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(A) Let the wall be $AB$ and the ground be $BC$. The ladder is $AC = 5 \ m$. Initially,$BC = 3 \ m$. In $	riangle ABC$,by Pythagoras theorem,$AB^2 +

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$1 \ m$

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(A) Let the wall be $AB$ and the ground be $BC$. The ladder is $AC = 5 \ m$. Initially,$BC = 3 \ m$. In $	riangle ABC$,by Pythagoras theorem,$AB^2 + BC^2 = AC^2$,so $AB^2 + 3^2 = 5^2$,which gives $AB^2 = 25 - 9 = 16$,so $AB = 4 \ m$.Now,the bottom of the ladder is pulled $1 \ m$ farther,so the new distance from the wall is $BC' = 3 + 1 = 4 \ m$. The ladder length remains $DC' = 5 \ m$. In $	riangle DBC'$,$BD^2 + BC'^2 = DC'^2$,so $BD^2 + 4^2 = 5^2$,which gives $BD^2 = 25 - 16 = 9$,so $BD = 3 \ m$.The distance the top of the ladder slides down is $AD = AB - BD = 4 - 3 = 1 \ m$.

$A$ ladder $5 \ m$ long leans against a vertical wall. The bottom of the ladder is $3 \ m$ from the wall. If the bottom of the ladder is pulled $1 \ m$ farther from the wall,how much does the top of the ladder slide down the wall?

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