Answer:
[tex]Solution,\\Here,\\Hypotenuse(h)=\frac{37}{3} =12.34yd\\Base(b)=4yd\\Perpendicular(p)=?\\Now,\\Using- pythagoras-theorem,\\h^{2} =p^{2} +b^{2} \\(12.34yd)^{2} =p^{2} +(4yd)^{2} \\152.11yd^{2} =p^{2} +16yd^{2} \\p^{2} =152.11yd^{2} -16yd^{2} \\p=\sqrt{136.11yd^{2} } \\p=11.667[/tex]
a≈11.667
Step-by-step explanation:
