Given:
The figure of a quadrilateral ABCD.
To find:
The perimeter of the quadrilateral ABCD.
Solution:
In an isosceles triangle, the two sides and base angles are congruent.
In triangle ABD,
[tex]\angle DAB\cong \angle ABD[/tex] [Given]
[tex]\Delta ABD[/tex] is an isosceles triangle [Base angle property]
[tex]AD=BD[/tex] [By definition of isosceles triangles]
[tex]8=BD[/tex] ...(i)
In triangle BCD,
[tex]\angle BCD\cong \angle CDB\cong \angle CBD[/tex] [Given]
All interior angles of the triangle BCD are congruent, so the triangle BCD is an equilateral triangle and all sides of the triangle area equal.
[tex]BC=CD=BD[/tex]
[tex]BC=CD=8[/tex] [Using (i)] ...(ii)
Now, the perimeter of quadrilateral ABCD is:
[tex]Perimeter=AB+BC+CD+AD[/tex]
[tex]Perimeter=11+8+8+8[/tex]
[tex]Perimeter=35[/tex]
Therefore, the perimeter of the quadrilateral ABCD is 35 units.
