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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –904.1 Rectangles and Squares — In-text Questions

Book page 89 Updated on2026-09-05

Q1.
In the earlier definition, we stated that a rectangle has (a) opposite sides of equal length, and (b) all angles equal to 90°. Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are 90°?
Answer

No, we would not be wrong. The shorter definition is enough.

Condition (a) is not an extra requirement at all — it follows from condition (b). Deduction 4 shows that once all four angles are 90°, the opposite sides are forced to be equal; there is no way to build a quadrilateral with four right angles and unequal opposite sides.

Why this is worth doing: a good definition should carry no dead weight. If part of a definition can be proved from the rest, it belongs among the properties, not in the definition. Keeping the definition minimal also makes it easier to test a figure — you only have to check four angles.
Tip: three right angles are already enough, since the angle sum of a quadrilateral is 360° and 360 – 270 = 90. The book uses all four in the definition because it is simpler to state.
Q2.
If you think that this definition is incomplete, try constructing a quadrilateral in which the angles are all 90° but the opposite sides are not equal. Are you able to construct such a quadrilateral?
Answer

No such quadrilateral exists — the construction always closes up into a rectangle, and you can see why while drawing it.

Try it: draw AB of any length. At B turn 90° and draw BC of any length. At C turn 90° and draw CD. To make the angle at D equal 90° as well, D must lie directly above A — and that forces

CD = AB   and   AD = BC

The moment you make CD longer or shorter than AB, the fourth corner refuses to be a right angle. You have only three free choices (two lengths and the starting direction); the fourth side is decided for you.

Why it happens: after three 90° turns you are facing back along your original direction. The figure can only close if the two sides you drew in that direction are equal, and likewise for the other pair. Deduction 4 turns this observation into a proof using congruent triangles.
Q3.
Join BD. ∆BAD and ∆DCB seem congruent. Can we justify this claim? Two equalities can be directly seen in the triangles. What can we say about ∠1 and ∠2?
Answer

∠1 = ∠2, and that is exactly the missing piece — the same argument as in Deduction 2.

Let ABCD have all four angles 90°, and join BD. Write ∠1 = ∠ADB, ∠2 = ∠CBD and ∠3 = ∠DBA.

Since ∠B = 90°:   ∠3 + ∠2 = 90°
In ∆ABD:   ∠3 + ∠1 + 90° = 180°, so ∠3 + ∠1 = 90°
Hence ∠1 = ∠2 = 90° – ∠3

Now compare ∆BAD and ∆DCB:

∠A = ∠C = 90°
∠1 = ∠2  (just proved)
BD = DB  (common side)
⇒ ∆BAD ≅ ∆DCB by the AAS condition
⇒ AD = CB and DC = BA

So a quadrilateral with all angles 90° automatically has equal opposite sides — it is a rectangle.

Why it happens: the diagonal BD splits each right angle into two parts. Whatever is taken from the corner at B is left over at D, because both angle sums (the corner and the triangle) come to the same 90°. Two triangles that share the diagonal and have matching angle pairs must be copies of each other.
Q4.
Is it wrong to write ∆BAD ≅ ∆CDB? Why?
Answer

Yes, it is wrong. In a congruence statement the letters must be written in corresponding order.

The correct statement is ∆BAD ≅ ∆DCB, which asserts:

∆BADmatches∆DCB
B↔D
A↔C
D↔B

Writing ∆BAD ≅ ∆CDB instead would claim B ↔ C, A ↔ D and D ↔ B, so it would claim BA = CD and ∠B = ∠C. But ∠B here is ∠DBA, one part of the corner, while ∠C is a full right angle — those are not equal in general.

Tip: read a congruence statement as a dictionary between the two triangles. Every equality you later quote — a side or an angle — comes straight off that dictionary, so scrambling the letters gives you false equalities.
Q5.
Are the opposite sides of a rectangle parallel?
Answer

Yes, and it can be proved from the right angles alone — no measurement needed.

Take AB as a transversal cutting AD and BC.

∠A + ∠B = 90° + 90° = 180°
∠A and ∠B are interior angles on the same side of the transversal AB
When such a pair adds to 180°, the two lines are parallel
⇒ AD ∥ BC
Why it happens: if AD and BC were not parallel they would meet somewhere, forming a triangle with AB. That triangle would already contain 90° + 90° = 180° at two of its vertices, leaving nothing for the third — impossible. So they never meet.

This gives Property 3: the opposite sides of a rectangle are parallel to each other.

Q6.
Can you similarly show that AB is parallel to DC (AB||DC)?
Answer

Yes — repeat the argument with a different transversal.

Take AD as the transversal cutting AB and DC
∠A + ∠D = 90° + 90° = 180°
These are interior angles on the same side of AD
⇒ AB ∥ DC

So both pairs of opposite sides of a rectangle are parallel. This is also why every rectangle is a parallelogram — it satisfies the parallelogram's definition and, on top of that, has all its angles 90°.

Tip: the same transversal argument works at any of the four sides, because every angle of a rectangle is 90°. You may pick whichever pair of angles is most convenient.
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