



 
Hi Wanda, The question was
I copied the diagram from my response in 2007, added one label, a line and changed the colouring. As you can see the triangle PQR is partitioned into three congruent triangles PQC, QRC and RPC. I am going to find the area of triangle PQC and multiply by 3 to determine the area of triangle PQR. To find the area of triangle PQC I cut it along the line MC and move the bottom half so that the line segment PM lies along QM. Since M is the midpoint of PQ these line segments will match. The labeling on this diagram is confusing since there are two points with the label C and one point with two labels P and Q o I am going to erase the label P and one of the Cs. The area of this triangle is the same as the area of triangle QCP in the original diagram. In my previous response I argued that measure of the angle QCM is 60^{o} and a similar argument shows angle MCP also has a measure of 60^{o}. Since the sum of the angles of a triangle 1s 180^{o} the triangle above is an equilateral triangle. We know that QC is a radius of the circle so QC = 3 and thus CM = 3/2 = 1 1/2. This is half the base of the triangle above and hence to determine its area (and hence the area of triangle QCP) I only need to find its height MQ. For this we can use Pythagoras theorem. Triangle QCM is a right triangle so
or
and thus MQ = 3 √3/2. Finally the area of triangle PQR is
I hope this helps,  


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