Subject: Compulsory Maths

#### Overview

Quadrilaterals are the polygons having four sides. Parallelogram, rectangle, square, rhombus, trapezium, and kite have some special properties.

#### Some special type of Quadrilaterals

Quadrilaterals are the polygons having four sides. Parallelogram, rectangle, square, rhombus, trapezium, and kite have some special properties. So, they are called special types of a quadrilateral. Some types of quadrilateral are as follows

1. Parallelogram

Its opposite sides are equal and parallel.
$\therefore$ AB= DC and AB//DC, AD = BC and AD//BC.
Its opposite angles are equal.
$\therefore$ $\angle$A = $\angle$C and $\angle$B = $\angle$D
Its diagonal bisect each other.
$\therefore$ Diagonals AC and BD bisect each other ar O.
i.e AO = OC and BO = OD.
2. Rectangle

Its opposite sides are equal and parallel.
$\therefore$ AB = DC and AB//DC, AD = BC and AD//BC.
Its all angles are equal andeach of them is 90°.
$\angle$A = $\angle$B = $\angle$C = $\angle$D = 90°.
Its diagonalsare equal and they bisect each other.
$\therefore$ Diagonals AC = BD, AO = OC and BO = OD.
3. Square

ts all sides are equal.
$\therefore$ AB = BC = CD = DA
Its all angles are equal and they are 90°.
$\therefore$ $\angle$A = $\angle$B = $\angle$C = $\angle$D = 90.
Its diagonal are equal and they bisect each other at right angle.
$\therefore$ AC = BD and they bisect each other at O at a right angle.
i.e AO = OC, BO = OD and BD⊥AC at O.
4. Rhombus
Its all sides are equal and opposite sides are parallel.

$\therefore$ AB = BC = CD = DA and AB//DC, AD//BC
Its opposite angles are equal.
$\therefore$ $\angle$A = $\angle$C and $\angle$B = $\angle$D
Its diagonals are not equal but they bisect each other at right angle.
$\therefore$ AC and BD bisect each other at O at right angle.
i.e AO = OC, BO = OD and BD⊥AC at O.
5. Trapezium

Its any one pair of opposite sides are parallel.
$\therefore$ AB//DC.
6. Kite

Its particular pairs of adjacent sides are equal.
AB = AD and BC = DC
The opposite angles formed by each pair of unequal adjacent sides are equal.
$\angle$ABC = $\angle$ADC
Diagonals intersect each other at O at a right angle.
i.e BD⊥AC at O.
##### Things to remember
• Opposite sides of a parallelogram are equal and parallel.
• The sides of rhombus are equal and opposite sides are parallel.
• The particular pairs of adjacent sides are equal.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Questions and Answers

Solution:

Let the angles of the quadrilateral be x°, 2x°, 3x° and 4x° respectively.

Here, x° + 2x° + 3x° + 4x° = 360°

or, 10x° = 360°

or, x° = $\frac{360°}{10}$ = 36°

$\therefore$ x° = 36°, 2 × 36° =72°, 3x° = 3 × 36° = 108° and 4x° = 4 × 36° = 144°.

Solution:

y° + 105° = 180° [Being the sum of a straight angle]

or, y° = 180° - 105° = 75°

Now, x° + (2x° + 10°) + 2x° + y° = 360° [The sum of the angles of a quadrilateral]

or, 5x° + 10° + 75° = 360°

or, 5x° = 360° - 85°

or x° = $\frac{275°}{5}$ = 55°

$\therefore$ x° = 55°, 2x° + 10° = 2×55° + 10° = 120°, 2x° = 2×55° = 110° and y° = 75°.

Solution:

Here, 2x° + 3x° + (2x+10)° + (x+30)° = 360° [Sum of the angles of quadrilateral]

or, (8x + 40)° = 360°

or, 8x° = 360° -40°

or, x = $\frac{320°}{8}$ = 40°

$\therefore$ 2x° = 2×40° = 80°, 3x° = 3×40° = 120°, (2x + 10)° = 2×40° + 10° = 90° and (x + 30)° = 40° + 30° = 70°

Solution:

Here,

w° = 25° [Being alternate angles]

x° = 20° [Being alternate angles]

Now, w° + 20° = 25° + 20° = 45°

x° + 25° = 20° + 25° = 45°

Again, y° + 45° = 180° [Being the sum of co-interior angles]

or, y° = 180° - 45°

or, y° = 135°

Also, z° = y° = 135° [Being the opposite angles of a parallelogram]

$\therefore$ w° = 25°, x° = 20°, y° = z° = 135°.

Some special types of quadrilaterals are as follows:

1. Parallelogram
2. Rectangle
3. Square
4. Rhombus
5. Trapezium
6. Kite