Problem Solving

Why couldn't I make the base go all the way across the circle? Is it because that would make it "on the circle" instead of just "at the center" of the circle?

AleksandrM wrote:Why couldn't I make the base go all the way across the circle? Is it because that would make it "on the circle" instead of just "at the center" of the circle?

Yes, that's right. The question states that one of the vertices lies at the center of the circle. If the base were to lie across the center of the circle, it would be impossible for the origin to be a vertice of the triangle.

What is the greatest possible area of a triangle region with one vertex at the center of a circle of radius 1 and other two vertices on the circle?

Pls find the attachment for the rough draw.

Assume a square is lying on a circle and the diagonal of the square in the circle's diameter. "The greatest possible area of a triangle region with one vertex at the center of a circle" means, its the diagonal of the triangle. If radius is 1, then diameter(diagonal) is, 2.

If the diagonal is going at the center of the circle, then the would definitely be the right angle triangle. Other two vertices are lying on the circle. (It doesn't mean that for right angle triangle the two sides should be equal.)

Both the vertices wouldn't go above the diagonal area. It should be below 2.

so, if i take both the vertices as 2 *(considering diagonal) 1/2 * 2 * 2 = 2, the area wouldn't go above 2 and wouldn't go below 1. so, it should be between 1 and 2. With this, we can eliminate A, B and C.

Take 1/2 * b * h = 1;

b=2/h;

b^2 + h^2 = 4

4/h^2 + h^2 = 4

4+ h ^ 4 - 4 h^2=0

h^4-4h^2+4=0 (take h^2 = m)

m^2 - 4m + 4=0

m=2

so, h = Sqroot(2).

b=2/Sqroot(2)

which is, Sqroot(2);

Guys, let me know if i'm wrong.[/img]

umaa, u have made two mistakes.

1) ur drawing shows that u haven't taken centre as vertex for one side. The triangle line is maybe passing thru centre but not is not a vertex. A vertex at centre means that it's a meeting point for two different sides of triangle.

2) you have already assumed the area to be 1. that's what we want to find out.

we know radius is one so to sides have length of 1. After that u can use any of the approach give above e.g Ian's approach to get the answer

hi ppl

I am a new member. I am not able to see the original diagram (JPG 2) as posted in the question.

Is there a trick to see it that I am missing out.

Regards

Can someone plz draw a diagram for what the triangle should look like??

Thanks

So I came across this question in my test and got it wrong..I assumed the equilateral triangle has the greatest area and marked root3/4
Now i see the logic..any triangle drawn by the above specifications will have two legs as the radius..we have to maximise the area so the third leg should be the largest.

However,is this some kind of a theoram/fact that we should be knowing?That to get the largest area of a triangle,the triangle has to be a right angle and not an equilateral one?

uptowngirl92 wrote:So I came across this question in my test and got it wrong..I assumed the equilateral triangle has the greatest area and marked root3/4
Now i see the logic..any triangle drawn by the above specifications will have two legs as the radius..we have to maximise the area so the third leg should be the largest.

However,is this some kind of a theoram/fact that we should be knowing?That to get the largest area of a triangle,the triangle has to be a right angle and not an equilateral one?

You can do it with calculus and set the derivative of area with respect to height to 0 and solve for the base in terms of height, which will be at maximum area. Turns out base = twice height which only occurs when its a 45,45,90 triangle, which with two legs of length 1 has an area of 1/2

Calculus is not tested..........