## Task 1 - 210721

a) A rectangular plot of land is divided into two rectangular fields by a path. The length of the plot, measured parallel to this path, is $105 \mathrm{~m}$. The width of the first part of the field is $270 \mathrm{~m}$, and the width of the second part of the field is $180 \mathrm{~m}$. The path is $3 \mathrm{~m}$ wide.

Determine the area of the first part of the field and the area of the second part of the field!

b) The entire plot is now combined into one large field by plowing over the path.

Determine the area of the resulting large field!

c) Determine how many meters of wire are needed for an electric fence if the entire field is to be completely enclosed with two wires! The sag and securing of the wire should be accounted for by increasing the double perimeter by one hundredth.

(Round to the nearest meter.)

Hint for a) and b): The areas should be given in hectares, rounded to two decimal places.
a) The first subfield has a length of $105 \mathrm{~m}$ and a width of $270 \mathrm{~m}$, so $105 \cdot 270=28350$ and thus an area of $28350 \mathrm{~m}^{2}$, i.e., rounded to 2.84 ha.

The second subfield has a length of $105 \mathrm{~m}$ and a width of $180 \mathrm{~m}$, so $105 \cdot 180=18900$ and thus an area of $18900 \mathrm{~m}^{2}$, i.e., 1.89 ha.

b) The entire plot has a length of $105 \mathrm{~m}$ and, because $270+3+180=453$, a width of $453 \mathrm{~m}$, so $105 \cdot 453=47565$ and thus an area of $47565 \mathrm{~m}^{2}$, i.e., rounded to 4.76 ha.

c) The entire plot has a perimeter of $2 \cdot(105+453)=2 \cdot 558=1116$ meters, so $1116 \mathrm{~m}$. For the fence, $2 \cdot 1116=2232$ and $2232: 100=22.32$ as well as $2232+22.32=2254.32$ are needed, thus rounded to $2254 \mathrm{~m}$ of wire.