Question

A ladder is leaning against the wall and reaches a window at a height 15ft.the ladder is 29m.find the distance between the wall and base of the ladder

Answer 1

Answer:

24 .82 m

Step-by-step explanation:

Let x be the distance between wall and base of ladder.

Using Pythagoras theorem,

29² = 15² + x²

x² = 29² - 15² = 616

x = 24.82 m

Answer 2

CORRECT QUESTION :

A ladder is leaning against the wall and reaches a window at a height 15 m.the ladder is 29 m.find the distance between the wall and base of the ladder.

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

$\green{\therefore{\text{Distance\:between\:wall\:and\:foot\:of\:ladder=24.81\:m}}}$

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

• In the given question information given about a ladder is leaning against the wall and reaches a window at a height 15ft.the ladder is 29m.

• Wh have to find the distance between the wall and base of the ladder

$\green{\underline \bold{Given :}} \\ : \implies \text{Length\:of\: ladder =29\:m} \\ \\ : \implies \text{Heigh\:of\:window= 15\:m} \\ \\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Distance\:between\:wall\:and\:foot\:of\: ladder=?}$

• Accroding to given question :

$\bold{In \: \triangle \: ABC} \\ : \implies {h}^{2} = {p}^{2} + {b}^{2} \: \: \: \: \: \: \: \: \text{(by \: phythagoras \: theoram}) \\ \\ : \implies {29}^{2} = {15}^{2} + {BC}^{2} \\ \\ : \implies (BC)^{2} =841-225\\ \\ : \implies {(BC)}^{2} =616 \\ \\ : \implies BC= \sqrt{616} \\ \\ \green{: \implies \text{BC=24.81\:m}}$