Ordinary Quadrilaterals
A quadrilateral is a two dimensional shape having four vertices & four sides.
Quadri (four) + Latus (Side) = Quadrilateral (four Sided figure)
Also known as Quadrangle or Tetragon.
Sum of the interior angles = 360^{0}
Area & Perimeter of the quadrilateral:
AREA
Area of the quadrilateral = ½ x diagonal x (sum of the perpendicular lengths
Area of the quadrilateral ABCD =(1/2) x 9 x (2+6) = 36 sq.cm
PERIMETER
Perimeter of the quadrilateral = sum of the sides of the quadrilateral.
Perimeter of the quadrilateral ABCD = 6 + 6 + 8 + 4 =24 cm
New Formula for finding the Area of a Quadrilateral:
Area of the quadrilateral = ½ x Product of diagonals x Sine of the angle between them.
Area of the quad ABCD = ½ x 9 x 8 x Sin 60^{0 }
= ½ x 9 x 8 x (√3/2)
= 18 sq.cm
Note: If the angle between the diagonals is 30º, then the formula is more simple
Area of the quad ABCD = (Product of the diagonals)/4
Quadrilaterals
Parallelogram
 1. Opposites sides are equal & parallel
 2. Opposite angles are equal
 3. Consecutive angles are supplementary.
 4. Diagonals that bisect each other.
Rectangle
 1. Opposites sides are equal & Parallel
 2. All angles are equal to 90°
 3. Diagonals are equal & bisect each other.
Trapezium
Only one pair of opposite sides is parallel.
Isosceles Trapezium
 1. Only one pair of opposite sides is parallel.
 2. Nonparallel sides are equal.
Square
 1. All sides are equal.
 2. All angles are equal to 90°
 3. Opposites sides are Parallel
 4. Diagonals are equal & bisect each other at right angles.
Rhombus
 1. All sides are equal.
 2. Opposites sides are Parallel
 3. Opposite angles are equal
 4. Diagonals bisect each other at right angles.
Kite
 1. Two pairs of adjacent sides are equal.
 2. One pair of opposite angles (that are obtuse) are equal
 3. Diagonals are perpendicular to each other
 4. The longer diagonal bisects the shorter diagonal.
Quadrilaterals  Dimensions  Area  Perimeter 

Parallelogram 
Breadth = b 
b x h 
2 (a + b) 
Rectangle 

l x b 
2 ( l + b) 
Square 
Side = a 
a^{2} 
4a 
Rhombus 
Length of the diagonals = d_{1} , d_{2} 
½ x d_{1 } x d_{2} 
4a 
Trapezium 
Length of Parallel Sides = a, b. 
½ x ( a + b) x h 
Sum of all sides 
Kite 
Length of the diagonals
= d_{1, }d_{2}

