Definition: A ‘quadrilateral’ is a geometric figure on a plane enclosed by 4 line segments.

Properties:

 4 Vertices 4  Sides 4 Angles 2 Diagonals A, B, C, D BC, CD,DA,AB ABC,  BCD, CDA, DAB BD,AC

Each diagonal divides the quadrilateral in to 2 triangles

(BAC and ADC with AC as common base and

BAD and BDC with BD as common base..

 Other terms Properties Adjacent Sides Sides have a common vertex. Opposite Sides Sides do not have a common vertex. Adjacent Angles Angles have a common side. Opposite Angles Angles do not have  a common side.

 Adjacent Sides ( Have common vertex) Opposite Sides ( Do not have common vertex) Adjacent Angles ( Have common side) Opposite Angles ( Do not have common side) (AB,BC) : B common vertex (BC,CD) : C common vertex (CD,DA) : D common vertex (DA,AB) : A common vertex (AB,CD) (AD,BC) ( ABC, BCD)  : BC common side  ( BCD, CDA)  : CD common side  ( CDA, DAB)  : DA common side  ( DAB,  ABC) : AB common side (ABC, ADC) (BAD , BCD)

We notice that the sum of all four interior angles in a quadrilateral is 3600. Can we prove this mathematically?

6.7.1 Problem 1:.The four angles of a quadrilateral are in the ratio of 4:5:6:9. Find all the angles.

Solution:

 The sum of all angles in a quadrilateral is 3600. Since the angles are in the ratio of 4:5:6:9, let they be 4x, 5x, 6x and 9x.  Sum of all angles = 4x+5x+6x+9x = 3600 i.e. 24x = 3600 x = 360/24 = 150 4x = 4*15 = 600 5x = 5*15 =750 6x = 6*15 =900 9x= 9*15 = 1350 The angles are 600, 750, 900 and 1350

6.7.1 Problem 2: If the angles of a quadrilateral are 3x, 3x+150, 3x+300 and 900.Find all the angles.

Solution:

 We know that  sum of interior angles of a quadrilateral is 360o 3x+3x+150+3x+300+900 = 3600: i.e. 9x+1350=3600 i.e.9x=2250 i.e. x=250 Therefore the angles are 3x = 750, 3x+150 =750+150=900, 3x+300 = 1050 Verify that the sum of these four angles = 3600

We have seen earlier that a triangle totally has six elements (3 sides and 3 angles). We have also learnt to construct triangles with just three elements given (3 sides, 1 side and 2 angles, 2 sides and 1 included angle).

Note: With just three angles we cannot construct a unique triangle.

Compared to a triangle, quadrilateral has 10 elements (4 sides, 4 angles, 2 diagonals). We cannot draw a quadrilateral uniquely with 4 elements. We need at least 5 elements to draw unique quadrilaterals. However, it will not be possible to construct a unique quadrilateral with 1 side and 4 angles.

We need any one group of the following set of elements, to construct a unique quadrilateral.

6.7.1 Table 1:

 No Given  no. of Sides Given no. of Diagonals Given no. of Angles No. of  elements required 1 2 2 1 5 2 2 1 2 5 3 4 1 - 5 4 4 - 1 5 5 3 - 2(Included) 5 6 3 2 - 5 7 2(adjacent) - 3 5

General method for construction of quadrilateral:

(The method is more or less the same for construction of quadrilateral, for any combinations of the data given in the above table)

Notes:

a) Step for drawing a side:

First draw a line. Mark a point on this line

From the above point, draw an arc of given radius (length of the side) to cut this line to get the required side.

b) Use protractor to construct angles of required measure, wherever required.

c) For drawing of any figure follow the procedure:

Step 1: Draw a rough diagram

Step 2: Follow the steps described above (a,b), to construct sides/angles of required measure.

1. When two sides, two diagonals and one angle are given

6.7.1 Problem 3: Construct a quadrilateral ABCD with AB=4cm, BC=2cm, AC=5cm and BD=4cm and DAB= 600

First draw a rough diagram

 Step Construction 1 Mark a point A and draw a line though A 2 With A as center, draw an arc of radius 4cm to cut above line at B (AB=4cm) 3 From B, dawn an arc of radius 2cm above AB 4 From A, draw an arc of radius 5cm to cut the above arc at C (AC=5cm,BC=2cm) 5 From A, draw a line at an angle 600 with AB 6 From B, draw an arc of radius 4cm to cut the above line at D (DAB= 600, BD=4cm). Join DC.

2. When two sides, one diagonal and two angles are given

6.7.1 Problem 4: Construct a quadrilateral ABCD with AB=4cm, BC=3cm, BD=5cm and ABC= 600, BCD= 650

First draw a rough diagram.

