+0

# Area question

+1
59
2
+110

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?

Feb 3, 2021

#1
+116049
+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

Feb 3, 2021
#2
+110
+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

Feb 3, 2021