+126 Determine an expression for the area of triangle QCA in terms of p.
I solved it as follows
Area of Triangle = 12∗b∗ht
base = √(0−2)2+(12−12)2=√4=2
ht = √(0−0)2+(p−12)2=√(p−12)2=(p−12)
Area = 12∗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?
+130477 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
