The goal area is 20 yards wide and 6 yards deep. It is the smaller of the two areas in front of the goal. The other area is called the penalty area.
The penalty area is 44 yards wide.
The dimensions for area are [L2]
All players, except the penalty taker must be at least 10 yards away from the penalty spot when the penalty is taken. As the Penalty Area extends 18 yards from the goal line, and the Penalty Spot is 12 yards from the goal line, the distance from the Penalty Spot to the edge of the Penalty Area can be as little as 6 Yards. The Arc - not half circle - marks the area outside the Penalty Area, that is within 10 yards of the Penalty Spot, and therefore outside which players must be when the penalty is taken
area is 2, volume is 3
If the new linear dimensions are k times the old dimensions, then the new area is k2 times the old area.
No. It doesn't matter where the goalkeeper is.The ballmust be in the penalty area (on the line is inside) for the keeper to touch it.
Goal width = 16.5 m Distance from goalpost to each edge = 16.5 m Width of penalty area = 3 * 16.5 = 49.5 m Depth of penalty area = 16.5 m Area of penalty area = 49.5 * 16.5 = 816.75 m2
The answer depends entirely on how the dimensions change. It is possible to change the dimensions without changing the perimeter. It is also possible to change the dimensions without changing the area. (And it is possible to change the area without changing the perimeter.)
It is an arc that is drawn outside of the penalty area with a radius of ten yards and the center being the penalty mark. (Note: the mark is only 6 yards from the edge of the penalty area area) Whenever a penalty kick is taken all players, except the kicker and the goal keeper, must be outside of the penalty area, at least ten yards from the mark, and not closer to the goal line than the mark. Other than the taking of penalty kicks, it has no other function.
Area has the dimensions of (length)2 .
If you know the dimensions of the missing triangle, then compute the area from those dimensions, then subtract that answer from the area of the full rectangle.