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.
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
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
∴ 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
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.