Answer:
Layout B
Explanation:
Since Shinji needs to put a length of fence around whichever layout he chooses, we find the perimeter of each of the figures.
Figure A
The vertices are P(-9,8), Q(-4,9), R(-1,-6) and S(-6,-7).
We use the distance formula to find the length of each of the sides.
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex][tex]\begin{gathered} PQ=\sqrt[]{(-9-(-4))^2+(8_{}-9_{})^2}=\sqrt[]{(-9+4)^2+(-1)^2} \\ =\sqrt[]{(-5)^2+(-1)^2} \\ =\sqrt[]{26} \end{gathered}[/tex]
Similarly:
[tex]\begin{gathered} PS=\sqrt[]{(-9-(-6))^2+(8_{}-(-7)_{})^2}=\sqrt[]{(-9+6)^2+(8+7)^2} \\ =\sqrt[]{(-3)^2+(15)^2} \\ =\sqrt[]{234} \end{gathered}[/tex]
Therefore, the perimeter of Figure A will be:
[tex]\begin{gathered} =2\sqrt[]{26}+2\sqrt[]{234} \\ =40.79\text{ units} \end{gathered}[/tex]
Figure B
• 9-2 = 7 units
,
• 8-(-3)=8+3 = 11 units
The dimension of Figure B is 7 units by 11 units.
Therefore, the perimeter of Figure B is:
[tex]\begin{gathered} =2(7+11) \\ =2\times18 \\ =36\text{ units} \end{gathered}[/tex]
The layout in Figure B has a lower perimeter, therefore it will need less fencing.
