NCERT Solutions for Class 7th Maths Chapter 6 Angle Bisection for a Design — In-text Questions
Book page 143 Updated on2026-09-19
Q1.
How do we construct this figure?
Fig. 6.5, page 143.
Answer
Draw the eight supporting rays first, then hang one petal on each ray.
Draw a line through a point O and construct the perpendicular at O. That gives 4 rays, at 90° to each other.
Bisect each of the four right angles. That gives 4 more rays in between. Now there are 8 rays, at 45° to each other.
With centre O and a fixed radius, cut all eight rays at the same distance from O. Call the eight cut points P₁ … P₈.
On the segment OP₁ construct a petal (an eye): find two centres on the perpendicular bisector of OP₁ and draw the two arcs from O to P₁.
Repeat with the same compass opening on the other seven rays. Trace the sixteen arcs and rub out the rays.
Why it happens: the figure looks even because all eight petals sit on rays that are equally spread. Equal spread means every gap is the same, and eight equal gaps around a point must each be 360° ÷ 8 = 45°. A right angle bisected once gives exactly that.
Tip: do not try to set 45° with a protractor. Bisecting a right angle is both faster and exact.
Q2.
What is the angle between two adjacent lines?
Answer
45°.
The 8 rays split the complete angle around O into 8 equal parts
Complete angle around a point = 360°
Angle between adjacent rays = 360° ÷ 8
= 45°
Why it happens: for the design to look the same after every turn, every gap must be equal. Eight equal gaps must share the full turn of 360° between them, so each gap is one-eighth of 360°.
Check it yourself: 8 × 45° = 360°. ✓ Also, two opposite rays make 4 × 45° = 180°, a straight line — which is exactly what the picture shows.
Q3.
How do we construct a 45° angle using only a ruler and a compass?
Answer
Construct a 90° angle, then bisect it.
Step 1: 90° at a point O on a line (page 141 method)
Step 2: bisect ∠AOX
90° ÷ 2 = 45°
At O construct the perpendicular OA, so ∠AOX = 90°.
With centre O and any radius, cut OA at P and OX at Q. So OP = OQ.
With the same (or any equal) radius, draw arcs from P and from Q crossing at R.
Join OR. Then ∠ROX = 45°.
Why it happens: OP = OQ and RP = RQ, with OR common, so ΔOPR ≅ ΔOQR by SSS. Corresponding angles give ∠POR = ∠QOR, and together they make the 90°. So each is half of it.