NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –934.1 A Special Rectangle — In-text Questions

Book page 90 Updated on2026-09-05

Q1.

In the quadrilaterals below, are there any non-rectangles?

Answer

No — all four are rectangles, including (iv).

Figure	Sides	All angles 90°?	Rectangle?
(i)	5 cm, 2 cm, 5 cm, 2 cm	Yes	Yes
(ii)	3.6 cm, 6 cm, 3.6 cm, 6 cm	Yes	Yes
(iii)	5 cm, 1 cm, 5 cm, 1 cm	Yes	Yes
(iv)	4 cm, 4 cm, 4 cm, 4 cm	Yes	Yes — and also a square

Every one of the four figures carries a right-angle mark at all four corners, and that is the only thing the definition asks for. Being tilted on the page (as (ii) and (iii) are) changes nothing — angles do not depend on how a figure is turned.

Why (iv) counts: a square passes the rectangle test — all its angles are 90°. It is a special rectangle, one whose sides happen to be all equal. So every square is a rectangle, but not every rectangle is a square, exactly as a Malayali is an Indian while not every Indian is a Malayali.

Square: a quadrilateral in which all the angles are equal to 90°, and all the sides are of equal length.

Q2.

Let us consider the Carpenter's Problem again. If the wooden strips have to be placed such that the thread passing through their endpoints forms a square, what must be done?

Answer

The strips must be equal in length, joined at their midpoints, and set at right angles to each other.

Equal + bisecting → all angles 90° (a rectangle)
+ crossing at 90° → all four sides equal as well
⇒ a square

The first two conditions were already needed for a rectangle. Making the angle exactly 90° is the one extra demand, and Deduction 5 shows it is the right one.

Try This: to build a square of diagonal 8 cm, draw an 8 cm segment, mark its midpoint O, draw the perpendicular to it at O, and cut off 4 cm on each side of O. Join the four endpoints in order.

Q3.

What more needs to be done to get equal sidelengths as well? Can this be achieved by properly choosing the angle between the diagonals?

Answer

Yes — set the angle between the diagonals to 90°. That single change is all that is needed.

Recall from Deduction 3 that the four half-angles at a vertex are a = 90 – ^x⁄₂ and b = ^x⁄₂, where x is the angle between the diagonals. The sides come out equal exactly when the four triangles round O are all congruent, and that happens when the four angles at O are all the same:

x = 180 – x ⇒ 2x = 180 ⇒ x = 90°

At x = 90° we also get a = b = 45°, so each diagonal cuts each corner into two 45° halves.

Why it happens: the four triangles AOB, BOC, COD, DOA all have two sides equal to half a diagonal. They differ only in the angle between those two sides. Make that angle the same in all four and the triangles become congruent by SAS — so their third sides, which are the four sides of the quadrilateral, become equal too.

Q4.

To find the angle formed by the diagonals, what are the two triangles we should consider for congruence? Can this be used to find the angles ∠BOA and ∠BOC formed by the diagonals?

Answer

Take ∆BOA and ∆BOC. They give the angles at once.

BA = BC (sides of the square)
OA = OC (the diagonals bisect each other)
BO = BO (common side)
⇒ ∆BOA ≅ ∆BOC by the SSS condition

So ∠BOA = ∠BOC (corresponding parts)
But ∠BOA + ∠BOC = 180° (they form a straight angle along AC)
⇒ ∠BOA = ∠BOC = 90°

This is Deduction 5: the diagonals of a square bisect each other at right angles.

Why it happens: B is the same distance from A and from C, so B lies on the perpendicular bisector of AC. O is the midpoint of AC. So BO is the perpendicular bisector of AC — which is another way of saying the diagonals meet at 90°.

Tip: the same argument works in any rhombus, which is why Property 6 on page 102 says the diagonals of a rhombus meet at 90°. A square is just a rhombus that is also a rectangle.

Q5.

Using this fact, construct a square with a diagonal of length 8 cm.

Answer

Steps.

Draw AC = 8 cm.
Construct the perpendicular bisector of AC with a compass — arcs of equal radius from A and from C on both sides, joined. Let it cut AC at O, so OA = OC = 4 cm.
On this perpendicular mark B and D with OB = OD = 4 cm, one on each side of AC.
Join AB, BC, CD and DA.

ABCD is the required square.

Diagonals: AC = BD = 8 cm (equal)
They bisect each other at O and meet at 90°
⇒ ABCD is a square with each side = √(4² + 4²) = 4√2 ≈ 5.66 cm

Why the perpendicular bisector does the whole job: it gives you both the “bisect each other” condition and the “at 90°” condition in one construction, and marking equal 4 cm lengths gives “equal diagonals”. No protractor is needed anywhere.

Q6.

Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well.

Answer

Both apply word for word, because neither deduction ever used anything except the right angles and the equal opposite sides — which a square has.

Deduction	What it used	Does a square have it?
1 — diagonals equal	AB = CD, ∠BAD = ∠CDA = 90°, AD common (SAS)	Yes
2 — diagonals bisect each other	AB = CD, vertically opposite angles, ∠1 = ∠2 (AAS)	Yes

So a square has equal diagonals that bisect each other — and, from Deduction 5, they also cross at 90°.

Why we get this for free: a square is a rectangle. Every statement proved about all rectangles is automatically true of every square; the square only adds properties, it never loses any. This is the practical payoff of the Venn-diagram picture.

Q7.

There is one more special property of a square. What are the measures of ∠1, ∠2, ∠3, and ∠4? See if you can reason and/or experiment to figure this out! ... Similarly, find ∠2 and ∠4.

Answer

All four are 45°.

In ∆ADC: ∠1 + ∠3 + 90 = 180 ⇒ ∠1 + ∠3 = 90
AD = DC (sides of a square) ⇒ ∠1 = ∠3
So 2∠1 = 90 ⇒ ∠1 = ∠3 = 45°

In ∆ABC: AB = BC and ∠B = 90°
⇒ ∠2 = ∠4 = 45°

So the diagonal cuts each 90° corner into two 45° halves. This is Property 5: the diagonals of a square bisect the angles of the square.

Why it happens: a diagonal of a square splits it into two isosceles right triangles. In an isosceles triangle the base angles are equal, and here they must share the 90° left over from the right angle at the vertex — so each gets exactly half.

Tip: this is the same 45° that appeared in Deduction 3 when x = 90: a = 90 – 45 = 45 and b = 45. The two routes agree, which is a good sign that both are right.

Was this helpful? Report an error