 Step Construction 1 Mark a point A and draw a line though A 2 Cut the above line by an arc of radius 4cm to cut at B (AB=4cm) 3 From B, draw a line at an angle 600 with AB 4 From B, dawn an arc of radius 3cm to cut the above line at C (BC=3cm, ABC=600) 5 From C, draw a line at an angle 650 with AB 6 From B, draw an arc of radius 5cm to cut the above line at D (BD=5cm)

3. When four sides and one angle are given

6.7.1 Problem 5: Construct a quadrilateral PQRS with PQ=4cm, QR=3cm, RS=2.5cm and PS=3.5cm and SPQ= 500

First draw a rough diagram

 Step Construction 1 Mark a point P and draw a line though P 2 Cut the above line by an arc of radius 4cm to cut at Q  (PQ=4cm) 3 From P, draw a line  at an angle 500 with PQ 4 From P, dawn an arc of radius 3.5cm to cut the above line at S(PS=3.5cm, SPQ= 500) 5 From S, dawn an arc of radius 2.5cm. 6 From Q, dawn an arc of radius 3cm to cut the above arc at R(SR=2.5cm,QR=3cm). Join SR and QR.

6.7. 1 Exercise:  In Table 6.7.1 we have listed seven  combinations of data required for construction of a quadrilateral, out of this we constructed quadrilateral of three combinations (green color). Construct quadrilaterals with the four remaining combinations of data (yellow color).

6.7.2 Area of a Triangle

 In the adjacent figure of triangle ABC, AD, BF, CE are altitudes to the bases BC, AC and AB respectively from the opposite vertices A,B and C respectively. Area of a triangle =(1/2)Base*height(altitude) Area of ABC = (1/2)BC*AD = (1/2)AC*BF =(1/2)AB*CE   We are going to prove this formula for calculation of area of triangle in section 6.8.7.

6.7.2 Problem 1: The base of a triangular field is three times its height. If the cost of cultivating field at the rate Rs36.72 per 100 sq mts is Rs 49,572, find its base and height

Solution:

 Total cost of cultivating the field = 49,572 Area of the fileld = 49572*100/36.72 = 135000 Let x be the height of the field  its base = 3x  Area of the field = (1/2)*base*height = (1/2)*3x*x = 3x2/2  =135000  x2  =135000*2/3 =90000   x  =300 Thus height of the triangle is 300m and base is 900m

Verification:

Area of triangle = 1/2*900*300 = 900*150 = 135000

Cost of cultivation = 135000*36.72/100 = 49572 which is as given in the problem

 We have seen that a diagonal of a quadrilateral cuts the quadrilateral in to 2 triangles. We shall use this Property to calculate the area of a quadrilateral, as we already know how to calculate the area of a triangle. Let PQRS be the quadrilateral. Draw the diagonal PR. Draw a perpendicular (QA= h1) to PR  from vertex Q. Draw another perpendicular (SB=h2) to PR from vertex S. h1 and h2 are altitudes of PQR and PRS. Area of PQR= ½(base*height) =1/2(PR* h1) Area of PRS = ½(base*height) = 1/2(PR* h2) Area of PQRS = Area of PQR + Area of PRS = ½(PR* h1)+ ½(PR* h1) = 1/2*PR* (h1+h2) sq units    Area of quadrilateral = 1/2 * diagonal * sum of altitudes of the two triangles with diagonal as base

Depending on the shape of quadrilateral, they are classified as follows:

Type

Figure

Relationship between Sides

Relationship    between  Angles

Relationship    between Diagonals

Parallelogram

Both pairs of opposite sides are parallel

1.Both pairs of opposite sides are parallel

2.Both pairs of opposite sides are equal

1.Oppopsite angles are     equal.

2. Sum of any two consecutive angles

= 1800

1.Diagonals divide the parallelogram in to two congruent triangles

2. Diagonals bisect each other

Trapezium

Only one pair of opposite sides are parallel

A pair of opposite sides are parallel

Pairs of consecutive angles at the end points of the two non parallel sides are supplimentary

Isosceles Trapezium

One pair of opposite sides are parallel

and non parallel sides are equal

1.A pair of opposite sides are parallel

2. Non parallel sides are equal.

1.Pairs of consecutive angles at the end points of the two non parallel sides are suplementary

2.Pairs of consecutive angles at the end points of the two  parallel sides are equal.

Diagonals are equal

Rectangle

Both pairs of opposite sides are parallel and

all angles are right angle

1.Opposite sides are equal

2.  Both pairs of opposite sides are parallel

All angles are equal and are right angles

1.Diagonals divide the rectangle in to two Equal triangles

2. Diagonals are equal

3 Diagonals  bisect each other

Rhombus

All sides are equal and

both pairs of opposite sides parallel

1. All sides are equal

2. Both pairs of opposite sides are parallel

1.Oppopsite angles are     equal.

