NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –854.1 Rectangles and Squares — In-text Questions

Book page 83 Updated on2026-09-05

Q1.

Are there other ways to define a rectangle?

Answer

Yes. The chapter finds two more, and proves that all three pick out exactly the same figures.

Definition	Where it comes from
All angles are 90° and opposite sides are equal	the familiar one, stated first
The diagonals are equal and bisect each other	the answer to the Carpenter's Problem
All angles are 90°	Deduction 4 — the “opposite sides equal” part turns out to follow on its own

A definition is a test: a figure is a rectangle exactly when it passes the test. Two different tests can be equally good, provided each one lets in the same figures and keeps out the same figures.

Tip: the third definition is the shortest, so the book finally adopts it — a rectangle is a quadrilateral in which the angles are all 90°. Everything else about a rectangle is then a theorem, not part of the definition.

Q2.

A carpenter needs to put together two thin strips of wood, as shown in Fig. 1, so that when a thread is passed through their endpoints, it forms a rectangle. She already has one 8 cm long strip. What should be the length of the other strip? Where should they both be joined?

Answer

The other strip must also be 8 cm, and the two strips must be joined at their midpoints — that is, 4 cm from each end.

Model the strips as line segments. Their four endpoints are the vertices of the quadrilateral the thread makes, so the strips are its diagonals. So the question becomes: what must be true of the diagonals of a rectangle?

Deduction 1 → the diagonals of a rectangle are equal
Deduction 2 → they bisect each other
So: second strip = 8 cm, joined at the midpoint of each

Why it happens: the angle between the two strips is not fixed. Deduction 3 shows that any angle at all gives a rectangle, as long as the strips are equal and cross at their midpoints. So the carpenter has one free choice — she can swing the strips to make the rectangle long and thin or nearly square — and the figure stays a rectangle.

Did you know? This is a real workshop method. Carpenters in Europe use it to square up a frame, and farmers in Mozambique use it to set out a rectangular base for a house.

Q3.

What is the length of the other diagonal?

Answer

8 cm — the same as the first, because the diagonals of a rectangle are always equal.

Deduction 1. In rectangle ABCD, compare ∆ADC and ∆DAB:

AB = CD (opposite sides of a rectangle)
∠BAD = ∠CDA = 90° (angles of a rectangle)
AD = DA (common side)
So ∆ADC ≅ ∆DAB by the SAS condition
Hence AC = BD (corresponding parts of congruent triangles)

Why it happens: each diagonal is the hypotenuse of a right triangle whose legs are a length and a breadth of the rectangle. Both diagonals sit on right triangles with the same two legs, so the two hypotenuses must match. The congruence argument says exactly this, without needing to compute any length.

Check it yourself: draw three very different rectangles and measure both diagonals of each. They agree every time — but measuring three cases is not a proof, which is why Deduction 1 is written out.

Q4.

What is the point of intersection of the two diagonals?

Answer

It is the midpoint of both diagonals — the diagonals bisect each other.

Deduction 2. Let the diagonals of rectangle ABCD meet at O, and compare ∆AOB and ∆COD.

AB = CD (opposite sides)
∠AOB = ∠COD (vertically opposite angles)
Now for ∠1 = ∠OAB and ∠2 = ∠OCD, write ∠3 = ∠DBC:
∠B = 90°, so ∠3 + ∠1 = 90°
In ∆BCD: ∠3 + ∠2 + 90° = 180°, so ∠3 + ∠2 = 90°
Hence ∠1 = ∠2 = 90° – ∠3
So ∆AOB ≅ ∆COD by the AAS condition
Therefore OA = OC and OB = OD

So O is the midpoint of AC and also the midpoint of BD.

Why it happens: ∠1 and ∠2 are both the “left-over” part of a right angle after taking away the same angle ∠3 — one inside the corner at B, the other inside triangle BCD. Two angles that are each 90° – ∠3 have to be equal. That single equality is what unlocks the congruence.

Tip: to bisect something is to cut it into two equal parts. Saying “the diagonals bisect each other” is shorter than saying “each diagonal passes through the midpoint of the other”, and means the same thing.

Q5.

What should the angle be between the diagonals?

Answer

Any angle at all. This is the surprising part of the Carpenter's Problem: the angle is not forced.

Deduction 3 takes two equal diagonals that bisect each other with an arbitrary angle x between them, and shows every angle of the resulting quadrilateral works out to 90°. So the carpenter may join the strips at any angle she likes and still get a rectangle — different angles simply give rectangles of different shapes.

x = 90° → all four sides equal → a square
x close to 0° or 180° → a long, thin rectangle

Why it happens: the four endpoints all lie at the same distance from the crossing point O (half of the common diagonal length), so they lie on a circle centred at O. Turning one strip about O just slides two of the points around that circle — the four points stay on the circle, and the figure stays a rectangle.

Q6.

Can the following equalities be used to establish that ∆AOD ≅ ∆COB? AO = CO (proved above); ∠AOB = ∠COD (vertically opposite angles); AD = CB

Answer

No. Two of the three facts do not belong to the triangles being compared.

AO = CO — usable. AO is a side of ∆AOD and CO is a side of ∆COB.
∠AOB = ∠COD — not usable. ∠AOB is not an angle of ∆AOD, and ∠COD is not an angle of ∆COB. The angles we need are ∠AOD and ∠COB (which are indeed equal, being vertically opposite).
AD = CB — usable, but it is a third side, not the second side next to the angle.

A correct set is:

AO = CO, ∠AOD = ∠COB (vertically opposite), OD = OB
⇒ ∆AOD ≅ ∆COB by SAS

Why it matters: a congruence condition is a rule about which parts match, not just about how many equalities you can list. SAS needs the angle to sit between the two sides. Listing an angle from somewhere else in the figure proves nothing about these two triangles.

Q7.

Let us check what quadrilateral we get if we draw the two diagonals such that their lengths are equal, they bisect each other and have an arbitrary angle, say 60°, between them. Can you find all the remaining angles?

Answer

Every angle in the figure is determined. Start with the four angles at the crossing point O.

Given ∠AOB = 60°
∠COD = 60° (vertically opposite)
∠AOD = 180° – 60° = 120° (linear pair)
∠BOC = 120° (vertically opposite)

Now use the four isosceles triangles. Because the diagonals are equal and bisect each other, OA = OB = OC = OD.

Triangle	Apex angle at O	Base angles
∆AOB	60°	60°, 60°
∆COD	60°	60°, 60°
∆AOD	120°	30°, 30°
∆BOC	120°	30°, 30°

So at every vertex of ABCD the two half-angles are 30° and 60°, giving 30° + 60° = 90°. All four angles of ABCD are right angles, so ABCD is a rectangle.

Why it happens: OA = OB makes ∆AOB isosceles, so its base angles are equal — that is the only reason the numbers come out. Each vertex angle of ABCD is one base angle from a 60° triangle plus one base angle from a 120° triangle, and 60 + 30 = 90 every time.

Q8.

In ∆AOB, since OA = OB, the angles opposite them are equal, say a. Can you find the value of a?

Answer

a = 60°.

In ∆AOB: a + a + 60 = 180 (interior angles of a triangle)
2a = 180 – 60 = 120
a = 60

So ∆AOB has all three angles equal to 60° — it is equilateral, which also tells us AB = OA = OB.

Why it happens: in a triangle, equal sides face equal angles. OA = OB forces the two angles opposite them to be the same value a, so the angle sum gives one equation in one unknown and pins a down completely.

Was this helpful? Report an error