Chapter 4: Quadrilaterals (Geometry)

A quadrilateral is a closed figure made of exactly four straight line segments (sides), each meeting the next only at its endpoints. Figs. (i)–(iii) satisfy this. Fig. (iv) has a curved side (or is not closed), and Fig. (v) has sides crossing each other or more/fewer than four sides, so they are not quadrilaterals.

Quadrilaterals are closed figures with exactly four straight sides that meet only at their ends; (iv) and (v) break this.

No. ∠AOB and ∠COD are not angles of ΔAOD and ΔCOB at all. Even the matching angles ∠AOD and ∠COB would not help: with AO = CO and AD = CB, the angle at O is not between these two sides, which is the SSA case, and SSA does not guarantee congruence.

No: the given angle is not in these triangles, and AO, AD with the angle at O would only be SSA.

It is not possible: Deduction 4 proves that if all the angles are 90°, the opposite sides must be equal.

Yes, it is wrong. The correspondence must match equal parts: B ↔ D, A ↔ C, D ↔ B, so the correct statement is $\Delta BAD\cong \Delta DCB$. Writing ΔCDB would match B with C and A with D.

Impossible (all right angles force equal opposite sides); the correct statement is ΔBAD ≅ ΔDCB, not ΔCDB.

Take AD as a transversal of AB and DC: $\angle A+\angle D=90°+90°=180°$. The interior angles on the same side add up to 180°, so AB ∥ DC.

∠A + ∠D = 180° with AD as transversal, so AB ∥ DC.

The diagonals of a square bisect each other at 90°. To construct: draw AC = 8 cm, draw its perpendicular bisector, and mark B and D on it 4 cm from O on each side; join ABCD.

All four angles between a diagonal and the sides are 45°: in ΔADC, $\angle 1=\angle 3$ (AD = DC) and $\angle 1+\angle 3=90°$; similarly $\angle 2=\angle 4=45°$ in ΔABC.

At 90°; ∠1 = ∠2 = ∠3 = ∠4 = 45°.

The diagonals of a rectangle are equal and bisect each other, so OA = OB = OC = OD and the four triangles are isosceles.

(i) $\angle CAD=90°-30°=60°$. Since OA = OB, $\angle ABD=30°$, so $\angle DBC=60°$; similarly $\angle ACD=30°$, $\angle ACB=60°$, $\angle BDC=30°$, $\angle ADB=60°$. At O: $\angle AOB=\angle COD=120°$, $\angle BOC=\angle AOD=60°$.

(ii) $\angle POS=110°$ (vertically opposite) and $\angle QOP=\angle ROS=70°$. In isosceles ΔQOR: $\angle OQR=\angle ORQ=(180°-110°)÷2=35°$. Then $\angle OQP=\angle ORS=90°-35°=55°$; in ΔPOQ (OP = OQ) $\angle OPQ=55°$, so $\angle OPS=35°$; likewise $\angle OSR=55°$ and $\angle OSP=35°$.

(i) 60°, 30°, 60°, 30°, 60°, 30°, 60° at the corners; 120° and 60° at O. (ii) 110°, 70°, 70° at O; 35° and 55° at each corner.

Draw AB = 8 cm and mark its midpoint O. Draw a line through O making the given angle with AB, and mark C and D on it with OC = OD = 4 cm. Join A, D, B, C in order.

Since the diagonals are equal and bisect each other, every such quadrilateral is a rectangle; in (iii) the diagonals are also perpendicular, so it is a square.

Rectangles in all four cases (a square for 90°).

Its diagonals PL and AM are both diameters, so they are equal, they bisect each other at O, and they are perpendicular. So APML is a square.

A square (equal diagonals bisecting each other at right angles).

Cross the two sticks so that they meet at their midpoints (find each midpoint by folding the thread to the stick's length). Pass the thread around the four ends. Since the "diagonals" are equal and bisect each other, the thread forms a rectangle, and each corner is exactly 90°.

Join the sticks at their midpoints; the thread through the four ends forms a rectangle, whose corners are 90°.

No. A parallelogram that is slanted (angles 60° and 120°, say) has parallel and equal opposite sides but no right angles. So this cannot be the definition of a rectangle.

No, a slanted parallelogram is a counterexample.

No. The angles of any quadrilateral add up to 360° (a diagonal splits it into two triangles: $180°+180°$). With three right angles, the fourth is $360°-270°=90°$.

No: 360° − 3 × 90° = 90°.

Angles: 30°, 150°, 30°, 150°; sides 4, 5, 4, 5 cm. The diagonals are not equal (one is long, one short), but they bisect each other.

Yes, both are wrong. The correct correspondences are ΔABD ≅ ΔCDB (A ↔ C, B ↔ D) and ΔAOE ≅ ΔYOS (A ↔ Y, E ↔ S). The written versions match vertices that are not corresponding.

Angles 30°, 150°, 30°, 150°; sides 4, 5, 4, 5 cm; diagonals unequal but bisecting each other; the correct statements are ΔABD ≅ ΔCDB and ΔAOE ≅ ΔYOS.

No, the angle depends on the parallelogram: it can be anything between 0° and 180°. It is 90° only when the parallelogram is a rhombus.

No particular angle (90° only for a rhombus).

ΔGAE ≅ ΔMAE by SSS: GA = MA, GE = ME (all sides of a rhombus are equal) and AE is common.

The diagonals of a rhombus are not equal in general (only when it is a square).

ΔGEO ≅ ΔMEO by SSS: GE = ME, GO = MO (diagonals bisect each other) and EO is common. So ∠GOE = ∠MOE = 90°.

Both by SSS; a rhombus's diagonals are generally not equal.

(i) $\angle E=\angle R=180°-40°=$ 140°, $\angle A=$ 40°.

(ii) $\angle Q=\angle S=$ 70°, $\angle R=$ 110°.

(iii) The diagonal XV bisects the angles: $\angle XVW=30°$, so $\angle V=\angle X=$ 60° (with $\angle UXV=\angle VXW=30°$), and $\angle U=\angle W=$ 120°.

(iv) $\angle OEI=20°$, so $\angle E=\angle O=$ 40° (with $\angle AOE=\angle EOI=20°$), and $\angle A=\angle I=$ 140°.

(i) 140°, 40°, 140° (ii) 70°, 110°, 70° (iii) 60° at X and V, 120° at U and W (iv) 40° at O and E, 140° at A and I

Draw AB = 7 cm with midpoint O. At O, draw a line at 140° to OB and mark C and D on it with OC = OD = 2.5 cm. Join A, D, B, C.

Diagonals bisecting each other (OA = OB = 3.5 cm, OC = OD = 2.5 cm) at 140° give the parallelogram.

Draw AB = 5 cm and its perpendicular bisector through the midpoint O; mark C and D on it with OC = OD = 2 cm; join A, D, B, C.

Perpendicular diagonals bisecting each other (5 cm and 4 cm) give the rhombus.

The diagonals are equal, bisect each other and are perpendicular, so the quadrilateral is a square.

After extending one diagonal by 2 cm at each end, the diagonals still bisect each other at right angles but are unequal, so it becomes a rhombus.

A square; then a rhombus.

(1) Joined along a side: all four sides are 8 cm, so it is a rhombus (angles 60°, 120°, 60°, 120°).

(2) Along the 6 cm side: sides 8, 8, 8, 8, a rhombus. Along an 8 cm side turned around: sides 8, 6, 8, 6 with opposite sides equal, a parallelogram. Along an 8 cm side as mirror images: sides 8, 8, 6, 6 with equal adjacent pairs, a kite.

(3) Each of the three sides can be shared in two ways:

• Turned around (half-turn): a parallelogram each time (sides 6, 9, 6, 9; 6, 12, 6, 12; 9, 12, 9, 12).
• As mirror images: a kite. Along the 12 cm side it is a convex kite (sides 6, 6, 9, 9). Along the 9 cm or 6 cm side, the obtuse angle (about 104.5°) is doubled to more than 180°, giving a concave kite (arrowhead) with sides 6, 6, 12, 12 or 9, 9, 12, 12.

Justification: in a half-turn join, each side meets its equal on the opposite side; in a mirror join, equal sides are next to each other.

(1) rhombus (2) rhombus, parallelogram or kite (3) three parallelograms and three kites (two of them concave).

ΔABD ≅ ΔCBD by SSS (AB = CB, AD = CD, BD common), so $\angle ABD=\angle CBD$ and $\angle ADB=\angle CDB$: BD bisects ∠B and ∠D.

Now ΔABO ≅ ΔCBO by SAS (AB = CB, ∠ABO = ∠CBO, BO common). So AO = CO, and $\angle AOB=\angle COB$; these add to 180°, so each is 90°.

ΔABD ≅ ΔCBD (SSS) gives the angle bisection; then ΔABO ≅ ΔCBO (SAS) gives AO = OC and ∠AOB = 90°.

XY and WZ are perpendicular to UV, and XWZY is a rectangle, so XY = WZ. Also UX = VW (equal sides) and $\angle XYU=\angle WZV=90°$. By RHS, ΔUXY ≅ ΔVWZ, so ∠U = ∠V.

RHS: right angles at Y and Z, hypotenuses UX = VW, and XY = WZ.

All four sides are 4 cm (a rhombus). Its angles are 60°, 120°, 60°, 120° (each 120° angle is made of two 60° angles).

Sides 4 cm each; angles 60°, 120°, 60°, 120°.

Draw PQ = 6 cm and its perpendicular bisector. On it mark S on one side and R on the other so that RS = 8 cm with the two parts unequal (e.g. TS = 2 cm, TR = 6 cm). Join P, S, Q, R.

(If the parts are equal, 4 cm each, the kite becomes a rhombus.)

One diagonal must be the perpendicular bisector of the other: e.g. TS = 2 cm, TR = 6 cm on the bisector of PQ = 6 cm.

(i) Co-interior angles add to 180°: the top angles are $180°-135°=$ 45° and $180°-105°=$ 75°.

(ii) The equal angles at the 100° end are both 100°, and the angles at the other parallel side are $180°-100°=$ 80° each.

(i) 45° and 75° (ii) 100°, 80°, 80°

The parallelogram set contains the rectangle and rhombus sets, which overlap in the squares. The kite set overlaps the parallelogram set exactly in the rhombuses (so the rhombus set, including squares, lies inside both).

(i) A rhombus (and so also a square).

(ii) Yes, a square: it is a rectangle with equal adjacent sides. No other rectangle is a kite.

(iii) No. Every rhombus is a kite, but a kite need not have all four sides equal.

Rhombuses are exactly the kites that are parallelograms; (i) rhombus (ii) only a square (iii) no, every rhombus is a kite but not conversely.

In ΔRIO, $\angle RIO=90°$, so $\angle ROI=180°-90°-30°=60°$. In rectangle RODS, $\angle ROD=90°$, so $\angle IOD=90°-60°=$ 30°.

30°

Draw AB = 6 cm. Construct its perpendicular bisector with a compass (arcs of equal radius from A and B), meeting AB at O. Mark C and D on it with OC = OD = 3 cm. Join A, C, B, D.

Use the compass perpendicular bisector of AB and mark 3 cm on each side of O.

Each corner triangle (like ΔCUV) has two equal sides (half the side of the square) and a right angle, so all four are congruent (SAS): the four sides of UXWV are equal. Each corner triangle is isosceles right-angled, so its base angles are 45°; at U, $45°+\angle U+45°=180°$ gives $\angle U=90°$, and similarly for the others. So UXWV is a square.

Other ways: mark points at the same distance from each corner, all going round in the same direction (Fig. (b)). The four corner right triangles are congruent (SAS), so the inner sides are equal, and the angles at each inner corner are $180°-(a+b)=90°$ (since $a+b=90°$ in each right triangle). So the inner figure is always a square.

UXWV is a square; any four points at equal distances from the corners (going round the same way) also give a square.

Yes. Four equal sides make it a rhombus, so it is a parallelogram: opposite angles are equal and adjacent angles add to 180°. With one angle 90°, all four are 90°. Equal sides and right angles: a square.

Yes: a rhombus with one right angle has all right angles.

Join a diagonal AC. ΔABC ≅ ΔCDA by SSS (AB = CD, BC = DA, AC common). So the alternate angles $\angle BAC=\angle DCA$ and $\angle BCA=\angle DAC$ are equal, which gives AB ∥ DC and BC ∥ AD. It is a parallelogram.

A parallelogram (SSS gives equal alternate angles, so opposite sides are parallel).

Yes. Join BD. It splits ABCD into ΔABD and ΔCBD, whose angles add to $180°+180°=360°$. These six angles together make up exactly the four angles of ABCD, with the reflex angle at D counted as the inside angle (more than 180°). Measuring a drawn figure confirms it (e.g. 40° + 30° + 30° + 260° = 360°).

Yes, 360°: the diagonal BD splits it into two triangles (the angle at D is reflex).

(i) False: it must be a rectangle; it is a square only if the diagonals are also perpendicular.

(ii) True: the fourth angle is $360°-270°=90°$.

(iii) True: the two pairs of opposite triangles are congruent (SAS), giving equal alternate angles, so opposite sides are parallel.

(iv) False: a kite also has perpendicular diagonals.

(v) True: if the angles are a, b, a, b then $2a+2b=360°$, so $a+b=180°$, and the co-interior angles show both pairs of opposite sides are parallel.

(vi) True: each angle is $360°÷4=90°$.

(vii) False: only one pair of sides is parallel (unless it is a rectangle).

(i) F (ii) T (iii) T (iv) F (v) T (vi) T (vii) F

The triangular crease is repeated in all four quarters by the folding, so the four sides of the shape are mirror images of each other and are equal: the creases form a rhombus (with the fold lines as its diagonals).

Nested rhombuses: make several triangular creases at the same corner, one after another, parallel to each other at different distances from the corner.

A square: make the crease cut off equal lengths along the two folded edges (a crease at 45° to the folds). Then the diagonals of the rhombus are equal, so it is a square.

A rhombus; parallel creases at different distances give nested rhombuses; a crease at 45° (equal lengths on both folds) gives a square.