Answer:
[tex]\text{The area is }140units^2[/tex]
Step-by-step explanation:
Given trapezoid ABCD, AB =13, CD = 14, BC = 5 and AD = 20 units.
we have to find the area of trapezoid ABCD
Quadrilateral BEFC is a rectangle, then
EF = BC = 5 units and BE = CF = y units
Let AE = x, then FD = AD - AE - EF = 20 - x - 5 = 15 - x.
By the Pythagorean theorem,
In ΔDFC
[tex]CD^2=CF^2+FD^2[/tex]
[tex]14^2=y^2+(15-x)^2[/tex] → (1)
In ΔAEB,
[tex]AB^2=BE^2+AE^2[/tex]
[tex]13^2=y^2+x^2[/tex] → (2)
Subtract equation 2 from 1
[tex]14^2-13^2=(15-x)^2-x^2[/tex]
[tex]196-169=225+x^2-30x-x^2[/tex]
[tex]30x=225-196+169=198[/tex]
[tex]x=6.6 units[/tex]
(2) ⇒ [tex]13^2=y^2+x^2[/tex]
[tex]169-43.56=y^2[/tex]
[tex]y=11.2units[/tex]
Hence, the height is 11.2 units
[tex]\text{Area of trapezium ABCD=}\frac{1}{2}\times (BC+AD)\times height[/tex]
[tex]=\frac{1}{2}\times (5+20)\times 11.2=140 units^2[/tex]
[tex]\text{Hence, the area is }140units^2[/tex]
