 In this video we provide the solution to question number seven from the practice final exam for math 1060 In which case we're asked to find the area of the provided triangle Notice that we have the three side lengths of the triangle two two and two times the square root of three In which case we need to find the area Huron's formula comes to mind Notice this diagram is not drawn to scale and since we have two and two it is in a Saucerle's triangle So we could try to use Like a 3060 90 triangle approach if we wanted to but we're just going to use Huron's formula in this in this situation So the area here is going to equal the square root of s times s minus a Times s minus b times s minus c where s is the semi perimeter It's the sum of the three sides divided by two So we end up with two plus two plus two times the square root of three over two This gives us four plus two root three over two or in other words two plus the square root of three That's our s value there and so when we look at all the possibilities We're going to end up with two plus the square root of three We're going to then get s minus in this case two that gives us a square root of three We're going to get s minus again This minus two why you're going to the square of three again, and then finally you're going to get s Minus here two root two that's going to give us two minus the square root of three like so now notice the square root of three times Itself is going to give us three and then if you take the square root Our two plus root three times it by two minus root three that's its conjugate that actually end up with four minus three This is still inside of the square root of course And so you end up with the square root of three as the area of this triangle and so we would select choice a