Mahaviracharya/Ganita Sara Samgraha/Chapter VII/180

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Mahaviracharya: Ganita Sara Samgraha Chapter $\text {VII}$: Measurement of Areas: Problem $180$

Two pillars are of known height: $p$ and $q$.

Two strings are tied, one to the top of each.

Each of these strings is stretched so as to touch the foot of the other pillar.

From the point where the two strings meet, another string is suspended vertically till it touches the ground.

What is the length $r$ of this suspended string?


Solution

$r = \dfrac {p q} {p + q}$

that is:

$\dfrac 1 r = \dfrac 1 p + \dfrac 1 q$


Proof

Pillars-problem.png

Let $AB$ and $CD$ be the pillars, standing vertically on the ground at $A$ and $C$.

Let $EF$ be the string hanging to the ground.

Let the height of $AB$ be $p$, and the height of $CD$ be $q$.

Let the length of $EF$ be $r$.

Let the distance of $AC$ be $d$.

Let the distance of $F$ from $C$ be $y$.

We have that:

$\triangle AEF$ and $\triangle ACD$ are similar
$\triangle ABC$ and $\triangle FEC$ are similar.

Thus:

\(\ds \dfrac {AB} {AC}\) \(=\) \(\ds \dfrac {EF} {FC}\)
\(\ds \dfrac {CD} {AC}\) \(=\) \(\ds \dfrac {EF} {AF}\)
\(\ds AF + FC\) \(=\) \(\ds AC\)

That is:

\(\ds \dfrac p d\) \(=\) \(\ds \dfrac r x\)
\(\ds \dfrac q d\) \(=\) \(\ds \dfrac r {d - x}\)
\(\ds \leadsto \ \ \) \(\ds p x\) \(=\) \(\ds r d\)
\(\ds q \paren {d - x}\) \(=\) \(\ds r d\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {r d} p\)
\(\ds \leadsto \ \ \) \(\ds q d - q \dfrac {r d} p\) \(=\) \(\ds r d\) substituting for $x$ and expanding
\(\ds \leadsto \ \ \) \(\ds p q - q r\) \(=\) \(\ds p r\) multiplying through by $\dfrac p d$
\(\ds \leadsto \ \ \) \(\ds p q\) \(=\) \(\ds r \paren {p + q}\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds \dfrac {p q} {p + q}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 r\) \(=\) \(\ds \dfrac {p + q} {p q}\)
\(\ds \) \(=\) \(\ds \dfrac 1 p + \dfrac 1 q\)

$\blacksquare$


Sources

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