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?

geoNewbie21 Feb 3, 2021

#1**+2 **

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

CPhill Feb 3, 2021

#2**+1 **

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

geoNewbie21 Feb 3, 2021