A 5, m long ladder is placed leaning against | vedclass

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(D) Let $AC$ be the ladder of length $5\,m$ and $BC = 4\,m$ be the height of the wall where the ladder is placed. If the foot of the ladder is moved $

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) Let $AC$ be the ladder of length $5\,m$ and $BC = 4\,m$ be the height of the wall where the ladder is placed. If the foot of the ladder is moved $1.6\,m$ towards the wall,i.e.,$AD = 1.6\,m$,then the ladder slides upward to a new position $ED$.In right-angled $	riangle ABC$,by Pythagoras theorem:$AC^2 = AB^2 + BC^2$$5^2 = AB^2 + 4^2$$25 = AB^2 + 16$$AB^2 = 9 \Rightarrow AB = 3\,m$.Now,the new distance of the foot from the wall is $DB = AB - AD = 3 - 1.6 = 1.4\,m$.In right-angled $	riangle EBD$,by Pythagoras theorem:$ED^2 = EB^2 + DB^2$$5^2 = EB^2 + (1.4)^2$$25 = EB^2 + 1.96$$EB^2 = 25 - 1.96 = 23.04$$EB = \sqrt{23.04} = 4.8\,m$.The distance by which the top of the ladder slides upwards is $EC = EB - BC = 4.8 - 4 = 0.8\,m$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$	$a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity.
$2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$	$b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity.
$3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$	$c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity.
$4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$	$d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.

$A$ $5\, m$ long ladder is placed leaning against a vertical wall such that it reaches the wall at a point $4 \,m$ high. If the foot of the ladder is moved $1.6 \,m$ towards the wall,find the distance by which the top of the ladder would slide upwards on the wall (in $m$).

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