A bridge

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

Correct result:

a =  80 ft

Solution:

$$\begin{gathered} r=10+x \\ {r}^2=x^2+(27+y{)}^2 \\ {r}^2=(9+x{)}^2+y^2 \\ (10+x{)}^2=x^2+(27+y{)}^2 \\ (10+x{)}^2=(9+x{)}^2+y^2 \\ 20x=y^2+54y+629 \\ 2x+19=y^2 \\ 20\ast (y^2-19)\mathrm{/}2=y^2+54y+629 \\ 20\cdot (y^2-19)\mathrm{/}2=y^2+54y+629 \\ 9y^2-54y-819=0 \\ a=9;b=-54;c=-819 \\ D=b^2-4ac=54^2-4\cdot 9\cdot (-819)=32400 \\ D>0 \\ {y}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{54\pm \sqrt{32400}}{18}} \\ {y}_{1,2}={\displaystyle \frac{54\pm 180}{18}} \\ {y}_{1,2}=3\pm 10 \\ {y}_1=13 \\ {y}_2=-7 \\ \text{ Factored form of the equation: } \\ 9(y-13)(y+7)=0 \\ x>0;y>0;r>0 \\ y=y_1=13=13\text{ ft } \\ x=(y^2-19)\mathrm{/}2=(13^2-19)\mathrm{/}2=75\text{ ft } \\ r=10+x=10+75=85\text{ ft } \\ a=2\cdot 27+2\cdot y=2\cdot 27+2\cdot 13=80\text{ ft} \end{gathered}$$

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Showing 2 comments:

Dr Math

THE BRIDGE OVER THE RIVER

2 years ago  Like

Dr Math

r = radius of circle of arc

x = distance between center of circle and water level (center of the circle is under ground)

y = distance in horiznotal of point on circle which is 9 ft above the water

2 years ago  Like

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