# Trapezoid

• Aug 17th 2010, 10:57 PM
Frenchie
Trapezoid
hello i have the following trapezoid bellow:
http://img691.imageshack.us/img691/5622/fsdfsdfv.png
the angle of @ is suppose to be 10 and it is asking me for the smaller base of trapezoid a and larger base of trapezoid b and the height of the trapezoid c Any help would be appreciated. It looks like the left and right triangle of that trapezoid should have angles of 90, 80, and 10? But after that i'm not sure how to go about solving for a and b

Would i get b by doing sin10=b/8? and since it's a trapezoid wouldn't b be 3x that and a=b?

EDIT: this is what i have so far

http://img251.imageshack.us/img251/28/fsdfsdfz.png

a=(sin10=a/8) or a=b=(sin10=b/8)
b=3(sin10=b/8)
c=(sin80=c/8)
• Aug 17th 2010, 11:45 PM
earboth
Quote:

Originally Posted by Frenchie
hello i have the following trapezoid bellow:

the angle of @ is suppose to be 10 and it is asking me for the smaller base of trapezoid a and larger base of trapezoid b. Any help would be appreciated. It looks like the left and right triangle of that trapezoid should have angles of 90, 80, and 10? But after that i'm not sure how to go about solving for a and b <=== according to your sketch a = 8' (?)

...

1. I've modified your sketch a little bit: See attachment.

2. As far as I understand the question the angle $\theta$ is outside the trapezoid, the small base a = 8' and the legs of the trapezoid have the length 8' too.

3. Then $b = 2x+a$.

Use the sine function to calculate x: $\sin(80^\circ)=\dfrac x8$

Use the cosine function to calculate c: $\cos(80^\circ)=\dfrac c8$
• Aug 17th 2010, 11:51 PM
Frenchie
ah you know what i think you're correct, I wasn't sure at first but yeah it is outside of the trapezoid. Thanks for your help btw =)

Oh your right also a is given! LOL i just needed b and c! thanks again =)
• Aug 18th 2010, 01:19 AM
Frenchie
How would i go about writing a general equation for the area as a function of theta? I know the area of a trapezoid is c*((a+b)/2) but i don't know how that relates to a general form.

f(@)=8cos(@)*(8+8sin(@)/2)

i don't even know if this is considered a "general equation" i mean it involves f(@) and it can be solved with @.
• Aug 18th 2010, 04:19 AM
sa-ri-ga-ma
Quote:

Originally Posted by Frenchie
How would i go about writing a general equation for the area as a function of theta? I know the area of a trapezoid is c*((a+b)/2) but i don't know how that relates to a general form.

f(@)=(cos(@)8)*((8+sin(@)/8)/2)

i don't even know if this is considered a "general equation" i mean it involves f(@) and it can be solved with @.

In general b can be written as

b = a + c*cot(α) + c*cot(β)
• Aug 18th 2010, 07:43 AM
earboth
Quote:

Originally Posted by Frenchie
How would i go about writing a general equation (?) for the area as a function of theta? I know the area of a trapezoid is c*((a+b)/2) <== I'm going to start here

1. I assume that your trapezoid is symmetric. I assume also that the base a and the length of the legs are given.

2. As you said the area of the trapezoid is calculated by:

$A=\dfrac{a+b}2 \cdot c$

3. Determine b depending on the length of one leg and the value of angle $\theta$:

$b = 2x+a$ . Now $x = 8 \cdot \cos(\theta)$. That means:

$b = 16 \cdot \cos(\theta) + a$

Determine c depending on the length of one leg and the value of angle $\theta$:

$c = 8 \cdot \sin(\theta)$

4. Plug in all terms substituting b and c:

$A=\dfrac{a+b}2 \cdot c~\implies~A=\dfrac{a+16 \cdot \cos(\theta) + a}2 \cdot 8 \cdot \sin(\theta)$

Simplify!