Q25. a ladder ap of length 10m is kept against a wall ab such that it reaches a window at a, 6m high from the ground. the same ladder reaches another window at c when it is rested against the wall opposite to ab. if the walls are 14 m apart. find the height of the window at c from the ground.

Step-by-step explanation:

Given a ladder AP of length = 10m is kept against a wall AB such that it reaches a window at A, such that AB=6m .

Consider a triangle ABP

By Pythagoras theorem,

Now,The same ladder reaches another window at C when it is rested against the wall opposite to AB .

Let D be a point which represent the foot of the second building CD on the ground which is 14 m away from B.

∴ BD = BP+ PD              

⇒ 14 = 8+PD

⇒PD =14-8 = 6m

Now consider another triangle CPD

Again by Pythagoras theorem,

Therefore, the height of the window at C from the ground CD=8m

step-by-step explanation:

compare

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl