Q. 2

# A ladder 26 m long reaches a window 24 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =24m and length of the ladder, BC = 26m

Let AB = x m be the distance of the foot of the ladder from the base of the wall.

In ∆CAB, using Pythagoras Theorm,

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AC)2 + (AB)2 = (BC)2

(24)2 + (AB)2 = (26)2

(AB)2 = (26)2 – (24)2

(AB)2 = (26 – 24)(26+24)

[ (a2 – b2)=(a+b)(a – b)]

(AB)2 = (2)(50)

(AB)2 = 100

AB = ±10

AB = 10 [taking positive square root]

Hence, the distance of the foot of the ladder from base of the wall is 10m

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Basic Proportionality Theorem42 mins
A Peep into Pythagoras Theorem43 mins
R.D Sharma | Solve Exercise -4.2 and 4.3FREE Class
NCERT | Strong Your Basics of Triangles39 mins
RD Sharma | Imp. Qs From Triangles41 mins
Quiz | Criterion of Similarity of Triangle45 mins
How to Ace Maths in NTSE 2020?36 mins
Know About Important Proofs in Triangles33 mins
Master BPT or Thales Theorem39 mins
Champ Quiz | Thales Theorem49 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses