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NCERT Solutions for Class 7th Maths Chapter 6 Related Constructions · 6-Pointed Star — In-text Questions

Book page 154 Updated on2026-09-19

Q1.
How will you construct 30° and 15° angles?
Answer

Bisect 60° to get 30°, then bisect 30° to get 15°.

60° ÷ 2 = 30°
30° ÷ 2 = 15°
  1. Construct ∠CAX = 60° with the equilateral-triangle method.
  2. With centre A cut both arms at P and Q (AP = AQ). With one equal radius cut arcs from P and Q meeting at R. Join AR — then ∠RAX = 30°.
  3. Bisect ∠RAX the same way: cut the two arms of the 30° angle, cross two equal arcs, join. That ray makes 15° with AX.
Why it happens: each bisection is justified by SSS congruence, so each really does halve the angle — no approximation creeps in. Halving 60° twice gives 15°, and halving again would give 7.5°.
Tip: 45° comes from halving 90°, and 15° from halving 30°. Put them together and you can also build 75° (60° + 15°), 105° (90° + 15°) and 135° (90° + 45°).
Q2.
Construct the following 6-pointed star. Note that it has a rotational symmetry. [Hint: figure with A, B, C, D, E, F outside and G, H, I, J, K, L inside] Do you see a hexagon here?
Answer

Yes — there is a hexagon, in fact two. Start from the outer regular hexagon ABCDEF, then join alternate vertices.

  1. Draw a circle with centre O and mark six points A, B, C, D, E, F on it by stepping the radius round the circle. ABCDEF is a regular hexagon.
  2. Join A to C, C to E, E to A — one equilateral triangle.
  3. Join B to D, D to F, F to B — the second equilateral triangle.
  4. The two triangles overlap in a small regular hexagon GHIJKL, and six points stick out. Ink the star outline and rub out the parts of the triangles inside the small hexagon.
Rotational symmetry: turning the figure about O by 360° ÷ 6 = 60°
carries A to B, B to C, … and leaves the star looking unchanged.
So the star has 6-fold rotational symmetry.
Why it happens: ΔACE joins alternate vertices of a regular hexagon, so its three sides are equal — it is equilateral. So is ΔBDF, and it is just ΔACE turned by 60°. Overlapping the two gives the familiar six-pointed star, and the region common to both is the inner regular hexagon.
Q3.
Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG — equilateral? Why? [Hint: Find the angles.]
Answer

Yes, all six are equilateral. Each has all three angles equal to 60°.

Take ΔAGH.

Angle at A. ∠GAH is the angle of the equilateral triangle ACE at A
  ⇒ ∠GAH = 60°

Angles at G and H. In the regular hexagon ABCDEF, the chords FB and EC
are both perpendicular to the diagonal AD, so FB ∥ EC.
G and H lie on FB; E and C are the other two vertices of ΔACE.
Corresponding angles for the transversal AE give ∠AGH = ∠AEC = 60°
Corresponding angles for the transversal AC give ∠AHG = ∠ACE = 60°

Angles of ΔAGH: 60° + 60° + 60° = 180° ✓
All angles equal ⇒ all sides equal ⇒ ΔAGH is equilateral

By the 60° rotational symmetry of the star, the same argument works for ΔBHI, ΔCIJ, ΔDJK, ΔELK and ΔFLG. So all six points of the star are congruent equilateral triangles.

Why it happens: the star is made of two equilateral triangles crossing each other. Wherever a side of one triangle cuts across a corner of the other, it slices off a small triangle whose angles are copies of the big triangle's 60° angles — cut by parallel lines. A triangle with three 60° angles has to be equilateral.
Check it yourself: the inner hexagon GHIJKL is regular too. Its angle at G is 180° − 60° = 120°, since A, G, L lie on one straight line (the side AE of ΔACE).
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