NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –874.1 Rectangles and Squares — In-text Questions

Book page 86 Updated on2026-09-05

Q1.

Can we now identify what type of quadrilateral ABCD is? Notice that its angles all add up to 90° (30° + 60°).

Answer

It is a rectangle.

At each vertex the diagonal splits the angle into a 30° part and a 60° part, and 30° + 60° = 90°. So all four angles of ABCD are 90°, which is exactly the condition for a rectangle.

∠A = ∠B = ∠C = ∠D = 90°
Check: 90 × 4 = 360° ✓ (the angle sum of any quadrilateral)

Tip: the figure was not drawn as a rectangle — it was built from two equal segments crossing at their midpoints at 60°. The rectangle is the conclusion, not the starting assumption.

Q2.

What can we say about its sides?

Answer

The opposite sides are equal: AB = CD and AD = CB.

∆AOB ≅ ∆COD ⇒ AB = CD
∆AOD ≅ ∆COB ⇒ AD = CB

Each congruence follows from SAS: OA = OC, OB = OD (the diagonals bisect each other), and the angle at O between them is the same in both triangles because vertically opposite angles are equal.

Why it happens: half of one diagonal, half of the other, and the angle between them completely fix a triangle. The two triangles on opposite sides of O are built from exactly the same three pieces, so the third sides — which are the opposite sides of ABCD — must match.

With all angles 90° and opposite sides equal, ABCD satisfies the book's first definition of a rectangle.

Q3.

Will ABCD remain a rectangle if the angles between the diagonals are changed? Can we generalise this?

Answer

Yes — it stays a rectangle for every angle. That is exactly what Deduction 3 generalises.

Let one of the angles between the diagonals be x. Then the four angles at O are x, x, 180 – x, 180 – x.

In isosceles ∆AOB (OA = OB), base angles a:
a + a + x = 180 ⇒ 2a = 180 – x ⇒ a = 90 – ^x⁄₂

In isosceles ∆AOD (OA = OD), base angles b:
b + b + (180 – x) = 180 ⇒ 2b = x ⇒ b = ^x⁄₂

Each angle of ABCD = a + b = (90 – ^x⁄₂) + ^x⁄₂ = 90

The x cancels. That is the whole point: the answer does not depend on x, so no matter what the angle between the diagonals is, the quadrilateral is a rectangle.

Why it happens: at each vertex, one half-angle grows exactly as fast as the other shrinks. Open the diagonals by a little and a falls by half that amount while b rises by the same half. Their sum can never move away from 90°.

Check it yourself: put x = 60 into the formulas: a = 90 – 30 = 60 and b = 30, matching the worked case on page 85 exactly.

Q4.

Since we know that ∆AOB is isosceles, we can denote the measures of both of its base angles by a. What is the value of a (in degrees) in terms of x?

Answer

a = 90 – ^x⁄₂ degrees.

a + a + x = 180 (sum of the interior angles of a triangle)
2a = 180 – x
a = ^{(180 – x)}⁄₂ = 90 – ^x⁄₂

x	a = 90 – x/2	b = x/2	a + b
40°	70°	20°	90°
60°	60°	30°	90°
90°	45°	45°	90°
140°	20°	70°	90°

Why algebra is used here: a numerical case such as x = 60° checks one rectangle. Writing the answer in terms of x checks all of them at once — the letter stands for every possible angle the two strips could make.

Q5.

What can we say about AB and CD, and AD and BC?

Answer

AB = CD and AD = CB — the opposite sides are equal, whatever the angle x is.

∆AOB ≅ ∆COD ⇒ AB = CD
∆AOD ≅ ∆COB ⇒ AD = CB

Putting this together with Q3: if two segments are equal and bisect each other, the quadrilateral formed by their endpoints has all angles 90° and equal opposite sides. So it is a rectangle — and the Carpenter's Problem is completely solved.

Tip: this is what makes the second definition legitimate — a rectangle is a quadrilateral whose diagonals are equal and bisect each other. Deductions 1 and 2 prove every rectangle passes this test; Deduction 3 proves everything that passes it is a rectangle.

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