# A bridge

A bridge over a river is in the shape of the arc of a circle with each base 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?

Result

a =  80 ft

#### Solution:

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

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Dr Math
THE BRIDGE OVER THE RIVER

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

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