# Quadrilaterals: A quadrilateral is a closed figure bounded by four line segments.

**Type of Quadrilaterals:**

**Parallelogram**: A parallelogram is a quadrilateral with opposite sides parallel. The region bounded by a parallelogram is also known as parallelogram.

**Rectangle**: A rectangle is a parallelogram with a right angle .The region bounded by a rectangle is a rectangular region.

**Rhombus**: A rhombus is a parallelogram with equal adjacent sides ,i.e. ,the opposite sides of rhombus are parallel and the lengths of four sides are equal. Region bounded by a rhombus is also called rhombus.

**Square**: A square is a rectangle with equal adjacent sides, .e., a square is a parallelogram with all sides equal and all right angles. The area bounded by square is also called a square.

**Trapezium**: A trapezium is a quadrilateral with a pair of parallel sides. The region bounded by trapezium is also called trapezium.

## (a)Area of trapezium

Construct DA∥CE at C.

∴AECD is a parallelogram. From the figure

area of trapezium=area of parallelogram

AECD + area of triangle CEB

=a × h +b – a × h

=b + a h

∴ Area of trapezium=average of the sum of two parallel sides ×height.

(b)**Area of Rhombus**: The diagonals of a rhombus bisect each other at right angles. If we know the lengths of two diagonals, we can find the area of rhombus.

Let the diagonals AC and BD of a rhombus

ABCD intersect each other at O.

Denote the lengths of two diagonals by

a and b respectively.

Area of rhombus= area of triangle ∆DAC + ∆BAC.

=1/2 a×1/2 b+1/2 a×1/2b

=1/2a×b

∴ Area of rhombus =half of the product of two diagonals.

**Ex: Prove that every angle of a rectangle is right angle.**

General enunciation: Prove that every angle of a rectangle is right angle.

Particular enunciation: Let ABCD is a rectangle in which ∠A=1 right angle.

We have to prove that every angle of a rectangle is right angle i.e. ∠A=∠B=∠C=∠D=1 Right angle

Proof: We know that, Rectangle is a parallelogram and opposite angles of a parallelogram equal to each other

∴∠A=∠C and ∠B=∠D —————— (i)

Again, since rectangle is parallelogram

∴AB∥CD, AD is their transversal.

∴∠A+∠D=2 right angle [sum of two interior angles on the same side

of a transversal is 2 right angle]

Or, ∠D=2 right angle – ∠A

=2 right angles – 1 right angle

∴∠D=1 right angle

From (i) we can write,

∠A=∠C= 1 right angle

And ∠B=∠D=1right angle

∴∠A=∠B=C=∠D=1 right angle (Proved)

**Ex:1.Prove that the diagonals of a square are equal and bisect each other.**

General enunciation: To prove that the diagonals of a square are equal and bisect each other.

Particular enunciation: Let ABCD is a square whose the diagonals AC and BD insect each other at O. We have to prove that (i)AC=BD (ii)AO=CO,BO=DO.

Proof: We know, square is a parallelogram and the diagonals of a parallelogram bisect each other

∴AO=CO, BO=DO

In ∆ABD and ∆ACD,

AB=DC [sides of equal to each other]

And AD=AD [common side]

Included ∠BAD =included ∠ADC [each of the angles is right angle]

∴∆ABD≌∆ACD [SAS theorem]

Therefore AC=BD (proved)

**Ex: a worker has made a rectangular concrete slab. In how many different ways can he be sure that the slab is really rectangular?**

Solution: In the following way the worker can be sure that the slab is really rectangular.

- If the opposite sides of the slab are equal and parallel.
- If every angle of the slab is right angle.
- If its two diagonals are equal and bisect each other.

### Ex: Find the area of trapezium by an alternate method.

Solution: Let ABCD is a trapezium in which the parallel sides are AB = a unit and DC =b unit respectively. We join A,C and draw a perpendicular CE on the side AB from the point C.

Therefore Area of the trapezium ABCD

=(Area of ∆region ABC)+(Area of ∆region ADC)

=(1/2 ×AB×CE)square unit +(1/2 ×DC×CE)square unit [the height of ∆ADC is CE]

=1/2 ×CE × (AB+DC) square unit

=1/2 h (a +b ) square unit.