A&H

Geek Question

Sheffields Finest

Maybe I'm foolish, maybe I'm blind!
Level 7 Referee
What is the total area of a standard Penalty Area including the arc of the D.....?? (and show your working out) :)
 
Last edited:
The Referee Store
745.12 yards squared I make it.

Penalty area =
18 x 40 (18 yards either side of goal posts and 8 yards between)
=720yds squared.

Penalty Arc
(3.14 x 16)/2 = 25.12 yards squared.

adding both together to arrive at final answer.

EDIT: my penalty arc area is wrong. I've assumed a circle but in fact you would have an oval with varying radius so might will have to work that bit out.
 
Incorrect so far!!!!!
Hmmmm... sure my calculations are right at the second attempt. The only thing I can think next would be to subtract the goal area, area?
I think that area is 120yds squared so maybe 703.4yds squared. After that I give up and leave it to someone who is actually good at Maths.
 
Not sure what I did to get to that first answer - blaming that on lack of caffeine!

Penalty area is 6480 sq ft
Arc area is 402.6 sq ft

6882.6 sq ft
 
Its the penalty arc thats getting me. I have worked it out wrong as the second radius wont be 5 will it? I give up... im supposed to be working lol :ninja:
 
Radius of the arc, r =4 yards and subtends an angle of 180 degreees (pi radians)

g = distance from centre of arc on the penalty area line, to the edge of the arc on the penalty area line

Right angle triangle therefore
g = (10^2 - 6^2)^0.5 = 8 yards

Area of ellipse = pi*r*g we've got half an ellipse therefire Area of arc = 0.5*3.14*4*8 = 50.27 sq yards

Area of penalty area =18*44 = 792 sq yds
Total Area
792 + 50.27 = 842 sq yds (to 3 sf)
 
The area of the D on the edge of the penalty area is 44.73 square yards (to 2 decimal places)

The D is a segment. To find the area of a segment, first find the area of the sector (the 2D cone shape) by using the angle to find what fraction of a whole circle it is. (Use a bit of trig. to find this angle). Then find the area of the triangle (use half base x height. Use Pythagoras to find the length of the base) Subtract one from the other to find the area of the segment. Please see my photo for working out.

Now write to your old maths teacher and apologise for asking "when will I ever need this in real life?"

Fantastic question, I'm sure my year 11s will look forward to solving it tomorrow morning.;)

penaltyAreaworkingoutMidsize.jpg
 
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