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**Types of Quadrilaterals**

Look at the different quadrilaterals drawn below:

**Trapezium:**One pair of opposite sides of quadrilateral are parallel. AB II CD.**Isosceles Trapezium:**If the two nonparallel sides of a trapezium are equal then it is called an isosceles trapezium.**Parallelograms:**Both pairs of opposite sides of quadrilaterals are parallel.**Rectangle:**A parallelogram one of whose angles is 90^{0}, is called a rectangle,**Square:**A parallelogram whose all sides are equal and one of whose angles is 90^{0}is called a square. Kite: A quadrilateral in which two pairs of adjacent sides are equal is known as a Kite.**Rhombus:**The parallelogram DEFG of Fig. (iv) has all sides equal. it is called a rhombus.**Square:**The parallelogram ABCD of Fig. (v) has ∠ A = 90° and all sides equal; it is called a square.

Proof: Let ABCD be a quadrilateral. Join AC.

Clearly, ∠1 + ∠2 = ∠A …… (i)

And, ∠3 + ∠4 = ∠C …… (ii)

We know that the sum of the angles of a triangle is 180°

Therefore, from ∆ABC, we have

∠2 + ∠4 + ∠B = 180° (Angle sum property of triangle)

From ∆ACD, we have

∠1 + ∠3 + ∠D = 180° (Angle sum property of triangle)

Adding the angles on either side, we get;

∠2 + ∠4 + ∠B + ∠1 + ∠3 + ∠D = 360°

⇒ (∠1 + ∠2) + ∠B + (∠3 + ∠4) + ∠D = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360° [using (i) and (ii)].

Hence, the sum of all the four angles of a quadrilateral is 360^{0}

Note that a square, rectangle and rhombus are all parallelograms.
A square is a rectangle and also a rhombus.
A parallelogram is a trapezium.
A kite is not a parallelogram.
A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram).
A rectangle or a rhombus is not a square.

A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram).
A rectangle or a rhombus is not a square

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Sum of angles of a quadrilateral = 360^{0}

Fourth Angle = 360^{0}-(110^{0}+82^{0}+68^{0}) =100^{0}

Therefore, fourth angle measures **100 ^{0}.**

Given, ABCD is a quadrilateral and AC and BD are the diagonals intersect at the point O.

We know that the sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

=> 2(AB + BC + CA + AD) > 2(AC + BD)

=> (AB + BC + CA + AD) > (AC + BD)

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