# Two Isosceles Trapezoids with identical Area and Angles

#### MathStuck

##### New member
I'm attempting to do a math problem, and I'm been going in circles with how to solve it 2 days. And help with the equations would be much appreciated!

I have an isosceles trapezoid, and know all of its dimensions. I would like to create another isosceles trapezoid with the same Area and angles as the first one, but with a different value for the small base.

How do I create an equation to solve this? From the equations I've found online, I need more known information to do this, but my intuition has me thinking its solvable. I've attempted to break it down into the triangles and squares, to use the slope, but my mind is moosh and could use some help. I've included a graphic I made, and some of my thinking below that runs me in circles.

My current logic, which doesnt work and I dont know why:

Variables:
A=area
b=small base
h=height
j= small angle
g=size of triangle base(not shown)

Breaking it up into 2 squares
A = b*h + h*g
g= h*tan(j) //Getting the small size from the height and angle
A = b*h + (h*(h*tan(j))) //equations combined
But everything after that gets weird.

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#### Dr.Peterson

##### Elite Member
What you have is a quadratic equation in h. Do you know the quadratic formula?

I think your tangent should be a cotangent, but otherwise the work so far looks okay.

#### MathStuck

##### New member
What you have is a quadratic equation in h. Do you know the quadratic formula?

I think your tangent should be a cotangent, but otherwise the work so far looks okay.
Thank you Dr. Peterson. Unknowingly I have been using an online tool which has been giving me the quadratic answer. Before your response, I did not realize thats what it was. Also, based on your response I watched a Khan Academy video and used the quadratic formula to produce this:
h=-(cot(j) + SQRT(cot(j) - 4*b*-A))/2*b

However, with both the online tools and this work by hand, the results do not appear to be correct compared to the answers I'm expecting. Is this because my initial approach which gave me the quadratic equation was wrong?

Here is the steps I took:
A = b*h + (h*(h*cot(j)))
0 = b*h + (h*(h*cot(j))) -A
0 = b*h + cot(j)h2 -A

h=-(cot(j) + SQRT(cot(j) - 4*b*-A))/2*b

#### MathStuck

##### New member
What you have is a quadratic equation in h. Do you know the quadratic formula?

I think your tangent should be a cotangent, but otherwise the work so far looks okay.
You were 100% correct! Thank you for your help. Turns out my issue had to do with giving degrees to my equations rather than radians, and vice-verse.

#### Dr.Peterson

##### Elite Member
You were 100% correct! Thank you for your help. Turns out my issue had to do with giving degrees to my equations rather than radians, and vice-verse.
That is one of the most common errors in problems like this! If you hadn't discovered it yourself, you could have shown us an example of your results and expected results, and we probably would have checked for this possibility first.