NCERT Solutions for Class 7th Maths Chapter 6 Regular Hexagon and Equilateral Triangles · Construction of a 60° angle — In-text Questions
Book page 152 – 153 Updated on2026-09-19
Q1.
Consider this figure. Will the 70° angle fit into the gap? What is the gap angle ∠AOI? We have, 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°. Use this to determine whether the 70° angle fits the gap.
The figure on page 152 — six angles placed edge to edge around the point O, with the 70° angle waiting below.
Answer
The gap angle is 50°, so the 70° angle does not fit — it is 20° too wide.
Complete angle around a point = 360°
Gap angle ∠AOI = 360° − 310° = 50°
70° > 50°, so the 70° angle overlaps the neighbouring pieces.
It is 70° − 50° = 20° too big.
Six angles use up 310° of the turn around O. Only 50° is left, so a 70° piece cannot be dropped in.
Why it happens: the pieces sit around a single point O, and one full turn about a point is 360°. Whatever is not used up by the six pieces is the gap. Since the gap is smaller than the piece we want to insert, the piece would have to ride over a neighbour — that is an overlap, not a tiling.
Check it yourself: 40 + 60 + 50 + 30 + 40 + 90 + 50 = 360 ✓. The right piece for this gap is a 50° angle.
Q2.
In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines?
Fig. 6.12, page 151.
Answer
Because the two angles at O on each side of the line add up to a straight angle, 180°.
Take AOD. At O, the angles on one side of it are
∠AOB + ∠BOC + ∠COD = 60° + 60° + 60° = 180°
Angles that add to 180° at a point on one side of a ray form a straight angle
⇒ OA and OD are opposite rays ⇒ A, O, D are collinear
Same for BOE: ∠BOC + ∠COD + ∠DOE = 60° + 60° + 60° = 180° ✓
Same for COF: ∠COD + ∠DOE + ∠EOF = 60° + 60° + 60° = 180° ✓
Why it happens: six equal 60° angles fill the 360° around O. Skipping three of them takes you exactly half way round — and half of a full turn is a straight angle. So each vertex has its "opposite" vertex directly across O, and the segment joining them passes straight through the centre.
Tip: this is why the three long diagonals of a regular hexagon are concurrent — they are not three separate diagonals but three straight lines through the centre.
Q3.
Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.
Answer
Six equilateral triangles of side 4 cm, packed around one point.
Mark a point O. With the compass set at 4 cm, draw a circle with centre O.
Mark any point A on the circle.
Without changing the compass opening, put the point at A and cut the circle at B. From B cut at C, from C at D, from D at E, from E at F.
F should land exactly 4 cm from A. Join A–B–C–D–E–F–A with the ruler.
Every step used the same radius, so
OA = OB = OC = OD = OE = OF = 4 cm (radii)
AB = BC = CD = DE = EF = FA = 4 cm (the compass steps)
So ΔOAB has all three sides 4 cm ⇒ equilateral ⇒ ∠AOB = 60°
Six such triangles: 6 × 60° = 360°, the complete turn ✓
Angle of the hexagon at each vertex = 60° + 60° = 120°
Why it happens: the radius of a circle steps exactly six times around its own circumference. That is not a coincidence — each step makes an equilateral triangle with the centre, each such triangle contributes 60° at the centre, and 6 × 60° = 360° closes the circle perfectly. The construction needs no protractor at all.
Check it yourself: if the sixth step misses A, the compass opening slipped. Re-set it to 4 cm and start again from A.
Q4.
How do we do it? (constructing a 120° angle using a ruler and a compass)
Answer
Construct a 60° angle at a point on a line — then the angle on the other side of it is automatically 120°.
Let the line be XAY with A between X and Y.
Construct ∠CAX = 60° on one side.
∠CAX and ∠CAY are angles on a straight line:
∠CAX + ∠CAY = 180°
60° + ∠CAY = 180°
∠CAY = 120°
Why it happens: we get 120° free of charge. Any ray drawn from a point on a straight line creates a pair of angles that add to 180°. Fix one of the pair at 60° and the other is forced to be 120°. Since the hexagon needs 120° at each corner, the 60° construction is all we really have to master.
Tip: 120° can also be built as 60° + 60°: construct one 60° angle, then construct another 60° on top of it using the same arc.
Q5.
How do we construct a 60° angle?
Answer
Build an equilateral triangle. Its angles are 60°, so the corner you make is the angle you want.
Suppose the 60° angle is needed at A on the segment AX. With centre A and any radius r, draw an arc cutting AX at B.
Without changing the radius, put the compass point at B and cut the first arc at C.
Join A to C. Then ∠CAX = 60°.
AB = r (first arc)
AC = r (C is on the first arc)
BC = r (second arc, same radius)
⇒ AB = AC = BC ⇒ ΔABC is equilateral
All angles of an equilateral triangle are equal and add to 180°
⇒ each = 180° ÷ 3 = 60° ⇒ ∠CAX = 60°
Why it happens: one compass opening, used three times, forces three equal sides. Equal sides force equal angles, and three equal angles in a triangle must each be a third of 180°. No measurement enters anywhere — this is why the construction is exact, unlike setting 60° on a protractor.
Q6.
Why is ∠CAX = 60°? Is there an equilateral triangle here?
Answer
Yes — ΔABC is equilateral, and that is exactly why the angle is 60°.
Both arcs were drawn with the same radius r:
AB = r (B is on the arc centred at A)
AC = r (C is on the arc centred at A)
BC = r (C is on the arc centred at B)
Three equal sides ⇒ ΔABC is equilateral
⇒ ∠BAC = ∠ABC = ∠ACB
⇒ 3 × ∠BAC = 180° ⇒ ∠BAC = 60°
B lies on AX, so ∠CAX = ∠CAB = 60°
Why it happens: the point C was chosen to be at distance r from A and at distance r from B. Being on two arcs of the same radius is what makes the third side equal to the other two. The equilateral triangle is hidden in the construction, and the 60° is its corner.
Check it yourself: join B to C with the ruler. Measure all three sides with the compass — they will all match the radius you started with.
Q7.
Construct a regular hexagon of sidelength 5 cm.
Answer
Same method as the 4 cm hexagon, with the compass set to 5 cm.
Draw a circle of radius 5 cm with centre O.
Mark a point A on it.
Keeping the compass at 5 cm, step round the circle: A → B → C → D → E → F. The sixth step returns to A.
Join the six points in order.
Side of the hexagon = radius of the circle = 5 cm
Perimeter = 6 × 5 = 30 cm
Each interior angle = 120°
Sum of interior angles = 6 × 120° = 720° ✓ (matches (6 − 2) × 180° = 720°)
Alternative, without a circle: draw AB = 5 cm; construct a 120° angle at B and cut BC = 5 cm; construct 120° at C and cut CD = 5 cm; carry on. After the sixth side the figure closes.
Why it happens: in a regular hexagon the distance from the centre to a vertex is equal to the side. So the circle you start with already carries the side length in its radius, and stepping the radius around the circle marks the vertices for you.