A bridge

The bridge over the river has the shape of an arc. The bridge is 10 feet above the water at the center of the river. At 27 feet from the river's edge, the bridge is 9 feet above the water. How wide is the river?

Final Answer:

a =  80 ft

Step-by-step explanation:

$$\begin{gathered} r=10+x \\ {r}^2=x^2+(27+y{)}^2 \\ {r}^2=(9+x{)}^2+y^2 \\ (10+x{)}^2=x^2+(27+y{)}^2 \\ (10+x{)}^2=(9+x{)}^2+y^2 \\ 20x=y^2+54y+629 \\ 2x+19=y^2 \\ 20\cdot (y^2-19)/2=y^2+54y+629 \\ 20\cdot (y^2-19)/2=y^2+54y+629 \\ 9y^2-54y-819=0 \\ 9=3^2 \\ 54=2\cdot 3^3 \\ 819=3^2\cdot 7\cdot 13 \\ \text{GCD}(9,54,819)=3^2=9 \\ {y}^2-6y-91=0 \\ a=1;b=-6;c=-91 \\ D=b^2-4ac=6^2-4\cdot 1\cdot (-91)=400 \\ D>0 \\ {y}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{6\pm \sqrt{400}}{2} \\ {y}_{1,2}=\frac{6\pm 20}{2} \\ {y}_{1,2}=3\pm 10 \\ {y}_1=13 \\ {y}_2=-7 \\ x>0;y>0;r>0 \\ y=y_1=13\text{ft} \\ x=(y^2-19)/2=(13^2-19)/2=75\text{ft} \\ r=10+x=10+75=85\text{ft} \\ a=2\cdot 27+2\cdot y=2\cdot 27+2\cdot 13=80\text{ft} \end{gathered}$$

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