# need help.

• Jun 20th 2006, 12:12 PM
bobby77
need help.
A long rope is pulled out between two opposite shores of a lake. It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
• Jun 20th 2006, 12:50 PM
Soroban
Hello, Bobby!

Quote:

A long rope is pulled out between two opposite shores of a lake.
It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
Code:

                C               * * *           *    :x    *         *        :  35  *     A * - - - - -+- - - - - * B         \      D:        /           \      :R-x  /           R \    :    / R               \  :  /                 \:/                 *                 O

The center of the earth is $\displaystyle O.$ .$\displaystyle OA = OB = OC = R$ (radius of the earth).

The 70-km rope is $\displaystyle AB.$ .We see that: $\displaystyle AD = DB = 35.$

Let $\displaystyle x = CD$ be the distance the rope is underwater at its center.
Then $\displaystyle DO = R - x.$

From right triangle $\displaystyle ODB:\;\;DO^2 + DB^2\:=\:OB^2$

So we have: .$\displaystyle (R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$

Quadratic Formula: .$\displaystyle x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$

Since $\displaystyle R = 6370$, we have: .$\displaystyle x \;= \;6370 \pm \sqrt{6370^2 - 1225}$

Therefore: .$\displaystyle x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}$

• Jun 20th 2006, 06:32 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, Bobby!

Code:

                C               * * *           *    :x    *         *        :  35  *     A * - - - - -+- - - - - * B         \      D:        /           \      :R-x  /           R \    :    / R               \  :  /                 \:/                 *                 O

The center of the earth is $\displaystyle O.$ .$\displaystyle OA = OB = OC = R$ (radius of the earth).

The 70-km rope is $\displaystyle AB.$ .We see that: $\displaystyle AD = DB = 35.$

Let $\displaystyle x = CD$ be the distance the rope is underwater at its center.
Then $\displaystyle DO = R - x.$

From right triangle $\displaystyle ODB:\;\;DO^2 + DB^2\:=\:OB^2$

So we have: .$\displaystyle (R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$

Quadratic Formula: .$\displaystyle x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$

Since $\displaystyle R = 6370$, we have: .$\displaystyle x \;= \;6370 \pm \sqrt{6370^2 - 1225}$

Therefore: .$\displaystyle x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}$

Hello Soroban.
We could do it in this way also.
Extend CO to meet the circle at E
Now,
AB and CE are two chords of a cicle intresecting at D
Hence,$\displaystyle AD.DB = CD.DE$
$\displaystyle 35.35 = x(DO + OE)$
$\displaystyle 1225 = x(2R - x)$

KeepSmiling

Malay
• Jun 20th 2006, 07:24 PM
Soroban
Very nice, Malay!

An excellent application of that intersecting-chords theorem
. . . which I obviously forgot.