Figure (i): every angle inside the rectangle, starting from the given ∠CAB = 30°.
Figure (ii) — rectangle PQRS with P bottom-left, Q top-left, R top-right, S bottom-right, both diagonals drawn, and ∠QOR = 110° at the crossing point O.
Why the isosceles triangles appear: the diagonals of a rectangle are equal and bisect each other, so all four half-diagonals OP, OQ, OR, OS have the same length. Every triangle formed at O therefore has two equal sides, and its base angles must be equal. That is what turns one given angle into all twelve.
Q2.
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of (i) 30° (ii) 40° (iii) 90° (iv) 140°
Answer
The construction is the same in every case; only the angle at O changes.
Draw AB = 8 cm and mark its midpoint O (so OA = OB = 4 cm).
At O draw a ray making the required angle with OB.
On this ray, and on its opposite ray, cut OC = OD = 4 cm.
Join AD, DB, BC and CA. ADBC is the required quadrilateral.
Every one of these four figures is a rectangle, because the diagonals are equal (8 cm each) and bisect each other — Deduction 3 says the angle between them makes no difference.
Angle at O
Half-angles at each vertex (90 – x/2, x/2)
Figure obtained
(i) 30°
75° and 15°
a long, thin rectangle
(ii) 40°
70° and 20°
rectangle
(iii) 90°
45° and 45°
a square
(iv) 140°
20° and 70°
rectangle (same shape as (ii), turned)
Why (ii) and (iv) give the same shape: 40° and 140° are supplementary, and the four angles at O are always x, x, 180 – x, 180 – x. Choosing 140° simply swaps which pair of angles is the smaller one, so you get the same rectangle standing the other way round.
Check it yourself: measure the four sides in case (iii). They should all come to about 5.7 cm, since each side is √(4² + 4²) = 4√2 cm.
Q3.
Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.
Answer
APML is a square.
The two diameters are the diagonals of the quadrilateral APML, so check them against the square test.
PL = AM — every diameter of a circle has the same length ⇒ diagonals equal
Both pass through the centre O, and O is the midpoint of each ⇒ diagonals bisect each other
PL ⊥ AM (given) ⇒ diagonals meet at 90°
Equal diagonals that bisect each other at right angles ⇒ the figure is a square.
Why it happens: all four vertices lie on the circle, so each is at distance r from O. The four triangles round O are therefore congruent right isosceles triangles with legs r, so all four sides of APML equal r√2 and all four angles equal 45° + 45° = 90°.
Try This: draw a circle of radius 3 cm and two perpendicular diameters. Measure a side of APML — it should be about 4.2 cm, since 3√2 ≈ 4.24.
Q4.
We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?
Answer
Use the Carpenter's Problem in reverse: build a rectangle first, and its corners are your right angles.
Let AB and CD be the two equal sticks. Find the midpoint of each by folding the thread along a stick and halving it.
Cross the sticks so that the two midpoints coincide at a point O. Fix them there.
Run the thread around the four ends A, C, B, D and back to A, pulling it taut.
ACBD is a rectangle, since its diagonals AB and CD are equal and bisect each other. So ∠C = ∠A = ∠B = ∠D = 90° — each corner of the thread outline is an exact right angle.
Why it is exact and not approximate: nothing here depends on judging an angle by eye. The two conditions you can control physically — sticks of equal length, crossing at their midpoints — are precisely the conditions Deduction 3 proves are enough to force all four angles to 90°.
Another way: use the isosceles triangle instead. Tie the thread to make a triangle with two equal sides on the sticks; the line from the apex to the midpoint of the base is perpendicular to the base, because it is the perpendicular bisector.
Q5.
We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle?
Answer
No. “Opposite sides parallel and equal” is a true property of rectangles, but it is not enough to define them.
A definition must let in every rectangle and keep out everything else. This one fails the second half: every parallelogram has opposite sides parallel and equal, and most parallelograms are not rectangles.
Parallelogram with ∠A = 30°, sides 4 cm and 5 cm
Opposite sides parallel ✓ opposite sides equal ✓
But ∠A = 30° ≠ 90° ⇒ not a rectangle
Why the angles must be mentioned: parallelism fixes the directions of the sides but not the angle between the two directions. Push the top of a rectangle sideways and the sides stay parallel and equal while every angle changes — you get a “leaning” parallelogram. Only the condition “all angles are 90°” stops this from happening.
Tip: compare with the definitions that do work: “all angles 90°”, or “diagonals equal and bisect each other”. In a general parallelogram the diagonals bisect each other but are not equal — which is exactly what the failed definition leaves out.