A badge is made of a circle centre O, with radius 5cm and a chord AB of length 6cm creating a blue minor segment at the bottom. Give all your answers to 1 decimal place.

Find:
The angle AOB at the centre
The length of the minor arc AB
The area of the triangle AOB.
The area of the yellow major segment.

How do I do this? Please show the steps!! Thank you, I am so confused

Here is the screen shot of the question (it has a diagram) Please help me. thanks - also i just typed in random numbers to get the grey answers on the bottom

mymathsradians | Flickr - Photo Sharing!

Cool. thanks i'll have a look now.

One way to get the "angle at the center" is to use the cosine law: $c^2= a^2+ b^2- 2ab cos(C)$ where a, b, and c are the lengths of the three sides of a triangle and C is the angle opposite the side "c". Here, the two sides a, and b are the radii of length 5 cm and c has length 6 cm.

Once you know that angle, the length of the arc is easy- the arc length is just the angle, in radians, times the radius.

You could get the area of the triangle, now that you know three sides by "Heron's formula" $A= \sqrt{s(s-a)(s-b)(s- c)}$ where s is the "semi-perimeter", (a+b+c)/2.

But since this is an isosceles triangle, it follows that the "height" bisects the opposite side. So you have two right triangles with hypotenuse of length 5 cm and one leg of length 3 cm.
Use the Pyhagorean theorem to find the length of the other leg, the height of the original triangle. And then use A= (1/2)bh.

The area of the entire circle is, of course, $\pi r^2= 25\pi$. The area of the "pie section", between the radii, is proportional to the part of $2\pi$ the angle, in radians, is, so its area is $25\pi\frac{\theta}{2\pi}= \frac{25}{2}\theta$ where $\theta$ is the angle you found in the first part of the problem. Subtract the area of the triangle, that you just found, to find the area of the blue section, then subtract that from the area of the entire triangle, $25\pi$ to find the area of the yellow part.