Answer:
solution given:
let's see only in a right-angled triangle Δ ACD.
AC=3 units
AD=5 units
since Δ ACD is a right-angled triangle. It satisfies Pythagoras law
[tex]AD^{2}=AC^{2}+CD^{2}[/tex]
25=9+[tex]CD^2[/tex]
[tex]CD^2[/tex]=25-9=16
[tex]CD=\sqrt{16} =4 units[/tex]
Now
Area of rectangle Δ ACD=[tex]\frac{1}{2} CD*AC=\frac{1}{2}*4*3=6\: square \:units[/tex]
similarly,
[tex]\frac{1}{2} AD*CE=6\: square \:units\\ 5*CE=6*2\\CE=\frac{12}{5}=2.4 units[/tex]
since AB=CE=2.4 units
Δ ACE is a right-angled triangle. It satisfies Pythagoras law.
[tex]AC^{2}=AE^{2}+CE^{2}\\3^2=AE^2+2.4^2\\9-5.76=AE^2\\AE and BC=\sqrt{1.24} =1.11 units[/tex]
now
area of trapezoid=area ofΔACD+Area of ΔABC
=6+[tex]\frac{1}{2}AB*BC=6+\frac{1}{2}*2.4*1.11=6+1.33 =7.33\: square \:units[/tex]
Step-by-step explanation:
