NCERT Solutions for Class 7th Maths Chapter 6 Related Constructions — Figure it Out

Book page 154 – 155 Updated on2026-09-19

Q1.

Construct the following figures: (a) An Inflexed Arc; (b) a flower of six petals around a circle — the fun part about this figure is that it can also be constructed using only a compass! Can you do it?; (c) a regular hexagon inscribed in a circle; (d) a ring of six equal circles; (e) a regular hexagon filled with a triangular star pattern.

(a) An Inflexed Arc

(b) A flower of six petals

(d) A ring of six circles

(e) A hexagon filled with a star pattern

The five figures to construct — pages 154 and 155.

Answer

Every one of these five figures comes out of the same two facts: the radius steps six times round a circle, and 60° comes from an equilateral triangle.

(a) The inflexed arc.

Draw the two vertical sides of the arch and mark the springing points P and Q at the top of them.
Draw a vertical axis midway between them (perpendicular bisector of PQ) and mark the apex T on it.
Draw an arc from P that curves outward and then, from a second centre on the other side, an arc that curves inward up to T. Mirror both arcs on the right-hand side.

(b) The six-petal flower — with a compass alone.

Draw a circle with centre O and radius r. Do not change the compass again.
Mark any point A on the circle. With centre A and the same radius r, draw a full circle. It cuts the first circle at two points.
Move the compass point to one of those crossings and draw again. Keep going round — after six circles you are back at A.
The six circles overlap the central circle in six petals. This is the classic 'flower of life' figure — no ruler needed at all.

(c) Regular hexagon in a circle. Draw the circle, step the radius round it six times, join the six points.

(d) Ring of six equal circles. Draw a circle of radius r with centre O and step the radius round it to mark six points. These six points are the centres of the six circles. Neighbouring centres are r apart, so draw each circle with radius r ÷ 2 — then neighbouring circles just touch, and the ring closes with a clear hole in the middle, exactly as in the picture.

(e) Patterned hexagon. Construct a regular hexagon; join its centre to the six vertices; find the midpoints of all the sides with perpendicular bisectors; join midpoints to each other and to the centre. Repeating this inside each small triangle gives the woven look.

Why the radius steps six times:
each step makes an equilateral triangle with the centre ⇒ 60° at the centre
6 × 60° = 360° = one full turn ✓

Why it happens: in figure (b) every new centre is at distance r from the old one and at distance r from O, so each new circle passes through O and through its neighbour's centre. That is what locks the six circles into a perfect ring and makes the petals identical.

Q2.

Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

The figure on page 155.

Answer

You see a bright white triangle sitting on top of the figure — but there is no triangle drawn at all. Only three notched circles and three bent corners are on the page.

What is actually drawn	What you seem to see
3 circles, each with a wedge cut out	A white triangle in front
3 bent line pairs (corners)	A second, outlined triangle behind
Nothing joining them	Three straight edges

Why it happens: the three wedges point at one another and their straight edges lie along the same three imaginary lines. The eye prefers one simple explanation — "a white triangle is lying on top and hiding parts of three circles" — over three separate odd shapes. So the brain invents the missing edges. These invented edges are called illusory contours; the picture is known as the Kanizsa triangle.

Check it yourself: cover one wedge with your finger. The illusion collapses at once — with only two corners, the brain has no reason to complete a triangle. To draw it, first construct an equilateral triangle lightly, put a circle at each vertex, cut out the 60° wedge that faces inward, then rub out the triangle.

Q3.

Construct this figure. [Hint: Find the angles in this figure.]

The figure on page 155.

Answer

Draw a regular hexagon, then draw the two equilateral triangles that join its alternate vertices. The star's six points land exactly on the hexagon's six corners.

Draw a circle of radius r, and step r around it to mark A, B, C, D, E, F. Join them — the regular hexagon.
Join A–C–E–A. Join B–D–F–B. Two equilateral triangles.
Ink the hexagon and the star; rub out the parts of the triangles inside the small hexagon.

Angles in the finished figure:
At a hexagon vertex (say A): the hexagon's angle ∠FAB = 120°
The star's point at A is ∠EAC = 60° (angle of equilateral ΔACE)
The two leftover slivers on either side:
(120° − 60°) ÷ 2 = 30° each

Each small triangle between the star and the hexagon side:
30° + 30° + 120° = 180° ✓

Star point angle = 60° · Inner hexagon angle = 120°

Why it happens: a side of the hexagon, say AB, together with the two star edges meeting at A and B, cuts off a small triangle. It has 30° at A, 30° at B and therefore 120° at the third corner. Six of these little triangles fill the space between the star and the hexagon.

Check it yourself: the six small triangles are congruent isosceles triangles, so the six sides of the star sit symmetrically inside the hexagon.

Q4.

Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P. [Hint: Find a line segment on l whose perpendicular bisector passes through P.]

Answer

Make P equidistant from two points of l. Then P is on their perpendicular bisector, which is the perpendicular you want.

With centre P and a radius big enough to reach the line, draw an arc cutting l at two points, X and Y. Now PX = PY.
So P lies on the perpendicular bisector of XY.
Find a second point of that bisector: with equal radii, draw arcs from X and from Y on the far side of l, meeting at Q.
Join P to Q. The line PQ is perpendicular to l, and it passes through P.

PX = PY (both radii of the arc from P)
QX = QY (both equal arcs from X and Y)
⇒ P and Q both lie on the perpendicular bisector of XY
⇒ line PQ is that perpendicular bisector
⇒ PQ ⊥ l, and P is on PQ ✓

Why it happens: we cannot bisect anything at P, because P is not on the line. So we turn the problem around: instead of hunting for a perpendicular, we manufacture a segment XY on l for which P is already an equidistant point. The perpendicular bisector of that segment is forced to go through P — and it is forced to be perpendicular to l.

Tip: the arc from P must be wide enough to cut the line twice. If it only touches the line at one point, open the compass a little more.

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