+126 Determine an expression for the area of triangle QCA in terms of p.
I solved it as follows
Area of Triangle = \(\frac{1}{2}*b*ht\)
base = \(\sqrt{(0-2)^2 + (12-12)^2} =\sqrt{4} = 2\)
ht = \(\sqrt{(0-0)^2+(p-12)^2} = \sqrt{(p-12)^2} = (p-12)\)
Area = \(\frac{1}{2}*2*(p-12) =(p-12)\)
But this was wrong... should this be (12-p) as p is smaller than 12 and area cannot be a -ve value?
+129899 QCA is a right triangle with legs QA and QC
QA = 2
QC =12 - p (i think this is the part where you went a little wrong, geoNewbie )
Area = product of the legs / 2 =
2 (12 - p) / 2 = 12 - p
+126 Ahh thats what i thought. But many times when I calculate distance I have use (y1-y2) but since it would be squared the answer comes out as a +ve number. But in this case, I guess it does make a difference. Thanks CPhill
