Types of Quadrilaterals
Look at the different quadrilaterals drawn below:
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 3600
Sum of angles of a quadrilateral = 3600
Fourth Angle = 3600-(1100+820+680) =1000
Therefore, fourth angle measures 1000.
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)