Key Concepts & Formulae at a Glance
- Quadrilateral: a 4-sided polygon.
- Parallelogram: a quadrilateral with both pairs of opposite sides parallel and equal.
Other properties: opposite angles are equal; diagonals bisect each other; adjacent angles are supplementary. - Rectangle: a parallelogram with one angle \(90^\circ\) (subsequently, all angles are \(90^\circ\)).
- Rhombus: a parallelogram with all sides equal.
Other properties: diagonals are perpendicular bisectors of each other; diagonals bisect the angles of the rhombus. - Square: a parallelogram with all sides equal and one angle \(90^\circ\) (inherits the properties of both rectangle and rhombus).
Other properties: diagonals are equal and bisect each other at \(90^\circ\). - Kite: a quadrilateral with two distinct pairs of adjacent equal sides (\(AB=BC\), \(CD=DA\)). One diagonal bisects the angles at its endpoints, bisects the other diagonal, and is perpendicular to it. (A rhombus/square happens to satisfy this too — see the note on Q4 below.)
- Trapezium: a quadrilateral (not a parallelogram) with at least one pair of opposite sides parallel; co-interior angles on the non-parallel-side transversals are supplementary. An isosceles trapezium (equal non-parallel sides) has equal base angles at each parallel side.
\[\angle A+\angle B+\angle C+\angle D=360^\circ \qquad \text{(any quadrilateral } ABCD\text{)}\]
Other Useful Properties of Lines and Angles
- Vertically Opposite Angles: equal.
- Linear Pair: angles on a straight line add up to \(180^\circ\).
- Corresponding and Alternate Angles: equal, when two parallel lines are cut by a transversal.
- Angle Sum Property: angles of a triangle sum to \(180^\circ\); angles of a quadrilateral sum to \(360^\circ\).
- Isosceles Triangle Property: angles opposite to equal sides are equal.
Rectangle and Rhombus are both parallelograms; Square sits below both since it inherits the properties of each. The Kite and Trapezium shown here are the typical look of those quadrilaterals (with unequal side-pairs) — but by this chapter's own definitions, a rhombus or square technically satisfies the kite condition too (see the note on Question 4 below for the precise, set-based picture).
Note: a "generic" kite (like the one drawn above) is never a parallelogram — but a rhombus is a special case where the kite's two adjacent-side-pairs happen to become equal to each other, which is also exactly what makes it a parallelogram. That's why Rhombus and Square end up belonging to both families at once.
The Carpenter's Problem
QA carpenter has two wooden strips, forming the diagonals of the quadrilateral joining their endpoints. One strip is 8 cm. What should the length of the other strip be, and how should the two strips be joined, so that the resulting quadrilateral is a rectangle?
A rectangle's diagonals are equal and bisect each other
\(\Rightarrow\) second strip must also be \(8\) cm
\(\Rightarrow\) the strips must be joined at their common midpoint
The angle between them can be anything — it doesn't affect the rectangle property
\(\Rightarrow\) second strip must also be \(8\) cm
\(\Rightarrow\) the strips must be joined at their common midpoint
The angle between them can be anything — it doesn't affect the rectangle property
The second strip must also be 8 cm, joined to the first at their common midpoint (any angle between them still gives a rectangle).
MTCan AO = CO (proved above), ∠AOB = ∠COD (vertically opposite angles), and AD = CB be used to establish △AOD ≅ △COB?
Not directly by SAS as written — the angle included between sides \(AO\) and \(AD\) in \(\triangle AOD\) is \(\angle AOD\), not \(\angle AOB\).
But \(B,O,D\) lie on straight diagonal \(BD\)
\(\Rightarrow \angle AOB+\angle AOD=180^\circ\) (linear pair)
Similarly \(A,O,C\) lie on diagonal \(AC\)
\(\Rightarrow \angle COD+\angle COB=180^\circ\) (Linear Pair)
Given \(\angle AOB=\angle COD\) (Vertically Opposite Angles)
\(\Rightarrow \angle AOD=180^\circ-\angle AOB=180^\circ-\angle COD=\angle COB\) (from the two linear pairs above)
Now in \(\triangle AOD\) and \(\triangle COB\):
\(AO=CO\)
\(\angle AOD=\angle COB\) (shown above)
\(AD=CB\)
\(\therefore \triangle AOD\cong\triangle COB\) (SAS)
But \(B,O,D\) lie on straight diagonal \(BD\)
\(\Rightarrow \angle AOB+\angle AOD=180^\circ\) (linear pair)
Similarly \(A,O,C\) lie on diagonal \(AC\)
\(\Rightarrow \angle COD+\angle COB=180^\circ\) (Linear Pair)
Given \(\angle AOB=\angle COD\) (Vertically Opposite Angles)
\(\Rightarrow \angle AOD=180^\circ-\angle AOB=180^\circ-\angle COD=\angle COB\) (from the two linear pairs above)
Now in \(\triangle AOD\) and \(\triangle COB\):
\(AO=CO\)
\(\angle AOD=\angle COB\) (shown above)
\(AD=CB\)
\(\therefore \triangle AOD\cong\triangle COB\) (SAS)
Yes — but only after using the straight-line (linear pair) relationship to first show ∠AOD = ∠COB; then SAS gives △AOD ≅ △COB.
Figure it Out — Rectangles & Squares
1Find all the other angles inside the following rectangles.
