- INTRODUCTION
- TYPES OF QUADRILATERALS
- PERIMETER AND AREA OF QUADRILATERAL
- POLYGONS
- TYPES OF POLYGONS
- PERIMETER AND AREA OF A POLYGON
Quadrilaterals and Other Polygons
A Quadrilateral is a closed figure bounded by four straight lines (see figure below). It has four vertices (A, B, C and D) such that no three of them are collinear. These vertices are the intersection points of four lines (AB, BC, CD and DA) known as the sides of the quadrilateral.
The line joining opposite vertices is known as diagonal. So in the given quadrilateral, line AC and BD are two diagonals.
To name a quadrilateral, the names of the vertices are written in clockwise or anticlockwise order.
For example, □ABCD or □ADCB.
Properties of a quadrilateral:
- Sum of the four interior angles is 360
,
i.e. ∠A + ∠B + ∠C + ∠D = 360°
- Sum of the four exterior angles is 360
(This is true for any polygon).
- A quadrilateral is said to be a cyclic quadrilateral if it is possible to draw a circle passing through all the four vertices of the quadrilateral.
- In a cyclic quadrilateral, opposite angles are supplementary and an external angle is equal to the interior opposite angle.
Quadrilaterals can be broadly classified into six major categories:
- Trapezium
- Parallelogram
- Rhombus
- Kite
- Rectangle
- Square
A quadrilateral in which at least one set of opposite sides is parallel is known as trapezium or trapezoid. The non-parallel sides if any (AD and BC in the given figure) are known as oblique sides. If length of oblique sides is equal, it is known as isosceles trapezium. Parallel sides (AB and DC) are generally known as bases and the perpendicular distance between bases is known as height.
In a trapezium, the length of the line joining the midpoints of the oblique sides is equal to the average of the length of the bases. Also a line (XY) drawn parallel to the bases will divide the oblique sides in equal ratio. The length of such a line dividing the oblique sides in the ratio a : b is given by
where B1 and B2 denote the length of the two bases AB and DC.
A quadrilateral in which both the sets of opposite sides are parallel to each other is known as a parallelogram.
Area (A) = b × h ; where b , h represent the base and height of the parallelogram respectively.
This area can also be calculated as twice the area of the triangle formed by joining any one of the diagonals.
Additional properties of a parallelogram:
In □ABCD;
- Opposite angles are equal. (m∠A = m∠C and m∠B = m∠D)
- Adjacent angles are supplementary. (m
A
+ mB
= mB
+ mC
= mC
+ mD
= mD
+ mA
= 180)
- Opposite sides are equal. (AB = DC and AD = BC)
- Any of the diagonals (AC or BD) divides the parallelogram into two congruent triangles. (ΔABC and ΔCDA are congruent. Similarly, ΔABD and ΔCDB are also congruent.)
- Both the diagonals bisect each other. (AO = CO and BO = DO)
- If we draw both the diagonals, we get four triangles.
- Vertically opposite triangles are congruent and the area of all four of these triangles is equal. (ΔABO and ΔCDO are congruent. Similarly, ΔBCO and ΔDAO are also congruent.)
- If we take a point anywhere inside the parallelogram and join it with all the four vertices, we get four triangles such that the sum of the area of the set of vertically opposite triangles is equal.
- If in a quadrilateral one set of opposite sides is parallel as well as equal, then the other set of opposite sides has to be equal and parallel, and such a quadrilateral will be a parallelogram.
A parallelogram in which all the sides are of equal length is called a rhombus. In a rhombus, the diagonals bisect each other at right angles. Every rhombus is a parallelogram but the converse is not true.
A quadrilateral in which the four sides can be divided into two groups each containing two adjacent equal sides is known as a kite. In a kite, the diagonals meet at right angles and one diagonal (i.e. BD) is bisected by the other diagonal (i.e. AC).
A quadrilateral in which each of the angles is a right angle is called a rectangle. In a rectangle, both the diagonals are equal.
A square is a parallelogram in which all the sides are equal and all the angles are right angles. A square possesses properties of both rhombus and rectangle. Both the diagonals are also equal and they bisect each other perpendicularly.
The sum of the lengths of the four sides of a quadrilateral is known as the perimeter. So, in a quadrilateral with sides of length a, b, c and d; the perimeter is given by
Perimeter (P) = a + b + c + d
Case I: A cyclic quadrilateral with sides of lengths a + b + c + d
|
Find the area of a cyclic quadrilateral with sides 25, 33, 39 and 65 cm.
Solution:
|
Case II: If the length of one diagonal and two offsets are given.
where d the diagonal and h1 and h2 are the offsets, i.e. perpendiculars drawn to the given diagonal from the other two vertices.
|
Find the area of a quadrilateral with one diagonal as 4 cm and the two offsets on this diagonal as 1 cm and 2 cm.
Solution:
|
Case III: If two diagonals and the included angle are given.
where d1 and d2 are the diagonals and θ is the angle between these diagonals.
Here, we can take either of the angles
between diagonals as 
as sin (180
)
= sin .
|
If area of a quadrilateral is 6
Solution:
|
cm2 with one of the diagonals as 4 cm and the angle between the
two diagonals as 120
Case IV: A trapezium
where b1 and b2 are the bases and h is the distance between these bases (known as height).
|
ABCD is a square with sides of length 10 units. OCD is an isosceles triangle with base CD. OC cuts AB at point Q and OD cuts AB at point P. The area of trapezoid PQCD is 80 square units. The altitude from O of the triangle OPQ is: [XAT 2009] (1) 12(2) 13(3) 14 (4) 15(5) None of the above.
Solution: The given information can be illustrated as follows:
Area of the trapezoid PQCD = 80 units
Now, OPQ and ODC are similar triangles. Let M be the point of contact of the altitude of ΔOCD and side AB. Then, let length of OM = x units.
Hence, option 4. |
Case V: A parallelogram
Area (A) = b × h
where b is the base and h is the distance between the two bases.
Case VI: A rhombus and a kite
where d1 and d2 are the diagonals.
|
Find the side of a rhombus with area 24 cm2 and one of its diagonals as 6 cm.
Solution: Length of one of the diagonals, d1 = 6 cm Area of the rhombus, A = 24 cm2
The diagonals bisect perpendicularly in a rhombus
By using Pythagoras theorem, with half of the diagonals i.e. 6/2 and 8/2 or with 3 and 4, the side of the rhombus will be 5 cm. |
Case VII: A rectangle
Area (A) = lb
where l is the length and b is the breadth.
Case VIII: A square
Area (A) = a2
where a is the side
|
Find the diagonal of a square with area 36 cm2.
Solution: A = a2
Now using Pythagoras theorem, we can find that
diagonal of a square with side a is given by a Hence, diagonal = a |
|
The diagonal of a square is 4 [SNAP 2009] (1) 8 cm (2) 8 
(3) 4 Solution: Let the length of the side of the original square be x.
2x2 = 32 x2 = 16 x = 4 Now, the area of the other square is twice that of the first square. So, the area of the second square = 2x2 = 32 If the length of the second square is y, then y2 = 32
Hence, option 1. |
|
A square, whose side is 2 metres, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon, in metres is [CAT 2001]
Solution:
Let the side of the octagon be x, which is also the hypotenuse of the triangle in the corner of the square.
Hence, option 2. |
|
Mohan was playing with a square cardboard of side 2 metres. While playing, he sliced off the corners of the cardboard in such a manner that a figure having all its sides equal was generated. The area of this eight sided figure is: [IIFT 2009]
Solution:
Refer to the following figure.
As all the sides of the figure are equal,
Area of the eight sided figure = Area of the square
– 4
Hence, option 4. |
Area of each small triangle
|
Any five points are taken inside or on a square of side 1. Let a be the smallest possible number with the property that it is always possible to select one pair of points from these five such that the distance between them is equal to or less than a. Then a is: [FMS 2010]
Solution: Two points on or inside the square will be at the
maximum distance when they are on two opposite vertices. Let us select 4
points on the vertices of the square. Then, the distance between any two
of them is 1 or
Now we select the fifth point such that it is at the
maximum possible distance from each of the other four points. Such a point
lies on the point of intersection of the diagonals and its distance from
each of the other four points is Hence, option 2. |
.
