NCERT Solutions Ganita Prakash (Part 1) Chapter 7 –159When do the two circles intersect? — In-text Questions
Book page 156 Updated on2026-09-05
Q1.
Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure?
Answer
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
Only Case 3 — where the circles cut each other — produces a third vertex.
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?
Answer
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?
Answer
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
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.
Answer
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.
Q5.
Frame a complete procedure that can be used to check the existence of a triangle.
Answer
A three-step routine that never fails.
Step 1: Make sure all three lengths are in the same unit.
Step 2: Pick out the largest length. Add the other two.
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.