Understanding Quadrilaterals

1.0Curve

A plane figure formed by joining a number of points without lifting a pencil from the paper and without retracing any portion of the drawing other than single points is called a curve. Examples:

Curves can be of different types - open, closed, simple closed etc.

2.0Region

Interior $\to$ Region inside the boundary of the closed curve. Exterior $\to$ Region outside the boundary of the closed curve.

3.0Polygons

A polygon is a simple closed curve made of many line segments. Examples of curves that are polygons $\to$

Examples of curves that are not polygons $\to$

4.0Classification Of Polygons

Polygons are classified by the number of their sides.

5.0Diagonals

A diagonal is a line segment connecting two non-consecutive vertices of a polygon.

Note:

• $0^{\circ }<$ Acute angle $<90^{\circ }$
• $90^{\circ }<$ Obtuse angle $<180^{\circ }$
• $180^{\circ }<$ Reflex angle < $360^{\circ }$

6.0Convex And Concave Polygons

Convex polygon: In convex polygon, the measure of each interior angle is less than $180^{\circ }$. Convex polygons have all their diagonals in the interior of the polygon. Example:

Concave polygon: In concave polygon, the measure of at least one of the interior angles is more than $180^{\circ }$. Concave polygons could have some of their diagonals in the exterior of the polygon. Example:

7.0Regular And Irregular Polygons

A Regular polygon is one which is both equilateral (all sides equal) and equiangular (all angles equal). Equilateral triangle, square $\to$ equilateral and equiangular $\to$ Regular

Rectangle $\to$ Equiangular but not equilateral $\to$ Irregular

• Q. Determine whether each figure is a polygon. If it is, classify the polygon and state whether it is regular. If it is not a polygon, explain why? (i)
  
  (ii)
  
  Explanation: (i) The figure has 6 equal sides and 6 equal angles. It is a regular hexagon. (ii) The figure is not a polygon since it has a curved side.

Note:

• Sum of all the three angles of a triangle is $180^{\circ }$.
• Angle sum property of a quadrilateral. (a polygon with 4 sides) $\to$ Sum of all the four angles of a quadrilateral $=360^{\circ }$

8.0Angle Sum Property

Proof: Let ABCD be a quadrilateral. Draw one of its diagonals, AC

• Q. The three angles of a quadrilateral are $75^{\circ },105^{\circ }$ and $85^{\circ }$. Find its fourth angle. Solution: Let the fourth angle of the quadrilateral be x . According to the angle sum property of a quadrilateral, $75^{\circ }+105^{\circ }+85^{\circ }+\mathrm{x}=360^{\circ }$ $\Rightarrow 265^{\circ }+\mathrm{x}=360^{\circ }\Rightarrow \mathrm{x}=360^{\circ }-265^{\circ }\Rightarrow \mathrm{x}=95^{\circ }$

Sum of the measure of the exterior angles of a polygon:

The sum of the exterior angles of any polygon is equal to $360^{\circ }$. Let us prove this for the case of a quadrilateral $ABCD$.

• Q. Find the number of sides of a regular polygon whose each exterior angle has a measure of $60^{\circ }$. Solution: The measure of all the exterior angles $=360^{\circ }$ Measure of each exterior angle $=60^{\circ }$ $\therefore$ Number of sides of the polygon $=\frac{360^{\circ }}{60^{\circ }}=6$ Thus, the polygon has 6 sides.

Note: Exterior angle of quadrilateral

9.0Quadrilaterals

A polygon with four sides and vertices is called a quadrilateral. Before learning the different types of quadrilaterals, we must know some important terms related to quadrilaterals. Vertex: A, B, C and D are vertices of the quadrilateral ABCD.

Sides: $\mathrm{AB},\mathrm{BC},\mathrm{CD}$ and DA are the sides of the quadrilateral ABCD . Angles: $\angle \mathrm{A},\angle \mathrm{B},\angle \mathrm{C}$ and $\angle \mathrm{D}$ are the angles of the quadrilateral ABCD . Adjacent sides: AB & BC (Common vertex B ), $\mathrm{BC}\&\mathrm{CD}$ (Common vertex C), CD & DA (Common vertex D) and DA & AB (common vertex A) are the four pairs of adjacent sides in the quadrilateral ABCD. Opposite sides: AB & DC and BC & AD are the two pairs of opposite sides of the quadrilateral ABCD. Adjacent angles: $\angle \mathrm{A}\&\angle \mathrm{B}$ (common side AB ), $\angle \mathrm{B}\&\angle \mathrm{C}$ (common side BC ), $\angle \mathrm{C}\&\angle \mathrm{D}$ (common side CD ) and $\angle \mathrm{D}\&\angle \mathrm{A}$ (common side AD ) are the four pairs of adjacent angles in the quadrilateral ABCD. Opposite Angles: $\angle \mathrm{A}\&\angle \mathrm{C}$ and $\angle \mathrm{B}\&\angle \mathrm{D}$ are the two pairs of opposite angles in the quadrilateral ABCD .

Note:

• Any four-sided closed shape is a quadrilateral, but the sides have to be straight and it has to be 2-dimensional (2-D).

