Question

25 metre long ladder is placed against a vertical wall such that the foot of the ladder is 7 metre from the feet of the wall if the top of the ladder slides down by 4 metre by how much distance will the foot of the ladder slide​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Foot\:slided\:distance=8\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ : \implies \text{Length \: of \: ladder = 25 \: m} \\ \\ : \implies \text{Distance \: between \: foot \: of \: ladder \: and \: wall = 7m} \\ \\ : \implies \text{Distance \: of \: sliding \: the \: ladder \: from \: top = 4 \: m} \\ \\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Distance \: of \: sliding \: of \: bottom \: of \: ladder = ?}$

• Accroding to given question :

$\bold{In \: \triangle \: ABC} \\ : \implies {h}^{2} = {p}^{2} + {b}^{2} \: \: \: \: \: \: \: \: \text{(by \: phythagoras \: theoram}) \\ \\ : \implies {25}^{2} = {AB}^{2} + {7}^{2} \\ \\ : \implies 625 = {AB}^{2} + 49 \\ \\ : \implies 625 - 49 = {AB}^{2} \\ \\ : \implies {AB}^{2} = 576 \\ \\ : \implies AB = \sqrt{576} \\ \\ \green{: \implies \text{AB = 24 \: m}} \\ \\ \bold{Slided \: height \: from \: top \: of \: ladder}\\ : \implies Final \: height = 24 - 4 \\ \\ : \implies \text{Final \: height =20 \: m} \\ \\ \bold{In \: \triangle \: DAE} \\ : \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ : \implies {25}^{2} = {20}^{2} + {AE}^{2} \\ \\ : \implies {AE }^{2} = 625 - 400 \\ \\ : \implies {AE }^{2} =225 \\ \\ : \implies {AE } = \sqrt{225} \\ \\ \green{: \implies \text{{AE }=15 \: m}} \\ \\ \green{\therefore \text{Distance \: of \: foot \: of \: ladder \: slide = 15 - 7 = 8 \: m}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Length of Ledder = 25 m
• Distance between ledder foot and wall = 7 m
• Sliding Distance of ladder = 4 m

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

Use Pythagoras Theorem :

$\small \bigstar {\boxed{\sf{(Hypotenuse)^2 \: = \: (Base)^2 \: + \: (Perpendicular)^2}}} \\ \\ \implies {\sf{(25)^2 \: = \: (7)^2 \: + \: P^2}} \\ \\ \implies {\sf{625 \: = \: 49 \: + \: P^2}} \\ \\ \implies {\sf{P^2 \: = \: 625 \: - \: 49}} \\ \\ \implies {\sf{P^2 \: = \: 576}} \\ \\ \implies {\sf{P \: = \: \sqrt{575}}} \\ \\ \implies {\boxed {\sf{Perpendicular \: = \: 24 \: m}}}$

$\rule{200}{2}$

Height After Sliding :

⇒Height (Sliding) = 24 - 4

⇒Height = 20 m

$\rule{100}{0.5}$

Again use Pythagoras Theorem :

$\implies {\sf{(25)^2 \: = \: (20)^2 \: + \: (B)^2}} \\ \\ \implies {\sf{625 \: = \: 400 \: + \: B^2}} \\ \\ \implies {\sf{B^2 \: = \: 625 \: - \: 400}} \\ \\ \implies {\sf{B^2 \: = \: 225}} \\ \\ \implies {\sf{B \: = \: \sqrt{225}}} \\ \\ \implies {\sf{B \: = \: 15}} \\ \\ \implies {\boxed{\sf{Base \: = \: 15 \: m}}}$

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Distance = 15 - 7 = 8 m