NCERT Solutions for Class 7th Maths Chapter 6 In-text Questions — Geometric Constructions
Book page 137 Updated on2026-09-19
Q1.
How do we find such A and B?
Answer
Draw arcs of the same radius from X and from Y — one pair above the line XY and one pair below it. The point where the upper arcs cross is A; the point where the lower arcs cross is B.
A is on the arc from X ⇒ AX = radius
A is on the arc from Y ⇒ AY = same radius
So AX = AY
In the same way BX = BY
Both pairs of arcs used the same radius, so AX = AY = BX = BY
Same radius from X and from Y. The arcs cross at A above XY and at B below it.
Why it happens: the compass keeps the radius fixed. Every point on the arc drawn from X is at that fixed distance from X, and every point on the arc drawn from Y is at the same fixed distance from Y. A crossing point belongs to both arcs, so it is at that one distance from X and from Y. That is exactly the condition AX = AY the eye needs.
Tip: make the radius clearly more than half of XY, otherwise the two arcs will not meet at all.
Q2.
In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the angle formed between them?
Fig. 6.1, page 137.
Answer
AB cuts XY at its midpoint, and the angle between them is 90°.
Let O be the point where AB meets XY.
Then OX = OY — O is the midpoint of XY
and ∠AOX = ∠AOY = 90°
Why it happens: A and B were both built to be equally far from X and from Y. Fold the picture along AB: X lands exactly on Y. A fold line that carries X onto Y must pass through the middle of XY and must stand square to it. A line that cuts a segment into two equal parts and is perpendicular to it is called the perpendicular bisector.
Check it yourself: measure OX and OY with a compass, not a scale — open the compass to OX and swing it across to Y. It should land exactly on Y.
Q3.
Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length?
Fig. 6.1, page 137 — the segment XY with arcs struck from X and from Y, meeting above at A and below at B.
Answer
Yes, always. The only thing we used was AX = AY = BX = BY, and that is true for every length of XY and every radius that lets the arcs meet.
Given: AX = AY, BX = BY, AB common Step 1. In ΔABX and ΔABY:
AX = AY, BX = BY, AB = AB
⇒ ΔABX ≅ ΔABY by SSS
⇒ ∠XAB = ∠YAB, i.e. ∠XAO = ∠YAO
Step 2. In ΔAOX and ΔAOY:
AX = AY, ∠XAO = ∠YAO, AO = AO
⇒ ΔAOX ≅ ΔAOY by SAS
⇒ OX = OY and ∠AOX = ∠AOY
Step 3. ∠AOX and ∠AOY together make a straight angle:
∠AOX + ∠AOY = 180°
2 × ∠AOX = 180°
∠AOX = 90°
Why it happens: notice that no number ever entered the argument — not the length of XY, not the radius of the arcs. Only the four equal distances were used. So the conclusion holds for every segment and every workable radius. That is what makes it a construction and not a lucky drawing.
Q4.
Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)?
Answer
ΔAOX and ΔAOY.
If ΔAOX ≅ ΔAOY, then by corresponding parts OX = OY ⇒ O is the midpoint of XY ∠AOX = ∠AOY
and since ∠AOX + ∠AOY = 180° (straight angle at O),
∠AOX = ∠AOY = 90° ⇒ AB ⊥ XY
Why it happens: "perpendicular bisector" is really two claims at once — equal halves, and a right angle. One congruence delivers both, because congruent triangles hand us equal sides and equal angles in a single step.
Tip: to get ΔAOX ≅ ΔAOY we still need ∠XAO = ∠YAO. That comes from the bigger pair ΔABX ≅ ΔABY (SSS), which uses AB as the common side.