geoNewbie21

I believe it was a mathcounts question. It was on a alcumus quiz and I got the answer wrong.

could you please show me the steps so I understand how to solve. Thanks

I am not sure about that because when we calculate the angle of a clock dont we use something like Angle = Minute *60/360. What I am not sure is do I assume it to be 3:45 and 2:50 or 9:15 and 10:10. The angle value I get when I assume the clock hands as 3:45 and 2:50 is 12.5 while 9:15 and 10:10 gives me the answer 57.5. and there is my confusion

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

Thanks. graphing the quad helped calculate the area as a sum of two triangles.

ht = sqrt((0-1.5)^2 + (1-2.5)^2) = 1.5sqrt(2)

base = sqrt((1-4)^2 + (3-0)^2) = 3sqrt(2)

Area of 1 triangle = 1/2 *1.5sqrt(2)*3*sqrt(2) = 4.5

Area of quadrialater = 2*area of triangle = 9

Yes, I am sorry I had it wrong and I think I wasted everyone's time. I was struggling with it and thought I was missing something, and now I know what it was the "-" sign.

No : the answer wa possible areas are 1,2,4,5, and 9 . So sum of all possible areas was 21

Thanks but the answer is not 14. But then why not 16 then? Could you please explain

Thankyou Guest and CPhill,

CPhill could you actually share a link on how to use desmos.You have made it easy to understand/solve many geometry related questions, but I am not sure on how to use desmos to add the  points A and O in this case or plot the circle :)

So then if I have y=4x-6, y=-2x+12 and y-axis can I calculate the area to be

1/2*b*h where b = 18, h = 3 from the graph so area = 27?

Am I approaching this the right way?

Hi Melody, Thank you and Will do!!