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Answer:

Let AC be the ladder, AB be the wall.

Then in △ABC,

[tex] {AC}^{2} = {AB}^{2} + {BC}^{2} (Pythagoras \: theorem)[/tex]

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

[tex] {BC}^{2} = {8}^{2} [/tex]

[tex]BC=8 m[/tex]

Hence, the foot of the ladder is 8 m from the wall.

Explaination:

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

[tex] \\ \\ [/tex]

To Find : Find the Distance of foot of ladder from the buliding

[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]

SolutioN :

[tex] \dag [/tex] Formula Used :

• [tex] {\underline{\boxed{\pmb{\sf{ { \bigg( Hypotenuse \bigg) }^{2} = { \bigg( Height \bigg) }^{2} + { \bigg( Base \bigg) }^{2} }}}}} [/tex]

[tex] \\ \\ [/tex]

[tex] \dag [/tex] Calculating the Distance :

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { {\bigg( Hypotenuse \bigg) }^{2} = { \bigg( Height \bigg) }^{2} + { \bigg( Base \bigg) }^{2} } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { { \bigg( 17 \bigg) }^{2} = { \bigg( 15 \bigg) }^{2} + { \bigg( Base \bigg) }^{2} } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { \sqrt{ { \bigg( 17 \bigg) }^{2} - { \bigg( 15 \bigg) }^{2} } = Base } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { \sqrt{280 - 225} = Base } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { \sqrt{64} = Base } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf { Base = 8 \; m }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered} [/tex]

[tex] \\ \\ [/tex]

[tex] \therefore \; [/tex] Distance of the ladder from the Building is 8 m .

[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]