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Construct a quadrilateral ABCD in which AB = 4 cm, AC = 5 cm , AD = 5.5 cm and ∠ABC = ∠ACD = 90°. Write the steps of construction

or draw a diagram.


By PGT find BC
BC= 3 CM
draw AB=4cm
taking B as center, make an angle of 90 degree
on ray B, mark an arc of 3 cm and name it C
Join AC (diagonal of quad) ( it will automatically be equal to 5 cm)
on line AC, taking C as center, draw again 90 degree angle. Because angle ABC= angle ACD = 90 degree
on ray C, from pt. A draw the line = to 5.5 cm n mark the point of intersection as D now, ABCD is the req. quad. 
2 5 2
in which hypotenuse square = sum of the square of other 2 sides i.e base and perpendicular
bc^2 = ab^2+ac^2 = 4+5= 9 =3 cm. am i right?
plz mark as best if my answer helped u.
i don't know if i drew the quad correctly.
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