In the right triangles whose legs are a, b and its hypotenuse is c
![a^2+b^2=c^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/fdnnfwrccw5g60jmi691r5gcz9ekxf8waa.png)
In triangle JLK
∵ LK and JL are the legs of the triangle
∵ KJ is the hypotenuse
![\therefore(LK)^2+(JL)^2=(KJ)^2](https://img.qammunity.org/2023/formulas/mathematics/college/hcwgmjzmwt7u6sfpvzhxum91fp0rfafl1b.png)
∵ KL = 14 m and KJ = 18 m
Substitute them in the rule above to find JL
![\begin{gathered} (14)^2+(JL)^2=(18)^2 \\ 196+(JL)^2=324 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6kz6lggcevt1gnapnemnqo3gw0uahrh723.png)
Subtract 196 from both sides
![\begin{gathered} 196-196+(JL)^2=324-196 \\ (JL)^2=128 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jaihfqe5gpb3fvwrb081l68nprsrlii9zc.png)
Take a square root for both sides
![\begin{gathered} \sqrt[]{(JL)^2}=\sqrt[]{128} \\ JL=11.3137085 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qd7gd4l5z9vq3mtawu3l1il5csamshx8em.png)
Round it to the nearest hundredth
![\therefore JL=11.31m](https://img.qammunity.org/2023/formulas/mathematics/college/z5avfbxsj49m425tepuwppkdvda682dfyh.png)
The answer is 11.31 m