(i) Diagonals of a rectangle are equal and bisect each other (Property of a Rectangle)
\(\Rightarrow\) all four triangles formed (\(\triangle AOB,\triangle BOC,\triangle COD,\triangle DOA\)) are isosceles
\(\angle ABD=30^\circ\) (given)
\(\angle ADB=60^\circ\) (Angle Sum Property in \(\triangle ABD\), since \(\angle DAB=90^\circ\): \(180^\circ-90^\circ-30^\circ\))
\(\angle DBC=90^\circ-30^\circ=60^\circ\) (\(\angle ABC=90^\circ\), corner of rectangle)
\(\angle ACB=60^\circ\) (Isosceles Triangle Property, \(OB=OC\Rightarrow\angle ACB=\angle DBC\))
\(\angle CAD=60^\circ\) (Alternate Angles, \(AD\parallel BC\), transversal \(AC\Rightarrow\angle CAD=\angle ACB\))
\(\angle BDC=30^\circ\) (Alternate Angles, \(AB\parallel DC\), transversal \(BD\Rightarrow\angle BDC=\angle ABD\))
\(\angle ACD=30^\circ\) (Isosceles Triangle Property, \(OC=OD\Rightarrow\angle ACD=\angle BDC\))
\(\Rightarrow\) all four triangles formed (\(\triangle AOB,\triangle BOC,\triangle COD,\triangle DOA\)) are isosceles
\(\angle ABD=30^\circ\) (given)
\(\angle ADB=60^\circ\) (Angle Sum Property in \(\triangle ABD\), since \(\angle DAB=90^\circ\): \(180^\circ-90^\circ-30^\circ\))
\(\angle DBC=90^\circ-30^\circ=60^\circ\) (\(\angle ABC=90^\circ\), corner of rectangle)
\(\angle ACB=60^\circ\) (Isosceles Triangle Property, \(OB=OC\Rightarrow\angle ACB=\angle DBC\))
\(\angle CAD=60^\circ\) (Alternate Angles, \(AD\parallel BC\), transversal \(AC\Rightarrow\angle CAD=\angle ACB\))
\(\angle BDC=30^\circ\) (Alternate Angles, \(AB\parallel DC\), transversal \(BD\Rightarrow\angle BDC=\angle ABD\))
\(\angle ACD=30^\circ\) (Isosceles Triangle Property, \(OC=OD\Rightarrow\angle ACD=\angle BDC\))
(ii) \(\angle POS=110^\circ\) (given)
\(\angle QOR=110^\circ\) (Vertically Opposite Angles to \(\angle POS\))
\(\angle QOP=\angle ROS=70^\circ\) (Linear Pair with \(\angle POS\))
\(\angle OQR=\angle ORQ=35^\circ\) (Isosceles Triangle Property in \(\triangle OQR\), \(OQ=OR\); Angle Sum Property: \(\frac{180^\circ-110^\circ}{2}\))
\(\angle OQP=\angle OPQ=55^\circ\) (Isosceles Triangle Property in \(\triangle OPQ\), \(OP=OQ\); Angle Sum Property: \(\frac{180^\circ-70^\circ}{2}\))
\(\angle ORS=\angle OSR=55^\circ\) (Isosceles Triangle Property in \(\triangle ORS\), \(OR=OS\); Angle Sum Property: \(\frac{180^\circ-70^\circ}{2}\))
\(\angle QOR=110^\circ\) (Vertically Opposite Angles to \(\angle POS\))
\(\angle QOP=\angle ROS=70^\circ\) (Linear Pair with \(\angle POS\))
\(\angle OQR=\angle ORQ=35^\circ\) (Isosceles Triangle Property in \(\triangle OQR\), \(OQ=OR\); Angle Sum Property: \(\frac{180^\circ-110^\circ}{2}\))
\(\angle OQP=\angle OPQ=55^\circ\) (Isosceles Triangle Property in \(\triangle OPQ\), \(OP=OQ\); Angle Sum Property: \(\frac{180^\circ-70^\circ}{2}\))
\(\angle ORS=\angle OSR=55^\circ\) (Isosceles Triangle Property in \(\triangle ORS\), \(OR=OS\); Angle Sum Property: \(\frac{180^\circ-70^\circ}{2}\))
(i) ∠ABD=30°, ∠CAD=60°, ∠ADB=60°, ∠BDC=30°, ∠ACD=30°, ∠ACB=60°. (ii) ∠QOP=∠ROS=70°; ∠OQR=∠ORQ=35°; ∠OQP=∠OPQ=∠ORS=∠OSR=55°.
2Draw a quadrilateral whose diagonals have equal lengths of 8 cm, bisect each other, and intersect at an angle of (i) 30° (ii) 40° (iii) 90° (iv) 140°.
For (i), 30°:
Draw \(AB=8\) cm
Mark midpoint \(O\)
Draw \(30^\circ\) angle at \(O\) on ray \(OB\)
Mark \(C\) on this ray with \(OC=4\) cm
Mark \(D\) on the opposite ray with \(OD=4\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required quadrilateral
Draw \(AB=8\) cm
Mark midpoint \(O\)
Draw \(30^\circ\) angle at \(O\) on ray \(OB\)
Mark \(C\) on this ray with \(OC=4\) cm
Mark \(D\) on the opposite ray with \(OD=4\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required quadrilateral
(ii) 40°, (iii) 90°, (iv) 140° are all drawn by the same process, just changing the angle at \(O\).
(i), (ii), (iv) each give a rectangle (equal diagonals bisecting each other). (iii) at 90° gives a square.
3Consider 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.
\(PL\) and \(AM\) are diameters of the same circle
\(\Rightarrow\) they bisect each other at \(O\), and \(PL=AM\)
Given: \(PL \perp AM\)
Equal diagonals, bisecting each other, at \(90^\circ\)
\(\Rightarrow\) this is the condition for a square
\(\Rightarrow\) they bisect each other at \(O\), and \(PL=AM\)
Given: \(PL \perp AM\)
Equal diagonals, bisecting each other, at \(90^\circ\)
\(\Rightarrow\) this is the condition for a square
APML is a square.
4Math Talk — 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?
The little square marks ∠C — that's a corner of quadrilateral ACBD, not the crossing angle at O (which can be anything).
Cross the two equal sticks \(AB\) and \(CD\) so their midpoints coincide at \(O\)
Join the four endpoints with thread \(\to\) quadrilateral \(ACBD\)
Diagonals \(AB\) and \(CD\) are equal, and bisect each other
\(\Rightarrow ACBD\) is a rectangle
\(\therefore \angle C=90^\circ\)
Join the four endpoints with thread \(\to\) quadrilateral \(ACBD\)
Diagonals \(AB\) and \(CD\) are equal, and bisect each other
\(\Rightarrow ACBD\) is a rectangle
\(\therefore \angle C=90^\circ\)
Cross the equal sticks at their common midpoint; the quadrilateral formed by their endpoints is a rectangle, so every corner is 90°.
5Try This — 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?
No — opposite sides parallel and equal is the definition of a parallelogram, which need not have right angles. A better (minimal) definition of a rectangle is: a parallelogram with one angle \(90^\circ\) — once one angle is \(90^\circ\), the adjacent angle must also be \(90^\circ\) (adjacent angles of a parallelogram are supplementary: \(180^\circ-90^\circ=90^\circ\)), and then the opposite angles match these, so all four angles automatically become \(90^\circ\).
No — that describes a parallelogram in general, not specifically a rectangle. A rectangle is a parallelogram with one angle 90° (which then forces all four angles to be 90°).
Figure it Out — Parallelograms & Rhombuses
1Find the remaining angles in the following quadrilaterals.
(i) Parallelogram, \(\angle P=40^\circ\):
\(\angle E=140^\circ\) (Adjacent Angles of a Parallelogram are Supplementary: \(180^\circ-40^\circ\))
\(\angle R=140^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle R=\angle E\))
\(\angle A=40^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle A=\angle P\))
\(\angle E=140^\circ\) (Adjacent Angles of a Parallelogram are Supplementary: \(180^\circ-40^\circ\))
\(\angle R=140^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle R=\angle E\))
\(\angle A=40^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle A=\angle P\))
(ii) Parallelogram, \(\angle P=110^\circ\):
\(\angle Q=70^\circ\) (Adjacent Angles of a Parallelogram are Supplementary: \(180^\circ-110^\circ\))
\(\angle S=70^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle S=\angle Q\))
\(\angle R=110^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle R=\angle P\))
\(\angle Q=70^\circ\) (Adjacent Angles of a Parallelogram are Supplementary: \(180^\circ-110^\circ\))
\(\angle S=70^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle S=\angle Q\))
\(\angle R=110^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle R=\angle P\))
(iii) Diagonal-split angle \(30^\circ\) at \(V\):
\(\angle XVU=\angle XVW=30^\circ\) (given)
\(\angle UVW=\angle XVU+\angle XVW=60^\circ\) (Angle Addition)
\(\angle WXU=60^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle WXU=\angle UVW\))
\(\angle U=180^\circ-60^\circ=120^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(\angle W=\angle U=120^\circ\) (Opposite Angles of a Parallelogram are Equal)
\(\angle XVU=\angle XVW=30^\circ\) (given)
\(\angle UVW=\angle XVU+\angle XVW=60^\circ\) (Angle Addition)
\(\angle WXU=60^\circ\) (Opposite Angles of a Parallelogram are Equal: \(\angle WXU=\angle UVW\))
\(\angle U=180^\circ-60^\circ=120^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(\angle W=\angle U=120^\circ\) (Opposite Angles of a Parallelogram are Equal)
(iv) Marked angle \(20^\circ\) at \(E\):
Diagonal \(OE\) splits the interior angle at \(E\) (between sides \(EI\) and \(EA\)) into \(\angle IEO\) and \(\angle OEA\), each \(20^\circ\) (given)
\(\angle IEA=20^\circ+20^\circ=40^\circ\) (Angle Addition)
\(\angle A=\angle I=180^\circ-40^\circ=140^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
Diagonal \(OE\) splits the interior angle at \(E\) (between sides \(EI\) and \(EA\)) into \(\angle IEO\) and \(\angle OEA\), each \(20^\circ\) (given)
\(\angle IEA=20^\circ+20^\circ=40^\circ\) (Angle Addition)
\(\angle A=\angle I=180^\circ-40^\circ=140^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
(i) ∠E=140°, ∠R=140°, ∠A=40°. (ii) ∠Q=70°, ∠S=70°, ∠R=110°. (iii) ∠U=∠W=120°. (iv) ∠A=∠I=140°.
2Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
Draw \(AB=7\) cm
Mark midpoint \(O\)
Through \(O\), draw a ray at \(140^\circ\) to \(AB\)
Mark \(C\) on this ray with \(OC=2.5\) cm
Mark \(D\) on the opposite ray with \(OD=2.5\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required parallelogram
Mark midpoint \(O\)
Through \(O\), draw a ray at \(140^\circ\) to \(AB\)
Mark \(C\) on this ray with \(OC=2.5\) cm
Mark \(D\) on the opposite ray with \(OD=2.5\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required parallelogram
Bisect a 7 cm and a 5 cm segment at the same point, crossing at 140°, and join the endpoints in order.
3Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Draw \(AB=5\) cm
Mark midpoint \(O\)
Draw a line perpendicular to \(AB\) through \(O\) (rhombus diagonals meet at \(90^\circ\))
Mark \(C\) on this perpendicular with \(OC=2\) cm
Mark \(D\) on the opposite ray with \(OD=2\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required rhombus
Mark midpoint \(O\)
Draw a line perpendicular to \(AB\) through \(O\) (rhombus diagonals meet at \(90^\circ\))
Mark \(C\) on this perpendicular with \(OC=2\) cm
Mark \(D\) on the opposite ray with \(OD=2\) cm
Join \(A,D,B,C\) in order
\(\therefore ADBC\) is the required rhombus
Bisect a 5 cm and a 4 cm segment perpendicularly at the same point, and join the endpoints in order.
