Prove the sum of quadrilateral is 360 degree
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We know,
Sum of Angles at vertices of any quadrilateral is [ ( n - 2 ) × 180 ]
N = Number of sides,
Applying formula,
sum of all angles = ( 4 - 2 ) × 180
sum of all angles = 2 × 180
sum of all angles = 360°
Sum of Angles at vertices of any quadrilateral is [ ( n - 2 ) × 180 ]
N = Number of sides,
Applying formula,
sum of all angles = ( 4 - 2 ) × 180
sum of all angles = 2 × 180
sum of all angles = 360°
Azikhan:
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Hola! Friend here's ur answer...
So ur question was.. Prove.. That
◆THE SUM OF ALL ANGLES OF A QUADRILATERAL IS 360°◆
so..as per I know..I'm solving it...
let ..all the angles of a quadrilateral be equal to 360°
so..now we will divide / separate the quadrilateral into two triangles..as shown in the figure..
as we know that all the angles of a triangle=180°[Euclid's axiom]..
so we can say that those two triangles would sum up to=360°
So ur question was.. Prove.. That
◆THE SUM OF ALL ANGLES OF A QUADRILATERAL IS 360°◆
so..as per I know..I'm solving it...
let ..all the angles of a quadrilateral be equal to 360°
so..now we will divide / separate the quadrilateral into two triangles..as shown in the figure..
as we know that all the angles of a triangle=180°[Euclid's axiom]..
so we can say that those two triangles would sum up to=360°
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