2.Sum of any two consecytive angles = 1800

1.Diagonals divide the rhombus in to two congruent triangles

2. Diagonals bisect each other.

3. Diagonals are | to each other.

Square

All sides are equal and

all angles are right angles

1 All sides are equal

2.Both pairs of opposite sides are parallel

All angles are  equal and right angles

1.Diagonals divide the sqaure in to two congruent triangles

2. Diagonals are equal.

3. Diagonals bisect each other.

4. Diagonals are | to each other.

Note : Why  mothers and  grandmothers  cut burfies in  the shape of parallelogram and not  in rectangles?  Because they could cut more pieces from the same spread of sweet preparation (Did they study Geometry?)

The reason we will learn later (Section6.8.2) is that the area of a rectangle (base*another side) is greater than the area of parallelogram (base*height) when their sides are of same measures.

The hierarchy of different types of quadrilaterals can be represented by the following chart:

6.7.4 Problem 1:  Two consecutive angles of a parallelogram have measures, (x+30) and (2x-60) respectively. Find the measures of all angles.

Solution:

 We know that in a parallelogram, sum of any two consecutive angles = 1800. (x+30)+(2x-60) = 1800. I e 3x-30 =180o  3x = 210  x=70 Hence the angles are 100,80,100 and 80

6.7.4 Problem 2:   In a parallelogram ABCD, DAB= 700, DBC = 800 Find CDB and ADB.

Solution:

 We know that in a parallelogram, sum of any two consecutive angles 1800. BAD + ABD+DBC = 1800 i.e. 700+ABD+800 = 1800  ABD = 1800-700-800 =300 Since BA || CD, corresponding angles are equal CDB = ABD and ADB =DBC  CDB = 300 and ADB = 800 ABC = ABD+DBC = 300+800 =1100 Since in a parallelogram opposite angles are equal, BCD = 700. So the angles are 700, 1100, 700 and 1100.

6.7.4 Problem 3: The ratio of two sides of a parallelogram is 3:5 and the perimeter is 48cm. Find the sides of the parallelogram.

Solution:

 We know that in a parallelogram Perimeter = sum of four sides = 2*(sum of any 2 adjacent sides) Since it is given that the perimeter = 48cm   Sum of any 2 adjacent sides= 24cm Since the ratio of adjacent sides = 3:5, let the sides be 3x and 5x  Sum of any two adjacent sides = 3x+5x = 24: i.e. 8x =24    x = 24/8 = 3 3x = 3*3 = 9cm, 5x = 5*3 = 15cm So the lengths of adjacent sides are 9cm and 15cm Hence the sides are 9cm, 15 cm, 9 cm and 15cm. Verification: Perimeter = sum of all sides = 9+15+9+15 = 48cm, which is as given in the problem and hence our solution is correct.

6.7 Summary of learning

 No Points to remember 1 Area of quadrilateral = 1/2 * diagonal * sum of altitudes of the 2 triangles with diagonal as base

Heron’s formula for calculation of area of a triangle:

 If a, b and c are measures of the three sides of a triangle then area can also be calculated by using the formula: Area = Where s = (a+b+c)/2 (semi perimeter of the circle)   Note: Though this has come to be known as Heron’s formula, ‘Bhaskara’ had provided the proof of this in his book (‘Leelavati’- shloka 169).

6.7.2 Problem 2:  Find the area of a triangular park whose sides are 18m, 24m and 30m. Also find the length of the altitude corresponding to the largest side of the triangle.

Solution:

Let a=18, b=24 and c=30.

s = (a+b+c)/2 = 36

Area = =  =  = 216

Since the largest base is 30, let the height on this base be h.

We have 216 = 1/2*30*h ( area of a triangle = (1/2)*base*height)

h = 216/15 = 14.4m

Verification:

Area of triangle = 1/2*30*14.4 = 216

Note: Heron’s formula is very helpful if we have to calculate the area of a quadrilateral given its sides and a diagonal, as in the problem given below.

6.7.2 Problem 3: A floral design on a floor is made up of 8 tiles which are quadrilaterals. Its sides are 9cm, 28cm, 9cm and 28cm and it’s diagonal is 35cm as shown in the adjoining figure. Find the cost of polishing the tiles at the rate of 50paisa per sq.cm.

Solution:

 Let us calculate the area of each tile Area of a tile = area of quadrilateral = 2*area of the triangle The sides of the triangle are a=28, b=9 and c=35.  s = (a+b+c)/2 = 36 Area of triangle =  =  =  = 88.2 Area of a tile = 2*area of triangle = 176.4 sq.cm. Total area to be polished = number of tiles * area of a tile = 8*176.4 = 1411.2 sq.cm. Total polishing charges = Rs 0.5*1411.2 = Rs 705.6