Any other point on or inside the square will be at a distance less than
|
Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three are not right angles.
What is the area of the plot? [CAT 2001]
(1) 768(2) 534 (3) 696.5(4) 684
Solution:
Let us draw DE, which is perpendicular to AC. Since ΔADC is isosceles, therefore DE will be its median and altitude both.
= 384 + 300 = 684 Hence, option 4. |
REMEMBER:
- A square has the maximum area, out of all the quadrilaterals with a given perimeter.
- A square has the minimum perimeter, out of all the quadrilaterals with a given area.
- Two quadrilaterals are similar if their corresponding angles are equal. Similar quadrilaterals have the ratio of any corresponding linear measurements equal to the ratio of their corresponding sides. Ratio of area of similar quadrilaterals will be equal to the square of the ratio of their sides.
- Two quadrilaterals are congruent if their corresponding sides are equal.
|
In the above diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle CEF and that of the rectangle? [CAT 2001]
Solution:
Hence, option 1. |
AB = 3 
|
Consider a square S which is inside a circle A such that the four corner points of the square touch the circumference of the circle. A second circle B is inside the square S so that its four sides touches the circumference of B. Then, the ratio of the areas of the circles A : B equals: [JMET 2010]
(1) (3)
Solution: Let the radius of circle A be r.
Diagonal of the square S, will be same as diameter of the circle A i.e. 2r Let the length of each side of square S be x x2 + x2 = (2r)2 2x2 = 4r2
From the figure it is evident that the length of the diameter of the circle will be equal to the length of the side of the square
Hence, option 2. |
(2) 2 : 1
(4) 
: 1
A polygon is a closed figure bounded by three or more straight lines, known as sides. It has as many vertices as the number of sides, with no three of them collinear.
A line joining two non adjacent vertices is known as the diagonal.
PROPERTIES OF POLYGONS
- Number of diagonals in an n sided polygon is given by
Explanation:
For an n sided polygon, there exist n vertices.
Now from one of the vertices i.e. say point A, we can draw (n – 3) diagonals.
Also (n – 3) diagonals can be drawn from one of the adjacent vertices i.e. say point B.
Now (n – 4) diagonals are drawn from the point C, (n – 5) diagonals are drawn from point D, (n – 6) diagonals are drawn from point E and so on.
∴ Total number of diagonals = (n − 3) + (n − 3) + (n − 4) + (n − 5) + … + 3 + 2 + 1
Therefore for an n-sided polygon, the number of diagonals is given by n(n − 3)/2.
The sum total of all the interior angles of any polygon = (n – 2)Ï€
Explanation:
Let m∠HOG = 1.
Then m∠OHG + m∠OGH = 180° −
1.
Similarly, we can do this for the other triangles.
∴ m∠OHG + m∠OGH + m∠OGF + m∠OFG + m∠OEF + m∠OFE + m∠OED + m∠ODE + m∠ODC + m∠OCD ….. n triangles
= (180° −1)
+ (180° − 2)
+ (180° − 3)
+ (180° − 4)
+ (180° − 5)
+ …… (180°− n)
= 180° × n – (1
+ 2
+ 3
+ 4
+ ……. n)
= n × 180° − (360°)
Here LHS represents the sum of all the interior angles for an n-sided polygon.
∴ Sum of interior angles of a polygon = n × Ï€ − 2Ï€ = (n − 2)Ï€.
The sum total of all the exterior angles of any polygon = 360°.
The measure of each exterior angle of a regular polygon = 360°/n
The ratio of the number of sides of a polygon to the diagonals of a polygon = 2 : (n − 3)
|
Find the number of sides of a regular polygon with
each interior angle of 150
Solution: Since the interior angle is 150 The measure of each exterior angle of a regular
polygon = 360
Thus, the number of sides of the regular polygon = 12 |
|
The interior angles of a polygon are in Arithmetic
Progression. If the smallest angle is 120 [IIFT 2008]
(1) 7(2) 8 (3) 9(4) None of the above
Solution: The interior angles are 120 Now the sum of exterior angles needs to be 360
Hence, option 3. |
REMEMBER:
- A polygon, in which all the interior angles are less than 180,
is known as a convex polygon.
