Question

A ladder 17m long reaches a window which is 8m above the ground, on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window at a height of 15 m. Find the width of the street.​

Answer 1

Answer:

23

Step-by-step explanation:

it is simple Pythagoras theorem

take width of the street as x

the hypotenuse is the ladder

and the height as one of the sides of the triangle

in first case,

h²=a²+b²

17²=8²+x²

x²=17²-8²

x²=289-64

x²=225

x=√225

x=15

in second case,

h²=a²+b²

17²=15²+x²

x²=17²-15²

x²=289-225

x²=64

x=√64

x=8

since we need total width of street, we add it up

15+8

23

Answer 2

$\huge\mathfrak{ Answer-}$

$\huge{\bf{\underline {\underline {\red{Taken-}}}}}$

[ Refer the attachment]

• 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.

$\huge{\bf{\underline {\underline {\red{To \: Find-}}}}}$

• Total width of the street.

$\huge{\bf{\underline {\underline {\red{Formula\: to \: be \: used}}}}}$

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

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

Where,

• h = hypotenuse
• b = base
• p = perpendicular

$\huge{\bf{\underline {\underline {\red{Solution-}}}}}$

From $\rm\pink {∆ DAC }$, we have :

• AC = base ( ? )
• AD = perpendicular ( 8m)
• CD = hypotenuse ( 17m)

$\rm\red { {AC}^{2} + {AD}^{2} = {CD}^{2} }$

$\rm\blue { {AC}^{2} = ( {CD}^{2} - {AD}^{2} )}$

$\rm\blue { = {17m}^{2} - {8m}^{2} = {AC}^{2}}$

$\rm\blue { = 289 - 64 = {AC}^{2} }$

$\rm\blue {= 225= {AC}^{2}}$

$\rm\blue { = \sqrt{225} = AC}$

$\rm\green { = 15m = AC}$

______...

From $\rm\pink { ∆ CBE}$, we have :

• BC = base ( ? )
• BE = perpendicular ( 15m )
• CE = hypotenuse ( 17m )

$\rm\blue { {CB}^{2} + {BE}^{2} = {CE}^{2} }$

$\rm\green { {CB}^{2} = ( {CE}^{2} - {BE}^{2} ) }$

$\rm\green {= {17m}^{2} - {15m}^{2} = {CB}^{2} }$

$\rm\green { = 289m - 225m = {CB}^{2}}$

$\rm\green {= 64m= {CB}^{2} }$

$\rm\green {= \sqrt{64m} = CB }$

$\rm\orange { = 8m = CB}$

Hence the width of street = AB = ( AC + CB ) = ( 15+8 ) m = 23 m.