Geoboard & Joining Triangles Activities
QPlace two rubber bands perpendicular to each other on a geoboard, forming diagonals of equal length. Join the ends. What is the quadrilateral that you get? Justify your answer.
Diagonals: equal length, perpendicular, bisecting each other (joined at centre)
Equal diagonals + bisecting each other \(\Rightarrow\) rectangle
Also perpendicular \(\Rightarrow\) all sides equal too
Equal diagonals + bisecting each other \(\Rightarrow\) rectangle
Also perpendicular \(\Rightarrow\) all sides equal too
A square — equal diagonals, bisecting each other, at 90°.
QExtend one of the diagonals on both sides by 2 cm. What quadrilateral will you get now? Justify your answer.
One diagonal now longer than the other (unequal)
Both diagonals still bisect each other at the same centre point
Still perpendicular (extending along the same line doesn't change the angle)
Both diagonals still bisect each other at the same centre point
Still perpendicular (extending along the same line doesn't change the angle)
A rhombus — diagonals unequal now, but still bisecting each other at 90°.
QTake two cardboard cutouts of an equilateral triangle of side 8 cm. Join them to get a quadrilateral. What type of quadrilateral is this? Justify your answer.
Joined along one full side (shared side becomes a diagonal)
All four outer sides = 8 cm (sides of the equilateral triangles)
Two opposite corners (triangle apexes) = \(60^\circ\) each (angle of an equilateral triangle)
Other two corners (where triangles meet) = \(60^\circ+60^\circ=120^\circ\) each (Angle Addition)
All four outer sides = 8 cm (sides of the equilateral triangles)
Two opposite corners (triangle apexes) = \(60^\circ\) each (angle of an equilateral triangle)
Other two corners (where triangles meet) = \(60^\circ+60^\circ=120^\circ\) each (Angle Addition)
A rhombus — all sides 8 cm, angles 60°, 120°, 60°, 120°.
QTake two cardboard cutouts of an isosceles triangle with sides 8 cm, 8 cm, 6 cm. What are the different ways they can be joined to get a quadrilateral? Identify each resulting quadrilateral.
Joined base-to-base (along the 6 cm side):
Adjacent sides become 8,8 and 8,8 — two pairs of adjacent equal sides
Joined along an 8 cm side instead:
Opposite sides become 8 cm and 6 cm — two pairs of equal, parallel opposite sides
Adjacent sides become 8,8 and 8,8 — two pairs of adjacent equal sides
Joined along an 8 cm side instead:
Opposite sides become 8 cm and 6 cm — two pairs of equal, parallel opposite sides
Base-to-base gives a kite. Joined along the 8 cm side gives a parallelogram.
QTake two cardboard cutouts of a scalene triangle with sides 6 cm, 9 cm, 12 cm. What are the different ways they can be joined to get a quadrilateral? Identify each resulting quadrilateral.
Joined along any one of the three sides (6, 9, or 12 cm):
Each way gives two pairs of adjacent equal sides (from the two identical triangle copies)
A parallelogram is not possible — a scalene triangle has no two equal sides to form equal opposite pairs
Each way gives two pairs of adjacent equal sides (from the two identical triangle copies)
A parallelogram is not possible — a scalene triangle has no two equal sides to form equal opposite pairs
All three ways give a kite; a parallelogram is never possible with scalene triangles.
Kite Diagonal Properties (Math Talk)
MTIn kite ABCD, show that diagonal BD (i) bisects ∠ABC and ∠ADC, and (ii) bisects diagonal AC and is perpendicular to it.
(i) In \(\triangle ABD\) and \(\triangle CBD\):
\(AB=CB\) (given, kite)
\(AD=CD\) (given, kite)
\(BD=BD\) (common)
\(\therefore \triangle ABD\cong\triangle CBD\) (SSS)
\(\Rightarrow \angle ABD=\angle CBD\) and \(\angle ADB=\angle CDB\)
\(\therefore BD\) bisects \(\angle ABC\) and \(\angle ADC\)
\(AB=CB\) (given, kite)
\(AD=CD\) (given, kite)
\(BD=BD\) (common)
\(\therefore \triangle ABD\cong\triangle CBD\) (SSS)
\(\Rightarrow \angle ABD=\angle CBD\) and \(\angle ADB=\angle CDB\)
\(\therefore BD\) bisects \(\angle ABC\) and \(\angle ADC\)
(ii) Let \(BD\) and \(AC\) intersect at \(O\).
In \(\triangle AOB\) and \(\triangle COB\):
\(AB=CB\) (given)
\(\angle ABO=\angle CBO\) (shown in part (i))
\(BO=BO\) (common)
\(\therefore \triangle AOB\cong\triangle COB\) (SAS)
\(\Rightarrow AO=CO\)
and \(\angle AOB=\angle COB\)
Since \(\angle AOB+\angle COB=180^\circ\) (linear pair)
\(\angle AOB=\angle COB=90^\circ\)
In \(\triangle AOB\) and \(\triangle COB\):
\(AB=CB\) (given)
\(\angle ABO=\angle CBO\) (shown in part (i))
\(BO=BO\) (common)
\(\therefore \triangle AOB\cong\triangle COB\) (SAS)
\(\Rightarrow AO=CO\)
and \(\angle AOB=\angle COB\)
Since \(\angle AOB+\angle COB=180^\circ\) (linear pair)
\(\angle AOB=\angle COB=90^\circ\)
△ABD ≅ △CBD (SSS) gives the angle bisection; △AOB ≅ △COB (SAS) gives AO = CO and BD ⊥ AC.
