P is a point on the segment joining the feet | vedclass

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(C) Let the two poles be $AD$ and $BC$ with heights $a$ and $b$ respectively,standing on a horizontal ground $AB$. $P$ is a point on $AB$.In $	riangl

Vedclass · Accepted Answer

$2(a^2+b^2)$

Vedclass · Answer

(C) Let the two poles be $AD$ and $BC$ with heights $a$ and $b$ respectively,standing on a horizontal ground $AB$. $P$ is a point on $AB$.In $	riangle APD$,$	an 45^{\circ} = \frac{AD}{AP} = \frac{a}{AP} \Rightarrow AP = a$.In $	riangle BPC$,$	an 45^{\circ} = \frac{BC}{BP} = \frac{b}{BP} \Rightarrow BP = b$.Draw a line $DE$ perpendicular to $BC$ at $E$. Then $DE = AB = AP + PB = a + b$ and $CE = BC - BE = BC - AD = b - a$.In right-angled $	riangle DEC$,by Pythagoras theorem:$DC^2 = DE^2 + CE^2$$DC^2 = (a+b)^2 + (b-a)^2$$DC^2 = (a^2 + 2ab + b^2) + (b^2 - 2ab + a^2)$$DC^2 = 2(a^2 + b^2)$.

$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then,the square of the distance between the tops of the poles is

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