SOLUTION: A cast iron disk (flat surface) has 5 holes drilled. Determine net and percent reduction in area. 4 holes have 1" diameter. 1 hole has 4" diameter. Have illustration b

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Question 1067676: A cast iron disk (flat surface) has 5 holes drilled. Determine net and percent reduction in area.
4 holes have 1" diameter.
1 hole has 4" diameter.
Have illustration but unable to upload.
My attempt:
Area = Pi * r^2
A = 3.1416 * 7^2 = 153.9384 = 154 sq. in. (gross area)
-----------------
A = 3.1416 * 0.5 ^ 2 * 4 = 3.1416 sq. in. (small holes)
A = 3.1416 * 2^2 = 12.5664 = 13 sq. in. (large hole)
137.8584 = 138 sq. in. (net area)

154 - 138 = 16 / 154 = 0.1038961 = 10.38 per cent reduction.
Where am I incorrect? Thanks.
Non-homework.
Found 2 solutions by Boreal, Alan3354:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
I'm assuming the main disk is 14 inches in diameter.
The Area is 49 pi, subtracting pi and also 4 pi
so 5 pi/49 pi is subtracted, or 10.20 per cent. The net area is 44 pi. By keeping pi as an irrational number, no rounding needs to be done until the end. Your work is otherwise appropriate, given what I am visualizing.

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
A cast iron disk (flat surface) has 5 holes drilled. Determine net and percent reduction in area.
4 holes have 1" diameter.
1 hole has 4" diameter.
Have illustration but unable to upload.
My attempt:
Area = Pi * r^2
A = 3.1416 * 7^2 = 153.9384 = 154 sq. in. (gross area)
---------
Radius of the disk is 7" ?
Area = 49pi
==============
A = 3.1416 * 0.5 ^ 2 * 4 = 3.1416 sq. in. (small holes)
A = pi
A = 3.1416 * 2^2 = 12.5664 = 13 sq. in. (large hole)
A = 4pi
137.8584 = 138 sq. in. (net area)
Lost = 5pi
154 - 138 = 16 / 154 = 0.1038961 = 10.38 per cent reduction.
================
Without rounding off, it's
5pi/49pi =~ 10.2%
===========
Area left = 44pi =~ 138.23 sq inches