Figure it Out — Kite & Trapezium
1Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
All sides = 4 cm (equilateral triangle sides)
Two opposite angles = \(60^\circ\) each (apex angles)
Other two angles = \(60^\circ+60^\circ=120^\circ\) each (angles combine where triangles join)
Two opposite angles = \(60^\circ\) each (apex angles)
Other two angles = \(60^\circ+60^\circ=120^\circ\) each (angles combine where triangles join)
All four sides = 4 cm; angles = 60°, 120°, 60°, 120° (a rhombus).
2Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Note how T bisects PQ evenly (3 cm + 3 cm), but is not centred on SR — only 3 cm up to S versus 5 cm down to R. A kite's perpendicular bisector belongs to just one diagonal; the other diagonal only needs to be perpendicular to it, not bisected by it.
Diagonals of a kite are perpendicular, and only one is bisected
Draw \(PQ=6\) cm
Construct its perpendicular bisector, meeting it at \(T\)
On this bisector, mark \(R\) and \(S\) (unequal distances from \(T\)) so that \(RS=8\) cm
Join \(P,R,Q,S\)
\(\therefore PRQS\) is the required kite
Draw \(PQ=6\) cm
Construct its perpendicular bisector, meeting it at \(T\)
On this bisector, mark \(R\) and \(S\) (unequal distances from \(T\)) so that \(RS=8\) cm
Join \(P,R,Q,S\)
\(\therefore PRQS\) is the required kite
Draw a 6 cm segment PQ; construct its perpendicular bisector; mark S and R on it, 8 cm apart (unequal distances from the midpoint); join PSQR.
3Find the remaining angles in the following trapeziums — (i) PQRS with PQ ∥ SR, ∠P = 135°, ∠Q = 105°; (ii) isosceles trapezium ABCD with AD = BC and ∠D = 100°.
(i) \(PQ \parallel SR\) \(\Rightarrow\) co-interior angles (on each leg) are supplementary
\(\angle S=180^\circ-135^\circ=45^\circ\) (Co-interior Angles, leg \(PS\))
\(\angle R=180^\circ-105^\circ=75^\circ\) (Co-interior Angles, leg \(QR\))
\(\angle S=180^\circ-135^\circ=45^\circ\) (Co-interior Angles, leg \(PS\))
\(\angle R=180^\circ-105^\circ=75^\circ\) (Co-interior Angles, leg \(QR\))
(ii) Isosceles trapezium, \(\angle D=100^\circ\):
\(\angle A=180^\circ-100^\circ=80^\circ\) (Co-interior Angles, leg \(AD\), since \(AB\parallel DC\))
\(\angle B=\angle A=80^\circ\) (Property of an Isosceles Trapezium: base angles at the equal side \(BC\) are equal)
\(\angle A=180^\circ-100^\circ=80^\circ\) (Co-interior Angles, leg \(AD\), since \(AB\parallel DC\))
\(\angle B=\angle A=80^\circ\) (Property of an Isosceles Trapezium: base angles at the equal side \(BC\) are equal)
(i) ∠S=45°, ∠R=75°. (ii) ∠A=80°, ∠B=80°.
4Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions —
(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?
Rectangle is drawn entirely inside Parallelogram (every rectangle is a parallelogram), but its left portion dips into the region Parallelogram shares with Kite. That shared Kite∩Parallelogram region is exactly Rhombus — and the part of it that's also inside Rectangle is exactly Square. So Kite and Square do share common area (the innermost zone), since every square is a rhombus and every rhombus is a kite.
Why isn't Kite ∩ Parallelogram empty? This chapter defines a kite as any quadrilateral \(ABCD\) with \(AB=BC\) and \(CD=DA\) — it doesn't require the two pairs to be different lengths from each other. A rhombus (all four sides equal) technically satisfies \(AB=BC\) and \(CD=DA\) too, so by this definition every rhombus is also a kite — which is exactly why part (i)'s answer is "Rhombus and Square." A general kite (with two visibly different adjacent-side lengths, like the one drawn in the Key Concepts diagram above) is indeed never a parallelogram, but the special case where a kite's sides are all equal (a rhombus) is simultaneously a parallelogram.
(i) Rhombus and Square. (ii) No. (iii) No. A rhombus is a kite whereas a kite need not be a rhombus.
(i) Rhombus and Square. (ii) No. (iii) No — every rhombus is a kite, but not every kite is a rhombus.
5If PAIR and RODS are two rectangles, find ∠IOD.
Rectangle PAIR and rectangle RODS share the vertex \(R\). \(I\) is a right-angle corner of PAIR, and \(O\) lies on ray \(IA\) (so \(\angle RIO=\angle RIA=90^\circ\)). \(\angle IRO=30^\circ\) is given at \(R\). \(O\) is a corner of RODS, so \(\angle ROD=90^\circ\).
In \(\triangle RIO\):
\(\angle RIO=90^\circ\) (\(O\) lies on ray \(IA\), and \(\angle RIA=90^\circ\) is a corner of rectangle PAIR)
\(\angle IRO=30^\circ\) (given)
\(\angle IRO+\angle RIO+\angle IOR=180^\circ\) (Angle Sum Property)
\(30^\circ+90^\circ+\angle IOR=180^\circ\)
\(\angle IOR=60^\circ\)
\(\angle ROD=90^\circ\) (corner of rectangle RODS)
\(\angle ROD=\angle IOR+\angle IOD\) (Angle Addition, since ray \(OI\) lies inside \(\angle ROD\))
\(90^\circ=60^\circ+\angle IOD\)
\(\angle IOD=30^\circ\)
\(\angle RIO=90^\circ\) (\(O\) lies on ray \(IA\), and \(\angle RIA=90^\circ\) is a corner of rectangle PAIR)
\(\angle IRO=30^\circ\) (given)
\(\angle IRO+\angle RIO+\angle IOR=180^\circ\) (Angle Sum Property)
\(30^\circ+90^\circ+\angle IOR=180^\circ\)
\(\angle IOR=60^\circ\)
\(\angle ROD=90^\circ\) (corner of rectangle RODS)
\(\angle ROD=\angle IOR+\angle IOD\) (Angle Addition, since ray \(OI\) lies inside \(\angle ROD\))
\(90^\circ=60^\circ+\angle IOD\)
\(\angle IOD=30^\circ\)
∠IOD = 30°.
6Construct a square with diagonal 6 cm without using a protractor.
Draw \(AB=6\) cm
Using a compass, construct the perpendicular bisector of \(AB\), meeting it at \(O\)
Mark \(C\) on this perpendicular with \(OC=3\) cm
Mark \(D\) on the opposite ray with \(OD=3\) cm
Join \(A,C,B,D\) in order
\(\therefore ACBD\) is the required square
Using a compass, construct the perpendicular bisector of \(AB\), meeting it at \(O\)
Mark \(C\) on this perpendicular with \(OC=3\) cm
Mark \(D\) on the opposite ray with \(OD=3\) cm
Join \(A,C,B,D\) in order
\(\therefore ACBD\) is the required square
Compass-construct the perpendicular bisector of a 6 cm segment, mark points 3 cm either side of the midpoint on it, and join the four endpoints in order.
7CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).
Let side of square \(=x\)
\(CU=CV=\frac{x}{2}\) (\(U,V\) are midpoints, given), and \(\angle C=90^\circ\) (angle of a square)
By Pythagoras in \(\triangle CUV\): \(UV=\dfrac{x}{\sqrt2}\)
Similarly \(VW=WX=XU=\dfrac{x}{\sqrt2}\) (by the same argument at each corner)
\(\therefore\) all sides of \(UVWX\) are equal
\(\triangle CUV\) is right-angled and isosceles \(\Rightarrow\) its two base angles \(=45^\circ\) each (Isosceles Triangle Property + Angle Sum Property)
Same pattern at every corner \(\Rightarrow\) each angle of \(UVWX=180^\circ-45^\circ-45^\circ=90^\circ\) (Angle Sum Property / Linear Pair)
All sides equal, all angles \(90^\circ\)
\(\therefore UVWX\) is a square
\(CU=CV=\frac{x}{2}\) (\(U,V\) are midpoints, given), and \(\angle C=90^\circ\) (angle of a square)
By Pythagoras in \(\triangle CUV\): \(UV=\dfrac{x}{\sqrt2}\)
Similarly \(VW=WX=XU=\dfrac{x}{\sqrt2}\) (by the same argument at each corner)
\(\therefore\) all sides of \(UVWX\) are equal
\(\triangle CUV\) is right-angled and isosceles \(\Rightarrow\) its two base angles \(=45^\circ\) each (Isosceles Triangle Property + Angle Sum Property)
Same pattern at every corner \(\Rightarrow\) each angle of \(UVWX=180^\circ-45^\circ-45^\circ=90^\circ\) (Angle Sum Property / Linear Pair)
All sides equal, all angles \(90^\circ\)
\(\therefore UVWX\) is a square
UVWX is a square. Marking equal offsets from each corner (going around consistently) on any square's sides always gives another inner square.
8If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Four equal sides \(\Rightarrow\) it is at least a rhombus (a rhombus is a parallelogram, so its angle properties apply)
One angle \(=90^\circ\) (given)
\(\Rightarrow\) adjacent angle \(=180^\circ-90^\circ=90^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(\Rightarrow\) opposite angles \(=90^\circ\) each too (Opposite Angles of a Parallelogram are Equal)
All angles \(90^\circ\) + all sides equal
\(\therefore\) it is a square
One angle \(=90^\circ\) (given)
\(\Rightarrow\) adjacent angle \(=180^\circ-90^\circ=90^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(\Rightarrow\) opposite angles \(=90^\circ\) each too (Opposite Angles of a Parallelogram are Equal)
All angles \(90^\circ\) + all sides equal
\(\therefore\) it is a square
Yes — a rhombus with one right angle has all four angles equal to 90°, making it a square.
9What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer.
In \(ABCD\) with \(AB=CD\), \(BC=AD\), draw diagonal \(AC\)
In \(\triangle ABC\) and \(\triangle CDA\):
\(AB=CD\) (given)
\(BC=AD\) (given)
\(AC=AC\) (Common Side)
\(\therefore \triangle ABC\cong\triangle CDA\) (SSS Congruence)
\(\Rightarrow \angle 1=\angle 4\) and \(\angle 2=\angle 3\) (CPCT)
\(\Rightarrow AB\parallel CD\) and \(BC\parallel AD\) (Alternate Angles are equal \(\Rightarrow\) lines are parallel)
In \(\triangle ABC\) and \(\triangle CDA\):
\(AB=CD\) (given)
\(BC=AD\) (given)
\(AC=AC\) (Common Side)
\(\therefore \triangle ABC\cong\triangle CDA\) (SSS Congruence)
\(\Rightarrow \angle 1=\angle 4\) and \(\angle 2=\angle 3\) (CPCT)
\(\Rightarrow AB\parallel CD\) and \(BC\parallel AD\) (Alternate Angles are equal \(\Rightarrow\) lines are parallel)
A quadrilateral with both pairs of opposite sides equal is always a parallelogram.
10Try This — Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
Join \(BD\)
\(\triangle ABD\) (angles 1, 2, 3): sum \(=180^\circ\) (Angle Sum Property)
\(\triangle BDC\) (angles 4, 5, 6): sum \(=180^\circ\) (Angle Sum Property)
Total \(=180^\circ+180^\circ=360^\circ\) (Addition)
Regrouping into the four angles of quadrilateral \(ADCB\): \(\angle A=1\), \(\angle B=2\), \(\angle C=6\), and reflex \(\angle D=3+5\) (Angle Addition at inward vertex \(D\))
\(\triangle ABD\) (angles 1, 2, 3): sum \(=180^\circ\) (Angle Sum Property)
\(\triangle BDC\) (angles 4, 5, 6): sum \(=180^\circ\) (Angle Sum Property)
Total \(=180^\circ+180^\circ=360^\circ\) (Addition)
Regrouping into the four angles of quadrilateral \(ADCB\): \(\angle A=1\), \(\angle B=2\), \(\angle C=6\), and reflex \(\angle D=3+5\) (Angle Addition at inward vertex \(D\))
Yes — still 360°, as long as the reflex angle at the inward vertex is measured correctly.
