Question

Hello
A ladder 25 m long just reaches tbe top of a building 24 m high from the ground. Find the distance of the foot of the ladder from the building ​

Answer 1

Hello dear!❤️

Solution:

Let AB be the building and CB be the ladder.

Then,

AB = 24m , CB = 25m and Angle CAB = 90°

By Pythagoras theorem

${cb}^{2} = {ab}^{2} + {ac}^{2}$

${ac}^{2} = {cb}^{2} - {ab}^{2}$

${ac}^{2} = ( {25}^{2} - {24}^{2} ) {m}^{2}$

${ac}^{2} =( 625 - 576) {m}^{2}$

${ac}^{2} = 49 {m}^{2}$

$ac = \sqrt{49}$

$ac = 7m$

Hence, the distance of the foot of the ladder from the building is 7 m

Hope this helps dear ☺️❤️

Answer 2

Step-by-step explanation:

According to the question: BC = 25 m AC = 24 m

.. In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB...

BC^2 = AC^2 + AB^2

AB^2 = BC^2 - AC^2

= 25^2 - 24^2

= 625 - 576 = 49

AB^2 = 49 = 7^2