10.0Different Types of Quadrilaterals

1. Trapezium: A quadrilateral in which a pair of opposite sides are parallel is called a trapezium.
   
   $\mathrm{AB}|\mid \mathrm{CD}$ $\Rightarrow ABCD$ is a trapezium. Other examples:
   
   (Parallel sides donated by arrows)

If the non-parallel sides of a trapezium are equal, then it is called an isosceles trapezium.

Properties of Trapezium: In a trapezium, the interior angles on the same side of each of the non-parallel sides are supplementary. $\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$ (as $\mathrm{AB}\Vert \mathrm{CD}$ and AD is a transversal).

2. Kite: Kite is a special type of Quadrilateral in which one pair of adjacent sides are equal to each other and the other pair of adjacent sides are equal to each other.
   

In the kite $\mathrm{ABCD},\mathrm{AB}=\mathrm{BC}$ and $\mathrm{CD}=\mathrm{AD}$

Properties of Kite

3. Parallelogram: A parallelogram is a quadrilateral in which both the pairs of opposite sides are parallel. Quadrilateral ABCD is a parallelogram where $AB\Vert CD$ and $BC\Vert AD$.
   

Properties of Parallelogram (i) The opposite sides of a parallelogram are equal in length. (ii) The opposite angles in a parallelogram are equal. (iii) The adjacent angles in a parallelogram are supplementary. (iv) The diagonals of a parallelogram bisect each other.

Proof: Consider a parallelogram ABCD. Draw any diagonal, say AC. Looking at the angles, $\angle 1=\angle 4$. (as $\mathrm{AB}\Vert \mathrm{CD}$ and AC is a transversal $\angle 1$ and $\angle 4$ are alternate interior angles.) Similarly, $\angle 2=\angle 3$

In triangles ADC and $\mathrm{ABC},\angle 1=\angle 4$ $\mathrm{AC}=\mathrm{AC}$ (common side), $\angle 2=\angle 3$ So, $△\mathrm{ADC}\cong △\mathrm{CBA}$ (ASA congruence rule) $\Rightarrow \mathrm{CD}=\mathrm{AB}$ and $\mathrm{AD}=\mathrm{BC}$ Hence proved.

(ii) The opposite angles in a parallelogram are equal. Proof: We have proved above that $△\mathrm{ADC}\cong △\mathrm{CBA}$

(iii) The adjacent angles in a parallelogram are supplementary. Proof : $\mathrm{AB}\Vert \mathrm{CD}$ and AD is a transversal, $\angle \mathrm{A}$ and $\angle \mathrm{D}$ are interior angles on the same side of the transversal. $\Rightarrow \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$

(iv) The diagonals of a parallelogram bisect each other. Proof: Let we draw the two diagonals AC and BD (meeting at 0) of the parallelogram ABCD. In triangles AOB and COD,

$\angle 1=\angle 2$ [alternate interior angle] $\angle 3=\angle 4$ [alternate interior angle] $\mathrm{CD}=\mathrm{AB}$ [opposite sides of parallelogram are equal] So, $△\mathrm{AOB}\cong △\mathrm{COD}$ (ASA congruency) $\Rightarrow \mathrm{AO}=\mathrm{CO}$ and $\mathrm{OB}=\mathrm{OD}$ Hence proved.

• Q. Draw a parallelogram. Label the congruent angles. Explanation
  
  ABCD is a parallelogram with $\mathrm{AB}\Vert \mathrm{CD}$ and $\mathrm{BC}\mid \Vert \mathrm{AD}$. We know, the opposite angles of a parallelogram are equal. $\Rightarrow \angle \mathrm{A}=\angle \mathrm{C}$ and $\angle \mathrm{B}=\angle \mathrm{D}$ The 2 set of congruent angles
• Q. In the given parallelogram $PQRS,OS=5\mathrm{cm}$ and $PR$ is $7cm$ more than $QS$. Find $OP$.
  
  Solution: $\mathrm{OS}=5\mathrm{cm}\Rightarrow \mathrm{OQ}=5\mathrm{cm}$ (Diagonals of a parallelogram bisect each other.) $\therefore \mathrm{QS}=10\mathrm{cm}$ Given that $\mathrm{PR}=\mathrm{QS}+7=10+7=17\mathrm{cm}$ $\Rightarrow \mathrm{OP}=\frac{1}{2}\times \mathrm{PR}=\frac{1}{2}\times 17=8.5\mathrm{cm}$.
• Q. Find the value of $x$ in each quadrilateral. (i) Kite ABCD
  
  (ii) Trapezium EFGH
  
  (iii) Parallelogram IJKL
  
  Solution: (i) We know, in a kite, the opposite angles that are between the sides of unequal length are equal. Here, $\mathrm{AB}=\mathrm{BC}$ and $\mathrm{AD}=\mathrm{CD}$ So, $\angle \mathrm{A}=\angle \mathrm{C}(\angle \mathrm{A}$ - between AB and $\mathrm{AD},\angle \mathrm{C}$ - between BC and CD$)$ $\Rightarrow \angle \mathrm{C}=\angle \mathrm{A}=\mathrm{x}=90^{\circ }$ (ii) We know, in a trapezium, the interior angles on the same side of each of the nonparallel sides are supplementary. Here, EH || FG $\Rightarrow \mathrm{x}+50^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=130^{\circ }$ (iii) We know, in a parallelogram, opposite angles are equal. So, $\angle \mathrm{I}=\angle \mathrm{K}\Rightarrow \mathrm{x}=30^{\circ }$.

