A $26$ m long ladder reached a window $24$ m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.

10

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Solution

The correct option is A $10$ m
In $△ P Q R$ , we have
${P R}^{2} = {P Q}^{2} + {Q R}^{2}$
(By Pythagoras theorem)
$({26)}^{2} = {(24)}^{2} + {Q R}^{2}$
${Q R}^{2} = 676 - 576 = 100$
$\Rightarrow Q R = \sqrt{100}$
$\Rightarrow Q R = 10$
The distance of the foot of the ladder fro the wall is $10$ m.

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A 26 m long ladder reached a window 24 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.

The correct option is A 10 mIn △PQR, we havePR2=PQ2+QR2(By Pythagoras theorem)(26)2=(24)2+QR2QR2=676−576=100⇒QR=√100⇒QR=10The distance of the foot of the ladder fro the wall is 10 m.

A $26$ m long ladder reached a window $24$ m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.