11State whether the following statements are true or false. Justify your answers.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
(ii) A quadrilateral having three right angles must be a rectangle.
(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
(vi) A quadrilateral in which all the angles are equal is a rectangle.
(vii) Isosceles trapeziums are parallelograms.
(i) False — gives a rectangle, not necessarily a square (also needs diagonals at \(90^\circ\)).
(ii) True — Angle Sum Property forces the fourth angle to \(90^\circ\) too.
(iii) True — using SAS (\(AO=CO\), \(\angle AOB=\angle COD\) Vertically Opposite Angles, \(OB=OD\)), \(\triangle AOB\cong\triangle COD\), giving \(AB\parallel CD\) (Alternate Angles); similarly \(AD\parallel BC\), so it is a parallelogram.
(iv) False — describes a kite in general, not necessarily a rhombus.
(v) True — let \(\angle A=\angle C=x\) and \(\angle B=\angle D=y\); then \(2x+2y=360^\circ\) (Angle Sum Property) \(\Rightarrow x+y=180^\circ\). So \(\angle A+\angle B=180^\circ\) and \(\angle B+\angle C=180^\circ\) — each pair of adjacent (co-interior) angles is supplementary, which forces \(AD\parallel BC\) and \(AB\parallel DC\) (Converse of Co-interior Angles), so \(ABCD\) is a parallelogram.
(vi) True — each angle \(=\frac{360^\circ}{4}=90^\circ\).
(vii) False — only one pair of sides is parallel by definition.
(i) False. (ii) True. (iii) True. (iv) False. (v) True. (vi) True. (vii) False.
Puzzle Time — Which Quad?
Paper-Folding Creases
Fold a square sheet in half, then in half again into a quarter. Make a single triangular (diagonal) crease at the corner that sits at the centre of the original sheet, then unfold completely.
1. What shape do the creases form? The single diagonal fold made on the quarter-sheet, once unfolded, leaves four crease segments radiating symmetrically from the centre to the midpoints of each side of the original square — connecting these midpoint creases traces out a rhombus (in fact a square, since the midpoints of a square's sides, joined in order, always form a smaller square rotated \(45^\circ\), by the same reasoning as Question 7 above).
2. Nested diamond creases: repeating the triangular corner-fold at progressively smaller distances from the centre (before unfolding) leaves a set of nested square/diamond creases, each one a scaled-down copy of the first, still centred and rotated the same way.
3. Folding to get a square directly: folding each corner of the quarter-sheet straight in so that it meets the centre fold-point exactly (rather than at an angled diagonal) produces the plain square crease pattern shown — since all four folds are symmetric and meet the same central point, the resulting crease outline is a square, using the same corner-to-centre logic as the diamond fold.
Practice Questions
The questions below are not from the NCERT textbook — they're original practice problems for extra revision, in the same style as the chapter, with full solutions.
1The diagonals of a rectangle intersect at an angle of 50°. Find all eight angles formed at the intersection and inside the rectangle.
Diagonals of a rectangle are equal and bisect each other
\(\Rightarrow\) all four half-diagonals are equal, every triangle formed is isosceles
One pair of vertical angles \(=50^\circ\)
Other pair (linear pair) \(=180^\circ-50^\circ=130^\circ\)
Base angles of triangle with apex \(50^\circ\): \(\dfrac{180^\circ-50^\circ}{2}=65^\circ\) each
Base angles of triangle with apex \(130^\circ\): \(\dfrac{180^\circ-130^\circ}{2}=25^\circ\) each
\(\Rightarrow\) all four half-diagonals are equal, every triangle formed is isosceles
One pair of vertical angles \(=50^\circ\)
Other pair (linear pair) \(=180^\circ-50^\circ=130^\circ\)
Base angles of triangle with apex \(50^\circ\): \(\dfrac{180^\circ-50^\circ}{2}=65^\circ\) each
Base angles of triangle with apex \(130^\circ\): \(\dfrac{180^\circ-130^\circ}{2}=25^\circ\) each
Angles at the centre: 50°, 130°, 50°, 130°. Base angles around the rectangle: 65°, 25°, 65°, 25°, 65°, 25°, 65°, 25° (alternating).
2In a parallelogram, one angle is 30° more than the adjacent angle. Find all four angles.
Let adjacent angles be \(x\) and \(x+30^\circ\)
\(x+(x+30^\circ)=180^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(2x=150^\circ\)
\(x=75^\circ\)
\(\therefore\) angles are \(75^\circ\) and \(105^\circ\)
Opposite angles repeat these values (Opposite Angles of a Parallelogram are Equal)
\(x+(x+30^\circ)=180^\circ\) (Adjacent Angles of a Parallelogram are Supplementary)
\(2x=150^\circ\)
\(x=75^\circ\)
\(\therefore\) angles are \(75^\circ\) and \(105^\circ\)
Opposite angles repeat these values (Opposite Angles of a Parallelogram are Equal)
75°, 105°, 75°, 105°.
3A rhombus has side 10 cm and one diagonal 12 cm. Find the length of the other diagonal.
Diagonals of a rhombus bisect each other at \(90^\circ\) (Property of a Rhombus)
\(\Rightarrow\) each side is the hypotenuse of a right triangle formed by the half-diagonals
Half of known diagonal \(=6\) cm (bisected)
Let other diagonal \(=d\), its half \(=\dfrac{d}{2}\)
\(6^2+\left(\dfrac{d}{2}\right)^2=10^2\) (Pythagoras Theorem)
\(\dfrac{d^2}{4}=100-36=64\)
\(d^2=256\)
\(d=16\)
\(\Rightarrow\) each side is the hypotenuse of a right triangle formed by the half-diagonals
Half of known diagonal \(=6\) cm (bisected)
Let other diagonal \(=d\), its half \(=\dfrac{d}{2}\)
\(6^2+\left(\dfrac{d}{2}\right)^2=10^2\) (Pythagoras Theorem)
\(\dfrac{d^2}{4}=100-36=64\)
\(d^2=256\)
\(d=16\)
The other diagonal is 16 cm.
4A kite has adjacent sides 13 cm, 13 cm and 20 cm, 20 cm. The diagonal that gets bisected (connecting the two "unequal" corners) measures 10 cm. Find the length of the other diagonal (the axis of symmetry).
Bisected diagonal splits into two 5 cm halves, each \(\perp\) to the other diagonal (Property of a Kite)
Using the 13 cm side: \(\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\) (Pythagoras Theorem)
Using the 20 cm side: \(\sqrt{20^2-5^2}=\sqrt{400-25}=\sqrt{375}\approx19.36\) (Pythagoras Theorem)
Using the 13 cm side: \(\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\) (Pythagoras Theorem)
Using the 20 cm side: \(\sqrt{20^2-5^2}=\sqrt{400-25}=\sqrt{375}\approx19.36\) (Pythagoras Theorem)
The other diagonal ≈ 12 + 19.36 = 31.36 cm.
5In an isosceles trapezium PQRS with PQ ∥ SR, ∠P = 70°. Find the other three angles.
Isosceles trapezium \(\Rightarrow\) base angles at each parallel side are equal (Property of an Isosceles Trapezium)
\(\angle Q=\angle P=70^\circ\)
\(\angle S=180^\circ-70^\circ=110^\circ\) (Co-interior Angles, leg \(PS\), since \(PQ\parallel SR\))
\(\angle R=\angle S=110^\circ\) (Property of an Isosceles Trapezium)
\(\angle Q=\angle P=70^\circ\)
\(\angle S=180^\circ-70^\circ=110^\circ\) (Co-interior Angles, leg \(PS\), since \(PQ\parallel SR\))
\(\angle R=\angle S=110^\circ\) (Property of an Isosceles Trapezium)
∠Q = 70°, ∠S = 110°, ∠R = 110°.
6A quadrilateral's diagonals bisect each other, but are neither equal nor perpendicular. What type of quadrilateral must it be — and can it be anything more specific?
Diagonals bisecting each other is exactly the defining test for a parallelogram. Since they're additionally not equal, it can't be a rectangle or square; since they're not perpendicular, it can't be a rhombus either. So it must be a "general" (non-special) parallelogram — no more specific classification applies.
It's a parallelogram, but not a rectangle, rhombus, or square.
7If a quadrilateral's diagonals are equal but do not bisect each other, must it be a rectangle? Justify your answer.
No. A rectangle needs equal diagonals and bisection together — equal diagonals alone isn't enough. For example, an isosceles trapezium has equal diagonals (by symmetry) but they don't bisect each other, and it clearly isn't a rectangle (only one pair of sides is parallel).
No — an isosceles trapezium is a counterexample: equal diagonals, but not bisecting, and not a rectangle.
8Construct a rhombus with side 5 cm and one angle 60°.
Draw \(AB=5\) cm
At \(A\), construct a \(60^\circ\) angle
Mark \(D\) on that ray with \(AD=5\) cm
From \(D\), draw an arc of radius 5 cm
From \(B\), draw an arc of radius 5 cm
Their intersection is \(C\)
Join \(BC\) and \(DC\)
\(\therefore ABCD\) is the required rhombus
At \(A\), construct a \(60^\circ\) angle
Mark \(D\) on that ray with \(AD=5\) cm
From \(D\), draw an arc of radius 5 cm
From \(B\), draw an arc of radius 5 cm
Their intersection is \(C\)
Join \(BC\) and \(DC\)
\(\therefore ABCD\) is the required rhombus
Draw a 60° angle at A with two 5 cm sides, then complete the fourth vertex by striking equal 5 cm arcs from the other two points.
Frequently Asked Questions
A rectangle can be defined just by "all angles equal to 90°" — equal opposite sides and parallel opposite sides follow automatically once all angles are right angles. Equivalently, a rectangle can be defined as a quadrilateral whose diagonals are equal and bisect each other.
A rhombus has all four sides equal, and is always a parallelogram. A kite has only two pairs of adjacent sides equal (not all four, and not opposite sides) and is generally not a parallelogram. Every rhombus is technically a kite, but a general kite is not a rhombus.
Draw one diagonal to split the quadrilateral into two triangles. Each triangle's angles add to 180°, and together the two triangles' six angles regroup exactly into the quadrilateral's four angles — so the total is 180° + 180° = 360°. This holds even for non-convex ("dart"-shaped) quadrilaterals, as long as the reflex angle at any inward-pointing vertex is measured correctly.
Every square is a rhombus (all four sides are equal), but not every rhombus is a square — a rhombus only needs equal sides, while a square additionally needs all angles to be 90°. A rhombus becomes a square exactly when one of its angles happens to be a right angle.
A parallelogram needs both pairs of opposite sides parallel. A trapezium only needs at least one pair parallel — so every parallelogram happens to satisfy the trapezium condition too, but a trapezium with just one parallel pair (like most trapeziums drawn in practice) is not a parallelogram, since its other pair of sides isn't parallel.
Continue with Chapter 5
Move on to Number Play next.