The question is: "A rectangle ABCD has A and D on the equal sides of an isosceles triangle and B and C on its base. If AB=2cm and BC=6cm, find the value of θ [bottom right corner of the triangle] that produces a triangle of minimum area."
I keep getting θ=1.89Rad, but it does not seem right. Can someone show me their process for solving this?
Thanks!
Posted Dec 20, 2012, 8:02 PM UTC
[deleted]
For the love of all that is holy, draw a picture first, then start setting up equations.
I first set the base of the triangle equal to 2b, so that the area equation becomes a simple A=bh. Let's also pretend that this isosceles triangle has its base on the x-axis, and its vertex on the y=axis, like in this picture. This lets us set up a coordinate system and define the two legs of the triangle as linear equations. Using point-slope form with the points (3,2), (0,h), and (b,0) we can get a relationship between b and h: b = 3h/(h-2)
This lets us get the area equation solely in terms of h, which we can then find a minimum value for. Check the picture I posted for the full details (minus the actual derivative calculations, but you can do that yourself.) If it looks like I got anything wrong, please let me know.
tl;dr I got a lower answer than you did, check this picture.
Hey, I did not think of trying to solve with respect for height. I've been trying like this, http://imgur.com/lGWuH , but typing it out I think I have messed up the trig.
Yup and there it is 2/tanθ =! 2cscθsecθ
I just noticed a mistake I made near the end of my calculations, at the end it should have said b=6 and tan(theta) = 2/3. This is the same answer I got from your trig method.
If you show your process, we can help point out where you may have gone wrong.
I'm trying to use like triangles to get the base and height. http://imgur.com/lGWuH