Let n be the sides,then (n−2)×180=2×360 n=6
Three angles of the quadrilaterals are in the ratio 2:3:4, then it will be 2x,3x,4x.
Now it is given
2x+4x=180
x=30
So three angles are 60°,90°,120°
Let y be the fourth angle,Now in a quadrilateral
Sum of all angles =360
y+60+90+120=360
y=90
So All the four angles are 60°,90°,120°,90°
let the interior angle be x,then exterior angle will be (180- x )
Now as per the question
(180−x)=15x
Solving this, we get
x= 150°
Therefore,Each interior angle is 150°
Now we know that
Sum of all interior angles is given by =(n−2)×180
Therefore
(n−2)×180=150n
Simplifying this we get
6n - 5n = 12
n = 12
So, Number of sides= 12
Adjacent angles are supplementary, So
x+80=180
x=100°
So angles are 80°,100°,80°, 100°
Let x be common ratio, then angles are 1x,2x,3x,4x
Sum of all angles =360
1x+2x+3x+4x=360
x=36
So angles are 36°, 72°, 108°,144°
Here AB=BC=CD=AD
Also AB=AC
Now in Triangle ABC,
AB=BC=AC
Hence equliateral triangle
∠B=60, ∠BAC=60 and ∠BCA=60
Now in Triangle ACD
AC=CD=AD
Hence equliateral triangle
∠D=60, ∠DAC=60 and ∠DCA=60
Now ∠A=∠BAC+∠DAC=120
∠C=∠BCA+∠DCA=120
Hence angles are 60°, 120°, 60°,120°
Since DC and AB are parallel , these angles are supplementary
x−22+x+42=180
2x+20=180
x=80
Since diagonals bisect each other and are Perpendicular,Side is given by pythagorus theorem as
a=12√d21+d22=8.5 cm
Oppossite angles are equal
So 6y=120, y=20
Now adjacent angles are supplementary
120+5x+10=180
x=10
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