NCERT Solutions Ganita Prakash (Part 1) Chapter 7 Triangle Inequality — Figure it Out

Book page 154 Updated on2026-09-05

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.

Answer

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.

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

Answer

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.

Answer

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

a ≤ c, and b is positive, so a < b + c ✓
b ≤ c, and a is positive, so b < a + c ✓
Only c < a + b is in doubt

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

Try This: Pick any three lengths at random and test all three comparisons. You will never manage to make two of them fail.

Was this helpful? Report an error