Question

In the rectangular trapezoid ABCD AC is driven by a CD (see figure).

Answer 1

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

$AD^{2}=AC^{2}+CD^{2}$

25=9+$CD^2$

$CD^2$=25-9=16

$CD=\sqrt{16} =4 units$

Now

Area of rectangle Δ ACD=$\frac{1}{2} CD*AC=\frac{1}{2}*4*3=6\: square \:units$

similarly,

$\frac{1}{2} AD*CE=6\: square \:units\\ 5*CE=6*2\\CE=\frac{12}{5}=2.4 units$

since AB=CE=2.4 units

Δ ACE is a right-angled triangle. It satisfies Pythagoras law.

$AC^{2}=AE^{2}+CE^{2}\\3^2=AE^2+2.4^2\\9-5.76=AE^2\\AE and BC=\sqrt{1.24} =1.11 units$

now

area of trapezoid=area ofΔACD+Area of ΔABC

=6+$\frac{1}{2}AB*BC=6+\frac{1}{2}*2.4*1.11=6+1.33 =7.33\: square \:units$