Given:
A figure of an isosceles trapezoid with bases 18 and 24, and the vertical height is 4.
To find:
The legs of the isosceles trapezoid.
Solution:
Draw another perpendicular and name the vertices as shown in the below figure.
From the figure it is clear that the AEFD is a rectangle. So,
[tex]EF=AD=18[/tex]
Since ABCD is an isosceles trapezoid, therefore in triangle ABE and DCF,
[tex]AB=DC[/tex] (Legs of isosceles trapezoid)
[tex]AE=DF[/tex] (Vertical height of isosceles trapezoid)
[tex]m\angle AEB=m\angle DFC[/tex] (Right angle)
[tex]\Delta ABE\cong \Delta DCF[/tex] (HL postulate)
[tex]BE=CF[/tex] (CPCTC)
Now,
[tex]BE+EF+FC=BC[/tex]
[tex]2BE+18=24[/tex]
[tex]2BE=24-18[/tex]
[tex]2BE=6[/tex]
[tex]BE=3[/tex]
Using Pythagoras theorem in triangle ABE, we get
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
[tex](AB)^2=(AE)^2+(BE)^2[/tex]
[tex](AB)^2=(4)^2+(3)^2[/tex]
[tex](AB)^2=16+9[/tex]
[tex](AB)^2=25[/tex]
Taking square root on both sides, we get
[tex]AB=\pm \sqrt{25}[/tex]
[tex]AB=\pm 5[/tex]
Side length cannot be negative. So, [tex]AB=5[/tex].
Therefore, the length of legs in the given isosceles trapezoid is 5 units.
