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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –1024.4 Quadrilaterals with Equal Sidelengths — In-text Questions

Book page 99 Updated on2026-09-05

Q1.
Are squares the only quadrilaterals that have equal sidelengths? Let us explore this question through construction.
Answer

No. A square is only one member of a much larger family — the rhombuses.

Draw AD and AB of the same length with any angle between them other than 90°, say 50°. Complete the figure so that all four sides match, and you get a quadrilateral with four equal sides whose angles are 50°, 130°, 50°, 130°. It is certainly not a square.

Rhombus: a quadrilateral in which all the sides have the same length.

Any angle less than 180° can be used in place of 50°, so there are infinitely many rhombuses for each side length; only the one with a 90° angle is a square.

Why four equal sides do not fix the shape: a four-bar linkage with equal rods is not rigid — you can push it over into a leaning shape without changing any side length. A triangle made of three rods cannot be deformed this way, which is why SSS is a congruence condition for triangles but there is no “SSSS” for quadrilaterals.
Q2.
Can we complete this quadrilateral so that all its sides are of the same length? Mark a point C whose distance from B and D is equal to AB (or AD).
Answer

Yes, and a compass finds the fourth vertex in one step.

  1. Open the compass to the length AB.
  2. Keeping this as radius, draw an arc centred at B and another centred at D.
  3. The arcs meet at C. Join BC and DC.
AB = AD  (drawn equal)
BC = AB  (radius of the arc from B)
DC = AB  (radius of the arc from D)
⇒ AB = BC = CD = DA — a rhombus
Why the arcs must meet: every point on the arc centred at B is at distance AB from B, and every point on the arc centred at D is at distance AB from D. Their crossing point is the one place that is the correct distance from both — so it is the only possible position for C on that side of BD.
Tip: the same construction with a starting angle of 90° gives a square, so the square really is just the rhombus you get for one particular choice of angle.
Q3.
What are the other angles of the rhombus ABCD that we have constructed? Reason and/or experiment to figure this out.
Answer

50°, 130°, 50° and 130°.

Deduction 9 first shows that in any rhombus a diagonal splits the two angles it meets into four equal parts. Apply that to ABCD, where ∠A = 50° and the diagonal BD makes four equal angles a.

In ∆ADB:   a + a + 50 = 180
2a = 130  ⇒ a = 65°

∠B = ∠D = a + a = 65 + 65 = 130°
∠C = ∠A = 50°
Check: 50 + 130 + 50 + 130 = 360° ✓

There is a second, quicker route. Every rhombus is a parallelogram, so adjacent angles are supplementary:

∠A = ∠C = 50°  and  ∠D = ∠B = 180 – 50 = 130°
Why both routes agree: they must — a rhombus is a parallelogram (its opposite sides are parallel, as the equal alternate angles in Deduction 9 show), so the parallelogram rules apply. Meeting the same answer by two independent arguments is a good check on both.
Q4.
It can be seen that ∆GAE ≅ ∆MAE (How?)
Answer

By the SSS condition, using the diagonal AE as the shared side.

GA = MA  (all sides of rhombus GAME are equal)
GE = ME  (same reason)
AE = AE  (common side)
⇒ ∆GAE ≅ ∆MAE by SSS

From the congruence, ∠G = ∠M and the angle at A is split the same way as the angle at E. Together with the two isosceles triangles GAE and MAE:

GE = GA ⇒ a = d    ME = MA ⇒ b = c
congruence ⇒ a = b and c = d
⇒ a = b = c = d
Why this gives Property 5: a = b says the diagonal AE cuts the angle at A into two equal halves, and c = d says the same at E. That is exactly what “the diagonals of a rhombus bisect its angles” means. It also gives EM ∥ GA and GE ∥ AM, since the equal parts are alternate angles — so every rhombus is a parallelogram.
Q5.
So a rhombus is a parallelogram, and a rectangle is also a parallelogram. How can this be represented using a Venn diagram? Where will the set of squares occur in this diagram?
Answer

Draw one large oval for parallelograms. Inside it, two overlapping ovals — rectangles and rhombuses. The squares sit exactly in the overlap.

Parallelograms Rectangles Rhombuses Squares
A square is exactly a quadrilateral that is both a rectangle and a rhombus.
RegionWhat lives there
Rectangle onlyall angles 90°, sides not all equal
Rhombus onlyall sides equal, angles not 90°
Overlapsquares — all angles 90° and all sides equal
Outside both, inside the big ovala plain leaning parallelogram
Why the squares fall in the overlap: a square is a rectangle (all angles 90°) and it is also a rhombus (all sides equal). Conversely, anything in the overlap has all angles 90° and all sides equal, which is precisely the definition of a square. So the overlap is not merely where squares happen to be — it is the set of squares.
Q6.
Are the diagonals of a rhombus equal?
Answer

No, not in general. They are equal only when the rhombus happens to be a square.

In the rhombus with angles 50° and 130°, the diagonal joining the two 130° corners is much shorter than the one joining the two 50° corners. Squash a rhombus flatter and one diagonal grows while the other shrinks, though all four sides stay the same length.

Diagonals equal + bisect each other ⇒ all angles 90°  (Deduction 3)
A rhombus with all angles 90° = a square
Why the sides cannot control the diagonals: a rhombus is a flexible linkage. The four rods fix the sides but leave one degree of freedom — the angle — and the two diagonals move in opposite directions as that angle changes. Only at 90° do they cross.
Tip: the properties a rhombus does have are Property 4 (diagonals bisect each other), Property 5 (they bisect the angles) and Property 6 (they meet at 90°). “Equal length” is not on the list.
Q7.
Do the diagonals of a rhombus intersect at any particular angle? In the rhombus GAME, we have ∆GEO ≅ ∆MEO (why?)
Answer

Yes — always at 90°.

Why ∆GEO ≅ ∆MEO: the diagonals bisect each other (every rhombus is a parallelogram), so O is the midpoint of GM.

GE = ME  (sides of the rhombus)
GO = MO  (O is the midpoint of GM)
EO = EO  (common side)
⇒ ∆GEO ≅ ∆MEO by SSS

Deduction 10. From the congruence, ∠GOE = ∠MOE. These two angles sit along the straight line GM, so they form a linear pair:

∠GOE + ∠MOE = 180°
2 × ∠GOE = 180°  ⇒ ∠GOE = 90°

This is Property 6: the diagonals of a rhombus intersect each other at an angle of 90°.

Why it happens: E is equidistant from G and M, and so is O. Two points that are both equidistant from G and M determine the perpendicular bisector of GM — and EO is that line. Perpendicular bisector means exactly “through the midpoint, at right angles”.
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