+0  
 
+1
117
2
avatar+122 

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
avatar+118588 
+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

 

 

cool cool cool

 Feb 3, 2021
 #2
avatar+122 
+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

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