ABC = 80°

→ only one symbol is possible

. com
m a s
Why it happens: the angle locks the hinge at B. Once the openingem is fixed, the
. o
cposition a l whole figure is decided.
gthe
em
of A and the position of C are both decided, so

g l as Three measurements — two arms and the angle between them — are enough to
a c
.
make an exact replica.

s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 4 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q5 Can you draw the symbol if it is known that AB = 4 cm, BC = 8 cm, and ∠ABC = 80°?

Yes. Here is the construction.

1. Draw BC = 8 cm with a ruler.
2. Place the protractor at B and mark 80° from BC. Draw a ray BX along that mark.
3. From B, cut off BA = 4 cm on the ray BX.
4. The bent line A–B–C is the required symbol.

A

4 cm
80°

B 8 cm C

Draw the longer arm first, then set the angle at B, then cut the shorter arm.

Check it yourself: ask a friend to draw the same three measurements on another
sheet. Cut out both and place one over the other — they will fit exactly. The two
symbols are congruent.

Page 5 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q6 If it is known that both symbols have the same arm lengths, can it be concluded
that the two symbols are congruent?

No. Equal arm lengths are not enough.
We have already seen several symbols with AB = 4 cm and BC = 8 cm that are not copies of each
other. They differ in the angle at B.

Same arms + different angle → not congruent

Same arms + same angle → congruent

Why it happens: fixing the angle fixes the shape and the size together. So to be sure
that two such symbols are congruent, check three things — arm AB, arm BC, and the
angle ∠ABC.

Figure it Out — Pages 3–4
Section 1.1 Geometric Twins

Q1 Check if the two figures are congruent.

The two figures printed on page 3, drawn to the sizes given in the book.

They are not congruent.
Each figure is two arms joined at a point. Measure both arms and the angle in each figure:

Page 6 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

FIGURE SHORTER ARM LONGER ARM ANGLE BETWEEN THE ARMS

First about 1.7 cm about 3.4 cm about 90°

Second about 1.7 cm about 3.4 cm about 70°

Equal arms, different opening

about 90° about 70°

The arms match, but the second figure is more closed at the corner.

Why it happens: the arm lengths agree, so the two figures pass the first test. But
the angle at the corner is not the same, and the angle is exactly what fixes the
shape. Trace the first figure and try to place it on the second — the arms will never
lie along each other at the same time.

Tip: measure the angles in your own book with a protractor. Printing can shift a line
slightly, so trust your own measurement.

Q2 Circle the pairs that appear congruent.

Two of the four pairs are congruent.

Page 7 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

PAIR CONGRUENT? REASON

The two teardrops Yes Same size and shape; the second one is only turned (rotated) so
its tip points left

The two clouds No The first cloud is bigger and has more bumps along its edge

The two bursts No Same shape, but one is clearly smaller than the other
(stars)

The two leaf pairs Yes The second is the mirror image (flip) of the first; each leaf
matches its partner

Why it happens: while checking congruence a figure may be rotated or flipped
before it is placed on the other. So the turned teardrop and the flipped leaf pair still
count as congruent. Being a smaller copy, however, is never allowed — congruent
figures must have the same size, not just the same shape.

Check it yourself: trace each left-hand figure on tracing paper. Slide, turn and even
flip the paper over. If it settles exactly on its partner, the pair is congruent.

Q3 What measurements would you take to create a figure congruent to a given: (a)
Circle (b) Rectangle. Using this, state how would you check if two — (a) Circles are
congruent? (b) Rectangles are congruent?

(a) Circle — one measurement is enough: the radius (or the diameter, which is twice the
radius).

Radius r → open the compass to r → draw the circle

Two circles are congruent ⟺ their radii are equal

(b) Rectangle — two measurements: the length and the breadth.

Length l and breadth b → draw l, turn 90°, draw b, and close up
Two rectangles are congruent ⟺ same length and same breadth

Page 8 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: a circle has nothing to fix except how far its boundary is from the

m l a se
centre, so one number decides everything. A rectangle already has all four angles
o g needs to be
.c so only the two side lengths are left free. Nothing else
a
m
equal to 90°,

l a se
ag
measured.

o m
c ag
Tip: for the rectangles, compare the longer side with the longer side and the shorter
.
m is congruent to a 5 cm × 3 cm one — the
s e
with the shorter. A 3 cm × 5 cm rectangle
a
agl
second is just standing on its side.

. com
m a s em Use this to
How would we check if two figures like the one below are congruent?
co whether each of the following pairs are congruent.gl
Q4

. a
em
identify

l a s
agANSWER
m
The figure is made of two straight strokes that meet at a point. Three arms come out of that
a s
meeting point, so measure:
m .co agl
l a se
g
the three arm lengths from the meeting point, and
the angles between the arms. a

co
If the arms and the angles match up in the same order, the figures are congruent — a figure
m
m .
e
may be turned or flipped first.
m l as
m .co a g
l a se
ag
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
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

arm 1 arm 2

meeting point

arm 4
arm 3
Measure every arm from the meeting point, and the angles between neighbouring arms.

The two pairs given in the book: measure them and you will find that in both pairs the arms
and the angles agree exactly. So both pairs are congruent — in each pair one figure is simply
the other shifted to a new place on the page.

Why it happens: sliding a figure to a new position changes nothing about it. Length
and angle do not depend on where the figure sits on the paper, so a shifted copy is
always congruent to the original.

Check it yourself: trace one figure of a pair and slide the tracing paper across —
without turning it — on to its partner. It fits exactly.

In-text Questions — Pages 5–8

Page 10 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Section 1.2 Congruence of Triangles · Measuring the Sidelengths · Conventions to Express Congruence

Q1 Meera and Rabia have been asked to make a cardboard cutout identical to a
triangular frame they have in school. They see that the frame is too big to be traced
on a paper and replicated. What do you think they can do?

They should measure the frame and then build the cutout from those measurements.
With a measuring tape they can take the three sidelengths of the triangular frame. Those three
numbers can be carried back to the desk, and a triangle with exactly those sides can be drawn
on cardboard with a ruler and a compass.

Why it happens: tracing needs the original and the copy to touch each other.
Measurements do not — a number written in a notebook carries the shape from the
playground to the desk. This is the whole idea behind the congruence conditions in
this chapter.

Q2 Using a measuring tape, the girls measure the sides of the triangle to be 40 cm, 60
cm, and 80 cm. Then, Rabia takes out her protractor to measure the angles. She is
stopped by Meera. Meera: The angles of the triangle are not required! With the side
lengths we have measured, we can create a triangle congruent to this one. Do you
agree with Meera?

Yes, Meera is right. The three sidelengths are enough.
Try to draw a triangle with sides 40 cm, 60 cm and 80 cm in more than one way. Draw the 60 cm
side first, then swing an arc of 40 cm from one end and an arc of 80 cm from the other. The arcs
meet at only two points, one above the line and one below — and those two triangles are mirror
images of each other, so they are congruent.

Three sides fixed → only one triangle possible (up to a flip)

This is the SSS condition

Page 11 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: once the longest side is drawn, the third corner has to be at a fixed
distance from each of its ends. Two distances from two fixed points leave only two
possible positions, and those two give congruent triangles. Nothing is left to choose,
so the angles cannot come out differently — measuring them would only repeat
information we already have.

Q3 Instead of the lengths being 40 cm, 60 cm, and 80 cm, suppose the sidelengths had
been 4 cm, 6 cm, 8 cm (this triangle can fit on our page). Is this information
sufficient to replicate the triangle with the same size and shape? If yes, can you do
so?

Yes, it is sufficient. Here is the construction, drawn small enough to fit on a page.

1. Draw AB = 6 cm with a ruler.
2. With A as centre and radius 4 cm, draw an arc.
3. With B as centre and radius 8 cm, draw a second arc cutting the first at E.
4. Join AE and BE. ∆ABE has sides 6 cm, 4 cm and 8 cm.

E

8 cm
4 cm

A 6 cm B

Page 12 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

The two arcs decide the third corner E. Only 4 cm, 6 cm and 8 cm were used.

Check it yourself: compare your triangle with your neighbour's. Cut them out and
place one on the other — they fit, perhaps after a flip.

Page 13 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Rabia: If I were to construct this triangle, I would first draw a line segment having
se
Q4

o m l a
one of the given lengths, say 6 cm, and then draw circles from each of its end points
gpoints, forming two
m .c 4 cm and 8 cm. But the circles would intersect at two
with radii a
l a se
triangles: ∆ABE and ∆ABF. Rabia: Do these two triangles have the same shape and
a g size? If not, then we will not be sure which of these would actually be congruent to
the original triangle we are trying to replicate.

co m
em . ag
g l as
a

co m
e m.
m l as
m .co E a g
ase
agl
m a s
A
m .co agl
ase
B

agl AB = 6 cm

co m
m .
m as e
.co a g l
sem
F
a
agl
se m
com g l a
m . a
ase
agl
The construction on page 5 — a 6 cm segment AB, a circle of radius 4 cm about A and a

m
circle of radius 8 cm about B, meeting at E and F.

. co
e m
c o m g l as
. a
sem
a
agl Yes, they do have the same shape and size. ∆ABE and ∆ABF are congruent, so Rabia's worry
c
m .
e
does not arise.

m a s
. co agl
Both triangles are built from the same three lengths:

e m
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

AE = AF = 4 cm (both on the circle centred at A)

BE = BF = 8 cm (both on the circle centred at B)

AB is the same side in both

→ same three sidelengths

Why it happens: the whole construction above the line AB is repeated below it in
exactly the same way. So AB acts as a line of symmetry: fold the paper along AB and
E falls on F. Whichever of the two triangles Rabia draws, it will be congruent to the
frame.

Q5 Examine whether ∆ABE and ∆ABF are congruent. For this, you could use one or more
of the following methods — tracing and comparing, taking a cutout and
superimposing, or observing that AB acts as a line of symmetry due to the
‘sameness’ of the act of construction above and below this line.

All three methods give the same answer: ∆ABE ≅ ∆ABF.
Tracing: trace ∆ABE, flip the tracing paper over about the line AB, and it lands exactly on
∆ABF.
Cutout: cut both triangles out. Turn one over and place it on the other — the three sides
match.
Symmetry: the arcs below AB were drawn with the same centres and the same radii as the
arcs above it. So the lower half is the mirror image of the upper half in the line AB.

Why it happens: a reflection does not change any length or any angle. It only turns
the figure over. Since congruence allows a figure to be flipped before superimposing,
a figure and its mirror image are always congruent.

Did you know? This is exactly why the SSS condition works. Any two triangles with
the same three sidelengths can be constructed by the same arcs, so they must be
congruent.

Page 15 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q6 The two triangles given below are congruent. How can these two triangles be
superimposed? Which vertices of ∆XYZ and ∆ABC should we overlap? This has to be
done so that the equal sides overlap. Figure out how.

Place A over X, B over Y and C over Z.

AB fits over XY

BC fits over YZ

AC fits over XZ

→ the triangles fit exactly, one over the other

Since the triangles now fit exactly, their angles also match:

∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z

Why it happens: superimposing is only allowed to put equal sides on equal sides. If
you tried to put A over Y, the side AB would land on YX; but AB and YX need not be
equal, so the two triangles would stick out beyond each other. Matching the equal
sides first automatically matches the vertices and the angles.

Tip: This matching is written as ∆ABC ≅ ∆XYZ. The order of the letters is the message
— first with first, second with second, third with third.

Q7 Are there other ways of overlapping the vertices so that the triangles fit exactly
over each other?

For these triangles, no — there is only one way.
The three sides of the triangle have three different lengths, so each side has exactly one partner
of the same length in the other triangle. That fixes the matching completely.

Page 16 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: a second way of overlapping would need two sides of the same
triangle to be equal. That happens only in special triangles:

Isosceles triangle — two equal sides, so two ways of overlapping.
Equilateral triangle — all three sides equal, so six ways of overlapping.

Q8 Can you identify a pair of congruent triangles below? Why are they congruent?

A B

D C

Fig. 1.1 (page 7) — rectangle ABCD with the diagonal BD drawn.

The diagonal BD cuts the rectangle into ∆ABD and ∆CDB, and these two are congruent.

AB = CD (opposite sides of a rectangle)

AD = CB (opposite sides of a rectangle)

BD = DB (the same side, common to both)

→ all three sides match → SSS condition

So ∆ABD and ∆CDB are congruent.

Page 17 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: the third side of each triangle is the diagonal itself. A side is always
equal to itself, so it costs nothing — that is why a common side is so useful in
congruence proofs. Two given equalities plus one free one complete the SSS
condition.

Q9 Can they be the following? ∆ABD ↔ ∆CDB with A ↔ C, B ↔ B, D ↔ D. Verify this by
superimposing paper cutouts of the triangles obtained from the rectangle ABCD
(Fig. 1.1).

A B

D C

Fig. 1.1 (page 7) — rectangle ABCD with the diagonal BD drawn.

No, this matching does not work. Cut the two triangles out of the rectangle and try it — they
will not settle on each other.

A over C, B over B, D over D

puts the side AB over the side CB

But AB and CB are the length and the breadth — they need not be equal

Page 18 of 55

ase
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: a correspondence is only correct if every pair of matched sides is a

m l a se
pair of equal sides. Here AB is matched with CB, and in a rectangle those two are
o
.c other, not opposite, so they are usually different. The
a gcutouts overlap
m
next to each
a e and congruence is not established.
spartly,
ag l
only

co m
ag
Tip: before writing a congruence, list the matched sides. If any pair on your list is not
.
s em
equal, the order of the letters is wrong.
a
agl

co m
m.
Q10 Identify the correct correspondence of vertices and express the congruence

m
between the two triangles.
as e
.co a g l
se m
g l a

a The correct matching is A ↔ C, B ↔ D, D ↔ B.
m a s
m .co agl
se
∆ABD ≅ ∆CDB

g l a
a
AB ↔ CD (equal, opposite sides)
co m
m .
e
BD ↔ DB (the common diagonal)
m l as
.co
DA ↔ BC (equal, opposite sides)
m a g
l a se
ag Every matched pair is a pair of equal sides, so this correspondence is correct.

se m
com move — lift ∆ABD, turn it half a turn about g l a
.
Why it happens: think of it as a physical
a
a s
the centre of the rectangle, andemset it down. A lands on C, B lands on D and D lands
l letters ∆ABD ≅ ∆CDB record.
agthe
on B. That is exactly what

co m
Tip: ∆ADB ≅ ∆CBD says the same thing, because the pairs A↔C, D↔B, B↔D are
m .
m as e
.co
unchanged. Only the order of listing has changed.
a g l
se m
g l a
a c
m .
Figure it Out — Pages 8–9
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Section 1.2 Congruence of Triangles — the SSS condition

Q1 Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways of expressing
this congruence.

The statement ∆HEN ≅ ∆BIG fixes three pairs of corresponding vertices:

H ↔ B, E ↔ I, N ↔ G

Any way of writing the same three pairs is correct. The three pairs can be listed in 6 orders, so
besides the given one there are 5 other correct ways:

WAY CONGRUENCE PAIRS IT RECORDS

Given ∆HEN ≅ ∆BIG H–B, E–I, N–G

1 ∆HNE ≅ ∆BGI H–B, N–G, E–I

2 ∆EHN ≅ ∆IBG E–I, H–B, N–G

3 ∆ENH ≅ ∆IGB E–I, N–G, H–B

4 ∆NHE ≅ ∆GBI N–G, H–B, E–I

5 ∆NEH ≅ ∆GIB N–G, E–I, H–B

Why it happens: the congruence sign does not care which vertex is written first. It
only demands that the first letter on the left goes with the first letter on the right,
and so on. As long as the three pairs stay together, the statement means the same
thing.

Tip: each of these can also be turned around, for example ∆BIG ≅ ∆HEN. Writing
∆HEN ≅ ∆BGI would be wrong — it pairs E with G and N with I.

Page 20 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q2 Determine whether the triangles are congruent. If yes, express the congruence.
(∆RED with RE = 3.5 cm, ED = 5 cm, RD = 6 cm; and ∆JMA with JA = 3.5 cm, AM = 5 cm,
JM = 6 cm)

Yes, they are congruent, by the SSS condition.
Match the equal sides:

LENGTH SIDE OF THE FIRST TRIANGLE SIDE OF THE SECOND TRIANGLE

3.5 cm RE JA

5 cm ED AM

6 cm RD JM

Reading the matching down the columns gives R ↔ J, E ↔ A, D ↔ M. So

∆RED ≅ ∆JAM (SSS condition)

Why it happens: the three sidelengths of the two triangles are the same set — 3.5
cm, 5 cm, 6 cm — so SSS applies and the triangles are congruent. To write the
congruence you must then find which vertex sits between which two sides. R lies
between the 3.5 cm and 6 cm sides; in the second triangle that vertex is J.

Tip: writing ∆RED ≅ ∆JMA would be wrong. It would pair ED (5 cm) with MA —
correct — but also RE (3.5 cm) with JM (6 cm), which is false.

Page 21 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q3 In the figure below, AB = AD, CB = CD. Can you identify any pair of congruent
triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD
into two equal parts? Give reasons.

A

D

B

C

The figure below Q3 on page 9 — AB = AD and CB = CD, with AC drawn dashed.

Join AC. It splits the figure into ∆ABC and ∆ADC, and these two are congruent.

Page 22 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

AB = AD (given)

CB = CD (given)

AC = AC (common side)

→ SSS condition → ∆ABC ≅ ∆ADC

Does AC divide the two angles equally? Yes.
Because the triangles are congruent, their corresponding angles are equal:

∠BAC = ∠DAC → AC cuts ∠BAD into two equal parts

∠BCA = ∠DCA → AC cuts ∠BCD into two equal parts

Why it happens: ∠BAC sits in ∆ABC at the vertex A, between the sides AB and AC. Its
partner in ∆ADC sits at A between AD and AC. Since the triangles fit exactly over each
other, these two angles must be equal — they are corresponding parts of congruent
triangles. The same argument at C gives the second pair.

Did you know? A four-sided figure with two pairs of equal neighbouring sides like
this is a kite. The diagonal AC is its line of symmetry — fold along AC and B lands on
D.

Page 23 of 55

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Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
In the figure below, are ∆DFE and ∆GED congruent to each other? It is given that DF
e
Q4

m l as
.co
= DG and FE = GE.

a g
se m
g l a
a D
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
em
.co E agl
l a s
ag

co m
m .
m F as e G
.co a g l
se m
g l a
a The figure below Q4 on page 9 — DF = DG and FE = GE.

se m
com g l a
m . a
ase
agl

The two triangles in the figure are congruent — but ∆DFE ≅ ∆GED is not the correct way to
m
write it.

. co
m
First, the congruence itself. Look at ∆DFE and ∆DGE:

m as e
.co a g l
a s em DF = DG (given)

agl
.c
FE = GE (given)

s e m
m a
agl
DE = DE (common side)

→ SSS condition → ∆DFE ≅ m . c o
a s e ∆DGE

agl
Now test the order given in the question. Writing ∆DFE ≅ ∆GED pairs the vertices like this:

co m
m .
m ase
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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

∆DFE ∆GED SIDES IT MATCHES ARE THEY EQUAL?

D G DF with GE Not given

F E FE with ED Not given

E D ED with DG Not given

None of these three pairs is known to be equal, so that correspondence is wrong.

Why it happens: the two triangles really do fit over each other, but only when D
stays on D and E stays on E, with F landing on G. Fold the figure along DE and the left
half falls on the right half. So the honest statement is ∆DFE ≅ ∆DGE.

Tip: ∆FDE ≅ ∆GDE and ∆EFD ≅ ∆EGD say the same thing. Always read the letters in
pairs before you accept a congruence.

In-text Questions — Pages 9–13
Measuring the Angles · Two Sides and the Included Angle (SAS) · A Non-included Angle (SSA) · Two Angles
and the Included Side (ASA)

Q1 Instead of measuring the three sidelengths of the triangular frame, if Meera and
Rabia measure the three angles, can they recreate the triangle exactly?

No. The three angles fix the shape but not the size.
Two triangles with the same three angles look alike, but one may be a big version of the other.
They can be placed one over the other only if they also happen to be the same size, and the
angles do not tell us that.

Why it happens: the angles say how steeply the sides lean, not how long they are.
You can keep the leaning the same and stretch every side to twice its length — the
angles do not change at all.

Page 25 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q2 Suppose the angles are 30°, 70°, and 80°. Can we create an exact copy of the frame
with this?

No. Many different triangles have these three angles.

30° + 70° + 80° = 180° ✓ — so such triangles do exist

But nothing in these numbers says how long a side must be

Draw a base of 4 cm and put 30° at one end and 70° at the other; the third angle comes out as
80°. Now do it again with a base of 8 cm. Both triangles have angles 30°, 70°, 80°, but the
second is twice as big as the first.

Why it happens: the two rays from the ends of the base meet at one point whatever
the base length is. Lengthening the base simply pushes that meeting point further
away, giving a larger triangle with the same three angles. So two triangles with the
same set of angles need not be congruent — AAA is not a congruence condition.

Q3 ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 5 cm, and ∠A =
∠X = 30°. Are they congruent?

Yes, they are congruent.
Here the equal angle lies between the two equal sides — ∠A is the angle between AB and AC,
and ∠X is the angle between XY and XZ.

AB = XY = 6 cm
∠A = ∠X = 30° (included angle)

AC = XZ = 5 cm

→ SAS condition → ∆ABC ≅ ∆XYZ

Page 26 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: start at the vertex A. Draw the 30° angle, mark 6 cm along one arm
to get B and 5 cm along the other to get C. Both B and C are now decided, so the
third side BC is decided too. Nobody following these instructions can end up with a
different triangle.

Q4 Construct a triangle having the above measurements. Compare it with the triangles
constructed by your classmates. Are the triangles all congruent? Explain why all
such triangles with these measurements are congruent.

Yes — every triangle in the class will be congruent (some may be flipped over).
The construction leaves no free choice:

1. Draw AB = 6 cm.
2. At A, draw a ray making 30° with AB. There is only one such ray on a given side of AB.
3. On that ray mark AC = 5 cm. There is only one such point C.
4. Join BC — there is only one segment joining two fixed points.

One choice for the angle + one choice for C → one triangle

Why it happens: each step is forced by the step before it. Because the given angle is
squeezed between the two given sides, both ends of the third side are pinned down
before that side is drawn. This is exactly what the SAS condition claims.

Check it yourself: cut out two classmates' triangles and place one on the other. If
one was drawn with the 30° opening to the left, turn it over first.

Q5 ΔABC and ΔXYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 4 cm, and ∠B =
∠Y = 30°. Are they congruent? Can there exist non-congruent triangles having these
measurements? Construct and find out.

Not necessarily. Two different triangles can be drawn from these measurements.
Here the equal angle is not between the two equal sides — ∠B lies at the end of AB, while AC
starts at A. This is the SSA case. Follow the construction:

1. Draw the base PQ = 6 cm.

Page 27 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

2. From P draw a line l making 30° with PQ.
3. From Q draw a long arc of radius 4 cm. It cuts the line l at two points, R and S.

Both ∆PQR and ∆PQS have a 6 cm side, a 4 cm side and a 30° angle at P — yet they are clearly
different triangles.

l

R

S

30°

P 6 cm Q

One arc, two crossing points — two triangles that are not congruent.

Why it happens: an arc can cut a straight line twice. Because the given angle is not
squeezed between the two given sides, the third vertex is not pinned down — it may
sit at either crossing. So the SSA condition does not guarantee congruence.

Page 28 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
How do we find the required triangle from this figure?
e
Q6

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

co m
e m . l
ag
g l as
a

co m
R
em.
m l as
m .co a g
l a se 30°
a g Q
P
6 cm
m a s
m .co agl
l a se
g
Step 3 of the construction on page 11 — base PQ = 6 cm, the line l at 30° to PQ, and an
a
arc of radius 4 cm drawn from Q cutting l at R and S.

co m
m .
m as e
.co a g l
a s emS — so the figure gives two triangles, ∆PQR and ∆PQS.
A point where the arc meets the line l gives the third vertex. But the arc meets l at two points, R

agl and

se m
com g l a
.
Both satisfy: PQ = 6 cm, QR = QS = 4 cm, ∠P = 30°
m a
ase
agl
But PR is longer than PS

→ the two triangles are not congruent

. com
m a s em point to pick, so
Why it happens: the measurements do not tell us which crossing
gl SSS, SAS or ASA the
o data is incomplete. This is what makes SSA unsafe:awith
.cthe
se m construction never offers such a choice.
g l a
a c
m .
m a s e
co agl
Tip: a rough diagram drawn before the actual construction shows at once whether

m .
e
the given angle lies between the two given sides. That decides whether you are in

g l as
the safe SAS case or the risky SSA case.
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q7 ∆ABC and ∆XYZ are two triangles with, BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z =
30°. Are they congruent? Can there exist non-congruent triangles having these
measurements? Construct and find out.

Yes, they are congruent, and no other triangle is possible with these measurements.
The equal side lies between the two equal angles, so this is the ASA case.

1. Draw BC = 5 cm.
2. At B draw a ray making 50° with BC.
3. At C draw a ray making 30° with CB, on the same side.
4. The two rays meet at exactly one point — that is A.

∠B = ∠Y = 50°

BC = YZ = 5 cm (included side)

∠C = ∠Z = 30°

→ ASA condition → ∆ABC ≅ ∆XYZ

Why it happens: two rays that are not parallel meet at exactly one point. Since both
ends of the base and both directions are fixed, the meeting point cannot move.
Everyone in the class gets the same triangle.

Tip: the third angle is decided too — 180° − 50° − 30° = 100°. That is a useful check
on your drawing.

Page 30 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q8 In the figure, Point O is the midpoint of AD and BC. What can one say about the
lengths AB and CD?

A
C

O

B
D

The figure on page 12 — AD and BC cross at O.

AB = CD. The two lengths are equal.
Look at ∆AOB and ∆DOC:

AO = OD (O is the midpoint of AD)

∠AOB = ∠DOC (vertically opposite angles)

BO = OC (O is the midpoint of BC)

→ SAS condition → ∆AOB ≅ ∆DOC

The correspondence is A ↔ D, O ↔ O, B ↔ C. So AB and DC are corresponding sides, and

AB = DC

Page 31 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: the two triangles are joined at O like a bow tie. Turning ∆AOB half a
turn about O sends A to D and B to C, because O is the midpoint of both segments.
The side AB is carried to the side DC, so the two must have the same length.

Q9 We have, AO = OD (as O is the midpoint of AD) and BO = OC (as O is the midpoint of
BC). Are there any other equal sides or angles?

Yes — ∠AOB = ∠DOC, because they are vertically opposite angles formed where the segments
AD and BC cross at O.

Two known sides: AO = OD and BO = OC

One known angle: ∠AOB = ∠DOC

The angle lies between the two sides → SAS

Why it happens: this is the piece of information that completes the condition.
Without it we would only have two pairs of equal sides, which is never enough. The
crossing point O supplies the third fact free of cost, just as a common side does.

Tip: whenever two straight lines cross in a figure, look at once for vertically opposite
angles. They are equal without being marked.

Figure it Out — Pages 13–14
Section 1.2 Congruence of Triangles — SAS and ASA

Q1 Identify whether the triangles below are congruent. What conditions did you use to
establish their congruence? Express the congruence. (∆ABC with AB = 7 cm, BC = 5
cm, ∠B = 47°; ∆XYZ with XZ = 7 cm, ZY = 5 cm, ∠Z = 47°)

Yes, they are congruent, by the SAS condition.
In each triangle the marked angle lies between the two marked sides:

Page 32 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

FIRST TRIANGLE SECOND TRIANGLE EQUAL?

BA = 7 cm ZX = 7 cm Yes

∠B = 47° (between BA and BC) ∠Z = 47° (between ZX and ZY) Yes

BC = 5 cm ZY = 5 cm Yes

So B corresponds to Z, A corresponds to X and C corresponds to Y:

∆ABC ≅ ∆XZY (SAS condition)

Why it happens: the vertex carrying the 47° angle must be matched with the vertex
carrying the 47° angle, so B goes with Z. After that, the 7 cm arm goes with the 7 cm
arm (A with X) and the 5 cm arm with the 5 cm arm (C with Y).

Tip: ∆ABC ≅ ∆XYZ would be wrong here. It would match ∠B with ∠Y, but the 47°
angle in the second triangle is at Z, not at Y.

Page 33 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Given that CD and AB are parallel, and AB = CD, what are the other equal parts in
se
Q2

o m l a
this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are
g
m .c resulting triangles congruent? If so, express the congruence.)
the two a
l a se
a g

co m
D e m . ag
g l as
a
C om
. c
s e m
m a
e m . co agl
g l as
a
m a s
m .co O agl
l a se
a g

co m
m .
m as e
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se m
g l a
a
A se m
com g l a
m . a
ase
agl B
co m
m .
o m l a separallel sides, and DB and
.c ag
The figure above Q3 on page 14 — DC and AB are the marked

a s em AC cross at O.

agl
.c
s e m
m a
. co agl

se m
l a
In the figure the segments DB and CA cross at O, making ∆ODC and ∆OBA.

ag DC and AB are parallel, so:
Step 1 — the equal angles.

co m
m .
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.co


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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

∠CDB = ∠ABD (alternate angles, transversal DB)

∠DCA = ∠BAC (alternate angles, transversal CA)

Step 2 — the congruence. Together with the given equal sides:

∠ODC = ∠OBA

DC = BA (given, the included side)

∠DCO = ∠BAO
→ ASA condition → ∆ODC ≅ ∆OBA
Step 3 — the other equal parts. From the congruence:

OD = OB
OC = OA
∠DOC = ∠BOA (also vertically opposite angles)

Why it happens: parallel lines hand us the two angles free of charge, and the given
equal sides sit exactly between them. Once the triangles are congruent, the
remaining sides must match too — which says that O is the midpoint of both DB and
CA.

Q3 Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two
triangles congruent?

Yes, the triangles are congruent, and that is what makes the two angles equal.
The segment BC is shared by ∆ABC (above it) and ∆DBC (below it).

∠ABC = ∠DBC (given)

BC = BC (common side, included between the two angles)
∠ACB = ∠DCB (given)

→ ASA condition → ∆ABC ≅ ∆DBC
The correspondence is A ↔ D, B ↔ B, C ↔ C. Hence the third pair of angles must also match:

Page 35 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

∠BAC = ∠BDC

Why it happens: the third angle of a triangle is decided by the other two, since all
three add to 180°. Both triangles have the same two angles at B and at C, so their
third angles must be equal as well. Congruence says the same thing in one step:
corresponding angles of congruent triangles are equal.

Tip: because the triangles are congruent we also get AB = DB and AC = DC. So ABDC
is a kite, with BC as its line of symmetry.

Q4 Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB =
∠DBC.

A D

B C

The figure below Q4 on page 14 — the single arcs mark ∠ABD and ∠DCA, the double arcs
mark ∠DBC and ∠ACB.

Step 1 — add the given angles. At B the angle ∠ABC is made of ∠ABD and ∠DBC. At C the
angle ∠DCB is made of ∠DCA and ∠ACB.

Page 36 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

∠ABC = ∠ABD + ∠DBC

∠DCB = ∠DCA + ∠ACB

Given ∠ABD = ∠DCA and ∠DBC = ∠ACB

→ ∠ABC = ∠DCB

Step 2 — find the congruent triangles. Look at ∆ABC and ∆DCB, which share the side BC:

∠ABC = ∠DCB (just proved)
BC = CB (common side)

∠ACB = ∠DBC (given)

→ ASA condition → ∆ABC ≅ ∆DCB
Step 3 — list the equal parts.

AB = DC and AC = DB (corresponding sides)
∠BAC = ∠CDB (corresponding angles)
∠ABC = ∠DCB, and of course BC = CB

Step 4 — the crossing point. Let O be the point where AC and DB cross. In ∆OBC, ∠OBC =
∠OCB (these are the given equal angles ∠DBC and ∠ACB), so ∆OBC is isosceles and

OB = OC, and therefore OA = OD

(since AC = DB, and OA = AC − OC, OD = DB − OB)

It also follows that ∆ABD ≅ ∆DCA by SSS: AB = DC, BD = CA and AD = DA.

Why it happens: the whole figure is symmetric about the vertical line through the
middle of BC. The two given equalities are exactly what forces that symmetry, and
every pair listed above is a pair of parts swapped by it.

In-text Questions — Pages 14–17

Page 37 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Measuring Two Angles and a Non-Included Side (AAS) · Measuring Two Sides in a Right Triangle (RHS)

Q1 The following triangles ∆ABC and ∆XYZ are such that ∠A = ∠X = 35°, ∠C = ∠Z = 75°,
and BC = YZ = 4 cm. Are the triangles congruent? Give a reason.

Yes, they are congruent.
The given side BC is not between the two given angles ∠A and ∠C, so ASA cannot be used
straight away. But the third angle can be found first.

∠B + 35° + 75° = 180°

∠B + 110° = 180°

∠B = 70°, and in the same way ∠Y = 70°

Now use ∠B, BC and ∠C, which do fit the ASA pattern:

∠B = ∠Y = 70°

BC = YZ = 4 cm (included side)

∠C = ∠Z = 75°

→ ASA condition → ∆ABC ≅ ∆XYZ

Why it happens: in a triangle the three angles always add to 180°, so knowing two
of them gives the third for free. That is why two angles and any one side are enough.
This shortcut is called the AAS condition.

Q2 What are the measures of ∠B and ∠Y?

Both are 70°.

Page 38 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
e m.
In ∆ABC: ∠A + ∠B + ∠C = 180°
m l as
.co
35° + ∠B + 75° = 180°
a g
a s em
a∠glB = 70°
∠B + 110° = 180°

co m
e m . ag
In ∆XYZ: 35° + ∠Y + 75° = 180°
g l as
∠Y = 70° a

co m
em.
as
So ∠B = ∠Y.
m l
.co a g
a s
Whyemit happens: the two triangles were given the same pair of angles, 35° and 75°.
a gl Since all three angles must total 180°, whatever is left over is the same in both —
70°.
m a s
m .co agl
l a se
Q3 ag
Does this help in showing that ΔABC and ΔXYZ are congruent?

co m
m .
e

m l as
.co g
Yes. Finding the third angle turns the problem into an ordinary ASA problem.

em a
a s
a gl ∠B = ∠Y = 70°
se m
com a
BC = YZ = 4 cm
. a g l
m
ase
∠C = ∠Z = 75°

a gl
Now the equal side BC lies between the two equal angles ∠B and ∠C, so ASA applies and

co m
m .
≅ ∆XYZ
as e
com
∆ABC
. a g l
m
ase
agl Why it happens: the side did not have to be the included one from the start. Once
c
m .
e
the third angle is worked out, every side of the triangle is included between some
m a s
. co agl
pair of known angles. So AAS always guarantees congruence.

se m
l a
g new rule. It is ASA with one extra line of arithmetic.
Tip: AAS is not reallyaa

co m
m .
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.co


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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Q4 ΔABC and ΔXYZ are right-angled triangles such that BC = YZ = 4 cm, ∠B = ∠Y = 90°
and AC = XZ = 5 cm. Are they congruent? Can there exist non-congruent triangles
having these measurements? Construct and find out.

Yes, they are congruent, and no different triangle can be drawn with these measurements.
Follow the construction with a rough diagram in front of you:

1. Draw the base QR = 4 cm.
2. At Q draw a line l perpendicular to QR.
3. From R cut an arc of radius 5 cm on the line l. Call the crossing point P.
4. Join PR. ∆PQR is the required triangle.

∠Q = 90°

PR = 5 cm (the hypotenuse, opposite the right angle)

QR = 4 cm (one other side)

→ RHS condition → ∆ABC ≅ ∆XYZ

Why it happens: this looks like the risky SSA case — two sides and an angle that is
not between them. The difference is that here the known angle is a right angle and
the known side opposite it is the hypotenuse, the longest side. The arc from R can
meet the perpendicular line above Q or below Q, and those two triangles are mirror
images of each other, so they are congruent. There is no third possibility.

Did you know? The third side comes out as 3 cm, since 3, 4, 5 is the smallest whole-
number right triangle.

Q5 Consider the downward extension of line l below QR. Would the arc from R meet
this line downwards as well (as in the case of triangle construction when the
sidelengths are given)? If so, would this lead to a triangle whose size and shape are
different from ΔPQR, and yet has the given measurements?

Yes, the arc does meet the line below QR — but no, the new triangle is not different.
Call the lower crossing point P′. Then ∆P′QR also has ∠Q = 90°, QR = 4 cm and P′R = 5 cm, exactly
the required measurements.

Page 40 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

P and P′ lie on the same line l, one above Q and one below

QP = QP′ (both are cut by the same 5 cm arc from R)

→ ∆PQR and ∆P′QR are mirror images in the line QR

Why it happens: fold the paper along QR. The perpendicular line l falls back on itself
and the arc from R falls back on itself, so P lands exactly on P′. A figure and its mirror
image are always congruent, so the second triangle is not a new triangle at all.

Q6 It can be seen that the other triangle we get below is also congruent to ΔPQR. Why?
Therefore, all triangles having these measurements will be congruent to each
other.

Because the two triangles share a side and match in two more parts:

QR = QR (common side)

∠PQR = ∠P′QR = 90° (both are right angles on the same perpendicular)

QP = QP′ (the same arc from R cuts off equal lengths on either side of Q)
→ SAS condition → ∆PQR ≅ ∆P′QR

So both possible triangles are congruent to each other, and therefore every triangle with these
measurements is congruent to ∆PQR.

Why it happens: QP = QP′ can also be seen from the right angle itself. In each
triangle the third side is decided by the other two, so equal hypotenuses and equal
bases force equal heights. This is what makes RHS safe while ordinary SSA is not.

In-text Questions — Pages 17–20

Page 41 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Section 1.3 Angles of Isosceles and Equilateral Triangles · Congruent Triangles in Real Life

Q1 ΔABC is isosceles with AB = AC, and ∠A = 80°. What can we say about ∠B and ∠C?

∠B = ∠C. The angles opposite the two equal sides are equal.
Drop the altitude from A to BC, meeting BC at D. Then compare ∆ADB and ∆ADC:

AB = AC (given)
∠ADB = ∠ADC = 90° (AD is the altitude)

AD = AD (common side)

→ RHS condition → ∆ADB ≅ ∆ADC

So the corresponding angles ∠B and ∠C are equal.

Why it happens: AB and AC are the hypotenuses of the two small right triangles,
and they are equal. AD is common. That is the right angle, the hypotenuse and one
more side — exactly RHS. This is a general result: in a triangle, angles opposite to
equal sides are equal.

Q2 Can you use this fact to find ∠B and ∠C?

Yes. Call each of them x.

∠A + ∠B + ∠C = 180°

80° + x + x = 180°

2x = 100°

x = 50°

∠B = ∠C = 50°

Page 42 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: once we know the two base angles are equal, the angle sum turns
into a single equation with one unknown. Whatever is left after taking 80° out of
180° has to be shared equally between ∠B and ∠C.

Check it yourself: 80° + 50° + 50° = 180° ✓

Q3 Equilateral triangles are those in which all the sides have equal lengths. What can
we say about their angles?

All three angles are equal.
Use the fact just discovered, twice over, in ∆ABC:

AB = AC → the angles opposite them are equal → ∠C = ∠B

AB = BC → the angles opposite them are equal → ∠C = ∠A

→ ∠A = ∠B = ∠C

Why it happens: "angles opposite equal sides are equal" can be applied to any pair
of equal sides. An equilateral triangle has three such pairs, so all three angles get
tied together. Just like the sides, the angles are all alike.

Q4 What could be their measures?

Each angle is 60°.

∠A + ∠B + ∠C = 180°

All three are equal, so

3 × (angle of an equilateral triangle) = 180°

angle = 180° ÷ 3 = 60°

Page 43 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the angle sum of a triangle is fixed at 180°, and here it has to be

m l a se
divided into three equal shares. Notice that we never measured anything —
o g in the world
.c alone was enough to prove that every equilateral triangle
a
m
congruence
se 60° angles.
hasathree
l
ag

co m
e m . ag
as
Q5 Verify this by construction.

a g l

co m
m.
Construct an equilateral triangle and measure its angles.

o m
1. Draw AB = 5 cm.
l a se
2. With A.cas centre and radius 5 cm, draw an arc above AB. a g
m
se B as centre and the same radius 5 cm, draw another arc cutting the first at C.
3.aWith
l
g
a 4. Join AC and BC.
m a s
agl
5. Measure ∠A, ∠B and ∠C with a protractor.

m .co
l a se
Each angle measures 60°
a g
Check: 60° + 60° + 60° = 180° ✓

com
m .
m as e
.co a g l
Check it yourself: repeat with a side of 3 cm and again with 8 cm. The triangles are

s e mdifferent sizes, but every angle still measures 60°. The size of an equilateral
of

agla triangle does not change its angles.
se m
com g l a
m . a
ase
agl
Q6 Congruent Triangles in Real Life: Congruent triangles can be seen in various
constructions and designs from ancient to modern times. Describe the congruent
triangles you see in each picture. (Louvre Museum, Pyramid of Giza, dome design,

co m
.
rangoli design, Rabindra Setu or Howrah Bridge)

em
m l as
.co a g
emCongruent triangles appear in every one of the pictures.

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

co m
m .
m ase
.co


a g l Page 44 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

PICTURE CONGRUENT TRIANGLES WHY THEY ARE CONGRUENT
YOU CAN SEE

Louvre Museum, The glass pyramid is covered Each pane is cut to the same size, so all panes
Paris with rows of small triangular in a row are copies of one another
panes

Pyramid of Giza The four slanting faces of the The base is a square and the apex is directly
pyramid above its centre, so the four faces have the
same sidelengths

Dome design Triangles repeating around the The same triangular piece is repeated after
curved surface turning it through a fixed angle

Rangoli design The triangular petals around the The pattern is made by turning one triangle
centre again and again about the centre point

Rabindra Setu The steel triangles in the criss- The bars are cut to fixed lengths, so each
(Howrah Bridge) cross girders triangle in the truss repeats along the bridge

Why it happens: repeating one congruent piece is cheap and strong. A builder
needs only one measurement set and one mould, and a triangle cannot be pushed
out of shape the way a rectangle can — its three sides fix it completely, by SSS. That
is why bridges and domes are built out of triangles.

Try This: look for congruent triangles around you — a folding chair, a kite, an ironing
board, or the pattern on a dupatta.

Figure it Out — Pages 20–21
Section 1.3 Angles of Isosceles and Equilateral Triangles

Q1 ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles.

Read the letters in the same order on both sides of the ≅ sign.

Page 45 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

CORRESPONDING VERTICES CORRESPONDING SIDES CORRESPONDING ANGLES

A and F AI and FL ∠A = ∠F

I and L IR and LY ∠I = ∠L
R and Y RA and YF ∠R = ∠Y

Why it happens: a side is named by the two vertices at its ends. Since A goes with F
and I goes with L, the side AI must go with the side FL. An angle is named by the
vertex it sits at, so ∠A goes with ∠F.

Tip: the same congruence could be written ∆IRA ≅ ∆LYF or ∆RAI ≅ ∆YFL. The pairs A–
F, I–L, R–Y never change.

Q2 Each of the following cases contains certain measurements taken from two
triangles. Identify the pairs in which the triangles are congruent to each other, with
reason. Express the congruence whenever they are congruent. (a) AB = DE, BC = EF,
CA = DF (b) AB = EF, ∠A = ∠E, AC = ED (c) AB = DF, ∠B = ∠D = 90°, AC = FE (d) ∠A = ∠D,
∠B = ∠E, AC = DF (e) AB = DF, ∠B = ∠F, AC = DE

Four of the five cases give congruent triangles.

CASE CONGRUENT? CONDITION CONGRUENCE

(a) Yes SSS — all three sides match ∆ABC ≅ ∆DEF

(b) Yes SAS — ∠A lies between AB and AC, ∠E lies between EF ∆ABC ≅ ∆EFD
and ED

(c) Yes RHS — a right angle, the hypotenuse and one more ∆ABC ≅ ∆FDE
side

(d) Yes AAS — two angles and a non-included side ∆ABC ≅ ∆DEF

(e) No SSA — the angle is not between the two sides —

(a) AB = DE, BC = EF, CA = FD. Three pairs of sides, so SSS gives ∆ABC ≅ ∆DEF.
(b) ∠A is the angle between the sides AB and AC. ∠E is the angle between EF and ED. Since AB =
EF and AC = ED, the equal angle is the included one in both — SAS gives ∆ABC ≅ ∆EFD.

Page 46 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

(c) ∠B = ∠D = 90°. In the first triangle the side opposite the right angle is AC; in the second it is
FE. These are the hypotenuses, and AC = FE. The extra pair AB = DF is one more side. RHS gives
∆ABC ≅ ∆FDE.
(d) ∠A = ∠D and ∠B = ∠E, with AC = DF. The side AC is not between ∠A and ∠B, so this is AAS.
Find the third angle: ∠C = 180° − ∠A − ∠B = 180° − ∠D − ∠E = ∠F. Now ∠A = ∠D, AC = DF, ∠C =
∠F is ASA, so ∆ABC ≅ ∆DEF.
(e) AB = DF, AC = DE and ∠B = ∠F. The angle ∠B sits at the end of AB, opposite the side AC — it is
not between the two given sides. This is the SSA case, which we have seen can give two different
triangles. The triangles need not be congruent.

Why it happens: the safe conditions all pin the third vertex down completely. SSA
does not, because an arc can cut a line at two points. Always check where the equal
angle sits before naming the condition.

Q3 It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a
transversal for these two lines. Are there any equal alternate angles?]

The segments AD and BC cross at O. Look at ∆AOB and ∆DOC.

OA = OD (given)

∠AOB = ∠DOC (vertically opposite angles)

OB = OC (given)

→ SAS condition → ∆AOB ≅ ∆DOC
Corresponding angles of congruent triangles are equal, so

∠OAB = ∠ODC, that is ∠DAB = ∠ADC

AD is a transversal cutting the two lines AB and DC. The angles ∠DAB and ∠ADC are alternate
angles for that transversal, and they are equal. Hence

AB ∥ CD

Page 47 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: equal alternate angles is the test for parallel lines. The congruence
supplies exactly that pair of equal angles. Notice how congruence is being used here
as a tool — we wanted a fact about parallel lines, and congruent triangles delivered
it.

Tip: the same congruence also gives AB = DC. So ABDC has one pair of sides that are
both equal and parallel.

Q4 ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent to ΔCDA? Give
more examples of two triangles where one triangle is congruent to the other in two
different ways, as in the case above. Can you give an example of two triangles
where one is congruent to the other in six different ways?

Part 1 — ∆ABC ≅ ∆ADC. The diagonal AC splits the square into two triangles.

AB = AD (sides of a square)

BC = DC (sides of a square)

AC = AC (common diagonal)

→ SSS condition → ∆ABC ≅ ∆ADC

Part 2 — is ∆ABC ≅ ∆CDA as well? Yes. Check the three pairs of that correspondence:
∆ABC ∆CDA SIDES MATCHED EQUAL?

A C AB with CD Yes — sides of the square

B D BC with DA Yes — sides of the square

C A CA with AC Yes — same segment

So ∆ABC ≅ ∆CDA also holds, again by SSS. The same pair of triangles is congruent in two
different ways.

Why it happens: ∆ABC is isosceles — AB = BC, since both are sides of the square. An
isosceles triangle can be laid on its partner in two ways: the ordinary way, and after
being flipped over its line of symmetry. Each way gives one correct congruence
statement.

Page 48 of 55

as e
Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

More examples of two ways: take any isosceles triangle ∆PQR with PQ = PR and a copy of it,
co m
∆XYZ with XY = XZ. Then both ∆PQR ≅ ∆XYZ and ∆PQR ≅ ∆XZY are correct.
e m.
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Six ways: take an equilateral triangle and a copy of it — say ∆PQR and ∆XYZ, each with all sides

se m
5 cm. Every one of the six matchings works:

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∆PQR ≅ ∆XYZ, ∆PQR ≅ ∆XZY, ∆PQR ≅ ∆YXZ

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∆PQR ≅ ∆YZX, ∆PQR ≅ ∆ZXY, ∆PQR ≅ ∆ZYX
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Tip: the number of ways is the number of ways of matching equal sides. Scalene → 1

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way, isosceles → 2 ways, equilateral → 6 ways.

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a s em
a glQ5 Find ∠B and ∠C, if A is the centre of the circle. (∠A = 120°)
m a s
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∠B = ∠C = 30°.
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aglcentre, so AB and AC are both radii:
B and C lie on the circle and A is its

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AB = AC (radii of the same circle)
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→ ∆ABC is isosceles
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Now use the angle sum, writing x for each of the equal angles:

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∠A + ∠B + ∠C = 180°

120° + x + x = 180°

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x = 30°
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Why it happens: every radius of a circle has the same length, so any triangle formed

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by the centre and two points of the circle is automatically isosceles. That single fact
does all the work here.
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co m
m .
m ase
.co


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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Check it yourself: 120° + 30° + 30° = 180° ✓

Q6 Find the missing angles. As per the convention that we have been following, all line
segments marked with a single ‘|’ are equal to each other and those marked with a
double ‘|’ are equal to each other, etc.

The figure is a rectangle ACDB cut into 16 triangles. Four points inside it are not named in the
book, so let us call them P, Q, S and T, as marked here.

C R V D

S F
Q
U

T P

A K L B

The rectangle with the four inner points named S, T, Q and P.

The marked equal sides. Single ‘|’: CR = RV = VD = CU = UA = US = UT = VQ. Triple ‘|||’: RQ = AT
= KL = LB. Double ‘||’: PF = PB = FB.
Working, step by step.

1. ∆CRU has CU = CR and ∠C = 90°, so it is isosceles: ∠CRU = ∠CUR = 45°.
2. ∆UTA has UA = UT and ∠UAT = 56° (given at A), so ∠UTA = 56° and ∠AUT = 180° − 112° = 68°.
3. ∆RVQ has VR = VQ and ∠RVQ = 68°, so ∠VRQ = ∠VQR = (180° − 68°) ÷ 2 = 56°.
4. C, R, V, D lie on one line, so at R: 45° + ∠URS + 34° + 56° = 180°, giving ∠URS = 45°.

Page 50 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

5. C, U, A lie on one line, so at U: 45° + ∠RUS + 34° + 68° = 180°, giving ∠RUS = 33°. Then in
∆RUS, ∠RSU = 180° − 45° − 33° = 102°.
6. In ∆RSQ: ∠RSQ = 180° − 34° − 44° = 102°.
7. ∆UST has US = UT and ∠SUT = 34°, so ∠UST = ∠UTS = 73°.
8. The four angles at S fill a full turn: ∠TSQ = 360° − 102° − 73° − 102° = 83°. Then in ∆STQ,
∠STQ = 180° − 83° − 46° = 51°.
9. In ∆TQP: ∠TQP = 90° and ∠QPT = 56°, so ∠QTP = 34°.
10. In ∆ATK: ∠AKT = 180° − 34° − 44° = 102°.
11. The six angles at T fill a full turn: ∠KTP = 360° − (34° + 51° + 73° + 56° + 44°) = 102°. Then in
∆TPK, ∠TKP = 180° − 102° − 30° = 48°.
12. A, K, L, B lie on one line, so at K: ∠PKL = 180° − 102° − 48° = 30°. In ∆KPL, ∠PLK = 90°, so
∠KPL = 60°.
13. ∆PLK and ∆PLB have LK = LB, ∠PLK = ∠PLB = 90° and PL common, so by SAS they are
congruent. Hence ∠PBL = 30°, ∠LPB = 60° and PK = PB.
14. ∆PBF has PF = PB = FB (all double marks), so it is equilateral: ∠BPF = ∠PFB = ∠PBF = 60°.
15. The six angles at P fill a full turn: ∠FPQ = 360° − (56° + 30° + 60° + 60° + 60°) = 94°.
16. D, F, B lie on one line, so at F: ∠QFP = 180° − 98° − 60° = 22°. Then in ∆PFQ, ∠PQF = 180° −
94° − 22° = 64°.
17. At V, ∠QVD = 180° − 68° = 112°. ∆QVD has VQ = VD, so ∠VQD = ∠VDQ = (180° − 112°) ÷ 2 =
34°.
18. At the corner D: ∠QDF = 90° − 34° = 56°. Then in ∆QFD, ∠DQF = 180° − 98° − 56° = 26°.

All the angles, triangle by triangle.

TRIANGLE ANGLES TRIANGLE ANGLES

∆CRU 90°, 45°, 45° ∆TPK 102° (at T), 30° (at P), 48° (at K)

∆RUS 45° (at R), 33° (at U), 102° (at S) ∆KPL 30°, 60°, 90°

∆UST 34° (at U), 73°, 73° ∆PLB 90° (at L), 60° (at P), 30° (at B)

∆UTA 68° (at U), 56°, 56° ∆PBF 60°, 60°, 60°

∆ATK 34° (at A), 44° (at T), 102° (at K) ∆PFQ 94° (at P), 22° (at F), 64° (at Q)

∆RSQ 34° (at R), 102° (at S), 44° (at Q) ∆QFD 98° (at F), 56° (at D), 26° (at Q)

∆STQ 83° (at S), 51° (at T), 46° (at Q) ∆QDV 112° (at V), 34°, 34°

∆TQP 34° (at T), 90° (at Q), 56° (at P) ∆QVR 68° (at V), 56°, 56°

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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Why it happens: only three tools are used again and again — angles opposite equal
sides are equal, the three angles of a triangle add to 180°, and the angles at a point
add to 360° (or 180° along a straight line). Each answer unlocks the next triangle, like
links in a chain.

Check it yourself: the seven angles around Q must total 360°. Adding them: 44° +
46° + 90° + 64° + 26° + 34° + 56° = 360° ✓ Notice also that ∆RSQ and ∆ATK have the
same three angles 34°, 44°, 102° and the equal sides RQ = AT, so ∆RSQ ≅ ∆AKT by
ASA.

Puzzle Time — Expression Engineer!
End-of-chapter puzzle

Q1 Draw lines and split the region consisting of white squares into 6 smaller congruent
regions.

The grid is 5 × 5, and the green square in the middle is not part of the region.

Total small squares = 5 × 5 = 25
White squares = 25 − 1 = 24

24 ÷ 6 = 4 squares in each region

So each of the six pieces must be made of 4 squares, and all six must be the same shape. An L-
shaped piece of 4 squares works. Here is one way to do it — the six pieces are shown in six
colours.

Page 52 of 55

Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

1 1 1 2 2

1 3 3 3 2

4 3 5 2

4 5 5 5 6

4 4 6 6 6

Six congruent L-shaped pieces of 4 squares each. The green square in the centre is left out.

Every piece is the same L of 4 squares — three in a row with one more square turned at the end.
The pieces are only rotated, never made bigger or smaller, so all six are congruent.

Why it happens: first count. Six equal pieces out of 24 squares means 4 squares
each, which cuts down the shapes worth trying. A straight piece of 4 will not work,
because the squares next to the hole cannot be reached by one; the L bends around
the hole neatly. Notice that the finished pattern has half-turn symmetry — turn the
picture upside down and piece 1 lands on piece 6, piece 2 on piece 4, and piece 3 on
piece 5.

Page 53 of 55

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Class 7 Maths Chapter 9 Geometric Twins
a g l AglaSem · NCERT Solutions

co m
m.
Try This: take out a corner square instead of the middle one and try again with six

m as e
l
pieces of 4 squares. Does an L still work?

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l a se
a g

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Chapter at a glance
em . ag
g l as
a
Two figures are congruent if they have the same shape and the same size. Tracing one and
placing it on the other, it fits exactly. The figure may be rotated or flipped first.
Arm lengths alone do not fix a figure. Two arms of 4 cm and 8 cm can be opened at any
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angle, so the angle between them must also be measured.

o m
c two angles with a side (ASA or AAS). g l a
For triangles, three sides are enough: SSS. So are two sides with the angle between them
(SAS),.and a
m
se conditions fail. AAA gives the same shape but any size, and SSA (two sides and a non-
g l a
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Two
included angle) can give two different triangles.

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In a right-angled triangle, the right angle with the hypotenuse and one more side is enough

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— the RHS condition.

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Congruence is a tool, not just a test. It proves that angles opposite equal sides are equal,
ofgan equilateral triangle is 60°.
and hence that every angle a

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m ase
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Class 7 Maths Chapter 9 Geometric Twins AglaSem · NCERT Solutions

Quick revision

CONDITION WHAT MUST BE GUARANTEES EXAMPLE FROM THE
EQUAL CONGRUENCE? CHAPTER

Congruent Same shape and same size They superimpose exactly Trace and place one on the
figures other

Rotate or flip Allowed before Still congruent The two teardrops on page
superimposing 3

Circle Radius only Yes Equal radii → congruent
circles

Rectangle Length and breadth Yes All four angles are already
90°

SSS Three pairs of sides Yes 4 cm, 6 cm, 8 cm

SAS Two sides and the included Yes 6 cm, 5 cm and ∠A = 30°
angle

SSA Two sides and a non- No 6 cm, 4 cm, 30° gives two
included angle triangles

AAA All three angles No 30°, 70°, 80° — same
shape, any size

ASA Two angles and the included Yes 50°, BC = 5 cm, 30°
side

AAS Two angles and a non- Yes 35°, 75°, BC = 4 cm
included side

RHS Right angle, hypotenuse and Yes ∠B = 90°, AC = 5 cm, BC = 4
one side cm

Isosceles Angles opposite equal sides They are equal AB = AC → ∠B = ∠C
triangle

Equilateral All three angles Each is 60° 3 × angle = 180°
triangle

Page 55 of 55