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NCERT Solutions Ganita Prakash (Part 1) Chapter 7 –160Conclusion — Figure it Out

Book page 159 Updated on2026-09-05

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
Answer

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

SetLargest vs sum of other twoTriangle?
(a) 1, 100, 100100 < 1 + 100 = 101Exists
(b) 3, 6, 99 = 3 + 6Does not exist
(c) 1, 1, 55 > 1 + 1 = 2Does not exist
(d) 5, 10, 1212 < 5 + 10 = 15Exists
Why it happens: In (b) the sum is exactly equal, not less. The two circles just touch, 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, so the triangle is real but extremely thin.
Q2.
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
Answer

Yes to both.

50 < 50 + 50 = 100 ✓
All three comparisons are the same, so the triangle inequality holds
An equilateral triangle of side 50 exists

For any sidelength s (a positive number):

Sum of the other two sides = s + s = 2s
Since s is positive, s < 2s is always true
So an equilateral triangle of any sidelength exists
Why it happens: For equal sides the inequality reduces to "s is less than twice s", which no positive number can break. That is why you can draw an equilateral triangle of 2 mm or of 2 metres with exactly the same two-arc construction.
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
Answer

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

GivenThird length must lie betweenFive possible values
(a) 1, 100100 – 1 = 99 and 100 + 1 = 10199.5, 100, 100.2, 100.7, 99.8
(b) 5, 50 and 101, 3, 4.9, 5, 9
(c) 3, 77 – 3 = 4 and 7 + 3 = 104.5, 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.
Answer

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

Given pairDifferenceSumAll possible third lengths
(a) 1, 10099101strictly between 99 and 101
(b) 5, 5010strictly between 0 and 10
(c) 3, 7410strictly 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.
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