Question

A ladder 17m long,reaches a window of a building 15m above the ground. Find the distance of the foot of the ladder from the building.
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Answer 1

Answer:

Let AC be the ladder, AB be the wall.

Then in △ABC,

${AC}^{2} = {AB}^{2} + {BC}^{2} (Pythagoras \: theorem)$

${17}^{2} = {15}^{2} + {BC}^{2}$

${BC}^{2} = {8}^{2}$

$BC=8 m$

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

Explaination:

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Answer 2

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

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

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SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ { \bigg( Hypotenuse \bigg) }^{2} = { \bigg( Height \bigg) }^{2} + { \bigg( Base \bigg) }^{2} }}}}}$

$\dag$ Calculating the Distance :

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

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

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

$\begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { \sqrt{280 - 225} = Base } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; \dashrightarrow \; \; \sf { \sqrt{64} = Base } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf { Base = 8 \; m }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered}$

$\therefore \;$ Distance of the ladder from the Building is 8 m .

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