NCERT Solutions for Class 7th Maths Chapter 7 Triangle Inequality — In-text Questions

Book page 152–153 Updated on2026-09-19

Q1.

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

Answer

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)?

Answer

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

The direct path (red) beats the roundabout path through the pole (yellow).

Why it happens: Between two points, the straight line is the shortest possible path. Any detour through a third point adds length. This single everyday fact is the whole idea behind the triangle inequality.

Q3.

Can this understanding be used to tell something about the existence of a triangle having sidelengths 10 cm, 15 cm and 30 cm?

Answer

Yes. Suppose such a ∆ABC existed, with BC = 10 cm, AB = 15 cm and CA = 30 cm. Test all three pairs of vertices.

Between	Direct path	Roundabout path	Direct shorter?
B and C	BC = 10 cm	BA + AC = 15 + 30 = 45 cm	Yes
A and B	AB = 15 cm	AC + CB = 30 + 10 = 40 cm	Yes
C and A	CA = 30 cm	CB + BA = 10 + 15 = 25 cm	No!

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.

Answer

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.

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?

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).

Answer

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:

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.

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