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NCERT Solutions for Class 7th Maths Chapter 6 Construction of Perpendicular Bisector — Figure it Out

Book page 140 Updated on2026-09-19

Q1.
When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer. [Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector. Hint 2: We can draw the whole line if any two of its points are known.]
Answer

No, it is not necessary. The upper pair of arcs may use one radius and the lower pair a completely different radius. You still get the perpendicular bisector.

Upper arcs, radius r₁ ⇒ crossing point A with AX = AY = r₁
Lower arcs, radius r₂ ⇒ crossing point B with BX = BY = r₂

A is equidistant from X and Y ⇒ A is on the perpendicular bisector
B is equidistant from X and Y ⇒ B is on the perpendicular bisector
Two points determine a line ⇒ AB is the perpendicular bisector
Why it happens: the proof never compared r₁ with r₂. Each point only had to be equally far from X and from Y — its own distance. Once A and B are both on the line, the ruler does the rest. What you must keep equal is the radius within each pair.
Check it yourself: draw the upper arcs with a small radius and the lower arcs with a much bigger one. The join AB still passes through the midpoint at a right angle — the picture just looks lopsided.
Q2.
Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.
Answer

Both pairs may be drawn on the same side — provided the two pairs use different radii, so that they cross at two different points.

Pair 1, radius r₁, above XY ⇒ point A, with AX = AY
Pair 2, radius r₂ (r₂ ≠ r₁), also above XY ⇒ point C, with CX = CY

A and C both lie on the perpendicular bisector
A ≠ C because r₁ ≠ r₂
⇒ the line AC is the perpendicular bisector of XY, extended down through XY
Why it happens: the bisector is a whole line, not just the bit between A and B. Any two of its points fix it. Points above XY are just as good as one above and one below — as long as they are two distinct points. If you used the same radius twice on the same side you would get the same point back, and one point cannot fix a line.
Tip: in practice A and B on opposite sides are placed far apart, so the ruler line is more accurate. Two points close together on one side make small drawing errors grow when you extend the line.
Q3.
While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them? Explore this through construction, and then justify your answer.
Answer

Yes — within one pair the two radii must be equal. This is the one thing you cannot change.

Arc from X with radius r₁, arc from Y with radius r₂, crossing at P
Then PX = r₁ and PY = r₂

If r₁ ≠ r₂ then PX ≠ PY
⇒ P is not equidistant from X and Y
⇒ P does not lie on the perpendicular bisector
Why it happens: the whole method rests on one property — the point must be equally far from X and from Y. Unequal radii destroy exactly that property. The crossing point then slides towards whichever endpoint used the smaller radius, and the line you draw tilts off the midpoint.
Check it yourself: draw an arc of 4 cm from X and 6 cm from Y. Join the two crossing points. Measure where the line meets XY — it will not be the midpoint, and the angle will not be 90°.
Q4.
Recreate this design using only a ruler and compass —
A four-petalled design made of eight arcs
The design to be recreated (page 140) — four petals meeting at one point.
After completing the above design, you can use a colour pencil with a ruler or compass to trace its boundary. This will make the design stand out from the supporting lines and arcs.
Answer

The design is four eyes sharing one centre — one pointing up, one down, one left, one right. Build the two supporting lines first, then draw an eye on each of the four half-lines.

  1. Draw a line and mark a point O on it. Construct the perpendicular to it at O (Q4 method, page 139). You now have two perpendicular lines through O.
  2. With centre O and a fixed radius, cut the four half-lines at P, Q, R and S. So OP = OQ = OR = OS.
  3. Take the segment OP. Construct its perpendicular bisector and mark two points C and D on that bisector, one on each side, at equal distances from the bisector's foot.
  4. With centre C draw the arc from O to P; with centre D draw the arc from O to P. These two arcs make the first petal (an eye with corners O and P).
  5. Repeat exactly the same for OQ, OR and OS, using the same compass opening every time. Four identical petals appear.
  6. Trace only the eight arcs with a colour pencil, then rub out the supporting lines.
Why it happens: each petal is symmetric about its own half-line because its two centres, C and D, are equally far from O and from P — they sit on the perpendicular bisector of OP. Using the same radius for all four petals makes the four petals congruent, so the whole design has the same shape whichever way you turn it by 90°.
Tip: keep the compass at one setting for all eight arcs. The moment you change it, one petal becomes fatter than its neighbour and the four-fold symmetry is lost.
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