Two vertical poles of heights 20 m and 80 m | vedclass

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(C) Let the two poles be $AB = 20 \ m$ and $CD = 80 \ m$ standing on a horizontal plane at a distance $x$ apart.Let the intersection point of the line

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Let the two poles be $AB = 20 \ m$ and $CD = 80 \ m$ standing on a horizontal plane at a distance $x$ apart.Let the intersection point of the lines joining the top of each pole to the foot of the other be $P$,and let its height from the ground be $h$.Let $P$ be at a distance $y$ from the foot of the first pole $(20 \ m)$ and $(x-y)$ from the foot of the second pole $(80 \ m)$.From similar triangles,we have:$\frac{h}{y} = \frac{20}{x} \implies \frac{y}{h} = \frac{x}{20} \implies \frac{1}{h} = \frac{x}{20y}$$\frac{h}{x-y} = \frac{80}{x} \implies \frac{x-y}{h} = \frac{x}{80} \implies \frac{1}{h} = \frac{x}{80(x-y)}$Adding the two relations:$\frac{1}{h} + \frac{1}{h} = \frac{1}{20} + \frac{1}{80}$ is incorrect; rather,from $\frac{h}{y} = \frac{20}{x}$ and $\frac{h}{x-y} = \frac{80}{x}$,we get $\frac{y}{h} = \frac{x}{20}$ and $\frac{x-y}{h} = \frac{x}{80}$.Adding these: $\frac{y + x - y}{h} = \frac{x}{20} + \frac{x}{80} \implies \frac{x}{h} = x \left( \frac{1}{20} + \frac{1}{80} \right)$.$\frac{1}{h} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16}$.Therefore,$h = 16 \ m$.

Two vertical poles of heights $20 \ m$ and $80 \ m$ stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other,from this horizontal plane is:

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