Question

A ladder of length 17 m reaches a window which is 8 m above the ground on
one side of a street and at the same point, it reaches a window 15 m high on a
wall on the opposite side. Find the width of the street.

Answer 1

• Diagram Available With Answer.
• [ Please Refer Attachment ]

TAKEN»

• Let AB be the street and C be the foot of the ladder.

• Let D and E be the windows at the heights of 8m and 15m respectively from the ground.

• Then CD and CE are the two positions of the ladder.

To Find›››

⠀⠀⠀⠀⠀⠀Total width of the street.

⠀⠀⠀⠀Using Formula:

$\tt \green{Pythagoras \: \: Property}$

$\huge \tt {h}^{2} = {b}^{2} + {p}^{2}$

• H = Hypotenuse
• B = Base
• P = Prepedicular

$\huge{ \mathcal{ \red{SOLUTION}}}$

$\tt From \: \blue{ ∆DAC}, We \: Have :$

1. AC = Base ( —?— )
2. AD = Prepedicular ( 8m )
3. CD = Hypotenuse ( 17m )

${ \huge{ \blue{ \overline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}}}$

$\sf \red{AC² + AD² = CD² } \\ \sf \blue{AC = ( CD² - AD² )} \\ \sf \blue{= 17m² - 8m² = AC²}\\ \blue{ = 289 - 64 = AC²} \\ \sf \blue{ = 225 = {AC}^{2} } \\ \sf \blue{\sqrt{225} = AC} \\ \\ \\ \sf \green{ 15m = AC}$

$\sf From \: \orange{∆CBE } \: We \: Have \implies$

• AC = Base ( —?— )
• AD = Prepedicular ( 15m )
• CD = Hypotenuse ( 17m )

$\sf \blue{ CB² + BE²=CE²} \\ \sf \green{CB² =(CE²-BE²)} \\ \sf \green{17m²-15m²=CB²} \\ \sf \green{286m-225m=CB²} \\ \sf \green{64m=CB²}\\ \sf \green{\sqrt{64m} =CB} \\ \\ \sf \orange{=8m=CB}$

Hence, The Wedth of Street = AB = ( AC + CB ) = ( 15 + 8 )m = 23m.