percent

**sri340** · January 26th 2010, 01:17 PM

A rectangular volleyball court measures 20 meters by 10 meters.
There is one line judge on each side of the volleyball net, standing
on opposite corners of the court. If each judge has accurate vision
up to 10 meters from his spot, what percent of the total court is
accurately seen by at least one of the two line judges? Express your
answer to the nearest whole number.

**e^(i*pi)** · January 26th 2010, 01:23 PM

Originally Posted by sri340

A rectangular volleyball court measures 20 meters by 10 meters.
There is one line judge on each side of the volleyball net, standing
on opposite corners of the court. If each judge has accurate vision
up to 10 meters from his spot, what percent of the total court is
accurately seen by at least one of the two line judges? Express your
answer to the nearest whole number.

Please don't overuse the BBCode, makes it very hard to quote.

Total Area = $20 \times 10 = 200\, m^2$

If someone is standing at the corner they can see in an arc of radius 10m and angle $\frac{\pi}{2}$ . Using the arc length formula $s = r \theta$ we get $s = \frac{10\pi}{2} = 5\pi \,m$

Since there are two in total they can see double this distance which is $10\pi \,m$

The percentage therefore is given by $\frac{10\pi}{200} \times 100\% = \frac{\pi}{20} \times 100\%$ .

I do not have a calculator to hand so you'll need to do the sum

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