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.


it's Isha
gm ​

Answer 1

Step-by-step explanation:

${ac}^{2} = {ab}^{2} + {bc}^{2} (pythagoras \: theorem)$

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

$289 = 225 + {bc}^{2}$

$289 - 225 = {bc}^{2}$

$\sqrt{64 } = bc$

$8 = bc$

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

Answer 2

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.

$~$

• AC = 17m, AB = 15m.

$~$

${\sf{\frak{\underline{As~we~know~that~:}}}}$

${\sf:\implies{\underline{Pythagoras~Theorem}}}$

• $\boxed{\sf\pink{\pmb{AB^2~=~AC^2~+~BC^2}}}$

$~$

${\bf{\underline{According~to~the~question~:}}}$

$~$

${\sf:\implies{BC^2~=~(17m)^2~+~(15m)^2}}$

${\sf:\implies{BC^2~=~289~-~225}}$

${\sf:\implies{BC^2~=~64m^2}}$

• (Here, both square's get cancelled)

$:\implies\boxed{\frak{\underline{\pmb{\pink{BC~=~8cm}}}}}$★

$~$

Hence,

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