- A polygon, in which at least one of the interior angles is more
than 180,
is known as a concave polygon.
- Unless otherwise specified, we take polygons as convex polygons.
- The sum of the interior angles of an n sided polygon is
(n – 2)180.
- The sum of the exterior angles is 360.
(This is true for any polygon.)
- If all the sides and hence all the angles of a polygon are equal, it is known as a regular polygon.
- A polygon is said to be a cyclic polygon if it is possible to draw a circle passing through all its vertices.
|
Each side of a given polygon is parallel to either
the X or the Y axis. A corner of such a polygon is said to be convex if
the internal angle is 90 [CAT 2003 Leaked Test]
(1) 20(2) 0 (3) 21(4) 22
Solution:
Consider the polygon as shown in the figure. Here 4 corners A, D, G and J are the concave corners and remaining 8 corners are convex corners. In general, for a polygon with sides parallel to either of the axes, if n is the number of concave corners and m is the number of convex corners, then we have, m
Hence, option 3. |
Based on the number of sides, polygons are categorized as shown below:
| Number of Sides | Name of the Polygon |
| Three | Triangle |
| Four | Quadrilateral |
| Five | Pentagon |
| Six | Hexagon |
| Seven | Heptagon |
| Eight | Octagon |
| Nine | Nonagon |
| Ten | Decagon |
PERIMETER OF A POLYGON (P):
The sum of the lengths of all the sides of a polygon is known as the perimeter. So, in an n sided polygon with sides a1, a2, a3, … an; the perimeter is given by
P = a1 + a2 + a3 +… + an
AREA OF A POLYGON (A)
Area of an n-sided regular polygon is given by,
where a is the side.
Explanation:
For polygons with five or more sides, area is calculated by dividing the polygon into triangles.
Let us take an n-sided regular polygon A1A2A3A4A5A6….. An.
A1A2 = A2A3 = A3A4 = A4A5 = ….An-1 An = a units
OB (inradius) = r and OA3 = OA2 (circumradius) = R
Also, from the figure, we see that r2 = R2 – (a/2)2
In n-gon, there are n such triangles = n × area of triangle
General formula for the area of an n-sided regular polygon is given
by:
AREA OF A REGULAR PENTAGON
A regular pentagon with side a can be divided into five triangles.
|
Find the area of a regular pentagon with side 10 cm.
Solution: Area of a pentagon with side 5 cm = 1.725 |
AREA OF A REGULAR HEXAGON
A regular hexagon with side a can be
divided into six equilateral triangles with side a. The area of each
equilateral triangle will be equal to 
a2/4
|
Find the perimeter of a regular hexagon with area
6
Solution:
Perimeter = 6 |
|
Find the inradius of a hexagon with each side of 4 units.
Solution: In case of a hexagon n = 6 Also here each side = 4 units
|
|
In the figure below, ABCDEF is a regular hexagon and
[CAT 2003 Leaked Test]
Solution:
We can see that triangles AMF, AMB, BMC, CMD, DME and EMF are all equilateral triangles. Also, AE, XY and BD bisect MF, AB, ED and CM.
Hence, option 1. |
|
Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF? [CAT 2003 Re-Test]
Solution: ABCDEF is a regular hexagon, join centre O with vertices A, C and E.
In quadrilateral AFEO, diagonal EA divides it into two equal area triangles. i.e. Area (∆AFE) = Area (∆AOE) Similarly, Area (∆ABC) = Area (∆AOC) And, Area (∆CDE) = Area (∆COE)
Hence, option 2. |
REMEMBER:
- For a fixed perimeter, the area of a polygon with higher number of sides will always be more than the area of a polygon with lesser number of sides.
- So, if an equilateral triangle, square, regular pentagon, regular hexagon have the same perimeter, then the order of their areas will be:
Equilateral triangle < Square < Regular pentagon < Regular hexagon
- For any fixed area, the perimeter of a regular polygon with lesser number of sides will always be more than that of a regular polygon with a greater number of sides.
- If the circumference of a circle is the same as the perimeter of a regular polygon, then the area of the circle will always be more than the area of the polygon.
- For a fixed area, the perimeter of a regular polygon will always be greater than the circumference of the circle.
