The line segment AB is a chord of a circle centre (2,-1) where A and B are (3,7) and (-5,3) respectively. AC is a diameter of the circle. Find the area of ABC?

Area ABC = 59.9998 The best way to tackle this problem is to draw the question on a graph and plot each point. Once you’ve done this you’ll notice that AO is the radius (or half a diameter). Using coordinates, AO is (-1,-8). In triangle form this means that one length of the triangle AO is 8 and the other is 1. By using a^2=b^2+c^2, AO is the square root value of 8^2 + 1^2 which is 8.06226. This means AC is 2 x AO = 16.1245. Looking at your graph, we can see the triangle ABC. Use the above method to work out the length of AB and BC: AB: (-8,-4) => square root of 8^2+4^2= 8.94427 AC: (6,-12) => square root of 6^2+12^2= 13.4164 Now we have the lengths of all three sides of the triangle ABC. *Remember that drawing your working out will always help*. There are multiple methods but the easiest in my opinion is using trigonometry by calculating at least one angle of the triangle. Using sinx=opposite/hypotenuse, sinACB=BC/AC, sinACB=8.94427/16.1245, ACB=33.6901. Using Area ABC=0.5 x BC x AC x sinC, Area ABC=0.5 x 13.4164 x 16.1245 x sin(33.6901) = 59.9998