11.0Some Special Parallelograms

1. Rhombus

A rhombus is a parallelogram with all sides equal. $ABCD$ is a rhombus where $AB\Vert CD$ and $BC\Vert AD$. Also, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$. Rhombus has all the properties of a parallelogram. Besides these, a rhombus has some other properties also.

Properties:

(ii) Opposite angles are equal (Property of parallelogram) $\angle \mathrm{A}=\angle \mathrm{C}$ $\angle \mathrm{B}=\angle \mathrm{D}$

(iii) Diagonals bisect each other at right angles. Proof: ABCD is a rhombus, therefore, it is a parallelogram. $\Rightarrow$ Its diagonals bisect each other. $\Rightarrow \mathrm{OA}=\mathrm{OC}$ and $\mathrm{OB}=0\mathrm{D}$ Now, in triangles AOB and COB

• All parallelograms are not rhombus, but all rhombuses are parallelogram. (i) $\angle \mathrm{DAB}=\angle \mathrm{DCB}$ (ii) $\angle \mathrm{ABD}=\angle \mathrm{BDC}$ (iii) $\mathrm{AB}|\mid \mathrm{CD}$ (iv) $BG=DG$
• Q. $ABCD$ is a Rhombus in which $AC$ and $BD$ are diagonals intersecting at 0 . If $\angle OAB=$ $55^{\circ }$, find $\angle \mathrm{CDO}$. Solution: The diagonals of a rhombus bisect each other at right angles. $\therefore \angle \mathrm{AOB}=90^{\circ }$
  
  Applying angle sum property in $△AOB$, $55^{\circ }+90^{\circ }+\angle \mathrm{OBA}=180^{\circ }\Rightarrow \angle \mathrm{OBA}=180^{\circ }-145^{\circ }=35^{\circ }$ We know that opposite sides of Rhombus are parallel to each other, $\angle OBA=\angle CDO(\mathrm{AB}||\mathrm{CD}$ and BD is transversal) $\Rightarrow \angle \mathrm{OBA}=\angle \mathrm{CDO}$ (Alternate angles) $\Rightarrow \angle \mathrm{OBA}=\angle \mathrm{CDO}=35^{\circ }$

2. Rectangle

A parallelogram which has all its angles as right angles is called a rectangle. MNOP is a rectangle.

Properties (i) Opposite sides are parallel and equal (Property of parallelogram) $$\begin{gathered} \mathrm{MN}\Vert \mathrm{OP},\mathrm{MN}=\mathrm{OP} \\ \mathrm{MP}\Vert \mathrm{NO},\mathrm{MP}=\mathrm{NO} \end{gathered}$$ (ii) All angles are equal and are right angles (Definition of rectangle). (iii) Diagonals bisect each other (Property of parallelogram). (iv) Diagonals are equal. Proof: In rectangle MNOP, let us draw its diagonals NP and OM.

In triangles POM and $\mathrm{OPN},\mathrm{PO}=\mathrm{OP}$ (common side) $\angle \mathrm{MPO}=\angle \mathrm{NOP}=90^{\circ }$ $\mathrm{PM}=\mathrm{ON}$ (opposite sides of rectangle) $\Rightarrow △\mathrm{POM}\cong △\mathrm{OPN}$ (SAS congruency) $\Rightarrow \mathrm{OM}=\mathrm{PN}$ Hence, proved. Diagonals are equal and bisect each other. $\Rightarrow \mathrm{MX}=\mathrm{XO}=\mathrm{NX}=\mathrm{XP}$

• Q. In the given rectangle $ABCD$, the diagonals $AC$ and $BD$ meet at 0 . Find $x$, if $0A=5x-1$ and $OD=4x+4$.
  
  Solution: $OA$ is half the diagonal AC and OD is half the diagonal BD . Since, the diagonals of a rectangle are equal, their halves must also be equal. Hence, $\mathrm{OA}=\mathrm{OD}$ $\Rightarrow 5\mathrm{x}-1=4\mathrm{x}+4$ $\Rightarrow 5\mathrm{x}-4\mathrm{x}=4+1\Rightarrow \mathrm{x}=5$

3. Square: A parallelogram with all sides and all angles equal is a square.
   

ABCD is a square. It has $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ and $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ }$

Properties (i) Opposite sides are parallel (property of parallelogram) (ii) All sides are equal (property of rhombus) (iii)All angles are right angles (equal) (property of rectangle) (iv) Diagonals are equal (property of rectangle) (iv) Diagonals bisect each other at right angles. (property of rhombus) Definition of square

Note:

• If all sides of quadrilateral are equal it is a rhombus always and sometimes a square.
• If all angles of quadrilateral are equal, it is a rectangle always and sometimes a square.
• Q. Classify the quadrilateral using the name that best describes it.