NCERT Solutions for Class 7th Maths Chapter 6 Eyes of different shapes · Construction of Perpendicular Bisector — In-text Questions
Book page 139 Updated on2026-09-19
Q1.
How do we get these different shapes? Try!The three eye shapes printed on page 138 — a long thin one, a broader one with rings, and a round one. Each is made only of arcs drawn with a compass.
Answer
Keep the same XY, but choose a different pair of centres on the perpendicular bisector. Pick points C and D with CX = CY = DX = DY, using a bigger or smaller radius than before, and draw the two arcs from C and D.
Bigger radius ⇒ the centres sit far from XY ⇒ arcs are flatter ⇒ a long, narrow eye
Smaller radius ⇒ the centres sit near XY ⇒ arcs are rounder ⇒ a wide, open eye
Why it happens: the eye is drawn by two arcs that both start at X and end at Y. Their shape depends only on how curved they are, and the curvature depends on the radius. A small radius bends the arc a lot; a large radius keeps it almost straight. The corners X and Y never move, so the eye stays symmetric whatever radius you choose.
Try This: draw four eyes on the same XY using four different radii. Colour only the boundaries and rub out the supporting arcs — you get four faces with four different expressions.
Q2.
Will C and D lie on the perpendicular bisector AB?
Answer
Yes. C and D must lie on the line AB.
C is at the same distance from X and Y (CX = CY)
D is at the same distance from X and Y (DX = DY)
By the earlier result, the line CD is a perpendicular bisector of XY
But a segment has only one perpendicular bisector, and that is AB
⇒ line CD is the line AB ⇒ C and D lie on AB
Why it happens: we proved that joining any two points equidistant from X and Y gives the perpendicular bisector. C and D are two such points. Since XY cannot have two different perpendicular bisectors, the line through C and D has to be the very same line AB.
Q3.
Justify the following statement using the facts that we have established. Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.
Answer
Let P be any point with PX = PY. We show P is on the perpendicular bisector AB.
Take a second point Q with QX = QY, on the other side of XY.
By the result already proved, the line PQ is the perpendicular bisector of XY.
XY has exactly one perpendicular bisector, namely AB.
So line PQ = line AB.
P lies on line PQ ⇒ P lies on AB.
Why it happens: a shorter way to see it. Join P to O, the midpoint of XY. In ΔPOX and ΔPOY we have PX = PY, OX = OY and PO common, so ΔPOX ≅ ΔPOY by SSS. Hence ∠POX = ∠POY, and since they add to 180°, each is 90°. So PO is perpendicular to XY at its midpoint — that is the perpendicular bisector, and P is on it.
Did you know? This one sentence is the whole engine of the chapter. Every construction that follows — the 90° angle, the angle bisector, the arch, the hexagon — is built by finding points at equal distances from two fixed points.
Q4.
Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?
Answer
Three steps, no measuring at all.
Open the compass to some fixed radius, more than half of XY. With centre X draw a long arc above XY; with the same radius and centre Y draw another long arc above XY. Call their crossing point A.
With the same radius, draw the matching pair of arcs below XY. Call their crossing point B.
Join A and B with the ruler. AB is the perpendicular bisector of XY.
Step 1 gives A, step 2 gives B, step 3 joins them. O is the midpoint and the angle at O is 90°.
Why it happens: A and B are each at equal distances from X and Y, so both lie on the perpendicular bisector. Two points fix a line, so the line through A and B is that perpendicular bisector.
Tip: this is a better way to find the midpoint of a segment than measuring with a scale. A scale can be read wrong by half a millimetre; the compass method has no reading in it at all.