Step-by-step explanation:

[tex] {ac}^{2} = {ab}^{2} + {bc}^{2} (pythagoras \: theorem)[/tex]

[tex] {17}^{2} = {15}^{2} + {bc}^{2} [/tex]

[tex]289 = 225 + {bc}^{2} [/tex]

[tex] 289 - 225 = {bc}^{2} [/tex]

[tex] \sqrt{64 } = bc[/tex]

[tex]8 = bc[/tex]

Hence, distance of the foot of the ladder from the building=8m

Verified answer

Given : A ladder 17m long, reaches a window of a building 15m above the ground.

To Find : The distance of the foot of the ladder from the building ?

______________

Solution : Let AC be the ladder and A be the position of the window.

[tex]~[/tex]

• AC = 17m, AB = 15m.

[tex]~[/tex]

[tex]{\sf{\frak{\underline{As~we~know~that~:}}}}[/tex]

[tex]{\sf:\implies{\underline{Pythagoras~Theorem}}}[/tex]

• [tex]\boxed{\sf\pink{\pmb{AB^2~=~AC^2~+~BC^2}}}[/tex]

[tex]~[/tex]

[tex]{\bf{\underline{According~to~the~question~:}}}[/tex]

[tex]~[/tex]

[tex]{\sf:\implies{BC^2~=~(17m)^2~+~(15m)^2}}[/tex]

[tex]{\sf:\implies{BC^2~=~289~-~225}}[/tex]

[tex]{\sf:\implies{BC^2~=~64m^2}}[/tex]

• (Here, both square's get cancelled)

[tex]:\implies\boxed{\frak{\underline{\pmb{\pink{BC~=~8cm}}}}}[/tex]★

[tex]~[/tex]

Hence,

• The distance of the foot of ladder from the building is 8cm.