Question

a ladder 50 m long reaches a window of building 40 metre above the ground determine the distance of the foot of the ladder from the building​

Answer 1

The distance of the foot of the ladder from the building is 30 metres.

Answer 2

AnswEr :

Let the P be the window, and PR be the Ladder reaches on the window of building.

･ Join QR, & it will form Right Angle Triangle

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{P}}\put(7.7,1){\large{Q}}\put(10.6,1){\large{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\mathsf{\large{40 m}}}\put(9,0.7){\matsf{\large{? m}}}\put(9.4,1.9){\mathsf{\large{50 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\rule{120}{2}$

• By Pythagoras theorem in ∆ PQR :

$:\implies\sf PR^2 = PQ^2 + QR^2\\\\\\:\implies\sf{(50 \:m)}^{2} = {(40 \:m)}^{2} +QR^2 \\\\\\:\implies\sf{(50 \:m)}^{2} - {(40 \:m)}^{2} =QR^2\\\\\qquad \scriptsize{\bf{\dag} \: ({a}^{2} - {b}^{2}) = (a + b)(a - b) }\\\\:\implies\sf(50 \:m + 40 \:m)(50 \:m - 40 \:m) = QR^2\\\\\\:\implies\sf 90 \:m \times 10 \:m= QR^2 \\\\\\:\implies\sf900 \:{m}^{2} = QR^2 \\\\\\:\implies\sf \sqrt{900 \: {m}^{2} } = QR\\\\\\:\implies\sf \sqrt{30 \:m \times 30 \:m } = QR\\\\\\:\implies \boxed{\sf QR = 30 \:m}$

⠀

∴ Hence, Distance of foot of the ladder from the building will be 30 metres.