A ladder is placed so as to reach a window 63 | vedclass

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(C) Let the ladder be $AC$ and $CE$,where $AC = CE = 65 \ cm$. Let $AB = 63 \ cm$ be the height of the window on one side and $ED = 56 \ cm$ be the he

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(C) Let the ladder be $AC$ and $CE$,where $AC = CE = 65 \ cm$. Let $AB = 63 \ cm$ be the height of the window on one side and $ED = 56 \ cm$ be the height of the point on the opposite side. In right-angled triangle $	riangle ABC$,by Pythagoras theorem: $BC^2 + AB^2 = AC^2$ $BC^2 + 63^2 = 65^2$ $BC^2 = 4225 - 3969 = 256$ $BC = \sqrt{256} = 16 \ cm$. In right-angled triangle $	riangle CDE$,by Pythagoras theorem: $CD^2 + ED^2 = CE^2$ $CD^2 + 56^2 = 65^2$ $CD^2 = 4225 - 3136 = 1089$ $CD = \sqrt{1089} = 33 \ cm$. Therefore,the width of the street is $BD = BC + CD = 16 + 33 = 49 \ cm$.

$A$ ladder is placed so as to reach a window $63 \ cm$ high. The ladder is then turned over to the opposite side of the street and is found to reach a point $56 \ cm$ high. If the ladder is $65 \ cm$ long,then the width of the street is........$cm$

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