A ladder resting against a vertical wall has its foot on the ground at a distance of 6ft. from the wall. A man on the ground climbs two - thirds of the ladder. What will be his distance from the wall now?

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Solution

Man climb $\frac{2}{3}$ of the ladder.
So, in $△ A D E$ & $△ A B C$
$∠ A = ∠ A$ (common)
$∠ A D E = ∠ A B C$ (corresponding Angle)
$∠ A E D = ∠ A C B$ (corresponding Angle)
$△ A D E \sim △ A B C$
$⟹ \frac{A D}{A B} = \frac{D E}{B C}$
$⟹ \frac{3}{5} = \frac{D E}{6} ⟹ D E = \frac{18}{5} = 3.6 f t$

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A ladder resting against a vertical wall has its foot on the ground at a distance of 6ft. from the wall. A man on the ground climbs two - thirds of the ladder. What will be his distance from the wall now?

Man climb 23 of the ladder. So, in △ADE & △ABC∠A=∠A (common)∠ADE=∠ABC (corresponding Angle)∠AED=∠ACB (corresponding Angle)△ADE∼△ABC⟹ADAB=DEBC⟹35=DE6⟹DE=185=3.6ft