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NCERT Solutions for Class 7th Maths Chapter 6 Angle Bisection for a Design — Figure it Out

Book page 144 – 145 Updated on2026-09-19

Q1.
Construct at least 4 different angles. Draw their bisectors.
Answer

Use the same three steps for every angle — the method does not care how big the angle is.

  1. Draw ∠AOB of any size (try one acute, one right, one obtuse, one reflex-looking wide angle).
  2. With centre O and one radius, cut both arms at P and Q, so OP = OQ.
  3. With one equal radius, cut arcs from P and from Q meeting at R.
  4. Join OR — it is the bisector.
Angle you drawEach halfCheck
70°35°35 + 35 = 70 ✓
90°45°45 + 45 = 90 ✓
120°60°60 + 60 = 120 ✓
150°75°75 + 75 = 150 ✓
Why it happens: the construction proves ΔOPR ≅ ΔOQR by SSS every single time, so the two halves are equal every single time. Measure with a protractor afterwards only to check your drawing, never to make it.
Check it yourself: after bisecting, put the compass point on R and check that RP = RQ. If they differ, one arc slipped.
Q2.
Construct the 8-petalled figure shown in Fig. 6.5.
Fig. 6.5, page 143.
Answer

Two right angles bisected twice give the eight rays; then eight identical petals.

  1. Mark O on a line and construct the perpendicular at O. Four rays, 90° apart.
  2. Bisect each right angle. Eight rays, 45° apart.
  3. With centre O and one fixed radius, cut all eight rays at P₁ … P₈, so OP₁ = OP₂ = … = OP₈.
  4. For OP₁: construct its perpendicular bisector, mark two centres C and D on it at equal distances from OP₁, and draw the arcs C→(O to P₁) and D→(O to P₁). That is one petal.
  5. Repeat with the same compass settings for the other seven rays.
  6. Ink the sixteen arcs; rub out the rays and the supporting bisectors.
Number of petals = 8
Angle between neighbouring petals = 360° ÷ 8 = 45°
Check: 8 × 45° = 360° ✓
Why it happens: equal rays plus equal petals give the figure its 8-fold rotational symmetry — turn it by 45° and it looks unchanged. The symmetry is a consequence of the construction, not of careful eyeballing.
Q3.
In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.
XBOACY
The construction shown on page 144 — the equal arcs from A and B are drawn on the far side of O and meet at C.
Answer

Yes. The line OC is still the bisector — but it now bisects the angle on the other side of O, and the bisecting ray of ∠AOB is the opposite ray of OC.

A and B are still marked with OA = OB
C is now on the far side, with CA = CB (equal radii again)

In ΔOAC and ΔOBC: OA = OB, AC = BC, OC common
⇒ ΔOAC ≅ ΔOBC by SSS
⇒ ∠AOC = ∠BOC

So the line OC is a line of symmetry for the two arms.
∠AOC = ∠BOC are the two angles outside ∠AOB,
and the ray opposite to OC divides ∠AOB into two equal parts.
Why it happens: the congruence only knows that O and C are both equidistant from A and B — so OC is the perpendicular bisector of the chord AB. A perpendicular bisector is a full line, and a full line through O splits the plane into two pairs of equal angles. Whichever side C falls on, the same line appears; only the ray you name changes.
Check it yourself: extend OC backwards through O with a ruler. The extension falls exactly inside ∠AOB and cuts it in half.
Q4.
What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle?
Answer

From 90° and 60°, repeated bisection (and adding or subtracting the pieces) gives a long list of angles — but 65.5° is not one of them.

Start fromBisect onceTwiceThree times
90°45°22.5°11.25°
60°30°15°7.5°
180°90°45°22.5°
Adding and subtracting these gives many more:
60° + 15° = 75°  ·  90° + 15° = 105°  ·  45° + 22.5° = 67.5°
60° − 7.5° = 52.5°  ·  30° + 7.5° = 37.5°  ·  90° + 60° = 150°

65.5°? Every angle built this way is a whole number of steps of 15°, or 7.5°, or 3.75°, … (each step being 15° halved again and again).
65.5 ÷ 15 = 4.366…   65.5 ÷ 7.5 = 8.733…   65.5 ÷ 3.75 = 17.466…
None of these is a whole number, and halving further never fixes it.
⇒ 65.5° cannot be constructed by bisection.
Why it happens: bisection can only halve. Starting from 90° and 60°, everything you can reach is made of pieces of size 15° cut in half some number of times. 65.5° is 65½°, and a half-degree simply never appears among those pieces. The nearest constructible angle is 67.5° (= 45° + 22.5°), which is 2° away.
Tip: if you need 65.5° for a drawing, use a protractor. Ruler-and-compass work is exact, but it cannot reach every angle.
Q5.
Come up with a method to construct the angle bisector using a rope.
Answer

Copy the compass steps, using the rope as both compass and ruler.

  1. Let the angle be ∠AOB with a peg at the corner O and the two arms marked on the ground.
  2. Take a rope and knot it at a fixed length r. Hold one end at O and stretch it along one arm — peg the far end as P. Stretch the same rope along the other arm — peg the far end as Q. Now OP = OQ = r.
  3. Take a second rope of some length s. Hold one end at P and swing the other end to sweep an arc. Do the same from Q. Peg the crossing point as R. Now PR = QR = s.
  4. Stretch a rope from O to R and mark the line. OR bisects ∠AOB.
In ΔOPR and ΔOQR:
  OP = OQ (= r), PR = QR (= s), OR common
  ⇒ ΔOPR ≅ ΔOQR by SSS
  ⇒ ∠POR = ∠QOR
Why it happens: a taut rope of fixed length does exactly what a compass does — it marks all points at one distance from a fixed peg. Since the proof of the bisector uses only equal lengths, it works just as well with rope on the ground as with a compass on paper. This is how large layouts, such as fire altars, were set out.
Q6.
Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square?
The four-petal design, page 145.
Answer

The four petals are largest when each arc is a semicircle drawn on a side of the square as diameter.

  1. Construct the square ABCD (four right angles, four equal sides).
  2. Construct the perpendicular bisector of each side. This marks the midpoints P, Q, R, S of the four sides — and the four bisectors meet at the centre O of the square.
  3. With centre P (midpoint of AB) and radius PA, draw the semicircle inside the square. It runs from A through O to B.
  4. Repeat from Q, R and S with the same radius. The four semicircles cut each other and four petals appear.
P S Q R O
Four semicircles on the four sides as diameters. Each passes through two corners and through the centre O, and the overlaps are the four petals.
Let the side of the square be s.
Radius of each arc = s ÷ 2
Distance from P (midpoint of AB) to A = s ÷ 2 ✓
Distance from P to the centre O = s ÷ 2 ✓
So the semicircle from P passes through A, O and B.
Why it happens: the arc must stay inside the square. Its centre is on a side and it has to reach the two nearest corners, which are s ÷ 2 away. Any larger radius spills outside the square; any smaller radius leaves the arc short of the corners and the petals shrink. So radius = s ÷ 2 is the largest that fits — the petals are then as big as they can be, with their four tips exactly at the four corners.
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