I know someone else asked this but when I emailed my teacher he said it wasn't right please help!!!!?

Question

Find the dimensions of the largest isosceles triangle that can be inscribed in a circle of radius 4.

Answer 1

Is the largest in perimeter or area although I think in both cases its the same anyway and that will be an equilateral triangle with sides equal to

4sin120/sin30 = 4√3

P = 12√3 and Area = 12√3

Answer 2

If the triangle is equilateral, then the center of the circle will coincide with the center of the triangle.
Therefore, the distance from the center to any corner of the triangle must be equal 4.
Given that there are 3 points around the circle, the angle between any of the 3 radii must be 360/3 which is 120.
Now you must envision a triangle with a 120 degree angle at the top and two 30 degree angles on the bottom corners. Draw a line down from the top to the middle of the bottom. (this will be from the center of the circle/triangle to the middle of the inscribed triangle's base.
We know that the angle of the lower corner is 30, the top is 60 and the intersection of the bottom line is 90. We also know the diagonal is 4. The tangent is the diagonal.

The ratio of half the bottom leg of the triangle to the radius is Cos(30). Therefore the bottom leg of the triangle must be 2 times that, 2Cos(30). Since the diagonal is a radius with distance = 4, the base must be 8Cos(30). All sides would be the same.

The hight from the bottom of the above triangle would be Sin(30). Sin(30) comes out evenly at .5. The hight of the equilateral would be the 2 + 8Cos(30)

Area will equal 1/2 base * hight.
( 8Cos(30) * (2+8Cos(30)) ) / 2
4Cos(30) * (1+4Cos(30) )

Answer 3

The largest isosceles triangle would actually be an equilateral triangle...

First inscribe an equilateral triangle in a circle. Now draw three lines from the center to each vertex, dividing the triangle into 3. Now divide these each in half with another line from the center, to the midpoint of each side of the triangle.

The final result is 6 triangles that are all 30-60-90.

Now to find the area. The hypotenuse is equal to 4 (the radius). Either via Pythagorean theorem, or remembering the ratios of a 30-60-90 triangle are 1 : 2 : sqrt(3), you should be able to figure that your triangle has sides that are 2 : 4 : 2*sqrt(3).

So the small triangles will have an area of 1/2 * 2 * 2 sqrt(3) = 2 sqrt(3). But you have six of them, so the actual area is:

Area = 12 sqrt (3) ≈ 20.785 sq. units.
Interestingly, the perimeter is also 6 * 2 sqrt(3)
Perimeter = 12 sqrt (3) ≈ 20.785 units.

If it helps you visualize this, I've attached a picture...

Answer 4

i am not sure but i think an Equilangular is the biggest with one side is 4 radical 3 and the high is 6

Answer 5

I got the perimeter as 20.85

and the area as 20.91 {unit} squared