# A man goes 15 m due west and then 8 m due north. How far is he from the starting point? Let AB = 15m and AC = 8m

In ∆CAB, using Pythagoras Theorm,

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

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

(8)2 + (15)2 = (BC)2

(BC)2 = 64 + 225

(BC)2 = 289

BC = ±17

BC = 17 [taking positive square root]

Hence, the man is 17m far from the starting point.

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  NCERT | Strong Your Basics of Triangles39 mins  RD Sharma | Imp. Qs From Triangles41 mins  R.D Sharma | Solve Exercise -4.2 and 4.3FREE Class  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 