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NCERT Solutions for Class 9th Maths Chapter 5 Think and Reflect — Symmetries of a Circle

Book page 94 Updated on2026-09-08

Q1.
What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
Answer

A regular polygon with n sides has rotational symmetry of order n and exactly n lines of reflection symmetry.

Square (n = 4): rotations of 90°, 180°, 270°, 360° — order 4; 4 lines of symmetry (2 diagonals, 2 through midpoints of opposite sides).
Regular pentagon (n = 5): rotations of 72°, 144°, 216°, 288°, 360° — order 5; 5 lines of symmetry (each joins a vertex to the midpoint of the opposite side).
Regular hexagon (n = 6): rotations of 60°, 120°, 180°, 240°, 300°, 360° — order 6; 6 lines of symmetry (3 long diagonals, 3 through midpoints of opposite sides).
Why it happens: the vertices of a regular n-gon sit on a circle, spaced 360°/n apart. Turning the figure through 360°/n sends each vertex to the next one, so the picture is unchanged. Doing this n times brings you back to the start — so there are exactly n rotations that work.
Did you know? As n grows the polygon looks more and more like a circle, and its symmetries multiply. The circle is the limit: it has every angle as a rotational symmetry and infinitely many lines of reflection symmetry — one for each diameter.
Q2.
What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
Answer

The longest chord is the diameter.

Longest chord = 2 × radius = 2 × 5 = 10 units

Is there a smallest chord? No. There is no chord of least length.

Why it happens: a chord at distance d from the centre has length 2√(25 − d²), and d can be any value with 0 ≤ d < 5. As d creeps up towards 5 the chord gets shorter and shorter — 2√(25 − 24) = 2, then 2√(25 − 24.99) = 0.2, and so on — but it never actually reaches 0, because at d = 5 the line only touches the circle at a single point and is no longer a chord. So chords can be made as short as you please, and there is no shortest one.
Tip: "longest" exists because d = 0 is allowed; "shortest" fails because d = 5 is not.
Q3.
The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points? (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)
Answer

The locus of points equidistant from two given points A and B is the perpendicular bisector of AB.

A locus claim always needs two statements proved, not one:

  1. Every point of the locus is on the line. Let P satisfy PA = PB, and let M be the midpoint of AB. Then in ΔPMA and ΔPMB: PA = PB, AM = BM, PM common. By SSS, ΔPMA ≅ ΔPMB, so ∠PMA = ∠PMB. These are angles on a line, so each is 90°. Hence PM ⊥ AB and passes through the midpoint — P lies on the perpendicular bisector.
  2. Every point of the line is in the locus. Let Q be any point on the perpendicular bisector, meeting AB at its midpoint M. In ΔQMA and ΔQMB: QM common, ∠QMA = ∠QMB = 90°, AM = BM. By SAS, ΔQMA ≅ ΔQMB, so QA = QB.
P is equidistant from A and B ⇔ P lies on the perpendicular bisector of AB
So the locus is exactly the perpendicular bisector of AB.
Why both halves matter: part 1 alone would allow the locus to be some part of the line; part 2 alone would allow extra points off the line. Only together do they pin the locus down exactly. The chapter uses this result immediately: the centres of all circles through A and B are precisely the points of this line.
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