Question

A man goes 10 m due east and then 24 m due north. Find the distance from the
starting point.​

Answer 1

Hello .................lol
I know this answer but I’m not in the mood to solve it sorry


OTHERS DINT REPORT MY ANSWER IF THAT GUY HAS ANY PROBLEM THEN HE WILL DO IT HIMSELF

Answer 2

Answer:

:

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

⠀

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

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

⠀

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