We need to compute the hypotenuse of 4 right triangles.
The Pythagorean theorem states:
[tex]c^2=a^2+b^2[/tex]where a and b are the legs and c is the hypotenuse of the right triangle.
In one of the triangles, the length of the legs are: 6 and 8 yards. Then the length of the hypotenuse is:
[tex]\begin{gathered} c^2_1=6^2+8^2 \\ c^2_1=36+64 \\ c_1=\sqrt[]{100} \\ c_1=10yd_{} \end{gathered}[/tex]In another triangle, the length of the legs are: 12 and 8 yards. Then the length of the hypotenuse is:
[tex]\begin{gathered} c^2_2=12^2+8^2 \\ c^2_2=144+64 \\ c_2=\sqrt[]{208} \\ c_2=4\sqrt[]{13}\text{ yd} \end{gathered}[/tex]In the triangle whose hypotenuse (c3) is 15 yd and one of its legs is 12 yd, the unknown is one of the legs, b, which can be computed as follows:
[tex]\begin{gathered} 15^2=12^2+b^2 \\ 225=144+b^2 \\ 225-144=b^2 \\ \sqrt[]{81}=b \\ 9=b \end{gathered}[/tex]The last triangle has legs of 9 yd and 6 yd. Its hypotenuse is:
[tex]\begin{gathered} c^2_4=9^2+6^2 \\ c^2_4=81+36 \\ c_4=\sqrt[]{117} \\ c_4=3\sqrt[]{13} \end{gathered}[/tex]Finally, the length of the walkway is:
[tex]\begin{gathered} c_1+c_2+c_3+c_4=10+4\sqrt[]{13}+15+3\sqrt[]{13}= \\ =(10+25)+(4\sqrt[]{13}+3\sqrt[]{13})= \\ =35+7\sqrt[]{13} \end{gathered}[/tex]This value is irrational because it includes and square root
