Question

A ladder $15$ m long reaches a window that is $9$ m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window $12$ m high. The width of the street is
(a) $27$ m
(b) $21$ m
(c) $24$ m
(d) $18$ m

Answer 1

(b) $21$ m

Let the ladder be at point $C$ on the ground.
At first, the ladder is placed towards the left side of the street and it reaches the window $AB$, which is $9$ m high from the ground.
Secondly, the ladder is placed towards the right side of the street and it reaches the window $DE$, which is $12$ m high from the ground.
Applying Pythagoras theorem in right-angled $△ABC$ and $△DEC$, we get,
$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow B{C}^2=A{C}^2-A{B}^2$

$\Rightarrow B{C}^2=152-92=225-81=144$

$\Rightarrow BC=144=12$ m

Also,

$D{C}^2=D{E}^2+C{E}^2$

$\Rightarrow C{E}^2=D{C}^2-D{E}^2$

$\Rightarrow C{E}^2=152-122=225-144=81$

$\Rightarrow CE=81=9$ m

Hence, total width of the street is,

$BE=BC+CE=12+9=21$ m