NCERT Solutions Class 7 Maths Chapter 7 a Tale of Three Intersecting Lines

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 7: A Tale of Three
Intersecting Lines

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

146 – 172 24 75 English

Solutions, notes, sample papers & more at 67 pages

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 7: A Tale of Three
Intersecting Lines
Chapter 7 of Ganita Prakash Part I is the story of the triangle — three points, three sides and three angles. We
construct triangles with a compass and a set square, discover the triangle inequality that decides when a
triangle can exist at all, and use parallel lines to prove the famous angle sum property.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) 146 – 172

SECTIONS QUESTIONS

24 75

MEDIUM

English

In-text Questions — Pages 146–149
Introduction · 7.1 Equilateral Triangles · 7.2 Constructing a Triangle When its Sides are Given

Q1 What happens when the three vertices lie on a straight line?

No triangle is formed at all. The figure collapses into a single line segment.

Three collinear points — no closed
shape

A B C

When A, B and C lie on one line, AB + BC = AC and the triangle has no inside.

Page 1 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: A triangle must enclose some space. If B lies on the segment AC,
then the "side" AB and the "side" BC together lie exactly along AC, so AB + BC = AC.
Nothing is enclosed, and the angle at B becomes 180° — a straight angle, not a
corner. Such a set of three points is called collinear.

Q2 Construct a triangle in which all the sides are of length 4 cm.

Use a compass — a ruler alone would need many trials.

1. Draw AB = 4 cm with a ruler.
2. With A as centre and radius 4 cm, draw a long arc above AB.
3. With B as centre and the same radius 4 cm, draw another arc cutting the first one at C.
4. Join AC and BC. ∆ABC is the required equilateral triangle.

C

4 cm 4 cm

A 4 cm B

The two arcs of radius 4 cm cross exactly at C, so AC = BC = AB = 4 cm.

Check it yourself: Measure the three angles with a protractor. Each one comes out
to be 60°.

Page 2 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q3 How did you construct this triangle and what tools did you use? Can this
construction be done only using a marked ruler (and a pencil)?

Yes, it can be done with only a marked ruler and a pencil — but it takes many trials.

Draw AB = 4 cm

Mark a point C with AC = 4 cm

Measure BC → it is usually not 4 cm

Shift C and measure again … and again

Why it happens: A ruler fixes one distance at a time. There are endlessly many
points C with AC = 4 cm, and only two of them also have BC = 4 cm. Hunting for them
by eye is slow. A compass fixes all the points at distance 4 cm from A in one sweep,
which is why it makes the job exact.

Tip: Tools used — ruler for the base, compass for the two arcs, pencil to join. A
protractor is not needed at all.

Q4 How do we make this construction more efficient?

Replace the guessing with two compass arcs.

Arc from A, radius 4 cm → every point on it is 4 cm from A

Arc from B, radius 4 cm → every point on it is 4 cm from B

Their meeting point C is 4 cm from both

Why it happens: This is exactly the trick used last year in Playing with Constructions
to fix the top point of a 'house'. One arc gives a whole family of correct points; the
second arc picks out the one point that satisfies the second condition too. No trial is
left.

Page 3 of 67

ase
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
m.
Let C be the point of intersection of the arcs. The construction ensures that both AC
e
Q5

m l as
.co
and BC are of length 4 cm. Can you see why?

a g
se m
g l a
a

Because C lies on both arcs at once.

. c om ag
C is on the arc centred at A of radius 4 cmm
a s e → AC = 4 cm

agl 4 cm → BC = 4 cm
C is on the arc centred at B of radius

m
And AB = 4 cm was drawn first
co
em.
as
So AB = BC = CA = 4 cm
m l
.co a g
a s emit happens: Every point of a circle (or arc) is at the same distance — the radius
a gl Why
— from its centre. So membership of an arc is a statement about distance. Being on

m a s
.co agl
two arcs means satisfying two distance conditions together.

se m
g l a
a
Q6 How do we construct triangles that are not equilateral?

co m
m .
o m l a se
g
.c two-arc method works — only the two radii are now different.
a
m
s1.e Choose any one of the three lengths as the base and draw it.
The same

g l a
a
se m
2. From one end, draw an arc whose radius is the second length.

com
3. From the other end, draw an arc whose radius is the third length.
g l a
m . a
ase
4. Join the crossing point to both ends.

a gl
Tip: Choosing the longest side as the base makes the two arcs cross at a
comfortable angle, so the third vertex is easy to mark sharply.
co m
m .
m as e
.co a g l
a s emQ7
gl
Construct a triangle of sidelength 4 cm, 5 cm and 6 cm.
a c
m .
m a s e
co agl

m .
e
Take AB = 4 cm as the base, AC = 5 cm and BC = 6 cm.

g l as
1. Step 1: Draw AB = 4 cm.
a
2. Step 2: With A as centre, draw a sufficiently long arc of radius 5 cm.

com
m .
m ase
.co


a g l Page 4 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

3. Step 3: With B as centre, draw an arc of radius 6 cm so that it cuts the first arc.
4. Step 4: Call the crossing point C. Join AC and BC to get ∆ABC.

C

6 cm
5 cm

A 4 cm B

Base 4 cm; the arc of radius 5 cm from A and the arc of radius 6 cm from B cross at C.

Check it yourself: 4 < 5 + 6, 5 < 4 + 6 and 6 < 4 + 5, so the arcs are certain to meet.
This is the triangle inequality you will meet a few pages later.

Q8 How do we construct this triangle more efficiently?

Do not draw the full circles — two short arcs are enough.

Points 5 cm from A → the circle centred at A, radius 5 cm

Points 6 cm from B → the circle centred at B, radius 6 cm

C = a point common to both → only the arcs near the crossing are needed

Page 5 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: The full circle carries far more information than we need. The
vertex C sits where the two circles cross, so we only have to draw the small pieces of
each circle around that region. Fewer lines means a cleaner, more accurate figure.

Did you know? The two circles actually cross at two points, one above AB and one
below. Both give a correct triangle — they are mirror images of each other in the line
AB.

Construct — Page 150
Section 7.2 Constructing a Triangle When its Sides are Given

Q1 Construct triangles having the following sidelengths (all the units are in cm): (a) 4,
4, 6 (b) 3, 4, 5 (c) 1, 5, 5 (d) 4, 6, 8 (e) 3.5, 3.5, 3.5

In every case draw the longest side as the base, then cut two arcs of the other two lengths.

SET BASE TO DRAW ARCS FROM THE TWO ENDS TYPE OF TRIANGLE

(a) 4, 4, 6 6 cm 4 cm and 4 cm Isosceles

(b) 3, 4, 5 5 cm 3 cm and 4 cm Scalene (right-angled)

(c) 1, 5, 5 5 cm 1 cm and 5 cm Isosceles (very thin)

(d) 4, 6, 8 8 cm 4 cm and 6 cm Scalene

(e) 3.5, 3.5, 3.5 3.5 cm 3.5 cm and 3.5 cm Equilateral

Every set works, because in each one the longest length is smaller than the sum of the other
two:

(a) 6 < 4 + 4 = 8 ✓

(b) 5 < 3 + 4 = 7 ✓

(c) 5 < 1 + 5 = 6 ✓

(d) 8 < 4 + 6 = 10 ✓

(e) 3.5 < 3.5 + 3.5 = 7 ✓

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Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: The two arcs meet only if together they can "reach across" the
base. In (c) the reach is 1 + 5 = 6 against a base of 5 — just one centimetre to spare
— so the triangle is long and thin, with a very small angle opposite the 1 cm side.

Check it yourself: Measure the angles of (b). You will get 37°, 53° and 90° — a right-
angled triangle.

Figure it Out — Pages 150–151
Section 7.2 Constructing a Triangle When its Sides are Given

Q1 Use the points on the circle and/or the centre to form isosceles triangles.

Take the centre O and any two points P and Q on the circle. ∆OPQ is isosceles.

O
r r

P Q

Page 7 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

OP and OQ are both radii, so ∆OPQ is always isosceles.

OP = radius

OQ = radius

So OP = OQ → ∆OPQ is isosceles

Why it happens: Every point of a circle is the same distance from the centre. So the
moment two of the three vertices are on the circle and the third is the centre, two
sides are forced to be equal. You can slide P and Q anywhere on the circle and the
triangle stays isosceles.

Try This: Mark P, Q, R, S … on the circle. ∆OPQ, ∆OQR, ∆ORS … are all isosceles — you
can make as many as you like. If you place P and Q so that PQ also equals the radius,
the triangle even becomes equilateral.

Q2 Use the points on the circles and/or their centres to form isosceles and equilateral
triangles. The circles are of the same size.

In both figures each circle passes through the centre of the other, so AB = radius.

Page 8 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
.co a g
a s em P
ag l

com
e m . ag
g l as
Aa B

co m
em.
m l as
m .co a g
l a se
ag
Q s
m a
m.c o agl
l a se
a g
P and Q are the crossing points of the two equal circles; ∆ABP and ∆ABQ are equilateral.

co m
Equilateral triangles
m .
m as e
.co a g l
s e m = radius (P is on circle A)
la
AP

ag BP = radius (P is on circle B)

se m
com
AB = radius (each circle passes through the other's centre)
g l a
m . a
ase
So AP = BP = AB → ∆ABP is equilateral (and so is ∆ABQ)

agl
Isosceles triangles

co m
m .
Take any point C on the circle with centre B. Then BC = radius = AB, so ∆ABC is isosceles.

m
Sliding C round that circle gives endlessly many.
o l a se
agAC = AD = radius, so ∆ACD is
.cTake any two points C and D on the circle centred at A. Then
m isosceles.
l a se
ag
.c
∆APQ and ∆BPQ are isosceles too, since AP = AQ and BP = BQ.

s e m
m a
co agl
In the second figure A, B and C are the three centres, and each of them lies on the other two
circles:
m .
as e
a g l

co m
m .
m ase
.co


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Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

AB = BC = CA = radius

So ∆ABC is equilateral

Why it happens: Equal circles fix one single length — the common radius. Every
segment you can draw from a centre to a point of its own circle equals that length.
Counting how many of a triangle's sides are such segments tells you at once
whether it is isosceles (two of them) or equilateral (all three).

In-text Questions — Page 151
Are Triangles Possible for any Lengths?

MATH TALK

Q1 Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are
you able to construct the triangle?

No. The two arcs never meet, so the third vertex simply does not exist.

Draw the base AB = 8 cm

Arc from A of radius 3 cm reaches only 3 cm along

Arc from B of radius 4 cm reaches only 4 cm back

3 + 4 = 7 cm, but the gap to be covered is 8 cm

7 < 8 → the arcs fall 1 cm short of each other

Page 10 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

gap — arcs never meet

3 cm 4 cm

A B
8 cm
With a base of 8 cm, arcs of 3 cm and 4 cm cannot reach each other.

Why it happens: The two shorter sides together have to bridge the base. Here they
add up to only 7 cm while the base is 8 cm, so however you tilt them they can never
join. No triangle exists with sides 3 cm, 4 cm and 8 cm.

Q2 Here is another set of lengths: 2 cm, 3 cm, and 6 cm. Check if a triangle is possible
for these sidelengths.

Not possible. The same thing happens, only worse.

Longest length = 6 cm

Sum of the other two = 2 + 3 = 5 cm

5 < 6 → the arcs fall short by 1 cm

No triangle exists

Check it yourself: Draw AB = 6 cm, then an arc of 2 cm from A and an arc of 3 cm
from B. A clear white gap is left between them.

Page 11 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q3 Try to find more sets of lengths for which a triangle construction is impossible. See
if you can find any pattern in them.

Here are five impossible sets, and the pattern behind all of them.

SET OF LENGTHS LONGEST SUM OF OTHER TWO TRIANGLE?

1, 2, 5 5 3 No

2, 4, 9 9 6 No

3, 5, 8 8 8 No (arcs only touch)

6, 7, 20 20 13 No

1, 1, 3 3 2 No

The pattern: a triangle is impossible exactly when

longest length ≥ sum of the other two lengths

Why it happens: The two smaller sides have to stretch from the two ends of the
longest side and meet somewhere above it. Their total length must be more than the
base — if it is exactly equal they flatten onto the base and give a straight line, and if
it is less they cannot meet at all.

Try This: Make an impossible set of your own by picking any two lengths and
making the third one bigger than their sum, for example 7, 9 and 20.

In-text Questions — Pages 152–153

Page 12 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Triangle Inequality

MATH TALK

Q1 Consider the lengths 10 cm, 15 cm and 30 cm. Does there exist a triangle having
these as sidelengths?

No such triangle exists.

Longest length = 30 cm

Sum of the other two = 10 + 15 = 25 cm

30 > 25 → the longest side is too long

So a triangle with sides 10 cm, 15 cm, 30 cm is impossible

Why it happens: Walking straight from one end of the 30 cm side to the other is a
30 cm journey. Walking round through the third vertex is 10 + 15 = 25 cm. That
would make the roundabout path shorter than the straight path — which can never
happen. The figure destroys itself.

Q2 Imagine you are at the entrance of the tent and want to go to the tree. Which is the
shorter path: (i) the straight-line path to the tree (the red path) or (ii) the straight-
line path from the tent to the pole, followed by the straight-line path from the pole
to the tree (the yellow path)?

The red path — the straight line from the tent to the tree — is shorter.

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Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
.co
pole
a g
se m
g l a
a
roundabout
com
e m . ag
g l as
a

co m
direct m.
se tree
m
tent
l a
m .co ag
l a se
ag The direct path (red) beats the roundabout path through the pole (yellow).

m a s
e
. c o
m the straight line is the shortest possible path. agl
s
Why it happens: Between two points,

a g la adds length. This single everyday fact is the whole
Any detour through a third point
idea behind the triangle inequality.

co m
m .
o m l a se
Q3 .cCan this understanding be used to tell something about
a gthe existence of a triangle
m
e having sidelengths 10 cm, 15 cm and 30 cm?
aglas
se m
om= 10 cm, AB = 15 cm and CA = 30 cm. Test all three
Yes. Suppose such a ∆ABC existed, with cBC g l a
. a
a s em
agl
pairs of vertices.

BETWEEN DIRECT PATH ROUNDABOUT PATH DIRECT SHORTER?

co m
m .
e
B and C BC = 10 cm BA + AC = 15 + 30 = 45 cm Yes

m AC + CB = 30 + 10 = 40 cm gl
as
.co a
em
A and B AB = 15 cm Yes

a s
agl C and A CA = 30 cm CB + BA = 10 + 15 = 25 cm No!

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

com
m .
m as e
.co


a g l Page 14 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

30 cm > 25 cm — the direct path is longer than the detour

That is absurd

So the triangle cannot exist

Why it happens: We assumed the triangle existed and reached an impossible
conclusion, so the assumption itself must be wrong. Notice we needed no ruler, no
compass and no construction — only reasoning. This is exactly how the properties of
parallel and intersecting lines were discovered earlier.

Q4 Can we say anything about the existence of a triangle having sidelengths 3 cm, 3
cm and 7 cm? Verify your answer by construction.

No triangle exists.

Direct path of 7 cm

Roundabout path = 3 + 3 = 6 cm

7 > 6 → the direct path would be longer than the detour — impossible

Verifying by construction: draw AB = 7 cm, then an arc of radius 3 cm from A and another of
radius 3 cm from B. The two arcs stop 1 cm apart and never cut each other, so there is no third
vertex.

Tip: Two sides of 3 cm can together bridge at most 6 cm. A 7 cm base is simply
beyond their reach.

Page 15 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q5 “In the rough diagram in Fig. 7.4, is it possible to assign lengths in a different order
such that the direct paths are always coming out to be shorter than the roundabout
paths? If this is possible, then a triangle might exist.” Is such rearrangement of
lengths possible in the triangle?

A

30
15

cm
cm

B
10 cm
C

Fig. 7.4, page 152 — the rough diagram of a supposed triangle with sidelengths 10 cm,
15 cm and 30 cm (not drawn to scale).

No. Rearranging the labels changes nothing.

Whichever side you call 30 cm, the other two are 10 cm and 15 cm

Their sum is always 10 + 15 = 25 cm

And 30 > 25 in every arrangement

There are only three genuinely different ways to place the labels, and all three fail in exactly the
same comparison:

Page 16 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

ARRANGEMENT FAILING COMPARISON TRIANGLE?

BC = 10, AB = 15, CA = 30 30 > 10 + 15 No

BC = 15, AB = 30, CA = 10 30 > 15 + 10 No

BC = 30, AB = 10, CA = 15 30 > 10 + 15 No

Why it happens: The comparison that decides everything only involves the largest
number and the sum of the other two. Names of vertices and the order of the sides
are just labels — they cannot change 10 + 15 into something bigger than 30.

Figure it Out — Page 154
Triangle Inequality

TRY THIS

Q1 We checked by construction that there are no triangles having sidelengths 3 cm, 4
cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without
trying to construct the triangle.

Yes — one comparison settles each case.

3, 4, 8 → 3 + 4 = 7 and 7 < 8 → no triangle

2, 3, 6 → 2 + 3 = 5 and 5 < 6 → no triangle

Why it happens: The direct path along the longest side must be shorter than the
roundabout path along the other two. Here the roundabout path (7 cm, 5 cm) is
shorter than the direct one (8 cm, 6 cm), which is impossible. So the drawing was
never going to work — the compass only confirmed what the numbers already said.

Tip: Always compare the largest length with the sum of the other two. That single
check is enough.

Page 17 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q2 Can we say anything about the existence of a triangle for each of the following sets
of lengths? (a) 10 km, 10 km and 25 km (b) 5 mm, 10 mm and 20 mm (c) 12 cm, 20 cm
and 40 cm

None of the three sets gives a triangle.

SET LONGEST SUM OF THE OTHER TWO COMPARISON TRIANGLE?

(a) 10, 10, 25 km 25 10 + 10 = 20 25 > 20 No

(b) 5, 10, 20 mm 20 5 + 10 = 15 20 > 15 No

(c) 12, 20, 40 cm 40 12 + 20 = 32 40 > 32 No

Why it happens: In each set the longest length beats the sum of the other two, so
the two shorter sides can never bridge it. The unit (km, mm, cm) makes no
difference at all — only the numbers matter, as long as all three are measured in the
same unit.

Q3 For each set of lengths seen so far, you might have noticed that in at least two of
the comparisons, the direct length was less than the sum of the other two (if not,
check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are
two comparisons where this happens: 10 < 15 + 30 and 15 < 10 + 30. But this doesn’t
happen for the third length: 30 > 10 + 15. Will this always happen? That is, for any
set of lengths, will there be at least two comparisons where the direct length is less
than the sum of the other two? Explore for different sets of lengths.

Yes, always. At most one comparison can ever fail.

SET COMPARISONS THAT HOLD COMPARISON THAT FAILS

7, 10, 15 7 < 25, 10 < 22, 15 < 17 none — triangle exists

12, 14, 18 12 < 32, 14 < 30, 18 < 26 none — triangle exists

2, 3, 9 2 < 12, 3 < 11 9>2+3

1, 4, 5 1 < 9, 4 < 6 5 = 1 + 4 (not less)

Arrange the lengths in increasing order as a ≤ b ≤ c. Then

Page 18 of 67

ase
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
e m.
a ≤ c, and b is positive, so a < b + c ✓
m l as
.co
b ≤ c, and a is positive, so b < a + c ✓
m a g
l a se
g
Only c < a + b is in doubt
a

. com
Why it happens: A smaller length is already no bigger than the largest one, so
ag
m sum bigger for certain. That is why the two
adding a third positive length makesethe
s
a
aglwhole existence question hangs on the single
smaller lengths always pass. The
comparison for the largest length.

co m
em.
com to make two of them fail. l
Try This: Pick any three lengths at random and test all three comparisons. You will
g as
never .manage a
a s em
a gl
m a s
.co agl
In-text Questions — Page 154
se m
l a
Triangle Inequality

a g
Further, for a given set of lengths, is it possible to identify which lengths will
m
Q1
immediately be less than the sum of the other two, without calculations? [Hint:
. co
e m
m as
Consider the direct lengths in the increasing order.]

.co a g l
a s em
gl

a Yes. Put the three lengths in increasing order — the two smaller ones pass automatically.

se m
com g l a
m . a
ase
Let a ≤ b ≤ c

agl
a ≤ c < b + c → a < b + c ✓ (no calculation needed)

b ≤ c < a + c → b < a + c ✓ (no calculation needed)
co m
m .
e
Only c < a + b has to be checked
m l as
.co a g
mFor example, in 6, 11, 14 you can say at once that 6 and 11 pass; only 14 vs 6 + 11 = 17 needs a
s e
agla look.
.c
s e m
m a
co agl
Why it happens: Comparing a small number with a sum that already contains the
m .
as e
largest number is no contest. The only tight comparison is the one where the largest

a
number stands alone on the left.
g l

co m
m .
m ase
.co


a g l Page 19 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q2 Given three sidelengths, what do we need to compare to check for the existence of
a triangle?

Compare the longest length with the sum of the other two. That is the only check needed.

If longest < sum of the other two → triangle exists

If longest = sum of the other two → the three points fall in a line, no triangle

If longest > sum of the other two → no triangle

When each length is smaller than the sum of the other two, the lengths are said to satisfy the
triangle inequality.

SET CHECK SATISFIES TRIANGLE INEQUALITY?

3, 4, 5 5<3+4=7 Yes

10, 15, 30 30 > 10 + 15 = 25 No

Tip: There are three comparisons in all, but two of them are free of charge. Do the
one that matters and save time.

In-text Questions — Page 155
Visualising the construction of circles

Q1 Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm? This satisfies the
triangle inequality: 8 < 4 + 5 = 9. Why do we not need to check the other two sides?

Because the other two comparisons are already guaranteed by the size order.

4 < 5 + 8 = 13 ✓ (4 is smaller than 8 alone)

5 < 4 + 8 = 12 ✓ (5 is smaller than 8 alone)

8 < 4 + 5 = 9 ✓ — the only comparison that could have failed

Page 20 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: Each of the smaller lengths is already less than the longest length
by itself; adding a third positive length can only widen the gap. So checking the
longest against the sum of the other two settles the whole question.

Q2 Now, suppose that a circle of radius 5 cm is constructed, centred at B. Can you draw
a rough diagram of the resulting figure?

4 cm
B
A
X

8 cm

Page 155 — the circle of radius 4 cm centred at A, drawn on the base AB = 8 cm; X is the
point where it cuts AB.

Take AB = 8 cm as the base. Draw the circle of radius 4 cm about A and the circle of radius 5 cm
about B — they overlap in a lens-shaped region.

Page 21 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

circle B, r = 5
circle A, r = 4

4 cm 4 cm

A X B

AB = 8 cm, AX = 4 cm so BX = 4 cm, which is less than the 5 cm radius of circle B.

Tip: Draw the base first and mark X, the point where circle A cuts AB. Then you can
see at a glance whether circle B swallows X or not.

Page 22 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q3 Note that in the figure below, AX = 4 cm and AB = 8 cm. So, what is BX? Does this
length help in visualising the resulting figure?

4 cm
B
A
X

8 cm

Page 155 — the circle of radius 4 cm centred at A, drawn on the base AB = 8 cm; X is the
point where it cuts AB.

BX = 4 cm, and it settles the whole picture.

BX = AB – AX

= 8 cm – 4 cm

= 4 cm

Radius of the circle centred at B = 5 cm

BX = 4 cm < 5 cm

So X lies inside circle B

Page 23 of 67

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Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: X is the point of circle A that is nearest to B. If even this nearest

m l a se
point is inside circle B, then part of circle A is inside circle B and part of it (the far
o
m .c — so the two circles must cross, and they cross atagexactly two points.
side) is outside

l a seof those points together with A and B gives a triangle. Hence a triangle with
g 4 cm, 5 cm and 8 cm does exist.
Either
asides

co m
e m . ag
g l as
Figure it Out — Page 156
a
Triangle Inequality

. com
m
Which of the following lengths can be the sidelengths of a triangle?
a s em Explain your
co Note that for each set, the three lengths have the l unit of measure.
Q1

. a gsame
em
answers.

a s
agl
(a) 2, 2, 5 (b) 3, 4, 6 (c) 2, 4, 8 (d) 5, 5, 8 (e) 10, 20, 25 (f) 10, 20, 35 (g) 24, 26, 28

s

m a
.co
Compare the longest length with the sum of the other two in each set.
em agl
a s
SET
agl OF OTHER TWO
LONGEST VS SUM SIDELENGTHS OF A TRIANGLE?

5>2+2=4
co m
(a) 2, 2, 5 No

m .
as e
com l
(b) 3, 4, 6 6<3+4=7 Yes

. a g
e m
as
(c) 2, 4, 8 8>2+4=6 No

agl (d) 5, 5, 8 8 < 5 + 5 = 10 Yes

se m
com g l a
. a
(e) 10, 20, 25 25 < 10 + 20 = 30 Yes

35 > 10 + 20 = 30as
em
agl
(f) 10, 20, 35 No

(g) 24, 26, 28 28 < 24 + 26 = 50 Yes

co m
m .
Writing out all three comparisons for the successful sets:
m as e
.co a g l
se m (b) 3 < 4 + 6, 4 < 3 + 6, 6 < 3 + 4 ✓
g l a
a c
(d) 5 < 5 + 8, 5 < 5 + 8, 8 < 5 + 5 ✓
m .
m a s e
. co agl
(e) 10 < 20 + 25, 20 < 10 + 25, 25 < 10 + 20 ✓
e m
l as
(g) 24 < 26 + 28, 26 < 24 + 28, 28 < 24 + 26 ✓
g
a

co m
m .
m ase
.co


a g l Page 24 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: Sets (a), (c) and (f) each have one length that is larger than the
other two put together, so those two sides can never reach across it. Sets (b), (d), (e)
and (g) satisfy the triangle inequality, so the two arcs are certain to cross and the
triangle certainly exists.

Tip: (d) 5, 5, 8 is isosceles and (g) 24, 26, 28 is scalene — the inequality decides
existence, not the type.

In-text Questions — Pages 156–159
When do the two circles intersect?

Q1 Will triangles always exist when a set of lengths satisfies the triangle inequality?
How can we be sure?

Yes — and we can be sure by looking at how the two circles can possibly sit.
Take the base AB = the longest length, and draw circles at A and B whose radii are the two
smaller lengths. Exactly three things can happen:

CASE PICTURE RELATION TRIANGLE?

Case 1 Circles touch at one point sum of the two radii = AB No (flat line)

Case 2 Circles do not meet sum of the two radii < AB No

Case 3 Circles cut each other sum of the two radii > AB Yes

Case 1: touch Case 2: apart Case 3: cut

Only Case 3 — where the circles cut each other — produces a third vertex.

Page 25 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: The triangle inequality says the sum of the two smaller lengths is
greater than the longest length. Since the two radii are the smaller lengths and AB is
the longest, this is exactly the condition of Case 3. So satisfying the inequality forces
the circles to cut, which forces the triangle to exist.

Q2 Case 2: Circles do not intersect internally. For this case to happen, what should be
the relation between the radii and AB?

The two radii together must fall short of AB.

sum of the two radii < AB

that is, sum of the two smaller lengths < longest length

In the figure, X is the point where circle A cuts AB. The gap XB is left uncovered because the
second circle is too small to reach X.

Why it happens: Circle A reaches only as far as its radius along AB; circle B reaches
back only as far as its radius. If those two reaches do not overlap, an empty strip is
left between them, and there is no point that belongs to both circles.

Q3 Can we use this analysis to tell if a triangle exists when the lengths satisfy the
triangle inequality?

Yes — the analysis closes the argument completely.

Triangle inequality holds

→ sum of the two smaller lengths > longest length
→ sum of the two radii > AB

→ we are in Case 3, the circles cut each other

→ a third vertex exists

→ the triangle exists

Page 26 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Conclusion: if a given set of three lengths satisfies the triangle inequality, a triangle exists
having those as sidelengths. If it does not, no such triangle exists.

Why it happens: Earlier we only knew that failing the inequality ruled a triangle out.
The three-case study supplies the missing half — passing the inequality rules a
triangle in. Together they give a complete test.

Q4 How will the two circles turn out for a set of lengths that do not satisfy the triangle
inequality? Find 3 examples of sets of lengths for which the circles: (a) touch each
other at a point, (b) do not intersect.

They land in Case 1 or Case 2.
(a) Circles touch at a point — sum of the two smaller lengths = longest length:

SET CHECK

3, 5, 8 3+5=8

2, 4, 6 2+4=6

5, 5, 10 5 + 5 = 10

(b) Circles do not intersect — sum of the two smaller lengths < longest length:

SET CHECK

2, 3, 8 2+3=5<8

1, 2, 10 1 + 2 = 3 < 10

4, 5, 12 4 + 5 = 9 < 12

Why it happens: In the touching case the meeting point lies on AB itself, so A, B and
it are collinear — a flattened triangle with no height. In the non-intersecting case
there is no common point at all.

Page 27 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q5 Frame a complete procedure that can be used to check the existence of a triangle.

A three-step routine that never fails.

1. Step 1: Make sure all three lengths are in the same unit.
2. Step 2: Pick out the largest length. Add the other two.
3. Step 3: Compare.

largest < sum of the other two → the triangle exists
largest = sum of the other two → the three points are collinear, no triangle
largest > sum of the other two → no triangle

Example: 7, 9, 15

Largest = 15, sum of others = 7 + 9 = 16
15 < 16 → triangle exists

Tip: No compass, no ruler, no drawing — one subtraction and one comparison
decide the answer.

Figure it Out — Pages 159–160
Conclusion — the triangle inequality test

TRY THIS

Q1 Check if a triangle exists for each of the following set of lengths: (a) 1, 100, 100 (b) 3,
6, 9 (c) 1, 1, 5 (d) 5, 10, 12

Compare the largest length with the sum of the other two.

Page 28 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
m.
SET LARGEST VS SUM OF OTHER TWO TRIANGLE?

m l a se
.co ag
(a) 1, 100, 100 100 < 1 + 100 = 101 Exists

se m
g l a 9=3+6
a
(b) 3, 6, 9 Does not exist

(c) 1, 1, 5 5>1+1=2 Does not exist

com
(d) 5, 10, 12 12 < 5 + 10 = 15
e m . Exists ag
g l as
a
Why it happens: In (b) the sum is exactly equal, not less. The two circles just touch,

c o m
so the third "vertex" lands on the base itself and the figure flattens into a straight
.
line — equality is not good enough. In (a) the margin is only 1 unit, somthe triangle is
m
real butoextremely l a se
.c
thin.
a g
se m
g l a
a
m
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there
a s
.co agl
Q2

m
exist an equilateral triangle of any sidelength? Justify your answer.

l a se
ANSWER a g
m
Yes to both.

. co
e m
m l as
.co g
50 < 50 + 50 = 100 ✓
m three comparisons are the same, so the triangle inequality holds
eAll a
a s
agl An equilateral triangle of side 50 exists

se m
com g l a
m . a
ase
For any sidelength s (a positive number):

a gl
Sum of the other two sides = s + s = 2s

co m
.
Since s is positive, s < 2s is always true
em
m l as
.co g
So an equilateral triangle of any sidelength exists
m a
l a se
ag Why it happens: For equal sides the inequality reduces to "s is less than twice s",
.c
which no positive number can break. That is why you can draw an equilateral
s e m
m a
. co
triangle of 2 mm or of 2 metres with exactly the same two-arc construction.
e m agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q3 For each of the following, give at least 5 possible values for the third length so there
exists a triangle having these as sidelengths (decimal values could also be chosen):
(a) 1, 100 (b) 5, 5 (c) 3, 7

The third length must be more than the difference and less than the sum of the two given
lengths.

GIVEN THIRD LENGTH MUST LIE BETWEEN FIVE POSSIBLE VALUES

(a) 1, 100 100 – 1 = 99 and 100 + 1 = 101 99.5, 100, 100.2, 100.7, 99.8

(b) 5, 5 0 and 10 1, 3, 4.9, 5, 9

(c) 3, 7 7 – 3 = 4 and 7 + 3 = 10 4.5, 5, 6.4, 7, 9

Check for (c), third length 5:

7<3+5=8✓

5 < 3 + 7 = 10 ✓

3 < 5 + 7 = 12 ✓

Why it happens: Two conditions act at once. The new length must not be so long
that it beats the other two together, and it must not be so short that the longer given
side beats it plus the shorter one. Together they trap it strictly between the
difference and the sum.

Q4 See if you can describe all possible lengths of the third side in each case, so that a
triangle exists with those sidelengths. For example, in case (a), all numbers strictly
between 99 and 101 would be possible.

In every case the answer is an open range: difference < third length < sum.

Page 30 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

GIVEN PAIR DIFFERENCE SUM ALL POSSIBLE THIRD LENGTHS

(a) 1, 100 99 101 strictly between 99 and 101

(b) 5, 5 0 10 strictly between 0 and 10

(c) 3, 7 4 10 strictly between 4 and 10

If the two given lengths are p and q with p ≤ q, the third length x must satisfy

q–p<x<q+p

Why it happens: x < p + q keeps x from being the over-long side; x > q – p (which is
the same as q < x + p) keeps q from being the over-long side. The end values are
excluded because equality flattens the triangle into a line.

Try This: For 5 and 5 the range starts at 0, because two equal sides can hold a third
side as tiny as you like — the triangle just becomes very flat.

In-text Questions — Page 160
7.3 Construction of Triangles When Some Sides and Angles are Given — Two Sides and the Included Angle

Q1 How do we construct a triangle if two sides and the angle included between them
are given?

Draw one given side as the base, build the given angle at one end, and cut off the second side
along the new arm.

1. Draw the first given side as the base.
2. At the end where the angle sits, use a protractor to draw the second arm of the angle.
3. Along that arm, mark off the second given length.
4. Join the marked point to the far end of the base.

Tip: The included angle is the angle between the two given sides — both sides are
arms of that angle. Measurements are usually written in the order side, angle, side,
as in 4 cm, 60°, 3 cm.

Page 31 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q2 Construct a triangle ABC with AB = 5 cm, AC = 4 cm and ∠A = 45°.

Take AB as the base, since the given angle sits at A.

1. Step 1: Construct a side AB of length 5 cm.
2. Step 2: Construct ∠A = 45° by drawing the other arm of the angle.
3. Step 3: Mark the point C on the other arm such that AC = 4 cm.
4. Step 4: Join BC to get the required triangle.

C

4 cm

45°

A 5 cm B

Base AB = 5 cm, ∠A = 45° drawn with a protractor, AC = 4 cm cut off along the new arm.

Why it happens: Once the angle at A is fixed, the whole arm AC is fixed as a
direction. Marking 4 cm along it fixes C exactly. There is no choice left, so this
construction gives exactly one triangle.

Figure it Out — Page 161

Page 32 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Two Sides and the Included Angle

MATH TALK

Q1 Construct triangles for the following measurements where the angle is included
between the sides: (a) 3 cm, 75°, 7 cm (b) 6 cm, 25°, 3 cm (c) 3 cm, 120°, 8 cm

In each case draw the longer given side as the base, make the angle at one end, and mark the
other side along the new arm.

SET BASE TO ANGLE AT THAT SECOND SIDE ALONG SHAPE YOU
DRAW END THE ARM GET

(a) 3 cm, 75°, 7 cm 75° 3 cm Scalene, acute-
7 cm angled

(b) 6 cm, 25°, 6 cm 25° 3 cm Scalene, thin
3 cm

(c) 3 cm, 120°, 8 cm 120° 3 cm Scalene, obtuse-
8 cm angled

For (a), for example:

1. Draw PQ = 7 cm.
2. At P draw an arm making 75° with PQ.
3. Mark R on that arm with PR = 3 cm.
4. Join QR. ∆PQR is the required triangle.

Check it yourself: Each set gives one and only one triangle. Two classmates who
draw (c) carefully will get triangles that fit exactly on top of each other.

Q2 We have seen that triangles do not exist for all sets of sidelengths. Is there a
combination of measurements in the case of two sides and the included angle
where a triangle is not possible? Justify your answer using what you observe during
construction.

Yes — but only when the given angle is not a genuine angle of a triangle, that is when it is 180°
or more.

Page 33 of 67

ase
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
m.
MEASUREMENTS WHAT HAPPENS ON PAPER TRIANGLE?

m as e
.co
3 cm, 180°, 7 cm
a g l
The two arms lie in one straight line; the third vertex falls on the No

a s em base

l
a4gcm, 210°, 6 cm The angle cannot even be drawn as an interior angle No

m
.co ag
4 cm, 179°, 6 cm An extremely flat but genuine triangle Yes

a sem
agl
Why it happens: As long as the included angle is more than 0° and less than 180°,
the two arms point in different directions, so their far ends can always be joined and

. c om
a triangle appears — the two sidelengths never matter. Only at 180° do the arms fall

s em here it is the
into one line and the triangle collapses. So, unlike the three-sides case,
m a
. co that decides.
angle alone
a gl
a s em
a gl Tip: Compare this with the triangle inequality. There the lengths could fail; here they

m a s
agl
never do.

m .co
l a se
a g
In-text Questions — Page 161
co m
.
Two Angles and the Included Side

se m
o m l a
g∠B = 80°.
Q1 .cConstruct a triangle ABC where AB = 5 cm, ∠A = 45° and
a
e m
las
ag ANSWER
se m
com
Take the given side AB as the base and grow both angles from it.
g l a
m . a
ase
agl
1. Step 1: Draw the base AB of length 5 cm.
2. Step 2: Draw ∠A and ∠B of measures 45° and 80° respectively, on the same side of AB.
3. Step 3: The point of intersection of the two new line segments is the third vertex C.

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

com
m .
m ase
.co


a g l Page 34 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

C

45° 80°

A 5 cm B

The two arms drawn at 45° and 80° meet at C, the third vertex.

Third angle ∠C = 180° – (45° + 80°)

= 180° – 125°

= 55°

Why it happens: The two arms lean towards each other because 45° + 80° = 125° is
less than 180°. Leaning arms are bound to meet, and where they meet is the only
possible third vertex — so again there is exactly one triangle.

Figure it Out — Page 162
Two Angles and the Included Side

Q1 Construct triangles for the following measurements: (a) 75°, 5 cm, 75° (b) 25°, 3 cm,
60° (c) 120°, 6 cm, 30°

Draw the given side as the base and set the two given angles at its two ends.

Page 35 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

SET BASE ANGLES AT SUM OF THE THIRD TYPE
THE ENDS TWO ANGLES ANGLE

(a) 75°, 5 5 cm 75° and 75° 150° 30° Isosceles, acute-
cm, 75° angled

(b) 25°, 3 3 cm 25° and 60° 85° 95° Scalene, obtuse-
cm, 60° angled

(c) 120°, 6 6 cm 120° and 30° 150° 30° Isosceles,
cm, 30° obtuse-angled

Third angle = 180° – (sum of the two given angles)

(a) 180° – 150° = 30°
(b) 180° – 85° = 95°

(c) 180° – 150° = 30°

Why it happens: Each sum is less than 180°, so in every case the two arms lean
towards each other and must meet. In (c) the third angle equals one of the given
angles (30°), which is why that triangle turns out isosceles.

Check it yourself: Measure the third angle with a protractor after construction and
compare it with the value in the table.

In-text Questions — Pages 162–163
Do triangles always exist?

MATH TALK

Q1 Do triangles exist for every combination of two angles and their included side?
Explore.

No. If the two given angles are too big, the two arms never meet.

Page 36 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

TWO ANGLES SUM WHAT THE ARMS DO TRIANGLE?

45°, 80° 125° Lean towards each other Yes

90°, 90° 180° Both go straight up — parallel No

100°, 95° 195° Lean away from each other No

Why it happens: Two arms meet only if together they turn inwards. If both angles
are 90° or more, each arm is already vertical or leaning outwards, so they can only
drift apart. Just as three lengths must pass a test, two angles must pass one too.

Q2 Find examples of measurements of two angles with the included side where a
triangle is not possible.

Any pair whose sum reaches 180° will do, whatever the length of the included side.

MEASUREMENTS SUM OF THE TWO ANGLES TRIANGLE?

90°, 5 cm, 90° 180° No — the arms are parallel

110°, 4 cm, 95° 205° No

40°, 6 cm, 150° 190° No

120°, 7 cm, 60° 180° No

Tip: If the two angles are each greater than or equal to a right angle (90°), a triangle
is clearly not possible.

Page 37 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q3 Now we make one of the base angles an acute angle, say 40°. What are the possible
values that the other angle should take so that the lines don’t meet? (a) Try to find
a possible ∠B (marked in the figure) for this to happen. (b) What could be smallest
value of ∠B for the lines to not meet?

C

l

40°

A B

Page 162 — the base AB with the ray l drawn from A at 40°.

With ∠A = 40°, the arm from B must bend far enough to the right.
(a) ∠B = 150° works — so do 145°, 160°, 170°. With any of these the line from B never meets the
line l drawn from A.
(b) The smallest such value is ∠B = 140°.

Page 38 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
em.
m l as
m .co a g meets
l a se l
a g
l
co m
em . ag
g l as m (∠B
a
40°
co m
e m.
m l aBs
.co g
A
m a
l a se
ag The line m through B parallel to l makes the smallest ∠B that still avoids l.

a s
. coBmtips to the right, the later it would cross l — agl
Why it happens: The more the arm at
s m
eparallel
g l a
a
until at one special tilt it becomes to l and never crosses at all. That parallel
position gives the smallest ∠B that fails, and every larger ∠B fails too.

co m
m .
m l a se that AB is the
Q4 .co
em transversal.]
Can you tell the actual value of
ag
∠B be in this case? [Hint: Note

l a s
ag
se m
com g l a
∠B = 140°.
m . a
ase
agl
Line m through B is parallel to line l through A

AB is the transversal
c o m
∠A and ∠B are interior angles on the same side of the transversalem
.
c o m g l as
m.So ∠A + ∠B = 180° a
l a se
ag 40° + ∠B = 180°
.c
∠B = 140° s e m
m a
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 39 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: This is the co-interior (allied) angle property of parallel lines learnt
in the chapter on lines and angles. Because the borderline case is exactly the parallel
case, the borderline value of ∠B is exactly 180° minus the other angle.

Q5 So, for what values of ∠B, does a triangle not exist? Does the length AB play any
part here?

A triangle does not exist when ∠B ≥ 140°, and the length AB plays no part at all.

∠A = 40°

Triangle exists when ∠A + ∠B < 180°, i.e. ∠B < 140°

Triangle fails when ∠B ≥ 140°

In general, for base angles ∠A and ∠B:

∠A + ∠B < 180° → triangle exists

∠A + ∠B ≥ 180° → triangle does not exist

Why it happens: Making AB longer or shorter only slides the arm at B sideways; it
does not change the direction the arm points in. Two lines meet or fail to meet
purely because of their directions, so only the angles decide.

Figure it Out — Page 163
Do triangles always exist?

Q1 For each of the following angles, find another angle for which a triangle is (a)
possible, (b) not possible. Find at least two different angles for each category: (a)
30° (b) 70° (c) 54° (d) 144°

The other angle must be less than 180° minus the given angle for a triangle to be possible.

Page 40 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

GIVEN LIMIT (180° – TRIANGLE POSSIBLE — TWO TRIANGLE NOT
ANGLE GIVEN) CHOICES (THIRD ANGLE) POSSIBLE — TWO
CHOICES

(a) 30° 150° 50° (third 100°), 90° (third 60°) 150°, 170°

(b) 70° 110° 60° (third 50°), 80° (third 30°) 120°, 140°

(c) 54° 126° 90° (third 36°), 64° (third 62°) 134°, 154°

(d) 144° 36° 20° (third 16°), 35° (third 1°) 36°, 90°

Sample check, (b) with 60°:

70° + 60° = 130° < 180° ✓

Third angle = 180° – 130° = 50°

Why it happens: The two chosen angles must leave something over for the third
angle, so their sum has to stay strictly under 180°. Angles at or above the limit leave
nothing (or less than nothing) for the third angle.

Tip: In (d), because 144° is already large, the partner angle has very little room —
anything from just above 0° up to just below 36°.

Q2 Determine which of the following pairs can be the angles of a triangle and which
cannot: (a) 35°, 150° (b) 70°, 30° (c) 90°, 85° (d) 50°, 150°

Add the pair and compare with 180°.

PAIR SUM ANGLES OF A TRIANGLE? THIRD ANGLE

(a) 35°, 150° 185° Cannot —

(b) 70°, 30° 100° Can 80°

(c) 90°, 85° 175° Can 5°

(d) 50°, 150° 200° Cannot —

Page 41 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: In (c) the sum 175° squeezes past 180° with only 5° to spare, so the
triangle is real but very thin and sharp. In (a) and (d) the sum already exceeds 180°,
leaving a negative amount for the third angle — impossible.

Q3 Like the triangle inequality, can you form a rule that describes the two angles for
which a triangle is possible? Can the sum of the two angles be used for framing this
rule?

Yes. The rule uses exactly the sum of the two angles.

For two angles ∠A and ∠B of a triangle:

0° < ∠A + ∠B < 180°
If the sum is less than 180°, a triangle with these two angles exists.
If the sum is 180° or more, no such triangle exists.

RULE FOR LENGTHS RULE FOR ANGLES

each length < sum of the other two sum of two angles < 180°

triangle inequality angle condition

Why it happens: The third angle is whatever is left over from 180°. A leftover exists
only when the first two angles have not already used up the full 180°. This is the first
hint of the angle sum property which is proved a page later.

In-text Questions — Page 164

Page 42 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Finding the third angle

MATH TALK

Q1 Let us take two angles, say 60° and 70°, whose sum is less than 180°. Let the
included side be 5 cm. What could the measure of the third angle be? Does this
measure change if the base length is changed to some other value, say 7 cm?
Construct and find out.

The third angle is 50°, and it does not change when the base changes.

Third angle = 180° – (60° + 70°)

= 180° – 130°

= 50°

BASE (INCLUDED SIDE) ANGLES AT THE ENDS THIRD ANGLE ON MEASURING

5 cm 60°, 70° 50°

7 cm 60°, 70° 50°

10 cm 60°, 70° 50°

Why it happens: Changing the base only makes the triangle bigger or smaller; the
arms keep the same directions, so the corner they form keeps the same opening.
Size changes, shape does not.

Q2 In general, once the two angles are fixed, does the third angle depend on the
included sidelength? Try with different pairs of angles and lengths.

No. The third angle depends only on the two given angles.

Page 43 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
m.
TWO ANGLES INCLUDED SIDE THIRD ANGLE

m ase
45°, 80°
.co
3 cm 55°
a g l
se m
g l a 8 cm
a
45°, 80° 55°

30°, 30° 4 cm 120°

co m
30°, 30° 9 cm
em . 120° ag
g l as
a
In every row: third angle = 180° – (sum of the two given angles)

co m
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m l as
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Tip: If your measurement differs by a degree or two, it is drawing error, not

se m
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mathematics. Sharpen the pencil and read the protractor from directly above.

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Try experimenting with different triangles to see if there is a relation between any
agl
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two angles and the third one. To find this relation, what data will you keep track of
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Keep a three-column table of the three angles, and add a fourth column for their sum.

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70° 50°
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60° 180°

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45° 80° 180°

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30° 30°

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120° 180°

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60°
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60° 60° 180°
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third angle = 180° – (sum of the other two)
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equivalently, ∠A + ∠B + ∠C = 180°
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a

co m
m .
m ase
.co


a g l Page 44 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Why it happens: Recording only two angles at a time hides the pattern; recording
the sum makes it jump out, because the last column is the same number every time.
Choosing what to tabulate is half the discovery.

Check it yourself: Draw five very different triangles — thin, fat, right-angled, obtuse
— and fill the table. The last column stays 180°.

Q4 Consider a triangle ABC with ∠B = 50° and ∠C = 70°. Let us suppose we construct a
line XY parallel to BC through vertex A. We can see new angles being formed here:
∠XAB, and ∠YAC. What are their values? Can we find ∠BAC from this?

∠XAB = 50°, ∠YAC = 70°, and therefore ∠BAC = 60°.

X A Y
50° 70°
60°

50° 70°

B C

XY is parallel to BC, so the two outer angles at A copy ∠B and ∠C.

Page 45 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

XY ∥ BC, with AB as a transversal → ∠XAB = ∠B = 50° (alternate angles)

XY ∥ BC, with AC as a transversal → ∠YAC = ∠C = 70° (alternate angles)

∠XAB + ∠BAC + ∠YAC = 180° (they form a straight angle at A)

50° + ∠BAC + 70° = 180°

120° + ∠BAC = 180°

Thus ∠BAC = 60°

Why it happens: The parallel line carries the two base angles up to the vertex A,
where they lie side by side with ∠BAC along a straight line. A straight angle is 180°,
so the three of them must add to 180° — and that is exactly the angle sum property.

Figure it Out — Page 165
Finding the third angle

TRY THIS

Q1 Find the third angle of a triangle (using a parallel line) when two of the angles are:
(a) 36°, 72° (b) 150°, 15° (c) 90°, 30° (d) 75°, 45°

Draw XY through A parallel to BC. The two base angles reappear at A as alternate angles, so the
three angles at A fill a straight angle.

∠XAB + ∠BAC + ∠YAC = 180°

∠B + ∠BAC + ∠C = 180°

Third angle = 180° – (sum of the other two)

Page 46 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

TWO ANGLES SUM THIRD ANGLE

(a) 36°, 72° 108° 72°

(b) 150°, 15° 165° 15°

(c) 90°, 30° 120° 60°

(d) 75°, 45° 120° 60°

Worked out fully for (a), as the book does:

∠XAB = ∠B = 36° and ∠YAC = ∠C = 72° (alternate angles)

∠XAB + ∠BAC + ∠YAC = 180°

36° + ∠BAC + 72° = 180°

108° + ∠BAC = 180°

∠BAC = 72°

Tip: (a) and (b) each give an isosceles triangle, because the third angle equals one of
the given ones.

Q2 Can you construct a triangle all of whose angles are equal to 70°? If two of the
angles are 70° what would the third angle be? If all the angles in a triangle have to
be equal, then what must its measure be? Explore and find out.

No, such a triangle is impossible.

70° + 70° + 70° = 210°

210° ≠ 180° → no such triangle

If two of the angles are 70°:

Page 47 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Third angle = 180° – (70° + 70°)

= 180° – 140°

= 40°

If all three angles are equal:

Let each angle be x

x + x + x = 180°

3x = 180°

x = 60°

Why it happens: The three angles have a fixed budget of 180° to share. Three
angles of 70° overspend it by 30°. Sharing the budget equally gives 180° ÷ 3 = 60°
each — which is exactly why every equilateral triangle has three 60° angles.

Q3 Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C?

∠B = ∠C = 65°.

Page 48 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
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co m
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m ase
.co


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Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

In-text Questions — Pages 165–167
Angle Sum Property · Exterior Angles

Q1 What can we say about the sum of the angles of any triangle? (Consider a triangle
ABC and construct a line through A that is parallel to BC.)

The sum is always 180° — for every triangle, big or small, thin or fat.

X A Y
∠XAB ∠YAC
∠A

∠B ∠C

B C

The straight angle at A is split into ∠XAB, ∠A and ∠YAC — copies of ∠B, ∠A and ∠C.

XY ∥ BC → ∠B = ∠XAB and ∠C = ∠YAC (alternate angles)

So ∠A + ∠B + ∠C = ∠A + ∠XAB + ∠YAC

= 180°, as together they form a straight angle

This result is called the angle sum property of triangles.

Why it happens: The single clever step is drawing a line through the top vertex
parallel to the base. It moves ∠B and ∠C up beside ∠A, where all three lie along one
straight line. This idea appears in The Elements, the famous book of the Greek
mathematician Euclid, who lived around 300 BCE.

Page 50 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q2 There is a convenient way of verifying the angle sum property by folding a
triangular cut-out of a paper. Do you see how this shows that the sum of the angles
in this triangle is 180°?

Yes — fold the three corners inwards and they line up along the base.

1. Cut out any paper triangle ABC and mark its three angles with different colours.
2. Fold the top vertex A down so that it lands exactly on the base BC.
3. Fold the corner at B inwards until it touches the same point, and do the same with the corner
at C.
4. The three coloured corners now sit side by side, with no gap and no overlap, along the
straight edge BC.

Three angles fitted along a straight line

Straight angle = 180°

So ∠A + ∠B + ∠C = 180°

Why it happens: The first fold is really the parallel line of the proof — the crease is
parallel to BC. Folding turns the printed proof into something you can hold in your
hand, which is why it convinces so quickly.

Try This: Repeat with a very obtuse triangle and with a right-angled one. The
corners still fit exactly along the straight edge.

Q3 Find ∠ACD, if ∠A = 50°, and ∠B = 60°.

∠ACD = 110°.

Page 51 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

A

50°

110°
60°

B C D
exterior angle
BC is extended to D; ∠ACD is the exterior angle at C.

From the angle sum property,

50° + 60° + ∠ACB = 180°

110° + ∠ACB = 180°

So ∠ACB = 70°

∠ACB and ∠ACD together form a straight angle

∠ACD = 180° – 70°

= 110°

Tip: An exterior angle is the angle between the extension of one side and the other
side meeting it. Every triangle has six of them, two at each vertex.

Q4 Find the exterior angle for different measures of ∠A and ∠B. Do you see any
relation between the exterior angle and these two angles? [Hint: From angle sum
property, we have ∠A + ∠B + ∠ACB = 180°.] We also have ∠ACD + ∠ACB = 180°, since
they form a straight angle. What does this show?

The exterior angle always equals the sum of the two opposite interior angles.

Page 52 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

∠A ∠B ∠ ACB ∠ ACD ∠A + ∠B
50° 60° 70° 110° 110°

40° 80° 60° 120° 120°

90° 30° 60° 120° 120°

35° 35° 110° 70° 70°

The two hints prove it in one line:

∠A + ∠B + ∠ACB = 180° … (angle sum property)

∠ACD + ∠ACB = 180° … (straight angle)

Both right sides are 180°, so the left sides are equal:

∠A + ∠B + ∠ACB = ∠ACD + ∠ACB

Take away ∠ACB from both sides:

∠ACD = ∠A + ∠B

Why it happens: Both expressions describe the same 180°, once through the
triangle and once along the straight line. The angle ∠ACB is shared by both, so
cancelling it leaves the exterior angle property: an exterior angle of a triangle equals
the sum of its two remote interior angles.

In-text Questions — Pages 167–169
7.4 Constructions Related to Altitudes of Triangles

Q1 Consider a triangle ABC. What is the height of the vertex A from its opposite side
BC, and how can it be measured?

Drop a perpendicular from A to BC and measure it. That perpendicular is the height.

Page 53 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
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Let AD ⟂ BC, with D on BC
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Length of AD = height of A from BC
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The segment AD is called an altitude of the triangle
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The other two altitudes are BE and CF — the perpendiculars from B and from C to their own
opposite sides.
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Why it happens: "Height" always means the shortest distance, and the shortest
distance from a point to a line is measured along the perpendicular. Any slanting
co m
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segment from A to BC would be longer than AD, so it would not describe how high A

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really is.
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whichever side is being taken as the base.
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BC is extended beyond B, and the perpendicular from A meets that extension at D.
a

com
m .
m ase
.co


a g l Page 54 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Extend BC beyond B

Drop the perpendicular from A to this extended line

Foot of the perpendicular = D

AD is the altitude from A to BC

Why it happens: The angle at B is obtuse, so A leans out past the end of the base.
The line containing BC still runs under A — only the segment BC does not. Altitudes
are measured to the line of the base, so the line is extended to receive the
perpendicular.

Q3 Cut out a paper triangle. Fix one of the sides as the base. Fold it in such a way that
the resulting crease is an altitude from the top vertex to the base. Justify why the
crease formed should be perpendicular to the base.

Fold so that the two halves of the base lie exactly on each other, with the crease passing through
the top vertex.

1. Cut out ∆ABC and take BC as the base.
2. Fold the paper so the crease passes through A and part of BC falls exactly on the rest of BC.
3. Open it out. The crease from A to BC is the altitude.

Why the crease is perpendicular:

Folding places the base line on itself

The two angles that the crease makes with BC land exactly on each other

So the two angles are equal

They are also on a straight line, so together they make 180°
Each angle = 180° ÷ 2 = 90°

Why it happens: A fold is a mirror. If one part of a line reflects onto the other part,
the crease must be its mirror line, and a mirror line always meets what it reflects at a
right angle.

Page 55 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

Q4 Construct an arbitrary triangle. Label the vertices A, B, C taking BC to be the base.
Construct the altitude from A to BC. Can you see how to do this?

Use a set square sliding along a ruler — a ruler alone cannot give an accurate 90°.

1. Step 1: Keep the ruler aligned to the base BC. Place the set square on the ruler so that one of
the edges of its right angle touches the ruler.
2. Step 2: Slide the set square along the ruler till the vertical edge of the set square touches the
vertex A.
3. Step 3: Draw the altitude to BC through A using the vertical edge of the set square.

Why it happens: The ruler holds the direction of BC fixed. The set square carries a
built-in right angle, so while it slides its vertical edge stays exactly perpendicular to
BC. Sliding only moves the edge sideways until it reaches A — the direction never
changes.

Tip: If the foot of the perpendicular falls outside the segment BC, simply extend BC
with the ruler first and then slide.

Q5 Does there exist a triangle in which a side is also an altitude? Visualise such a
triangle and draw a rough diagram.

Yes — in a right-angled triangle. If ∠B = 90°, then AB is itself the altitude from A to BC.

Page 56 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

A

altitude
= side AB

B C

With ∠B = 90°, the side AB is already perpendicular to BC, so it is the altitude from A.

∠B = 90° → AB ⟂ BC

So the altitude from A to BC is AB itself

Similarly, CB is the altitude from C to AB

Triangles having one right angle are called right-angled triangles, or simply right triangles.

Why it happens: An altitude is just a perpendicular from a vertex to the opposite
side. In a right-angled triangle the two arms of the right angle are already
perpendicular to each other, so two of the three altitudes are sides of the triangle.
Only the third altitude, from the right-angle vertex to the longest side, has to be
drawn.

In-text Questions — Page 170

Page 57 of 67

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines AglaSem · NCERT Solutions

7.5 Types of Triangles

MATH TALK

Q1 Did you spot any other type of triangle?

Yes. Besides equilateral, isosceles, scalene and right-angled triangles, the chapter also throws
up acute-angled and obtuse-angled triangles.

WHERE IT TURNED UP TRIANGLE TYPE

Construct, page 150 3, 4, 5 Scalene, right-angled

Construct, page 150 4, 4, 6 Isosceles, obtuse-angled

Figure it Out, page 161 3 cm, 120°, 8 cm Scalene, obtuse-angled

Figure it Out, page 162 75°, 5 cm, 75° Isosceles, acute-angled

Tip: Sorting by sides gives equilateral / isosceles / scalene. Sorting by angles gives
acute-angled / right-angled / obtuse-angled. Every triangle has one label from each
list.

Q2 What are the other types of triangles based on angle measures?

Three types altogether — acute-angled, right-angled and obtuse-angled.

TYPE CONDITION ON THE ANGLES EXAMPLE

Acute-angled All three angles less than 90° 60°, 60°, 60°

Right-angled One angle exactly 90° 90°, 60°, 30°

Obtuse-angled One angle greater than 90° 120°, 30°, 30°

Why it happens: A triangle can hold at most one angle of 90° or more, because two
such angles would already use up 180° and leave nothing for the third. So looking at
the biggest angle alone is enough to name the type.

Page 58 of 67

as e
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines
a g l AglaSem · NCERT Solutions

co m
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What could an acute-angled triangle be? Can we define it as a triangle with one
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Q3

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acute angle? Why not?

a g
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An acute-angled triangle is one in which all three angles are acute. Defining it as "a triangle

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with one acute angle" would be useless.
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Right-angled example: 90°, 60°, 30° — has two acute angles

Obtuse-angled example: 120°, 40°, 20° — has two acute angles

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So "having an acute angle" does not single out anything

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a s emit happens: Every triangle already has at least two acute angles, because at
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a most one angle can be 90° or more. A definition must separate one type from the

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that all three angles be acute is what makes the definition useful.

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sem it Out — Pages 170–171
la Types of Triangles
Figure
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Q1 Construct a triangle ABC with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude
from A to BC.

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First check the triangle inequality, then build it and drop the perpendicular.

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Longest side = 6 cm

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6 < 5 + 5 = 10 ✓ → the triangle exists

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1. Draw BC = 5 cm.
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2. With B as centre, draw an arc of radius 6 cm (this is AB).

co m
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m ase
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a g l Page 59 of 67