Area of a trapezoid using variables/angles

[sd.1991]

Find the area of the trapezoidal cross-section of the irrigation canal shown below. Your answer will be in terms of h, w, and θ. Use theta for θ.


I am unable to find base 2. So far, I have .5(w+ ?)h
I know I can use the alternate interior angles for theta, but don't know what to do after that. Thank you to whoever attempts to help me.

 

[burakaltr]

Find the area of the trapezoidal cross-section of the irrigation canal shown below. Your answer will be in terms of h, w, and θ. Use theta for θ.
View attachment 1345

I am unable to find base 2. So far, I have .5(w+ ?)h
I know I can use the alternate interior angles for theta, but don't know what to do after that. Thank you to whoever attempts to help me.

Call L one of the two congruent legs,

where Sin(theta)=h/L

but Area of trapezoid is = w*L*Sin(180-theta)=w*L*sin(theta)

Substitute for Sin(theta) to obtain

Area=w*h*Sin(theta)

 

[burakaltr]

To show whar Buraka is telling you:

A  a  E         w          F  a  B


      h                    h


      D         w          C

ABCD is the trapezoid; E,F are on AB such that ED and FC are vertical.
So we have 2 congruent right triangles AED and BFC; ok?
Let x = theta, and a = AE = BF. Let's use right triangle AED.
AngleEAD = x ; see that?
So angleADE = 90 - x
Using Law of Sines:
a / SIN(90 - x) = h / SIN(x)
a = hSIN(90 - x) / SIN(x)

Buraka is telling you all that in condensed form

I liked it to be called " Buraka ". Btw my name is Burak Alkan. ))

Thanks, Denis.

Have a Great Day !

 

[burakaltr]

Mine is just as bad: Denis Alfred Borris

Yes, I'm Bad, I'm Nation-Wide

 

[Deleted member 4993]

To show whar Buraka is telling you:

A  a  E         w          F  a  B


      h                    h


      D         w          C

ABCD is the trapezoid; E,F are on AB such that ED and FC are vertical.
So we have 2 congruent right triangles AED and BFC; ok?
Let x = theta, and a = AE = BF. Let's use right triangle AED.
AngleEAD = x ; see that?
So angleADE = 90 - x
Using Law of Sines:
a / SIN(90 - x) = h / SIN(x)
a = hSIN(90 - x) / SIN(x)

Buraka is telling you all that in condensed form

Stealing Denis's drawing and derivation:

a = hSIN(90 - Θ) / SIN(Θ) = h* cot(Θ)

Area = 1/2 * [w + (w+2a)] * h = [w + h*cot(Θ)] * h