½ x d_{1 } x d_{2} 
2 (a +b) 
Theorems
Theorem # 1
Theorem # 2
Theorem # 3
Cyclic Quadrilaterals
A quadrilateral is called Cyclic quadrilateral if its all vertices lie on the circumference of a circle.
If ABCD is a cyclic quadrilateral, then the points A,B,C,D are called as concyclic points.
Theorem # 1
Theorem # 2
Theorem # 3
Theorem # 4
Theorem # 5
Theorem # 6
1. For a parallelogram to be cyclic or inscribed in a circle, the opposite angles of that parallelogram should be supplementary.2. A cyclic square/rectangle can be formed by taking the centre as the point of intersection of diagonals.
Theorem # 7
Hence, Points A, B, C, D are concyclic.
Theorem # 8
Ptolemy’s Theorem
The product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.
If ABCD is a cyclic quadrilateral with AC & BD as diagonals, then (AB x CD) + ( BC x AD) = AC x BD
AC x BD = 11x10 = 110
(AB x CD) + (BC x AD)= (8 x 5) + (7 x 10) = 110
Brahmagupta Formula
(for finding Area of Cyclic Quadrilateral
Ifa, b, cand dare the sides of a cyclic quadrilateral, then its area is given by:
Where s is the semi perimeter s = 1/2(a+b+c+d)
Diagonals of Cyclic Quadrilaterals
Suppose a, b, c and d are the sides of a cyclic quadrilateral and di & d2 are the diagonals, then we can find the diagonals of it using the below given formulas:
Radius of Cyclic Quadrilateral
(for finding Area of Cyclic Quadrilateral
Ifa, b, cand dare the sides of a cyclic quadrilateral, then its area is given by:
Where s is the semi perimeter s = 1/2(a+b+c+d)
Brahmagupta’s theorem
In a cyclic quadrilateral if the diagonals intersect each other at right angles, then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.
Author’s Creation in Quadrilaterals
1. Question:
In quadrilateral ABCD, AC & BD meet at O; AB = AD; ∠ABD = 40°; ∠CBD = 30°; and ∠BDC = 20°. Find the measurement of ∠AOD.
Question created by
Dr.M.Raja Climax
Founder Chairman, CEOA
2. Question:
ABCD is a square. E & F are points on AB & BC respectively such that ∠EDF=45º. Prove that ED & FD are the bisectors of ∠AEF & ∠CFE
3. Question:
ABCD is a square. E & F are points on DC& BD respectively such that ∠EAF =45°& ∠AFB =70° as shown in the figure. Ifa perpendicular is drawn from Ato EF meeting EF at M, then find ∠DMB
4. Question:
Research done by
Dr.M.Raja Climax
Prof G. LAKSHMANA MURTHY.
5. Question:
In the above figure, ‘O’ is the orthocentre of Δ ABC. Identify & write the cyclic quadrilaterals figuring inside ΔABC.
6. Question:
Modified Rider :
The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the centroid of ∆APB and H be the orthocentre of ∆CPD. Show that the points H,P,O are collinear.
7. Question:
ABCD is a Cyclic quadrilateral with AB = AD and CB = CD. M and N are points on AB and AD respectively such that ∠MCN = ∠ABD. Prove that MN = MB + ND.
8. Question:
Given a quadrilateral ABCD where BD bisects ∠B , P is a point on BC such that PD bisects ∠APC. Show that ∠BDP + ∠PAD = 90°
Click here to view the problem in GeoGebra.
https://www.geogebra.org/m/yanv3fgjAuthor’s Solution for Challenging Problems in Quadrilaterals
1. Question:
ABCD is a quadrilateral. Its diagonals AC & BD measuring 6 cm & 5 cm respectively cut each other at O and ∠AOD is 30°. Find the area of the quadrilateral ABCD.
Question created by
Dr.M.Raja Climax
Founder Chairman, CEOA
2. Question:
Find the value of θ, if ∠PAB = ∠ PBA = 15º
The question sender was taken aback on viewing two simple solutions for such a challenging problem.
3. Question:
Find tanθ in the given figure
4. Question:
P is a point on side AB of square ABCD such that DP =5cm. DQ is the angle bisector of ∠PDC where Q is a point on side BC. Then find the length of (CQ+AP)
5. Question:
If ABCD is a square with side 5 cm, E, F, G are the midpoints of the sides AB, BC, CD respectively. Find the area of the shaded region
6. Question:
ABCD is a cyclic quadrilateral. AC & BD meet at P. O is the circumcentre of ∆ APB. Prove that, PZ is an altitude of ∆ CPD and there by prove that O, P and the orthocentre of ∆CPD are collinear,
7. Question:
In the square ABCD, E is the interior point such that ∠AED is 90º. F is point on DE such that ∠CFD is 90º. AF meets CD at G. CE meets DA at H. Lines GH and CF meets at N, GH and AE meet at M. Then prove that GN=HM
8. Question:
ABCD is a square. E & F are midpoints of sides AD & BC respectively. Line segments AC,BD,CE & DF intersect in the interior of the square to form a quadrilateral PQRS as shown in the figure. What is the ratio of area PQRS to that of ABCD?