Since the triangle is a right triangle we can draw the triangle like so, calling x the distance corresponding to the base of the triangle and x+5 the height oh the triangle.
now use the formula of the area
![A=(bh)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/xo42jv4osrwdvc7knat7xohbopzvkw08x9.png)
we now the area and the corresponding variables for the base and height, replace them on the formula
![52=(x\cdot(x+5))/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/b1g2lqhphwtkftn6y8v4kugaf5vjrfohjb.png)
clear for the x
![104=x^2+5x](https://img.qammunity.org/2023/formulas/mathematics/college/4llid48in3fectto7dtmeq5lf66xpv9bkp.png)
equal the equation to 0 and solve the quadratic equation or by factorization
![x^2+5x-104=0](https://img.qammunity.org/2023/formulas/mathematics/college/iakb1s8r44bxoeg6n7xfyummgadk8ntlj3.png)
![(x-8)\cdot(x+13)=0](https://img.qammunity.org/2023/formulas/mathematics/college/v34ab87pzvltr6o7bh5y4qntb563j2cqp0.png)
by factorization we know that roots are x=-13 and x=8, since we are talking about a distance the base of the triangle must be 8 ft.
check the answer with the formula of the area
![\begin{gathered} A=(bh)/(2) \\ A=(8\cdot(8+5))/(2) \\ A=(8\cdot13)/(2) \\ A=52 \\ 52=52 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3031fmqwdhwn89d73sseua4vlh5c72q5ul.png)
The procedure is correct, the base of the triangle is 8 ft and the height is 13 ft.