4. Quadrilaterals

A quadrilateral is a closed figure made of four line segments (sides).

The other figures are either open, have curved sides, or do not have exactly four straight sides. So they are not quadrilaterals.

No. The angle ∠AOB = ∠COD is not an angle of ∆AOD or ∆COB.

Also, the equal angle must be between the two equal sides (SAS). So these equalities cannot be used.

Yes, it is wrong. In congruence, the order of letters must match the corresponding vertices.

Here B ↔ D, A ↔ C and D ↔ B. So the correct statement is ∆BAD ≅ ∆DCB.

AD is a transversal to AB and DC. ∠A + ∠D = 90° + 90° = 180°.

The interior angles on the same side of the transversal add up to 180°. So AB || DC.

The strips must be of equal length, joined at their midpoints, and placed at right angles (90°) to each other.

1. Draw AC = 8 cm and mark its midpoint O.
2. Draw a line through O perpendicular to AC.
3. On this line, mark B and D such that OB = OD = 4 cm.
4. Join AB, BC, CD and DA. ABCD is the required square.

In each right triangle formed by a diagonal, the two sides are equal. So the two base angles are equal.

Each angle = (180° − 90°) ÷ 2 = 45°. So ∠1 = ∠2 = ∠3 = ∠4 = 45°.

1. Draw AC = 8 cm. Mark its midpoint O.
2. At O, draw a line making the given angle (30°, 40°, 90° or 140°) with OC.
3. On this line, mark B and D on opposite sides of O such that OB = OD = 4 cm.
4. Join AB, BC, CD and DA.

In each case we get a rectangle. In (iii), with 90°, the rectangle is a square.

PL and AM are diameters, so they are equal and bisect each other at O. They are also perpendicular.

Diagonals that are equal and bisect each other at 90° form a square. So APML is a square.

Cross the two sticks at their midpoints. Pass the thread around their four ends to make a quadrilateral.

The diagonals are equal and bisect each other, so the figure is a rectangle. Each corner of the thread gives an exact 90°.

No. A parallelogram that is not a rectangle (for example, with angles 60° and 120°) has opposite sides parallel and equal, but its angles are not 90°.

So this cannot be chosen as the definition of a rectangle.

Yes. The vertices must correspond. A ↔ C, B ↔ D and D ↔ B. So the correct statement is ∆ABD ≅ ∆CDB.

No. In a parallelogram with angles 30° and 150°, one diagonal is clearly longer than the other. They are equal only when the parallelogram is a rectangle.

Yes. A corresponds to Y and E corresponds to S. So the correct statement is ∆AOE ≅ ∆YOS.

No. They can meet at any angle. They meet at 90° only in a rhombus (or square).

GE = ME, GA = MA (all sides of a rhombus are equal), and AE is common.

So ∆GAE ≅ ∆MAE by the SSS condition.

Draw a big oval for parallelograms. Inside it, draw two overlapping ovals for rectangles and rhombuses.

Squares are in the overlapping part, because a square is both a rectangle and a rhombus.

No, not always. They are equal only when the rhombus is a square.

GE = ME (sides of a rhombus), GO = MO (diagonals bisect each other) and EO is common.

So ∆GEO ≅ ∆MEO by the SSS condition.

1. Draw AC = 7 cm. Mark its midpoint O.
2. At O, draw a line making an angle of 140° with OC.
3. On this line, mark B and D on opposite sides of O such that OB = OD = 2.5 cm.
4. Join AB, BC, CD and DA. ABCD is the required parallelogram.

1. Draw AC = 5 cm. Mark its midpoint O.
2. Draw a line through O perpendicular to AC.
3. On it, mark B and D on opposite sides of O such that OB = OD = 2 cm.
4. Join AB, BC, CD and DA. ABCD is the required rhombus.

First we get a square, because the diagonals are equal and bisect each other at 90°.

After extending one diagonal by 2 cm on both sides, the diagonals still bisect each other at 90° but are not equal. So we get a rhombus.

(i) In ∆ABD and ∆CBD: AB = CB, AD = CD and BD is common. So ∆ABD ≅ ∆CBD (SSS).

Hence ∠ABD = ∠CBD and ∠ADB = ∠CDB. So BD bisects ∠ABC and ∠ADC.

(ii) In ∆AOB and ∆COB: AB = CB, ∠ABO = ∠CBO and BO is common. So ∆AOB ≅ ∆COB (SAS).

Hence AO = CO, and ∠AOB = ∠COB. These two angles form a linear pair, so each is 90°. So BD bisects AC at right angles.

UX = VW (equal sides), XY = WZ (opposite sides of rectangle XWZY) and ∠XYU = ∠WZV = 90°.

So ∆UXY ≅ ∆VWZ by the RHS condition. Hence ∠U = ∠V.

All four sides are 4 cm. So it is a rhombus.

Two opposite angles are 60° each. The other two are 60° + 60° = 120° each.

1. Draw AC = 6 cm and mark its midpoint O.
2. Draw a line through O perpendicular to AC.
3. On this line, mark B on one side with OB = 2 cm and D on the other side with OD = 6 cm (so BD = 8 cm).
4. Join AB, BC, CD and DA. ABCD is the required kite (AB = CB and AD = CD).

Draw two overlapping ovals: Parallelograms and Kites. Their overlap is Rhombuses. Inside Parallelograms, draw Rectangles overlapping Rhombuses. The common part of Rectangles and Rhombuses is Squares.

In ∆ROI, ∠RIO = 90° and ∠ORI = 30°. So ∠ROI = 180° − 90° − 30° = 60°.

∠ROD = 90° (angle of rectangle RODS).

So ∠IOD = 90° − 60° = 30°

1. Draw AC = 6 cm.
2. With a compass, draw the perpendicular bisector of AC. Let it cut AC at O.
3. On it, mark B and D such that OB = OD = 3 cm (use the compass).
4. Join AB, BC, CD and DA. ABCD is the required square.

The four corner triangles (∆CUV, ∆AXU, ∆SWX, ∆EVW) are congruent by SAS (two half-sides and a 90° angle). So UV = UX = XW = WV.

Each corner triangle is isosceles right-angled, so its base angles are 45°. At U, ∠CUV + ∠VUX + ∠AUX = 180°, so ∠VUX = 180° − 45° − 45° = 90°. Similarly, all angles are 90°.

So UVWX is a square.

Other ways: Mark points on the four sides at the same distance from the corners, going round in the same direction (for example, 1 cm from C along CA, 1 cm from A along AS, and so on). Joining them always gives a square.

Yes. Four equal sides make it a rhombus, which is a parallelogram.

Adjacent angles add up to 180°, so the next angle is 90°. Opposite angles are equal, so all angles are 90°.

All sides equal and all angles 90° means it is a square.

Let ABCD have AB = CD and BC = DA. Draw diagonal AC.

∆ABC ≅ ∆CDA by SSS. So ∠BAC = ∠DCA and ∠BCA = ∠DAC.

These are alternate angles. So AB || DC and BC || AD.

Hence ABCD is a parallelogram.

Yes. Join BD. It divides ABCD into ∆ABD and ∆CBD.

The angles of both triangles together make up all the angles of ABCD (the angle at D is the reflex angle inside the figure).

So the sum = 180° + 180° = 360°

The creases form a rhombus. The four parts are mirror images of one another, so all four sides are equal.

Make several triangular creases at the same corner, one after another, each parallel to the first and a little farther from the corner. On opening, we get rhombuses one inside another.

Make the triangular crease so that it cuts off equal lengths on both folded edges (the crease is at 45° to each edge). On opening, the creases form a square.