Question

a man goes 10 m due east and then 24 m due north . find the distance from starting point

Answer 1

let the distance travelled in the east be =AB=10m
let the distance travelled in the north =BC=24m
according to pythagorus theorem
ABsq +BCsq=ACsq
10sq+24sq=ACsq
100+576=ACsq
676=ACsq
AC=26
thus distance from the starting point is 26 km

Answer 2

Step-by-step explanation:

Answer:

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(10.6,2.9){\large\sf{C}}\put(7.7,1){\large\sf{A}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(10.5,1){\line(0,2){1.9}}\qbezier(8,1)(8.5,1.4)(10.5,2.9)\put(9,0.7){\sf{\large{10 m}}}\put(10.7,1.9){\sf{\large{24 m}}}\put(10.3,1){\line(0,1){0.2}}\put(10.3,1.2){\line(3,0){0.2}}\put(8,1){\circle*{.15}}\put(7.3,2){\sf{\large{West}}}\put(11.7,2){\sf{\large{East}}}\put(10,0){\sf{\large{South}}}\put(10,3.8){\sf{\large{North}}}\end{picture}$

⠀

$\underline{\bigstar\:\textsf{By Pythagoras Theorem :}}$

$:\implies\sf (Hypotenuse)^2 = (Perpendicular)^2+(Base)^2\\\\\\:\implies\sf (AC)^2=(BC)^2+(AB)^2\\\\\\:\implies\sf (AC)^2=(24 \:m)^2+(10 \:m)^2\\\\\\:\implies\sf (AC)^2= 576\:m^2+100\:m^2\\\\\\:\implies\sf (AC)^2=676 \:m^2\\\\\\:\implies\sf AC= \sqrt{676 \:m^2}\\\\\\:\implies\sf AC = \sqrt{26 \:m \times 26 \:m}\\\\\\:\implies\underline{\boxed{\sf AC = 26 \:m}}$

⠀

$\therefore\:\underline{\textsf{Man is \textbf{26 m} away from starting point}}.$